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RS Agarwal Solution | Class 6th | Chapter-4 | Integers | Edugrown

Exercise 4A

Page No 63:

Question 1:

Write the opposite of each of the following:
(i) An increase of 8
(ii) A loss of Rs 7
(iii) Gaining a weight of 5 kg
(iv) 10 km above sea level
(v) 5°C below the freezing point
(vi) A deposit of Rs 100
(vii) Earning Rs 500
(viii) Going 6 m to the east
(ix) 24
(x) −34

ANSWER:

(i) A decrease of 8
(ii) A gain of Rs 7
(iii) Losing a weight of 5 kg
(iv) 10 km below the sea level
(v) 5oC above the freezing point
(vi) A withdrawal of Rs 100
(vii) Spending Rs 500
(viii) Going 6 m to the west
(ix) The opposite of 24 is -24.
(x) The opposite of -34 is 34.

Page No 63:

Question 2:

Indicate the following using ‘+’ or ‘−’ sign:
(i) A gain of Rs 600
(ii) A loss of Rs 800
(iii) 7°C below the freezing point
(iv) Decrease of 9
(v) 2 km above sea level
(vi) 3 km below sea level
(vii) A deposit of Rs 200
(viii) A withdrawal of Rs 300

ANSWER:

(i) +Rs 600  
(ii) -Rs 800
(iii) -7oC
(iv) -9
(v) +2 km
(vi) -3 km
(vii) + Rs 200
(viii) -Rs 300

Page No 64:

Question 3:

Mark the following integers on a number line:
(i) −5
(ii) −2
(iii) 0
(iv) −7
(v) −13

ANSWER:

(i) -5


(ii) -2

(iii) 0 


(iv) 7



Page No 64:

Question 4:

Which number is larger in each of the following pairs?
(i) 0, −2
(ii) −3, −5
(iii) −5, 2
(iv) −16, 8
(v) −365, −913
(vi) −888, 8

ANSWER:

(i)0, -2
  0 > -2
This is because zero is greater than every negative integer.

(ii) -3, -5
  -3 > -5
Since 3 is smaller than 5, -3 is greater than -5.

(iii) -5, 2
2 > -5
This is because every positive integer is greater than every negative integer.

(iv) -16, 8
8 > -16
This is because every positive integer is greater than every negative integer.
v) -365, -913
-365 > -913
Since 365 is smaller than 913,  -365 is greater than -913.
vi) -888, 8
8 > -888
This is because every positive integer is greater than every negative integer.

Page No 64:

Question 5:

Which number is smaller in each of the following pairs?
(i) 6, −7
(ii) 0, −1
(iii) −13, −27
(iv) −26, 17
(v) −317, −603
(vi) −777, 7

ANSWER:

i) -7 < 6
This is because every positive integer is greater than every negative integer.
ii) -1 < 0
This is because zero is greater than every negative integer.
iii) -27 < -13
Since 27 is greater than 13, -27 is smaller than -13.
iv) -26 < 17
This is because every positive integer is greater than every negative integer.
v) -603 < -317
Since 603 is greater than 317, -603 is smaller than -317.
vi) -777 < 7
This is because every positive integer is greater than every negative integer.

Page No 64:

Question 6:

Write all integers between
(i) 0 and 6
(ii) −5 and 0
(iii) −3 and 3
(iv) −7 and −5

ANSWER:

i) 1, 2, 3, 4, 5

ii) -4, -3, -2, -1

iii) -2, -1, 0, 1, 2

iv) -6

Page No 64:

Question 7:

Fill in the blanks by appropriate symbol > or <:
(i) 0 …… 7
(ii) 0 …… −3
(iii) −5 …… −2
(iv) −15 …… 13
(v) −231 …… −132
(vi) −6 … … 6

ANSWER:

i) 0 < 7
This is because 0 is less than any positive integer.
ii) 0 > -3
This is because 0 is greater than any negative integer.
iii) -5 < -2
Since 5 is greater than 2, -5 is smaller than -2.
iv) -15 < 13
This is because every positive integer is greater than every negative integer.
v) -231 < -132
Since 231 is greater than 132, -231 is smaller than -132.
vi) -6 < 6
This is because every positive integer is greater than every negative integer.

Page No 64:

Question 8:

Write the following integers in the increasing order:
(i) 5, −7, −2, 0, 8
(ii) −23, 12, 0, −6, −100, −1
(iii) −17, 15, −363, −501, 165
(iv) 21, −106, −16, 16, 0, −2, −81

ANSWER:

i) -7 < -2 < 0 < 5 < 8
ii) -100 < -23 < -6 < -1 < 0 < 12
iii) -501 < -363 < -17 < 15 < 165
iv) -106 < -81 < -16 < -2 < 0 < 16 < 21

Page No 64:

Question 9:

Write the following integers in the decreasing order:
(i) 0, 7, −3, −9, −132, 36
(ii) 51, −53, −8, 0, −2
(iii) −71, −81, 36, 0, −5
(iv) −365, −515, 102, 413, −7

ANSWER:

i) 36 > 7 > 0 > -3 > -9 > -132
ii) 51 > 0 > -2 > -8 > -53
iii) 36 > 0 > -5 > -71 > -81
iv) 413 > 102 > -7 > -365 > -515

Page No 64:

Question 10:

Using the number line, write the integer which is
(i) 4 more than 6
(ii) 5 more than −6
(iii) 6 less than 2
(iv) 2 les than −3

ANSWER:

i) 4 more than 6
We want an integer that is 4 more than 6. So, we will start from 6 and proceed 4 steps to the right to obtain 10.

 
ii) 5 more than -6
We want an integer that is 5 more than -6. So, we will start from -6 and proceed 5 steps to the right to obtain -1.



iii) 6 less than 2
We want an integer that is 6 less than 2. So, we will start from 2 and proceed 6 steps to the left to obtain -4.





iv) 2 less than -3
We want an integer that is 2 less than -3. So, we will start from -3 and proceed 2 steps to the left to obtain -5

Page No 64:

Question 11:

For each of the following statements, write (T) for true and (F) for false:
(i) The smallest integer is zero.
(ii) Zero is not an integer.
(iii) The opposite of zero is zero.
(iv) −10 is greater than −6.
(v) The absolute value of an integer is always greater than the integer.
(vi) 0 is larger than every negative integer.
(vii) Every negative integer is less than every natural number.
(viii) The successor of −187 is −188.
(ix) The predecessor of −215 is −214.

ANSWER:

i) False
This is because 0 is greater than every negative integer.

ii) False
0 is an integer as we know that every whole number is an integer and 0 is a whole number.

iii) True
0 is an integer that is neither positive nor negative. So, the opposite of zero is zero.

iv) False
Since 10 is greater than 6, -10 is smaller than -6.

v) True
This is because an absolute value is a positive number. For example, -2 is an integer, but its absolute value is 2 and it is greater than -2.

vi) True
This is because all negative integers are to the left of 0.

vii) True
This is because natural numbers are positive and every positive integer is greater than every negative integer.

viii) False
This is because the successor of -187 is equal to -186 (-186 + 1). In succession, we move from the left to the right along a number line.

ix) False
This is because the predecessor of -215 is -216 (-216 – 1). To find the predecessor, we move from the right to the left along a number line.

Page No 64:

Question 12:

Find the value of
(i) |−9|
(ii) |−36|
(iii) |0|
(iv) |15|
(v) −|−3|
(vi) 7 + |−3|
(vii) |7−4|
(viii) 8 −|−7|

ANSWER:

i) The value of |-9| is 9
ii) The value of |-36| is 36
iii) The value of |0| is 0
iv) The value of |15| is 15
v) The value of |-3| is 3
∴∴ -|-3| = -3
vi) 7 + |-3|
= 7 + 3          (The value of |-3| is 3)
= 10
vii) |7 – 4|
= |3| 
= 3                 (The value of |3| is 3)

viii) 8 – |-7|
= 8 – 7           (The value of |-7| is 7)
= 1

Page No 64:

Question 13:

(i) Write five negative integers greater than −7.
(ii) Write five negative integers less than −20.

ANSWER:

i) Every negative integer that is to the right of -7 is greater than -7.
So, five negative integers that are greater than -7 are -6, -5, -4, -3, -2 and -1.
ii) Every negative integer that is to the left of -20 is less than -20.
So, five negative integers that are less than -20 are -21, -22, -23, -24 and -25.

Page No 68:

Exercise 4B

Question 1:

Use the number line and add the following integers:
(i) 9 + (−6)
(ii) (−3) + 7
(iii) 8 + (−8)
(iv) (−1) + (−3)
(v) (−4) + (−7)
(vi) (−2) + (−8)
(vii) 3 + (−2) + (−4)
(viii) (−1) + (−2) + (−3)
(ix) 5 + (−2) + (−6)

ANSWER:

i) On the number line, we start from 0 and move 9 steps to the right to reach a point A. Now, starting from A, we move 6 steps to the left to reach point B.


B represents the integer 3.
∴∴ 9 + (−6) = 3

(ii) On the number line, we start from 0 and move 3 steps to the left to reach point A. Now, starting from A, we move 7 steps to the right to reach point B.
B represents the integer 4.
∴∴ (−3) + 7 = 4

                 

(iii) On the number line, we start from 0 and move 8 steps to the right to reach point A. Now, starting from A, we move 8 steps to the left to reach point B.
B represents the integer 0.
∴∴ 8 + (−8) = 0


(iv) On the number line, we start from 0 and move 1 step to the left to reach point A. Now, starting from A, we move 3 steps to the left to reach point B.
B represents the integer −4.
∴∴ (−1) + (−3) = −4
                                

(v) On the number line, we start from 0 and move 4 steps to the left to reach point A. Now, starting from A, we move 7 steps to the left to reach point B.
B represents the integer −11.
∴∴ (−4) + (−7) = −11



(vi) On the number line, we start from 0 and move 2 steps to the left to reach point A. Now, starting from A, we move 8 steps to the left to reach point B.
B represents the integer −10.
∴∴ (−2) + (−8) = −10


(vii) On the number line, we start from 0 and move 3 steps to the right to reach point A. Now, starting from A, we move 2 steps to the left to reach point B. Again, starting from B, we move 4 steps to the left to reach point C.
C represents the integer −3.
∴∴ 3 + (−2) + (−4) = −3



(viii) On the number line, we start from 0 and move 1 step to the left to reach point A. Now, starting from A, we move 2 steps to the left to reach point B. Again, starting from B, we move 3 steps to the left to reach point C.
C represents the integer −6.
∴∴ (−1) + (−2) + (−3) = −6

              

(ix) On the number line, we start from 0 and move 5 steps to the right to reach point A. Now, starting from A, we move 2 steps to the left to reach point B. Again, starting from B, we move 6 steps to the left to reach point C.
C represents the integer −3.
∴∴ 5 + (−2) + (−6) = −3

               

Page No 68:

Question 2:

Fill in the blanks:
(i) (−3) + (−9) = …….
(ii) (−7) + (−8) = …….
(iii) (−9) + 16 = …….
(iv) (−13) + 25 = …….
(v) 8 + (−17) = …….
(vi) 2 + (−12) = …….

ANSWER:

(i)
(−3) + (−9)
= −3 − 9
= −12

(ii)
(−7) + (−8)
= −7 − 8
= −15

(iii)
(−9) + 16
= −9 + 16
= 7

(iv)
(−13) + 25 
= −13 + 25
= 12

(v)
8 + (−17)
= 8 − 17
= −9

(v)
2 + (−12)
= 2 − 12
= −10

Page No 68:

Question 3:

Add:
(i) −365  −87   -365  -87   

(ii)−73  −687   -73  -687   

(iii)−1065  −987   -1065  -987   

(iv) −3596  −1089   -3596  -1089   

ANSWER:

(i) 
−365   −87−452−365   −87−452
−365
−365  −87 −365  −87-365 and -87 are both negative integers. So, we add 365 and 87, and put the negative sign before the sum. 

(ii) 
−687 −73−760−687 −73−760
-687 and -73 are both negative integers. So, we add 365 and 87, and put the negative sign before the sum. 

(iii)
 −1065 −987−2052−1065 −987−2052
-1065 and -987 are both negative integers. So, we add 1065 and 987, and put the negative sign before the sum.

(iv)
 −3596−1089−4685−3596−1089−4685
-3596 and -1089 are both negative integers. So, we add 3596 and 1089, and put the negative sign before the sum.

Page No 68:

Question 4:

Add:
(i) −206  +98   -206  +98   

(ii)+178  −69   +178  -69   

(iii)−103  +312   -103  +312   

(iv) −493  +289   -493  +289   

ANSWER:

i)
−206   +98−108  Since we are adding a negative number with a positive number,we shall subtract the smaller number, i.e. 98  from the greater number, i.e. 206206 − 98 = 108        Since the greater number is negative, the sign of the result will be negative. So, the answer will be −108−206   +98−108  Since we are adding a negative number with a positive number,we shall subtract the smaller number, i.e. 98  from the greater number, i.e. 206206 – 98 = 108        Since the greater number is negative, the sign of the result will be negative. So, the answer will be -108

ii) 
     178     -69  109       178     -69  109  
Since we are adding a negative number with a positive number,we shall subtract the smaller number, i.e. 69, from the greater number, i.e. 178178 – 69 = 109     Since the greater number is positive, the sign of the result will be positive.    So, the answer will be 109Since we are adding a negative number with a positive number,we shall subtract the smaller number, i.e. 69, from the greater number, i.e. 178178 – 69 = 109     Since the greater number is positive, the sign of the result will be positive.    So, the answer will be 109

(iii) 

       312    -103  209           312    -103  209    
Since we are adding a negative number with a positive number,
we shall subtract the smaller number, i.e. -103, from the greater number, i.e. 312
312 – 103 = 209
Since the greater number is positive, the sign of the result will be positive.
So, the answer will be 209

(iv) −493+289−204(iv) −493+289−204

Since we are adding a negative number with a positive number,
we shall subtract the smaller number, i.e. 289, from the greater number, i.e. 493.
493 – 289 = 204
Since the greater number is negative, the sign of the result will be negative.
So, the answer will be -204

Page No 68:

Question 5:

Find the sum of
(i) 137 and −354
(ii) 1001 and −13
(iii) −3057 and 199
(iv) −36 and 1027
(v) −389 and −1032
(vi) −36 and 100
(vii) 3002 and −888
(viii) −18, + 25 and −37
(ix) −312, 37 and 192
(x) −51, −203, 36 and −28

ANSWER:

(i) 137 and −354−354+137−217(ii) 1001 and −131001−13988(iii) −3057 and 199−3057   199−2858(iv) −36 and 10273057−363021(v) −389 and −1032−1032−389−1421(vi) −36 and 100100−3664(vii) 3002 and −8883002−8882114(i) 137 and −354−354+137−217(ii) 1001 and −131001−13988(iii) −3057 and 199−3057   199−2858(iv) −36 and 10273057−363021(v) −389 and −1032−1032−389−1421(vi) −36 and 100100−3664(vii) 3002 and −8883002−8882114
(viii) −18, + 25 and −37
 25 + (−18) + (−37)
= 25 – (18 + 37)
= 25 – 55
= –30
 
(ix) −312, 39 and 192
  39 + 192 + (−312)
= 39 + 192 – 312
= 231 −312
= −81
(x) −51, −203, 36 and −28
  36 + (−51) + (−203) + (−28)
= 36 − (51 + 203 + 28)
= 36 – 282
= −246

Page No 68:

Question 6:

Find the additive inverse of
(i) −57
(ii) 183
(iii) 0
(iv) −1001
(v) 2054

ANSWER:

(i) −57 + 57 = 0
So, the additive inverse of −57 is 57.

(ii) 183 − 183 = 0
So, the additive inverse of 183 is −183.

(iii) 0 + 0 = 0
So, the additive inverse of 0 is 0.

(iv) −1001 + 1001 = 0
So, the additive inverse of​ −1001 is 1001.

(v) 2054 − 2054 = 0
So, the additive inverse of​ 2054 is −2054

Page No 68:

Question 7:

Write the successor of each one of the following:
(i) 201
(ii) 70
(iii) −5
(iv) −99
(v) −500

ANSWER:

(i) The successor of 201:
201 + 1 = 202
(ii) The successor of 70:
70 + 1 = 71
(iii) The successor of −5:
−5 + 1 = −4
(iv) The successor of −99:
−99 + 1 = −98
(v) The successor of −500:
−500 + 1 = −499

Page No 68:

Question 8:

Write the predecessor of each one of the following:
(i) 120
(ii) 79
(iii) −8
(iv) −141
(v) −300

ANSWER:

(i) The predecessor of 120:
120 − 1 = 119
(ii) The predecessor of 79:
79 − 1 = 78
(iii) The predecessor of −8:
−8 − 1 = −9
(iv) The predecessor of −141:
−141 − 1 = −142
​(v) The predecessor of −300:
−300 − 1 = −301

Page No 69:

Question 9:

Simplify:
(i) (−7) + (−9) + 12 + (−16)
(ii) 37 + (−23) + (−65) + 9 + (−12)
(iii) (−145) + 79 + (−265) + (−41) + 2
(iv) 1056 + (−798) + (−38) + 44 + (−1)

ANSWER:

(i) (−7) + (−9) + 12 + (−16)
  = 12 − (7 + 9 + 16)
  = 12 − 32
  = −20

(ii)  37 + (−23) + (−65) + 9 + (−12)
      = 37 + 9 − (23 + 65 + 12) 
      = 46-100
      = −54

​(iii) (−145) + 79 + (−265) + (−41) + 2
     = 79 +2 − ( 145 + 265 + 41)
     = 81 − 451
     = −370

(iv) 1056 + (−798) + (−38) + 44 + (−1)
     = 1056 + 44 − (798 + 38 + 1)
     = 1100 − 837
     = −263

Page No 69:

Question 10:

A car travelled 60 km to the north of Patna and then 90 km to the south from there. How far from Patna was the car finally?

ANSWER:

Let the distance covered in the direction of north be positive and that in the direction of south be negative.

Distance travelled to the north of Patna = 60 km
Distance travelled to the south of Patna = -90 km
Total distance travelled by the car = 60 + (​-90)
                                                   = -30 km
The car was 30 km south of Patna.

Page No 69:

Question 11:

A man bought some pencils for Rs 30 and some pens for Rs 90. The next day, he again bought some pencils for Rs 25. Then, he sold all the pencils for Rs 20 and the pens for Rs 70. What was his net gain or loss?

ANSWER:

Total cost price  = Price of pencils + Price of pens
                         = 30 + 90 + 25
                         = Rs 145

Total amount sold = Price of pen + Price of pencils
                           = 20 + 70
                           = 90
Selling price – costing price = 90 −- 145
                                          = −-55 
The negative sign implies loss.
Hence, his net loss was Rs 55. 

Page No 69:

Question 12:

For each of the following statements write (T) for true and (F) for false:
(i) The sum of two negative integers is always a negative integer.
(ii) The sum of a negative integer and a positive integer is always a negative integer.
(iii) The sum of an integer and its negative is zero.
(iv) The sum of three different integers can never be zero.
(v) | −5| < |−3|
(vi) |8 − 5| = |8| + |−5|

ANSWER:

(i) True
For example: – 2 + (-1) = -3
 
(ii) False
It can be negative or positive.
For example: -2 + 3 = 1 gives a positive integer, but -5 + 2 = -3 gives a negative integer.

(iii) True
For example: 100 + (-100) = 0

(iv) False
For example: (-5) + 2 + 3 = 0

(v) False
|-5| = 5  and | -3 | = 3, 5 > 3

(vi) False
|8 − 5| = 3
|8| + |−5| = 8 + 5
= 13
∴∴ |8 − 5|≠≠|8| + |−5|

Page No 69:

Question 13:

Find an integer a such that
(i) a + 6 = 0
(ii) 5 + a = 0
(iii) a + (−4) = 0
(iv) −8 + a = 0

ANSWER:

(i) a + 6 = 0
 => a = 0 − 6
=> a = − 6

(ii) 5 + a = 0
=> a = 0 − 5

(iii) a + (−4) = 0
=> a = 0 − (−4)
=> a = 4

(iv) −8 + a = 0
   => a = 0 + 8
  => a = 8

Page No 70:

Exercise 4C

Question 1:

Subtract:
(i) 18 from −34
(ii) −15 from 25
(iii) −28 from −43
(iv) 68 from −37
(v) 219 from 0
(vi) −92 from 0
(vii) −135 from −250
(viii) −2768 from −287
(ix) 6240 from −271
(x) −3012 from 6250

ANSWER:

(i) −34 − 18
 = −52

(ii) 25 − (−15)
 = 25 + 15
 = 40
(iii) −28 from −43
= −43 − (−28)
= −43 + 28
​= −15

(iv) 68 from −37
 = −37 − 68
 = −105
​(v)  219 from 0
 =  0 − 219 
 = −219

(vi) −92 from 0
= 0 − (−92)
= 0 + 92
= 92

(vii) −135 from −250
= −250 − (−135)
​= −250 + 135
= −115

(viii) −2768 from −287
= −287 − (−2768)
​= 2768 −​ 287
= 2481

(ix) 6240 from −271
= −271 − (6240)
= −271 − 6240
= −6511

(x) −3012 from 6250
= 6250 − (−3012)
= 6250 + 3012
​= 9262

Page No 70:

Question 2:

Subtract the sum of −1050 and 813 from −23.

ANSWER:

Sum of −1050 and 813:
−1050 + 813
= −237
Subtracting the sum of −1050 and 813 from −23:
−23 − (−237)
= −23 +237
= 214

Page No 70:

Question 3:

Subtract the sum of −250 and 138 from the sum of 136 and −272.

ANSWER:

Sum of 138 and −250:
138 + (−250)
= 138 − 250
= −112
Sum of 136 and −272:
= 136 + (−272)
= 136 − 272
 = −136
Subtracting the sum of −250 and 138 from the sum of 136 and −272:
−136 − (−112​)  
= −136 + 112​
= −24
 

Page No 70:

Question 4:

From the sum of 33 and −47, subtract −84.

ANSWER:

Adding 33 and −47:
​33 + (−47)
= 33 − 47
= −14

Subtracting −84 from −14:
−14 − (−84)
= −14 + 84
= 70 

Page No 70:

Question 5:

Add −36 to the difference of −8 and −68.

ANSWER:

Difference of −8 and −68:
−8 − (−68)
​= −8 + 68
= 60

Adding -36 to 60:
−36 + 60
= 24 

Page No 70:

Question 6:

Simplify:
(i) [37 − (−8)] + [11 − (−30)]
(ii) [−13 − (−17) + [−22 − (−40)]

ANSWER:

(i) [37 − (−8)] + [11 − (−30)]
= (37 + 8) + (11 + 30)
= 45 + 41
= 86

(ii) [−13 − (−17) + [−22 − (−40)]
=  (−13 +17) + (-22 + 40)
= 4 + 18
= 22

Page No 70:

Question 7:

Find 34 − (−72) and (−72) − 34. Are they equal?

ANSWER:

No, they are not equal.

34 − (−72) 
= 34 + 72
​= 106

(−72) − 34
= −72 − 34
= −106

Since 106 is not equal to −106, the two expressions are not equal.

Page No 70:

Question 8:

The sum of two integers is −13. If one of the numbers is 170, find the other.

ANSWER:

Let the other integer be x.
According to question, we have:
x + 170 =  −13
=> x = −13 − 170
=>  x = −183
Thus, the other integer is −183.

Page No 70:

Question 9:

The sum of two integers is 65. If one of the integers is −47, find the other.

ANSWER:

Let the other integer be x.
According to question, we have:
x + (−47) = 65
=> x − 47 = 65
=>  x = 65 + 47
=> x = 112 
Thus, the other integer is 112.

Page No 70:

Question 10:

Which of the following statements are true and which are false?
(i) The sum of two integers is always an integer.
(ii) The difference of two integers is always an integer.
(iii) −14 > −8 − (−7)
(iv) −5 − 2 > −8
(v) (−7) − 3 = (−3) − (−7)

ANSWER:

(i) True
An integer added to an integer gives an integer.

(ii) True
An integer subtracted from an integer gives an integer.

iii) False
 −8 − (−7)
 = −8 + 7
 = −1       
Since 14 is greater than 1, −1 is greater than −14.

iv) True
−5 − 2 = −7 
Since 8 is greater than 7, −7 is greater than −8.
                − 7 > −8

​v) False
L.H.S.
(−7) − 3 = −10
R.H.S.
(−3) − (−7)
= (−3) + 7
= 4
∴∴ L.H.S. ≠≠ R.H.S.

Page No 71:

Question 11:

The point A is on a mountain which is 5700 metres above sea level and the point B is in a mine which is 39600 metres below sea level. Find the vertical distance between A and B.
Figure

ANSWER:

Let us consider the height above the sea level as positive and that below the sea level as negative.
∴∴ Height of point A from sea level = 5700 m
    Depth of point B from sea level = -39600 m
Vertical distance between A and B = Distance of point A from sea level – Distance of point B from sea level  
= 5700 – (​-39600) 
= 45300 m                                                            

Page No 71:

Question 12:

On a day in Srinagar, the temperature at 6 p.m. was 1°C but at midnight that day, it dropped to −4°C. By how many degrees Celsius did the temperature fall?

ANSWER:

Initial temperature of Srinagar at 6 p.m. = 1°C
Final temperature of Srinagar at midnight = −4°C
Change in temperature = Final temperature – Initial temperature
                                      ​= (−4 − 1)°C
                                      = −5°C
So, the temperature has changed by −5°C.
The negative sign indicates that the temperature has fallen.
So, the temperature has fallen by 5°C.

Page No 72:

Exercise 4D

Question 1:

Multiply:
(i) 15 by 9
(ii) 18 by −7
(iii) 29 by −11
(iv) −18 by 13
(v) −56 by 16
(vi) 32 by −21
(vii) −57 by 0
(viii) 0 by −31
(ix) −12 by −9
(x) −746 by −8
(xi) 118 by −7
(xii) −238 by −143

ANSWER:

(i) 15 by 9
    = 15 × 9
    = 135

(ii) 18 by −7
= –(18 × 7)
      = –126

(iii) 29 by –11
= –(29 × 11)
= –319

(iv) –18 by 13
= –(18 × 13)
       = –234

(v) –56 by 16
= –(56 × 16)
      = –896

(vi) 32 by –21
= –(32 × 21)
= –672

(vii) –57 by 0
= –(57 × 0)
        = 0

(viii) 0 by –31
= –(0 × 31)
= 0

(ix) –12 by –9
      = (–12) × (– 9)
      = 108


(x) (–​746) by (–8)
      = (–746) × (–8)
      = 5968

(xi) 118 by −7
​= 118 × (-7)
= –826

(xii) −238 by −143
= (−238) × (−143)
= 34034


 

Page No 73:

Question 2:

Find the products:
(i) (−2) × 3 × (−4)
(ii) 2 × (−5) × (−6)
(iii) (−8) × 3 × 5
(iv) 8 × 7 × (−10)
(v) (−3) × (−7) × (−6)
(vi) (−8) × (−3) × (−9)

ANSWER:

(i)  (–2) × 3 × (–4)
   = [(–2) × 3] × (–4)
   = (–6) × (–4)
  = 24

(ii) 2 × (–5) × (–6)
     = [2 × (–5)] × (–6)
     = (–10) × (–6)
     = 60

(iii) (–8) × 3 × 5
     = [(–8) × 3] × 5
     = (–24) × 5
     = –120
(iv) 8 × 7 × (–10)
     = [8 × 7] × (–10)
     = 56 × (–10)
     = –560
(v)  (–3) × (–7) × (–6)
     = [(–3) × (–7)] × (–6)
     = 21 × (–6)
     = –126
(vi) (–8) × (–3) × (–9)
     = [(–8) × (–3)] × (–9)
     = 24 × (–9)
     = –216

Page No 73:

Question 3:

Use convenient groupings and find the values of
(i) 18 × (−27) × 30
(ii) (−8) × (−63) × 9
(iii) (−17) × (−23) × 41
(iv) (−51) × (−47) × (−19)

ANSWER:

(i) 18 × (–27) × 30
            = (–27) × [18 × 30]
            = (–27) × 540
            = –14580

(ii) (–8) × (–63) × 9
            = [(–8) × (–63)] × 9
            = 504 × 9
            = 4536

(iii) (–17) × (–23) × 41
            = [(–17) × (–23)] × 41
            = 391 × 41
            = 16031

(iv) (–51) × (–47) × (–19)
            = [(–51) × (–47)] × (–19)
            = 2397 × (–19)
            = – 45543

Page No 73:

Question 4:

Verify the following:
(i) 18 × [9 + (−7)] = 18 × 9 + 18 × (−7)
(ii) (−13) × [(−6) + (−19)] = (−13) × (−6) + (−13) × (−19)

ANSWER:

(i)
L.H.S.                                             
=18 × [9 + (–7)]                                
= 18 × [9 – 7]                                      
= 18 × 2                                              
= 36 
   R.H.S.
=18 × 9 + 18 × (–7)
= 162 – (18 × 7)  
= 162 – 126
= 36
∴∴ L.H.S = R.H.S
Hence, verified.

(ii) (–13) × [(–6) + (–19)] = (–13) × (–6) + (–13) × (–19)
L.H.S.                                                             
= (–13) × [(–6) + (–19)]                                
= (–13) × [–6 – 19]                          
= (–13) × (–25)                                              
= 325
R.H.S.
= (–13) × (–6) + (–13) × (–19)
= 78 + 247
= 325

∴∴ L.H.S = R.H.S
Hence, verified.

Page No 73:

Question 5:

Complete the following multiplication table:

x−3−2−10123
−3       
−2       
−1       
0       
1       
2       
3       

ANSWER:

×–3–2–10123
–39630–3–6–9
–26420–2–4–6
–13210–1–2–3
00000000
1–3–2–10123
2–6–4–20246
3–9–6–30369

Page No 73:

Question 6:

Which of the following statements are true and which are fals?
(i) The product of a positive integer and a negative integer is negative.
(ii) The product of two negative integers is a negative integer.
(iii) The product of three negative integers is a negative integer.
(iv) Every integer when multiplied with −1 gives its multiplicative inverse.

ANSWER:

(i) The product of a positive integer and a negative integer is negative.
            True

(ii) The product of two negative integers is a negative integer.
            False
The product of two negative integers is always a positive integer.

(iii) The product of three negative integers is a negative integer.
            True

(iv) Every integer when multiplied by (–1) gives its multiplicative inverse.
          False

Every integer when multiplied by (1) gives its multiplicative inverse.

Page No 73:

Question 7:

Simplify:
(i) (−9) × 6 + (−9) × 4
(ii) 8 × (−12) + 7 × (−12)
(iii) 30 × (−22) + 30 × (14)
(iv) (−15) × (−14) + (−15) × (−6)
(v) 43 × (−33) + 43 × (−17)
(vi) (−36) × (72) + (−36) × 28
(vii) (−27) × (−16) + (−27) × (−14)

ANSWER:

(i) (–9) × 6 + (–9) × 4
Solution:
Using the distributive law:
(–9) × 6 + (–9) × 4
= (–9) × (6+9)
= (–9) × 10
= –90

(ii) 8 × (–12) + 7 × (–12)
Solution:
Using the distributive law:
8 × (–12) + 7 × (–12)
= (–12) × (8+7)
= (–12) × 15
= –180

(iii) 30 × (–22) + 30 × (14)
Solution:
Using the distributive law:
30 × (–22) + 30 × (14)
= 30 × [(–22) + 14]
= 30 × [–22 + 14]
= 30 × (–8)
= –240

(iv) (–15) × (–14) + (–15) × (–6)
Solution:
(–15) × (–14) + (–15) × (–6)
Using the distributive law:
= (–15) × [ (–14) + (–6)]
= (–15) × [–14 – 6]
= (–15) × (–20)
= 300

(v) 43 × (–33) + 43 × (–17)
Solution:
43 × (–33) + 43 × (–17)
Using the distributive law:
= (43 ) × [–(33) + (–17)]
= (43 ) × [–33 – 17]
= 43 × (–50)
= –2150

(vi)  (–36) × (72) + (–36) × 28
Solution
(–36) × (72) + (–36) × 28
Using the distributive law:
 = (–36) × (72 + 28 )
 = (–36) × 100
 = –3600 

(vii) (–27) × (–16) + (–27) × (–14)
Solution:
(–27) × (–16) + (–27) × (–14)
Using the distributive law:
= (–27) × [(–16) + (–14)]
= (–27) × [–16 –14]
= (–27) × [–30]
= 810

Page No 75:

Exercise 4E

Question 1:

Divide:
(i) 85 by −17
(ii) −72 by 18
(iii) −80 by 16
(iv) −121 by 11
(v) 108 by −12
(vi) −161 by 23
(vii) −76 by −19
(viii) −147 by −21
(ix) −639 by −71
(x) −15625 by −125
(xi) 2067 by −1
(xii) 1765 by −1765
(xiii) 0 by −278
(xiv) 3000 by −100

ANSWER:

(i) 85 by –17

= −8517−8517
= –5

(ii) –72 by 18

=−7218−7218
= –4
(iii) –80 by 16

= −8016−8016
= –5
(iv) –121 by 11

=−12111−12111
 = –11

 (v) 108 by –12

=  108−12108−12
 = –9
(vi)  –161 by 23
= −16123−16123
= –7
(vii) –76 by –19

=−76−19−76−19
= 4
(viii) –147 by –21

= −147−21−147−21
= 7
(ix) –639 by –71

=−639−71=9=−639−71=9
(x) –639 by –71

=−639−71=9=−639−71=9
(x) –15625 by –125

=−15625−125=125=−15625−125=125


(xi) 2067 by –1

=2067−1=−2067=2067−1=−2067

(xii) 1765 by –1765

=1765−1765=−1×17651765= −1×1= −1=1765−1765=−1×17651765= -1×1= -1

(xiii) 0 by –278

 =0−278=0=0−278=0

(xiv) 3000 by –100

=3000−100=−30=3000−100=−30

Page No 75:

Question 2:

Fill in the blanks:
(i) 80 ÷ (……) = −5
(ii) (−84) + (……) = −7
(iii) (……) ÷ (−5) = 25
(iv) (……) ÷ 372 = 0
(v) (……) ÷ 1 = −186
(vi) (……) ÷ 17 = −2
(vii) (……) ÷ 165 = −1
(viii) (……) ÷ (−1) = 73
(ix) 1 ÷ (……) = −1

ANSWER:

(i) 80 ÷ (–16) = –5
(ii) (–84) ÷ (12) = –7
(iii) (–125) ÷ (–5) = 25
(iv) (0) ÷ (372) = 0
(v) (–186) ÷ 1 = –186
(vi) (–34) ÷ 17 = –2
(vii) (–165) ÷ 165 = –1
(viii) (–73) ÷ –1 = 73
(ix) 1 ÷ (–1) = –1

Page No 75:

Question 3:

Write (T) for true and (F) for false for each of the following statements:
(i) 0 ÷ (−6) = 0
(ii) (−8) ÷ 0 = 0
(iii) 15 ÷ (−1) = −15
(iv) (−16) ÷ (−4) = −4
(v) (−7) ÷ (−1) = 7
(vi) (−18) ÷ 9 = −2
(vii) 20 ÷ (−5) = −4
(viii) (−10) ÷ 1 = −10
(ix) (−1) ÷ (−1) = −1

ANSWER:

(i) True
(ii) False
This is because we cannot divide any integer by 0. If we do so, we get the quotient as infinity.
(iii) True
(iv) False
This is because the division of any two negative integers always gives a positive quotient.
(v) True
(vi) True
(vii) True
(viii) True
(ix) False
This is because the division of any two negative integers always gives a positive quotient.

Page No 75:

Question 1:

Which of the following is a true statement?
(a) −4 > −3
(b) −4 < −3
(c) −4 and −3 are non-comparable

ANSWER:

(b) –4 < –3
Since 4 is greater than 3, –4 is less than –3.

Page No 75:

Question 2:

2 less than −3 is
(a) −1
(b) 1
(c) −5
(d) 5

ANSWER:

(c) –5

2 less than –3 means the following:
= –3 – 2
= –5

Page No 76:

Question 3:

4 more than −5 is
(a) 9
(b) −9
(c) −1
(d) 1

ANSWER:

c) –1

4 more than –5 means the following:
= –5 + 4
= –1

Page No 76:

Question 4:

2 less than −7 is
(a) −9
(b) −5
(c) 5
(d) none of these

ANSWER:

(a) –9

2 less than −7 means the following:
= −7 − 2
= −9

Page No 76:

Question 5:

7 + |−3| = ?
(a) 4
(b) 10
(c) −10
(d) none of these

ANSWER:

(b) 10
7 + |-3|
= 7 + (+ 3)   (The absolute value of −3 is 3.)
= 7 + 3 
= 10

Page No 76:

Question 6:

(−42) + (−35) = ?
(a) −7
(b) 7
(c) −77
(d) none of these

ANSWER:

(c) –77
(−42) + (−35) 
= −42 − 35
= −77

Page No 76:

Question 7:

(−37) + 6 = ?
(a) −43
(b) −31
(c) 31
(d) none of these

ANSWER:

(b) –31
(−37) + 6
= −37 + 6
= −31

Page No 76:

Question 8:

49 + (−27) = ?
(a) −73
(b) 73
(c) 22
(d) none of these

ANSWER:

(c) 22
49 + (−27)
= 49 − 27
​= 22 

Page No 76:

Question 9:

The successor of −18 is
(a) −19
(b) 17
(c) −17
(d) 19

ANSWER:

(c) –17

In succession, we move from the left to the right of the number line.

Page No 76:

Question 10:

The predecessor of −16 is
(a) −15
(b) −17
(c) 15
(d) 17

ANSWER:

(b) –17
To find the predecessor of a number, we move from the right to the left of a number line. 

Page No 76:

Question 11:

The additive inverse of −5 is
(a) 5
(b) 0
(c) −4
(d) −6

ANSWER:

(a) 5
If we add the additive inverse of a number to the number, we get 0.
−5 + 5 = 0

Page No 76:

Question 12:

−12 − (−5) = ?
(a) −17
(b) −7
(c) 7
(d) none of these

ANSWER:

(b) –7
−12 − (−5) 
= −12 + 5
= −7

Page No 76:

Question 13:

−5 − (−8) = ?
(a) 3
(b) 13
(c) −3
(d) none of these

ANSWER:

(b) 13.5 − (−8) 
= 5 + 8
= 13

Page No 76:

Question 14:

The sum of two integers is −25. If one of them is 30 then the other is
(a) 55
(b) 5
(c) −55
(d) none of these

ANSWER:

 (c) –55
Let x be the other integer.
x + 30 = –25
⇒⇒ x = –25 – 30
⇒⇒ x = –55

Page No 76:

Question 15:

The sum of two integers is 20. If one of them is −5 then the other is
(a) 25
(b) −25
(c) 15
(d) none of these

ANSWER:

(a) 25

Let the other integer be x
x + (-5) = 20
⇒⇒x – 5 = 20
⇒⇒x = 25

​

Page No 76:

Question 16:

The sum of two integers is −13. If one of them is 8 then the other is
(a) −5
(b) −21
(c) 21
(d) none of these

ANSWER:

(b) −21

Let the other integer be x.
x + 8 = −13
=> x  = −13 − 8
=> x = −21

Page No 76:

Question 17:

On subtracting −8 from 0, we get
(a) −8
(b) 8
(c) none of these

ANSWER:

 (b) 8

0 − (−8)
= 0 + 8
= 8

Page No 76:

Question 18:

8 + (−8) = ?
(a) 16
(b) −16
(c) 0
(d) none of these

ANSWER:

(c) 0

8 + (−8) 
= 8 − 8 
= 0

Page No 76:

Question 19:

(−6) + 4 − (−3) = ?
(a) −5
(b) −1
(c) 1
(d) none of these

ANSWER:

(c) 1

(−6) + 4 − (−3)
= −6 + 4 + 3
= −6 + 7
= 1

Page No 76:

Question 20:

6 − (−4) = ?
(a) 2
(b) −10
(c) 10
(d) none of these

ANSWER:

(c) 10
6 − (−4) 
= 6 + 4 
= 10

Page No 76:

Question 21:

(−7) + (−9) + 12 + (−16) = ?
(a) −20
(b) 20
(c) −12
(d) none of these

ANSWER:

(a) –20
(−7) + (−9) + 12 + (−16)
= −7 − 9 + 12 −16 
= −20

Page No 76:

Question 22:

On subtracting 8 from −4, we get
(a) 4
(b) 12
(c) −12
(d) none of these

ANSWER:

(c) –12
–​4 –​ 8
= –​12

Page No 76:

Question 23:

On subtracting −9 from −6, we get
(a) −15
(b) −3
(c) 3
(d) none of these

ANSWER:

(c) 3

We have:
−6 − (−9)
= −6 + 9
= 3

Page No 76:

Question 24:

On subtracting −5 from 10, we get
(a) 5
(b) −15
(c) 15
(d) none of these

ANSWER:

(c) 15

We have:
 10  − (−5)
​= 10 + 5
= 15 

Page No 77:

Question 25:

(−6) × 9 = ?
(a) 54
(b) −54
(c) none of these

ANSWER:

(b) –54
We have:
(−6) × 9 
= −(6 × 9​)
= −54 

Page No 77:

Question 26:

(−9) × 6 + (−9) × 4 = ?
(a) −90
(b) 90
(c) −18
(d) 18

ANSWER:

(a) –90

(−9) × 6 + (−9) × 4 
Using distributive law:
 (−9) × (6 + 4)
= (−9) × (10)
= −90

Page No 77:

Question 27:

36 ÷ (−9) = ?
(a) 4
(b) −4
(c) none of these

ANSWER:

(b) –4

36 ÷ (−9) 

 36−9=369×(−1)= 1(−1)×369= −1 ×4= −436-9=369×(-1)= 1(-1)×369= -1 ×4= -4

Page No 78:

Exercise 4F

Question 1:

What are integers? Write all integers from −5 to 5.

ANSWER:

The numbers …–4, –3, –2, –1, 0, 1, 2, 3, 4… are integers.
The group of positive and negative numbers including 0 is called integers.

–5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5

Page No 78:

Question 2:

In each of the pairs given below, find the larger integer.
(i) 0, −3
(ii) −4, −6
(iii) −99, 9
(iv) −385, −615

ANSWER:

(i) 0, –3
0
This is because 0 is greater than any negative integer.

(ii) –4, –6
–4
Since 6 is greater than 4, –4 is greater than –6.

(iii) –99, 9
9
This is because every positive integer is greater than any negative integer.

(iv) –385, –615
–385
Since 615 is greater than 385, –385 is greater than –615.

Page No 78:

Question 3:

Write the following integers in increasing order:
−18, 16, 0, −5, 8, −36, −1, 1

ANSWER:

We can arrange the given integers in the increasing order in the following manner:
 –36, –18, –5, –1, 0, 1, 8, 16

Page No 78:

Question 4:

Find the value of:
(i) 9 − |−6|
(ii) 6 + |−4|
(iii) −8 − |−3|

ANSWER:

(i) 9 – |–6|
= 9 – (6)
= 9 – 6
= 3

(ii) 6 + |–4|
= 6 + (4)
= 6 + 4
= 10

(iii) –8 – |–3|
= –8 – 3
= –11

Page No 78:

Question 5:

Write four integers less than −6 and four integers greater tha −6.

ANSWER:

Four integers less than –6 (i.e. four negative integers that lie to the left of –6) are –7, –8, –9 and –10.
Four integers greater than –6 (i.e. four negative integers that lie to the right of –6 ) are –5, –4, –3 and –2.

Page No 78:

Question 6:

Evaluate:
(i) 8 + (−16)
(ii) (−5) + (−6)
(iii) (−6) × (−8)
(iv) (−36) ÷ 6
(v) 30 − (−50)
(vi) (−40) ÷ (−10)
(vii) 8 × (−5)
(viii) (−30) − 15

ANSWER:

(i) 8 + (–16)
= 8 – 16
= –8

(ii) (–5) + (–6)
= –5 – 6
= –11

(iii) (–6) × (–8)
       = (6 × 8)
       = 48

(iv) (–36) ÷ 6
    −366= (−1)×366= −6-366= (-1)×366= -6
(v) 
30 – (–50)
= 30 + 50
= 80

(vi) (–40) ÷ (–10)
=−40−10=(−1)×40(−1)×10=4=-40-10=(-1)×40(-1)×10=4

(vii) 8 × (–5)
= –(8 × 5)
= –40

(viii) (–30) – 15
= –30 – 15
= –45

Page No 78:

Question 7:

The sum of two integers is −12. If one of them is 34, find the other.

ANSWER:

Let the integer be x.
∴∴ 34 + x = –12
or x = –12 – 34
or x = –46
Therefore, the other integer is –46.

Page No 78:

Question 8:

Simplify:
(i) (−24) × (68) + (−24) × 32
(ii) (−9) × 18 − (−9) × 8
(iii) (−147) ÷ (−21)
(iv) 16 ÷ (−1)

ANSWER:

(i) (–24) × (68) + (–24) × 32
= –(24) × (68+32)
= –24 × 100
= –2400

(ii) (–9) × 18 – (–9) × 8
= –(9 ) × [18 – 8]
= –9 × 10
= –90

(iii) (–147) ÷ (–21)

=−147−21=(−1)×147(−1)×21=(−1)(−1)×14721=7=-147-21=(-1)×147(-1)×21=(-1)(-1)×14721=7

(iv) 16 ÷ (–1)

=16−1=16×(−1)(−1)×(−1)=16×(−1)=−16=16-1=16×(-1)(-1)×(-1)=16×(-1)=-16   {Multiplying the numerator and the denominator by (–1)}

Page No 78:

Question 9:

The successor of −89 is
(a) −90
(b) −88
(c) 90
(d) 88

ANSWER:

(b) −88
The successor of −89 is ​−88. The successor of a number lies towards its right on a number line. ​ 
−88 lies to the right of ​−89.  

Page No 78:

Question 10:

The predecessor of −99 is
(a) −98
(b) −100
(c) 98
(d) 100

ANSWER:

(b) ​−100
The predecessor of a number lies to the left of the number.
​​−100 lies to the left of −​99. Hence, ​​−100 is a predecessor of −​99.

Page No 78:

Question 11:

Additive inverse of −23 is
(a) −123-123
(b) 123123
(c) 23
(d) −23

ANSWER:

(c) ​23
Additive inverse of a number added to the number gives 0.
 −23 + 23 = 0
Hence, 23 is the additive inverse of  −23. 

Page No 78:

Question 12:

If (−13 + 6)     −25 − (−9)(-13 + 6)     -25 – (-9), then the correct symbol in the place holder is
(a) <
(b) >
(c) =
(d) none of these

ANSWER:

(b) > 


Here, L.H.S. = (−13 + 6) 
                   = −7


R.H.S. = −25 − (−9)
            = −25 + 9
​            = −16
  
−7 > −16

L.H.S. > R.H.S.

Page No 78:

Question 13:

? + (−8) = 12
(a) −4
(−20)
(c) 20
(d) 4

ANSWER:

 (c) 20 

x + (−8) = 12
=> x − 8 = 12
=> x = 12 + 8
=> x = 20

Page No 78:

Question 14:

The integer which is 5 more that (−7) is
(a) −12
(b) 12
(c) −2
(d) 2

ANSWER:

(c) -2

5 more than (−7) means 5 added to (−7).
   5 + (−7)
= 5 − 7
= −2

Page No 78:

Question 15:

What should be added to 16 to get (−31)?
(a) 15
(b) −15
(c) 47
(d) −47

ANSWER:

(d) −47
Let the number to be added to 16 be x.
x + 16 = (−31)
=> x = (−31)−16
=> x = −47

Page No 78:

Question 16:

When 34 is subtracted from −36, we get
(a) 2
(b) −2
(c) 70
(d) −70

ANSWER:

(d) −70
−36 ​− 34
= −70

Page No 78:

Question 17:

Fill in the blanks.
(i) −23 − (?) = 15.
(ii) The largest negative integer is …… .
(iii) The smallest positive integer is …… .
(iv) (−8) + (−6) − (−3) = …… .
(v) The predecessor of −200 is …… .

ANSWER:

(i)
Let the required number be x.
   −23 − x = 15
=> −23 = 15 + x
=> 15 + x = −23
=> x = −15 −23
=> x = −38 

(ii) 
The largest negative integer is -1.
 
(iii) 
The smallest positive integer is 1.

(iv) 
(−8) + (−6) − (−3) 
= (−8) + (−6) +3
= −8 ​−6 + 3
= −11

(v) 
The predecessor of −200:
(−200 − 1)
= −201 

Page No 79:

Question 18:

Write ‘T’ for true and ‘F’ for false in each of the following:
(i) 0 is neither positive nor negative.
(ii) −(−36) − 1 = − 37.
(iii) On the number line −10 lies to the right of −6.
(iv) 0 is an integer.
(v) −|−15| = −15.
(vi) |−40| + 40 = 0.

ANSWER:

(i) T
(ii) F 

 −(−36) − 1 
= 36  − 1​ 
= 35

(iii) F
This is because −10 is less than −6.

(iv) T

(v) T

−|−15| 
= ​−(15)
= −15

​(vi) F

|−40| + 40
= 40 + 40
= 80

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RS Agarwal Solution | Class 6th | Chapter-5 | Fractions | Edugrown

Exercise 5A

Question 1:

Write the fraction representing the shaded portion:
(i) Figure
(ii) Figure
(iii) Figure
(iv) Figure
(v) Figure
(vi) Figure

ANSWER:

(i) The shaded portion is 3 parts of the whole figure
     ∴∴ 3434            
(ii) The shaded portion is 1 parts of the whole figure
     ∴∴ 1414      
(iii) The shaded portion is 2 parts of the whole figure.
     ∴∴ 2323        
(iv) The shaded portion is 3 parts of the whole figure.
     ∴∴310310              
(v)The shaded portion is 4 parts of the whole figure.
     ∴∴4949          
(vi) The shaded portion is 3 parts of the whole figure.
     ∴∴ 3838

Page No 82:

Question 2:

Shade 4949 of the given figure.
Figure

ANSWER:

Page No 82:

Question 3:

In the given figure, if we say that the shaded region is 1414, then identify the error in it.
Figure

ANSWER:

The given rectangle is not divided into four equal parts.

Thus, the shaded region is not equal to 1414 of the whole.

Page No 82:

Question 4:

Write a fraction for each of the following:
(i) three-fourths
(ii) four-sevenths
(iii) two-fifths
(iv) three-tenths
(v) one-eighth
(vi) five-sixths
(vii) eight-ninths
(viii) seven-twelfths

ANSWER:

(i) 3434        (ii) 4747             (iii) 2525           (iv) 310310           (v) 1818
(vi) 5656             (vii)8989              (viii) 712712

Page No 83:

Question 5:

Write down the numerator and the denominator of each of the fractions given below:
(i) 4949
(ii) 611611
(iii) 815815
(iv) 12171217
(v) 5151

ANSWER:

     Numerator        Denominator
(i) 4                         9
(ii) 6                       11
(iii) 8                      15
(iv) 12                     17
(v) 5                        1

Page No 83:

Question 6:

Write down the fraction in which
(i) numerator = 3, denominator = 8
(ii) numerator = 5, denominator = 12
(iii) numerator = 7, denominator = 16
(iv) numerator = 8, denominator = 15

ANSWER:

(i)3838        (ii) 512512          (iii)716716             (iv) 815815

Page No 83:

Question 7:

Write down the fractional number for each of the following:
(i) 2323
(ii) 4949
(iii) 2525
(iv) 710710
(v) 1313
(vi) 3434
(vii) 3838
(viii) 914914
(ix) 511511
(x) 615615

ANSWER:

(i) two-thirds
(ii) four−-ninths
(iii) two−-fifths
(iv) seven−-tenths
(v) one−-thirds
(vi) three−-fourths
(vii) three−-eighths
(viii) nine−-fourteenths 
(ix) five−-elevenths
(x) six−-fifteenths

Page No 83:

Question 8:

What fraction of an hour is 24 minutes?

ANSWER:

We know: 1 hour = 60 minutes
∴ The required fraction = 2460=252460=25  

Page No 83:

Question 9:

How many natural numbers are there from 2 to 10? What fraction of them are prime numbers?

ANSWER:

There are total 9 natural numbers from 2 to 10. They are 2, 3, 4, 5, 6, 7, 8, 9, 10.
Out of these natural numbers, 2, 3, 5, 7 are the prime numbers.
∴ The required fraction = 4949.

Page No 83:

Question 10:

Determine:
(i) 2323 of 15 pens
(ii) 2323 of 27 balls
(iii) 2323 of 36 balloons

ANSWER:

(i) 2323 of 15 pens = (231×1551) = 10 pens231×1551 = 10 pens
(ii) 2323 of 27 balls = (231×2791) = 18 balls231×2791 = 18 balls
(iii) 2323 of 36 balloons = ​(231×36121) = 24 balloons231×36121 = 24 balloons

Page No 83:

Question 11:

Determine:
(i) 3434 of 16 cups
(ii) 3434 of 28 rackets
(iii) 3434 of 32 books

ANSWER:

(i) 3434 of 16 cups = (341 × 1641) = 12 cups341 × 1641 = 12 cups
(ii) 3434 of 28 rackets = (341 × 2871) = 21 rackets341 × 2871 = 21 rackets
(iii) 3434 of 32 books = (341 × 3281) = 24 books341 × 3281 = 24 books

Page No 83:

Question 12:

Neelam has 25 pencils. She gives 4545 of them to Meena. How many pencils does Meena get? How many pencils are left with Neelam?

ANSWER:

Neelam gives 4545 of 25 pencils to Meena.
 (451 × 2551) = 20 Pencils 451 × 2551 = 20 Pencils
Thus, Meena gets 20 pencils.
∴ Number of pencils left with Neelam = 25 −- 20 = 5 pencils
Thus, 5 pencils are left with Neelam.

Page No 83:

Question 13:

Represent each of the following fractions on the number line:
(i) 3838
(ii) 5959
(iii) 4747
(iv) 2525
(v) 1414

ANSWER:

Draw a 0 to 1 on a number line. Label point 1 as A and mark the starting point as 0.

(i) Divide the number line from 0 to 1 into 8 equal parts and take out 3 parts from it to reach point P.



(ii) Divide the number line from 0 to 1 into 9 equal parts and take out 5 parts from it to reach point P

.

(iii) Divide the number line from 0 to 1 into 7 equal parts and take out 4 parts from it to reach point P.



(Iv) Divide the number line from 0 to 1 into 5 equal parts and take out 2 parts from it to reach point P.



(v) Divide the number line from 0 to 1 into 4 equal parts and take out 1 part from it to reach point P.

Page No 85:

Exercise 5B

Question 1:

Which of the following are proper fractions?
12,35,10774, 2, 158,1616,1011,231012,35,10774, 2, 158,1616,1011,2310

ANSWER:

12, 35, 101112, 35, 1011

Page No 85:

Question 2:

Which of the following are improper fractions?
32,56,94,88, 3, 2716,2331,1918,1013,262632,56,94,88, 3, 2716,2331,1918,1013,2626

ANSWER:

A fraction whose numerator is greater than or equal to its denominator is called an improper fraction. Hence, 32, 94, 88, 2716, 1918 and 262632, 94, 88, 2716, 1918 and 2626 are improper fractions.

Page No 85:

Question 3:

Write six improper fractions with denominator 5.

ANSWER:

Clearly, 65, 75, 85, 95, 115and 12565, 75, 85, 95, 115and 125 are improper fractions, each with 5 as the denominator.

Page No 85:

Question 4:

Write six improper fractions with numerator 13.

ANSWER:

Clearly, 132, 133, 134, 135, 136, 137132, 133, 134, 135, 136, 137 are improper fractions, each with 13 as the numerator.

Page No 85:

Question 5:

Convert each of the following into an improper fraction:
(i) 557557
(ii) 938938
(iii) 63106310
(iv) 35113511
(v) 1091410914
(vi) 1271512715
(vii) 88138813
(viii) 51235123

ANSWER:

We have:
(i) 557 = (5 × 7) + 57 = 407557 = (5 × 7) + 57 = 407

(ii) 938 = (9 × 8) + 38 = 758938 = (9 × 8) + 38 = 758

(iii) 6310 = (6 × 10) + 310 = 63106310 = (6 × 10) + 310 = 6310

(iv) 3511 = (3 × 11) + 511 = 38113511 = (3 × 11) + 511 = 3811

(v) 10914 = (10 × 14) + 914 = 1491410914 = (10 × 14) + 914 = 14914

(vi) 12715 = (12 × 15) + 715 = 1871512715 = (12 × 15) + 715 = 18715

(vii) 8813 = (8 × 13) + 813 = 112138813 = (8 × 13) + 813 = 11213

(viii) 5123 = (51 × 3) + 23 = 15535123 = (51 × 3) + 23 = 1553

Page No 85:

Question 6:

Convert each of the following into a mixed fraction:
(i) 175175
(ii) 627627
(iii) 10181018
(iv) 95139513
(v) 81118111
(vi) 87168716
(vii) 1031210312
(viii) 1172011720

ANSWER:

(i) On dividing 17 by 5, we get:
    Quotient = 3
    Remainder = 2
   ∴ 175 =  3 +25 = 325  175 =  3 +25 = 325  

(ii) On dividing 62 by 7, we get:
    Quotient = 8
    Remainder = 6
   ∴ 627 =  8 +67 = 867  627 =  8 +67 = 867  

(iii) On dividing 101 by 8, we get:
    Quotient = 12
    Remainder = 5
   ∴ 1018 =  12 +58 = 1258  1018 =  12 +58 = 1258  

(iv) On dividing 95 by 13, we get:
    Quotient = 7
    Remainder = 4
   ∴ 9513 =  7 +413 = 7413  9513 =  7 +413 = 7413  

(v) On dividing 81 by 11, we get:
    Quotient = 7
    Remainder = 4
   ∴ 8111 =  7 +411 = 7411  8111 =  7 +411 = 7411  

(vi) On dividing 87 by 16, we get:
    Quotient = 5
    Remainder = 7
   ∴ 8716 =  5 +716 = 5716  8716 =  5 +716 = 5716  

(vii) On dividing 103 by 12, we get:
    Quotient = 8
    Remainder = 7
   ∴ 10312 =  8 +712 = 8712  10312 =  8 +712 = 8712  

(viii) On dividing 117 by 20, we get:
    Quotient = 5
    Remainder = 17
   ∴ 11720 =  5 +1720 = 51720  11720 =  5 +1720 = 51720  

Page No 85:

Question 7:

Fill up the blanks with ‘>’, ‘<‘ or ‘=’:
(i) 12      112      1
(ii) 34      134      1
(iii) 1      671      67
(iv) 66      166      1
(v) 30163016      130163016      1
(vi) 115      1115      1

ANSWER:

An improper fraction is greater than 1. Hence, it is always greater than a proper fraction, which is less than 1.
(i) 12  <  112  <  1

(ii) 34  <  134  <  1

(iii) 1  >  671  >  67

(iv) 66  =  166  =  1

(v) 30163016  =  130163016  =  1

(vi) 115  >  1115  >  1

Page No 86:

Question 8:

Draw number lines and locate the following points:
(i) 14, 12, 34, 4414, 12, 34, 44
(ii) 18, 28, 38, 58, 7818, 28, 38, 58, 78
(iii) 25, 35, 45, 8525, 35, 45, 85

ANSWER:

(i) Draw a number line. Mark 0 as the starting point and 1 as the ending point.
Then, divide 0 to 1 in four equal parts, where each part is equal to 1/4.
Show the consecutive parts as 1/4, 1/2, 3/4 and at 1 show 4/4 = 1.



(ii) Draw 0 to 1 on a number line. Divide the segment into 8 equal parts, each part corresponds to 1/8. Show the consecutive parts as 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8 and 8/8. Highlight the required ones only.



(iii) Draw 0 to 2 on a number line. Divide the segment between 0 and 1 into 5 equal parts, where each part is equal to 1/5.
Show 2/5, 3/5, 4/5 and 8/5 3 parts away from 1 towards 2. (1 < 8/5 < 2)

Page No 89:

Exercise 5C

Question 1:

Write five fractions equivalent to each of the following:
(i) 2323
(ii) 4545
(iii) 5858
(iv) 710710
(v) 3737
(vi) 611611
(vii) 7979
(viii) 512512

ANSWER:

(i) 23 =2×23×2 =  2×33×3=  2×43×4= 2×53×5 = 2×63×623 =2×23×2 =  2×33×3=  2×43×4= 2×53×5 = 2×63×6

   ∴ 23 = 46 = 69 = 812 = 1015 = 121823 = 46 = 69 = 812 = 1015 = 1218

Hence, the five fractions equivalent to 2323 are  46, 69, 812, 1015 and 1218 46, 69, 812, 1015 and 1218.


(ii) ​ 45 =4×25×2 =  4×35×3=  4×45×4= 4×55×5 = 4×65×645 =4×25×2 =  4×35×3=  4×45×4= 4×55×5 = 4×65×6

   ∴ 45 = 810 = 1215 = 1620 = 2025 = 243045 = 810 = 1215 = 1620 = 2025 = 2430

Hence, the five fractions equivalent to 4545 are  810, 1215, 1620, 2025 and 2430 810, 1215, 1620, 2025 and 2430.


(iii) ​ 58 =5×28×2 =  5×38×3=  5×48×4= 5×58×5 = 5×68×658 =5×28×2 =  5×38×3=  5×48×4= 5×58×5 = 5×68×6

   ∴ 58 = 1016 = 1524 = 2032 = 2540 = 304858 = 1016 = 1524 = 2032 = 2540 = 3048

Hence, the five fractions equivalent to 5858 are   1016, 1524, 2032, 2540 and 3048  1016, 1524, 2032, 2540 and 3048.



(iv) ​ 710 =7×210×2 =  7×310×3=  7×410×4= 7×510×5 = 7×610×6710 =7×210×2 =  7×310×3=  7×410×4= 7×510×5 = 7×610×6

   ∴ 710 = 1420 = 2130 = 2840 =  3550= 4260710 = 1420 = 2130 = 2840 =  3550= 4260

Hence, the five fractions equivalent to 710710 are  1420, 2130, 2840, 3550 and 4260 1420, 2130, 2840, 3550 and 4260.


(v) ​​ 37 =3×27×2 =  3×37×3=  3×47×4= 3×57×5 = 3×67×637 =3×27×2 =  3×37×3=  3×47×4= 3×57×5 = 3×67×6

   ∴ 37 = 614 = 921 = 1228 =  1535= 184237 = 614 = 921 = 1228 =  1535= 1842

Hence, the five fractions equivalent to 3737 are 614, 921, 1228,1535 and 1842614, 921, 1228,1535 and 1842.


(vi)  ​ 611 =6×211×2 =  6×311×3=  6×411×4= 6×511×5 = 6×611×6611 =6×211×2 =  6×311×3=  6×411×4= 6×511×5 = 6×611×6

   ∴ 611 = 1222 = 1833 = 2444 =  3055= 3666611 = 1222 = 1833 = 2444 =  3055= 3666

Hence, the five fractions equivalent to 611611 are  1222, 1833, 2444, 3055 and 3666 1222, 1833, 2444, 3055 and 3666.


(vii)  79 =7×29×2 =  7×39×3=  7×49×4= 7×59×5 = 7×69×679 =7×29×2 =  7×39×3=  7×49×4= 7×59×5 = 7×69×6

   ∴ 79 = 1418 = 2127 = 2836 =  3545= 425479 = 1418 = 2127 = 2836 =  3545= 4254

Hence, the five fractions equivalent to 7979 are  1418, 2127, 2836, 3545 and 4254 1418, 2127, 2836, 3545 and 4254.


(viii)  512 =5×212×2 =  5×312×3=  5×412×4= 5×512×5 = 5×612×6512 =5×212×2 =  5×312×3=  5×412×4= 5×512×5 = 5×612×6

   ∴ 512 = 1024 = 1536 = 2048 =  2560= 3072512 = 1024 = 1536 = 2048 =  2560= 3072

Hence, the five fractions equivalent to 512512 are 1024, 1536, 2048,2560 and 30721024, 1536, 2048,2560 and 3072.

Page No 89:

Question 2:

Which of the following are the pairs of equivalent fractions?
(i) 56 and 202456 and 2024
(ii) 38 and 154038 and 1540
(iii) 47 and 162147 and 1621
(iv) 29 and 146329 and 1463
(v) 13 and 92413 and 924
(vi) 23 and 332223 and 3322

ANSWER:

The pairs of equivalent fractions are as follows:
(i) 56 and 2024                         (2024 = 5×46×4)56 and 2024                         2024 = 5×46×4
(ii) 38 and 1540                         (1540 = 3×58×5)38 and 1540                         1540 = 3×58×5
(iv) 29 and 1463                         (1463 = 2×79×7)29 and 1463                         1463 = 2×79×7

Page No 89:

Question 3:

Find the equivalent fraction of 3535 having
(i) denominator 30
(ii) numerator 24

ANSWER:

(i) Let 35 = 30 35 = 30 
Clearly, 30 = 5 ×× 6
So, we multiply the numerator by 6.

∴ ​35 = 3×65×6= 183035 = 3×65×6= 1830
Hence, the required fraction is 18301830.
(ii)  ​Let 35 = 24 35 = 24 
   Clearly, 24 = 3 ×× 8
   So, we multiply the denominator by 8.

∴ ​35 = 3×85×8= 244035 = 3×85×8= 2440
Hence, the required fraction is 24402440.

Page No 89:

Question 4:

Find the equivalent fraction of 5959 having
(i) denominator 54
(ii) numerator 35

ANSWER:

(i) Let 59 = 54 59 = 54 
Clearly, 54 = 9 ×× 6
So, we multiply the numerator by 6.
∴ ​59 = 5×69×6= 305459 = 5×69×6= 3054
Hence, the required fraction is 30543054.
(ii)  ​Let 59 = 35 59 = 35 
   Clearly, 35 = 5 ×× 7
   So, we multiply the denominator by 7.
∴ ​59 = 5×79×7= 356359 = 5×79×7= 3563
Hence, the required fraction is 35633563.

Page No 89:

Question 5:

Find the equivalent fraction of 611611 having
(i) denominator 77
(ii) numerator 60

ANSWER:

(i) Let 611 = 77 611 = 77 
   Clearly, 77 = 11 ×× 7
   So, we multiply the numerator by 7.

∴ ​611 = 6×711×7= 4277611 = 6×711×7= 4277
Hence, the required fraction is 42774277.
(ii)  ​Let 611 = 60 611 = 60 
   Clearly, 60 = 6 ×× 10
   So, we multiply the denominator by 10.

∴ ​611 = 6×1011×10= 60110611 = 6×1011×10= 60110
Hence, the required fraction is 6011060110.

Page No 89:

Question 6:

Find the equivalent fraction of 24302430 having numerator 4.

ANSWER:

   Let 2430 = 4 2430 = 4 
   Clearly, 4 = 24 ÷÷ 6
   So, we divide the denominator by 6.
 ∴ ​2430 = 24÷630÷6= 452430 = 24÷630÷6= 45
  Hence, the required fraction is 4545.

Page No 89:

Question 7:

Find the equivalent fraction of 36483648 with
(i) numerator 9
(ii) denominator 4

ANSWER:

(i) Let 3648 = 9 3648 = 9 
   Clearly, 9 = 36 ÷÷ 4
   So, we divide the denominator by 4.
∴ ​3648 = 36÷448÷4= 9123648 = 36÷448÷4= 912
Hence, the required fraction is 912912.
(ii)  ​Let 3648 = 4 3648 = 4 
   Clearly, 4 = 48 ÷÷ 12
   So, we divide the numerator by 12.
∴ ​3648 = 36÷1248÷12= 343648 = 36÷1248÷12= 34
Hence, the required fraction is 3434.

Page No 89:

Question 8:

Find the equivalent fraction of 56705670 with
(i) numerator 4
(ii) denominator 10

ANSWER:

(i) Let 5670 = 4 5670 = 4 
   Clearly, 4 = 56 ÷÷ 14
   So, we divide the denominator by 14.
  ∴ ​5670 = 56÷1470÷14= 455670 = 56÷1470÷14= 45
  Hence, the required fraction is 4545.
(ii)  ​Let 5670 = 10 5670 = 10 
     Clearly, 10 = 70 ÷÷ 7
     So, we divide the numerator by 7.
   ∴ ​5670 = 56×770×7= 8105670 = 56×770×7= 810
   Hence, the required fraction is 810810.

Page No 89:

Question 9:

Reduce each of the following fractions into its simplest form:
(i) 915915
(ii) 48604860
(iii) 84988498
(iv) 1506015060
(v) 72907290

ANSWER:

(i) Here, numerator = 9 and denominator = 15
Factors of 9 are 1, 3 and 9.
Factors of 15 are 1, 3, 5 and 15.
Common factors of 9 and 15 are 1 and 3.
H.C.F. of 9 and 15 is 3.
∴ 915 =9÷315÷3 = 35915 =9÷315÷3 = 35
Hence, the simplest form of 915 is 35915 is 35.

(ii) Here, numerator = 48 and denominator = 60
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12.
H.C.F. of 48 and 60 is 12.
∴ 4860 =48÷1260÷12 = 454860 =48÷1260÷12 = 45
Hence, the simplest form of 4860 is 454860 is 45.

(iii) Here, numerator = 84 and denominator = 98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 42 and 84.
Factors of 98 are 1, 2, 7, 14, 49 and 98.
Common factors of 84 and 98 are 1, 2, 7 and 14.
H.C.F. of 84 and 98 is 14.
∴ 8498 =84÷1498÷14 = 678498 =84÷1498÷14 = 67
Hence, the simplest form of 8498 is 678498 is 67.

(iv) Here, numerator = 150 and denominator = 60
Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 75 and 150.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors of 150 and 60 are 1, 2, 3, 5, 6, 10, 15 and 30.
H.C.F. of 150 and 60 is 30.
∴ 15060 =150÷3060÷30 = 5215060 =150÷3060÷30 = 52
Hence, the simplest form of 15060 is 5215060 is 52.

(v) ​Here, numerator = 72 and denominator = 90
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.
Common factors of 72 and 90 are 1, 2, 3, 6, 9 and 18.
H.C.F. of 72 and 90 is 18.
∴ 7290 =72÷1890÷18 = 457290 =72÷1890÷18 = 45
Hence, the simplest form of 7290 is 457290 is 45.

Page No 89:

Question 10:

Show that each of the following fractions is in the simplest form:
(i) 811811
(ii) 914914
(iii) 25362536
(iv) 815815
(v) 21102110

ANSWER:

(i) Here, numerator = 8 and denominator = 11
    Factors of 8 are 1, 2, 4 and 8.
    Factors of 11 are 1 and 11.    Common factor of 8 and 11 is 1.
   Thus, H.C.F. of 8 and 11 is 1.
   Hence, 811811 is the simplest form.

(ii) Here, numerator = 9 and denominator = 14
    Factors of 9 are 1, 3 and 9.
    Factors of 14 are 1, 2, 7 and 14.   Common factor of 9 and 14 is 1.
   Thus, H.C.F. of 9 and 14 is 1.
   Hence, 914914 is the simplest form.

(iii) Here, numerator = 25 and denominator = 36
     Factors of 25 are 1, 5 and 25.
     Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.    Common factor of 25 and 36 is 1.
    Thus, H.C.F. of 25 and 36 is 1.
   Hence, 25362536 is the simplest form.

(iv) Here, numerator = 8 and denominator = 15
      Factors of 8 are 1, 2, 4 and 8.
      Factors of 15 are 1, 3, 5 and 15.      Common factor of 8 and 15 is 1.
     Thus, H.C.F. of 8 and 15 is 1.
     Hence, 815815 is the simplest form.
(v) Here, numerator = 21 and denominator = 10
     Factors of 21 are 1, 3, 7 and 21.
     Factors of 10 are 1, 2, 5 and 10.     Common factor of 21 and 10 is 1.
    Thus, H.C.F. of 21 and 10 is 1.
    Hence, 21102110 is the simplest form.

Page No 90:

Question 11:

Replace          by the correct number in each of the following:
(i) 27=8    27=8    
(ii) 35=    3535=    35
(iii) 58=20    58=20    
(iv) 4560=9    4560=9    
(v) 4056=    74056=    7
(vi) 4254=7    4254=7    

ANSWER:

(i) 28            (27 = 2×47×4 = 828)27 = 2×47×4 = 828
(ii) 21           (35 = 3×75×7 = 2135)35 = 3×75×7 = 2135
(iii) 32          (58 = 5×48×4 = 2032)58 = 5×48×4 = 2032
(iv) 12          (4560 = 45÷560÷5 = 912)4560 = 45÷560÷5 = 912
(v) 5             (4056 = 40÷856÷8 = 57)4056 = 40÷856÷8 = 57 
(vi) 9              (4254 = 42÷654÷6 = 79)4254 = 42÷654÷6 = 79

Page No 93:

Exercise 5D

Question 1:

Define like and unlike fractions and give five examples of each.

ANSWER:

Like fractions:
Fractions having the same denominator are called like fractions.
Examples: 311, 511, 711, 911, 1011311, 511, 711, 911, 1011

Unlike fractions:
Fractions having different denominators are called unlike fractions.
Examples: 34, 45, 67, 911, 21334, 45, 67, 911, 213

Page No 93:

Question 2:

Convert 35, 710, 815 and 113035, 710, 815 and 1130 into like fractions.

ANSWER:

The given fractions are 35, 710, 815 and 1130.35, 710, 815 and 1130.


L.C.M. of 5, 10, 15 and 30 = (5 ×× 2 ×× 3) = 30
So, we convert the given fractions into equivalent fractions with 30 as the denominator.
(But, one of the fractions already has 30 as its denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:
35 = 3×65×6 = 1830; 710 = 7×310×3 = 2130 ; 815 = 8×215×2 = 163035 = 3×65×6 = 1830; 710 = 7×310×3 = 2130 ; 815 = 8×215×2 = 1630

Hence, the required like fractions are 1830, 2130, 1630 and 1130.1830, 2130, 1630 and 1130.

Page No 93:

Question 3:

Convert 14, 58, 712 and 132414, 58, 712 and 1324 into like fractions.

ANSWER:

The given fractions are 14, 58, 712 and 1324 .14, 58, 712 and 1324 .
L.C.M. of 4, 8, 12 and 24 = (4 ×× 2 ×× 3) = 24
So, we convert the given fractions into equivalent fractions with 24 as the denominator.
(But one of the fractions already has 24 as the denominator. So, there is no need to convert it into an equivalent fraction.)
Thus, we have:
14 = 1×64×6 = 624; 58 = 5×38×3 = 1524 ; 712 = 7×212×2 = 142414 = 1×64×6 = 624; 58 = 5×38×3 = 1524 ; 712 = 7×212×2 = 1424

Hence, the required like fractions are 624, 1524, 1424 and 1324.624, 1524, 1424 and 1324.

Page No 93:

Question 4:

Fill in the place holders with the correct symbol > or <:
(i) 89      5989      59
(ii) 910      710910      710
(iii) 37      6737      67
(iv) 1115      8151115      815
(v) 611      511611      511
(vi) 1120      17201120      1720

ANSWER:

Between two fractions with the same denominator, the one with the greater numerator is the greater of the two.

(i) >
(ii) >
(iii) <
(iv) >
(v) >
(vi) <

Page No 93:

Question 5:

Fill in the place holders with the correct symbol > or <:
(i) 34      3534      35
(ii) 78      71078      710
(iii) 411      49411      49
(iv) 811      813811      813
(v) 512      58512      58
(vi) 1114      11151114      1115

ANSWER:

Between two fractions with the same numerator, the one with the smaller denominator is the greater of the two.

(i) >
(ii) >
(iii)<
(iv) >
(v) <
(vi) >

Page No 93:

Question 6:

Compare the fractions given below:
45, 5745, 57

ANSWER:

45,  5745,  57
By cross multiplying:
5 ×× 5 = 25 and 4 ×× 7 = 28     
Clearly, 28 > 25
∴∴ 45 > 5745 > 57

Page No 93:

Question 7:

Compare the fractions given below:
38, 5638, 56

ANSWER:

38,  5638,  56
By cross multiplying:
3 ×× 6 = 18 and 5 ×× 8 = 40        
Clearly, 18 < 40
∴∴ 38  <   5638  <   56

Page No 93:

Question 8:

Compare the fractions given below:
711, 67711, 67

ANSWER:

711 , 67711 , 67

By cross multiplying:
7 ×× 7 = 49 and 11 ×× 6 = 66        
Clearly, 49 < 66
∴∴ 711 <  67711 <  67

Page No 93:

Question 9:

Compare the fractions given below:
56, 91156, 911

ANSWER:

711 , 67711 , 67
By cross multiplying:
5 ×× 11 = 55 and 9 ×× 6 = 54         
Clearly, 55 > 54
∴∴ 56  >  91156  >  911

Page No 93:

Question 10:

Compare the fractions given below:
23, 4923, 49

ANSWER:

711 , 67711 , 67
By cross multiplying:
2 ×× 9 = 18 and 4 ×× 3 = 12     
Clearly, 18 > 12
∴∴ 23 > 4923 > 49

Page No 93:

Question 11:

Compare the fractions given below:
613, 34613, 34

ANSWER:

613 , 34613 , 34
By cross multiplying:
6 ×× 4 = 24 and 13 ×× 3 = 39      
Clearly, 24 < 39
∴∴ 613 < 34613 < 34

Page No 93:

Question 12:

Compare the fractions given below:
34, 5634, 56

ANSWER:

613, 34613, 34
By cross multiplying:
3 ×× 6 = 18 and 4 ×× 5 = 20     
Clearly, 18 < 20
∴∴ 34 < 5634 < 56

Page No 93:

Question 13:

Compare the fractions given below:
58, 71258, 712

ANSWER:

58 ,71258 ,712
By cross multiplying:
5 ×× 12 = 60 and 8 ×× 7 = 56     
Clearly, 60 > 56
∴∴ 58  > 71258  > 712

Page No 93:

Question 14:

Compare the fractions given below:
49, 5649, 56

ANSWER:

L.C.M. of 9 and 6 = (3 ×× 3 ×× 2) = 18
Now, we convert 49 and 5649 and 56 into equivalent fractions having 18 as the denominator. 
∴​ 49  = 4×29×2  = 818 and   56 = 5×36×3 = 151849  = 4×29×2  = 818 and   56 = 5×36×3 = 15184949
         
Clearly, 818 < 1518818 < 1518
∴∴ 49 < 5649 < 56

Page No 93:

Question 15:

Compare the fractions given below:
45, 71045, 710

ANSWER:

L.C.M. of 5 and 10 = (5 ×× 2) = 10
Now, we convert 45 45  into an equivalent fraction having 10 as the denominator as the other fraction has already 10 as its denominator.
∴​ 45  = 4×25×2  = 810 45  = 4×25×2  = 810 4949
         
Clearly, 810 > 710810 > 710
∴∴ 45 > 71045 > 710

Page No 93:

Question 16:

Compare the fractions given below:
78, 91078, 910

ANSWER:

L.C.M. of 8 and 10 = (2 ×× 5 ×× 2 ×× 2) = 40
Now, we convert 78 and 91078 and 910 into equivalent fractions having 40 as the denominator.
∴​ 78  = 7×58×5  = 3540 and 910  = 9×410×4  = 3640 78  = 7×58×5  = 3540 and 910  = 9×410×4  = 3640 4949
         
Clearly, 3540 < 36403540 < 3640
∴∴ 78 < 91078 < 910

Page No 93:

Question 17:

Compare the fractions given below:
1112, 13151112, 1315

ANSWER:

L.C.M. of 12 and 15 = (2 ×× 2 ×× 3 ×× 5) = 60
Now, we convert 1112 and 13151112 and 1315 into equivalent fractions having 60 as the denominator.
∴​ 1112  = 11×512×5  = 5560 and 1315  = 13×415×4  = 5260 1112  = 11×512×5  = 5560 and 1315  = 13×415×4  = 5260 4949
         
Clearly, 5560 > 52605560 > 5260
∴∴ 1112 > 13151112 > 1315

Page No 93:

Question 18:

Arrange the following fractions in ascending order:
12, 34, 56 and 7812, 34, 56 and 78

ANSWER:



The given fractions are 12, 34, 56 and 7812, 34, 56 and 78.
L.C.M. of 2, 4, 6 and 8 = (2 ×× 2 ×× 2 ×× 3) = 24
We convert each of the given fractions into an equivalent fraction with denominator 24.
Now, we have:
 12 = 1×122×12 = 1224; 34 = 3×64×6 = 182456 = 5×46×4 = 2024; 78 = 7×38×3 = 212412 = 1×122×12 = 1224; 34 = 3×64×6 = 182456 = 5×46×4 = 2024; 78 = 7×38×3 = 2124

Clearly, 1224 <1824 <2024 <21241224 <1824 <2024 <2124

∴ ​12 <34 <56 <7812 <34 <56 <78
Hence, the given fractions can be arranged in the ascending order as follows:
12, 34, 56, 7812, 34, 56, 78​

Page No 93:

Question 19:

Arrange the following fractions in ascending order:
23, 56, 79 and 111823, 56, 79 and 1118

ANSWER:

The given fractions are 23, 56, 79 and 1118.23, 56, 79 and 1118.


L.C.M. of 3, 6, 9 and 18 = (3 ×× 2  ×× 3) = 18
So, we convert each of the fractions whose denominator is not equal to 18 into an equivalent fraction with denominator 18.
Now, we have:
23 = 2×63×6 = 1218; 56 = 5×36×3 = 1518; 79 = 7×29×2 = 141823 = 2×63×6 = 1218; 56 = 5×36×3 = 1518; 79 = 7×29×2 = 1418
Clearly, 1118 <1218 <1418 <15181118 <1218 <1418 <1518
∴ ​1118 <23 <79 <561118 <23 <79 <56

Hence, the given fractions can be arranged in the ascending order as follows:
1118 ,23 ,79 ,561118 ,23 ,79 ,56

Page No 93:

Question 20:

Arrange the following fractions in ascending order:
25, 710, 1115 and 173025, 710, 1115 and 1730

ANSWER:

The given fractions are 25,710, 1115 and 1730.25,710, 1115 and 1730.
L.C.M. of 5, 10, 15 and 30 = (2 ×× 5 ×× 3) = 30  

                                         
So, we convert each of the fractions whose denominator is not equal to 30 into an equivalent fraction with denominator 30.
Now, we have:
25 = 2×65×6 = 1230; 710 = 7×310×3 = 2130; 1115 = 11×215×2 = 223025 = 2×65×6 = 1230; 710 = 7×310×3 = 2130; 1115 = 11×215×2 = 2230
Clearly, 1230 <1730 <2130 <22301230 <1730 <2130 <2230
∴ ​25 <1730 <710 <111525 <1730 <710 <1115

Hence, the given fractions can be arranged in the ascending order as follows:
25, 1730, 710, 111525, 1730, 710, 1115 

Page No 93:

Question 21:

Arrange the following fractions in ascending order:
34, 78, 1116 and 233234, 78, 1116 and 2332

ANSWER:

The given fractions are 34, 78, 1116 and 2332.34, 78, 1116 and 2332.
L.C.M. of 4, 8, 16 and 32 = (2 ⨯ 2 ⨯ 2 ⨯ 2 ⨯ 2) = 32    

                                      
So, we convert each of the fractions whose denominator is not equal to 32 into an equivalent fraction with denominator 32.
Now, we have:
34 = 3×84×8 = 2432; 78 = 7×48×4 = 2832; 1116 = 11×216×2 = 223234 = 3×84×8 = 2432; 78 = 7×48×4 = 2832; 1116 = 11×216×2 = 2232
Clearly, 2232 <2332 <2432 <28322232 <2332 <2432 <2832
∴ ​1116 <2332 <34 <781116 <2332 <34 <78

Hence, the given fractions can be arranged in the ascending order as follows:
1116, 2332, 34, 781116, 2332, 34, 78

Page No 93:

Question 22:

Arrange the following fractions in descending order:
34, 58, 1112 and 172434, 58, 1112 and 1724

ANSWER:

The given fractions are 34, 58, 1112 and 1724.34, 58, 1112 and 1724.
L.C.M. of 4, 8, 12 and 24 = (2 ⨯ 2 ⨯ 2 ⨯ 3) = 24            

                              
So, we convert each of the fractions whose denominator is not equal to 24 into an equivalent fraction with denominator 24.
Thus, we have;
34 = 3×64×6 = 1824; 58 = 5×38×3 = 1524; 1112 = 11×212×2 = 222434 = 3×64×6 = 1824; 58 = 5×38×3 = 1524; 1112 = 11×212×2 = 2224
Clearly, 2224 >1824 >1724 >15242224 >1824 >1724 >1524

∴ ​1112 >34 >1724 >581112 >34 >1724 >58

Hence, the given fractions can be arranged in the descending order as follows:
1112, 34, 1724, 581112, 34, 1724, 58

Page No 93:

Question 23:

Arrange the following fractions in descending order:
79, 512, 1118 and 173679, 512, 1118 and 1736

ANSWER:

The given fractions are 79, 512, 1118 and 1736.79, 512, 1118 and 1736.
L.C.M. of 9, 12, 18 and 36 = (3 ⨯ 3 ⨯ 2 ⨯ 2) = 36      

                                    
We convert each of the fractions whose denominator is not equal to 36 into an equivalent fraction with denominator 36.
Thus, we have:
79 = 7×49×4 = 2836; 512 = 5×312×3 = 1536; 1118 = 11×218×2 = 223679 = 7×49×4 = 2836; 512 = 5×312×3 = 1536; 1118 = 11×218×2 = 2236
Clearly, 2836 >2236 >1736 >15362836 >2236 >1736 >1536

∴ ​79 >1118 >1736 >51279 >1118 >1736 >512

Hence, the given fractions can be arranged in the descending order as follows:
79 ,1118,1736,51279 ,1118,1736,512

Page No 93:

Question 24:

Arrange the following fractions in descending order:
23, 35, 710 and 81523, 35, 710 and 815

ANSWER:

The given fractions are 23, 35, 710 and 815.23, 35, 710 and 815.
L.C.M. of 3, 5,10 and 15 = (2 ⨯ 3 ⨯ 5) = 30  

                                        
So, we convert each of the fractions into an equivalent fraction with denominator 30.
Thus, we have:
23 = 2×103×10 = 2030; 35 = 3×65×6 = 1830; 710 = 7×310×3 = 2130; 815 = 8×215×2 = 163023 = 2×103×10 = 2030; 35 = 3×65×6 = 1830; 710 = 7×310×3 = 2130; 815 = 8×215×2 = 1630
Clearly, 2130 >2030 >1830 >16302130 >2030 >1830 >1630
∴ ​710 >23 >35 >815710 >23 >35 >815

Hence, the given fractions can be arranged in the descending order as follows:
710 ,23 ,35 ,815710 ,23 ,35 ,815

Page No 93:

Question 25:

Arrange the following fractions in descending order:
57, 914, 1721 and 314257, 914, 1721 and 3142

ANSWER:

The given fractions are 57, 914, 1721 and 3142.57, 914, 1721 and 3142.
L.C.M. of 7, 14, 21 and 42 = (2 ⨯ 3 ⨯ 7) = 42                                          


We convert each one of the fractions whose denominator is not equal to 42 into an equivalent fraction with denominator 42.
Thus, we have:
57 = 5×67×6 = 3042; 914 = 9×314×3 = 2742; 1721 = 17×221×2 = 344257 = 5×67×6 = 3042; 914 = 9×314×3 = 2742; 1721 = 17×221×2 = 3442
Clearly, 3442 >3142 >3042 >27423442 >3142 >3042 >2742
∴ ​1721 >3142 >57 >9141721 >3142 >57 >914
Hence, the given fractions can be arranged in the descending order as follows:
1721,3142,57, 9141721,3142,57, 914

Page No 93:

Question 26:

Arrange the following fractions in descending order:
112, 123, 17, 19, 117, 150112, 123, 17, 19, 117, 150

ANSWER:

The given fractions are 112,123, 17, 19 , 117 and 150.112,123, 17, 19 , 117 and 150.
As the fractions have the same numerator, we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.
Clearly, 17 >19 >112 >117>123>15017 >19 >112 >117>123>150
Hence, the given fractions can be arranged in the descending order as follows:
17, 19, 112, 117, 123, 15017, 19, 112, 117, 123, 150

Page No 93:

Question 27:

Arrange the following fractions in descending order:
37, 311, 35, 313, 34, 31737, 311, 35, 313, 34, 317

ANSWER:

The given fractions are 37, 311, 35, 313, 34 and 317.37, 311, 35, 313, 34 and 317.
As the fractions have the same numerator, so we can follow the rule for the comparison of such fractions.
This rule states that when two fractions have the same numerator, the fraction having the smaller denominator is the greater one.

Clearly, 34 >35 >37 >311>313>31734 >35 >37 >311>313>317
Hence, the given fractions can be arranged in the descending order as follows:
34, 35, 37, 311, 313, 31734, 35, 37, 311, 313, 317 

Page No 94:

Question 28:

Lalita read 30 pages of a book containing 100 pages while Sarita read 2525 of the book. Who read more?

ANSWER:

Lalita read 30 pages of a book having 100 pages.
Sarita read 2525 of the same book.
 2525 of 100 pages = ​25 × 100 = 2005 = 40 pages25 × 100 = 2005 = 40 pages
Hence, Sarita read more pages than Lalita as 40 is greater than 30.

Page No 94:

Question 29:

Rafiq exercised for 2323 hour, while Rohit exercise for 3434 hour. Who exercised for a longer time?

ANSWER:

To know who exercised for a longer time, we have to compare 23 hour with 34 hour 23 hour with 34 hour .
On cross multiplying:
4 ×× 2 = 8 and 3 ×× 3 = 9
Clearly, 8 < 9
∴∴ 23 hour < 34 hour23 hour < 34 hour
Hence, Rohit exercised for a longer time.

Page No 94:

Question 30:

In a school 20 students out of 25 passed in VI A, while 24 out of 30 passed in VI B. Which section gave better result?

ANSWER:

Fraction of students who passed in VI A = 2025 = 20÷525÷5 = 452025 = 20÷525÷5 = 45

Fraction of students who passed in VI B = 2430 = 24÷630÷6 = 452430 = 24÷630÷6 = 45
In both the sections, the fraction of students who passed is the same, so both the sections have the same result.

Page No 96:

Exercise 5E

Question 1:

Find the sum:
58+1858+18

ANSWER:

The given fractions are like fractions.
We know:
Sum of like fractions  = Sum of the numeratorsCommon denominatorSum of the numeratorsCommon denominator
Thus, we have:
58 + 18 = (5+1) 8 = 6 384 = 3458 + 18 = 5+1 8 = 6 384 = 34

Page No 96:

Question 2:

Find the sum:
49+8949+89

ANSWER:

The given fractions are like fractions.
We know:
Sum of like fractions  = Sum of the numeratosCommon denominatorSum of the numeratosCommon denominator
Thus, we have:
49 + 89 = (4+8) 9 = 12493 = 43 = 11349 + 89 = 4+8 9 = 12493 = 43 = 113

Page No 96:

Question 3:

Find the sum:
135+245135+245

ANSWER:

The given fractions are like fractions.
We know:
Sum of like fractions  = Sum of the numeratorsCommon denominatorSum of the numeratorsCommon denominator
Thus, we have:
135 + 245 = 85 + 145 = (8+14) 5 = 225 = 425 135 + 245 = 85 + 145 = 8+14 5 = 225 = 425 

Page No 96:

Question 4:

Find the sum:
29+5629+56

ANSWER:

L.C.M. of 9 and 6 = (2 ×× 3 ×× 3) = 18                                   


Now, we have:

     29 = 2 × 29 × 2 = 418; 56 = 5 × 36 × 3 = 1518∴ 29 + 56 = 418 + 1518 = (4 + 15)18 = 1918 = 1118                                                                                                                                                                                            29 = 2 × 29 × 2 = 418; 56 = 5 × 36 × 3 = 1518∴ 29 + 56 = 418 + 1518 = 4 + 1518 = 1918 = 1118                                                                                                                                                                                       

Page No 96:

Question 5:

Find the sum:
712+916712+916

ANSWER:

L.C.M. of 12 and 16 = (2 ×× 2 ×× 2 ×× 2 ×× 3) = 48                                   


Now, we have:

     712 = 7 × 412 × 4 = 2848; 916 = 9 × 316 × 3 = 2748∴ 712 + 916 = 2848 + 2748 = (28 + 27)48 = 5548 = 1748                                                                                                                                                                                            712 = 7 × 412 × 4 = 2848; 916 = 9 × 316 × 3 = 2748∴ 712 + 916 = 2848 + 2748 = 28 + 2748 = 5548 = 1748                                                                                                                                                                                       

Page No 96:

Question 6:

Find the sum:
415+1720415+1720

ANSWER:

L.C.M. of 15 and 20 = (3 ×× 5 ×× 2 ×× 2) = 60                                   


     ∴ 415 + 1720 = (16 + 51)60     {[60 ÷ 15 = 4, 4 × 4 = 16] and [60 ÷ 20 = 3, 17 × 3 = 51]}                       = 6760 = 1760                                                                                                                                                                                            ∴ 415 + 1720 = 16 + 5160     60 ÷ 15 = 4, 4 × 4 = 16 and 60 ÷ 20 = 3, 17 × 3 = 51                       = 6760 = 1760                                                                                                                                                                                       

Page No 96:

Question 7:

Find the sum:
234+556234+556

ANSWER:

We have:                                              

                                                                                      
     234 + 556 = 114 + 356                                L.C.M. of 4 and 6 = (2 × 2 × 3) = 12  = (66 + 140)24                                   {[24 ÷ 4 = 6, 6 × 11 = 66] and [24 ÷ 6 = 4, 4 × 35 = 140]}    = 2061032412 = 10312 = 8712     234 + 556 = 114 + 356                                L.C.M. of 4 and 6 = (2 × 2 × 3) = 12  = 66 + 14024                                   24 ÷ 4 = 6, 6 × 11 = 66 and 24 ÷ 6 = 4, 4 × 35 = 140    = 2061032412 = 10312 = 8712
234+556

Page No 96:

Question 8:

Find the sum:
318+1512318+1512

ANSWER:

We have:
                                                                                                                                                       

     318 + 1512 = 258 + 1712                                L.C.M. of 8 and 12 = (2 × 2 × 2 × 3) = 24  = (75 + 34)24                                   {[24 ÷ 8 = 3, 3 × 25 = 75] and [24 ÷ 12 = 2, 2 × 17 = 34]}    = 10924  = 41324     318 + 1512 = 258 + 1712                                L.C.M. of 8 and 12 = (2 × 2 × 2 × 3) = 24  = 75 + 3424                                   24 ÷ 8 = 3, 3 × 25 = 75 and 24 ÷ 12 = 2, 2 × 17 = 34    = 10924  = 41324
234+556

Page No 96:

Question 9:

Find the sum:
2710+38152710+3815

ANSWER:

We have:

                                                                                                                                                  
     2710 + 3815 = 2710 + 5315                                L.C.M. of 10 and 15 = (2 × 3 × 5) = 30  = (81 + 106)30                                   {[30 ÷10 = 3, 3 × 27 = 81] and [30 ÷ 15 = 2, 2 × 53 = 106]}    = 18730  = 6730     2710 + 3815 = 2710 + 5315                                L.C.M. of 10 and 15 = (2 × 3 × 5) = 30  = 81 + 10630                                   30 ÷10 = 3, 3 × 27 = 81 and 30 ÷ 15 = 2, 2 × 53 = 106    = 18730  = 6730
234+556

Page No 96:

Question 10:

Find the sum:
323+156+2323+156+2

ANSWER:

We have:


                                                                                                                                                       
     323 + 156 + 2 = 113 + 116 + 21                             L.C.M. of 3 and 6 = (2 × 3) = 6  = (22 + 11 + 12)6                                   {[6 ÷ 3 = 2, 2 × 11 = 22], [6 ÷ 6 =1, 1 × 11 = 11] and [6 ÷ 1 = 6, 6 × 2 = 12]}    = 451562  = 152 = 712     323 + 156 + 2 = 113 + 116 + 21                             L.C.M. of 3 and 6 = (2 × 3) = 6  = 22 + 11 + 126                                   6 ÷ 3 = 2, 2 × 11 = 22, 6 ÷ 6 =1, 1 × 11 = 11 and 6 ÷ 1 = 6, 6 × 2 = 12    = 451562  = 152 = 712
234+556

Page No 96:

Question 11:

Find the sum:
3+1415+13203+1415+1320

ANSWER:

We have:

                                                                                                                                                       
     3 + 1415 + 1320  = 31 + 1915 + 2320                              L.C.M. of 15 and 20 = (2 × 2 × 3 × 5) = 60  = (180 + 76 + 69)60                                   {[60 ÷ 1 = 60, 60 × 3 = 180], [60 ÷ 15 = 4, 4 × 19 = 76] and [60 ÷ 20 =3, 3 × 23 = 69]}    =  325656012  = 6512 = 5512     3 + 1415 + 1320  = 31 + 1915 + 2320                              L.C.M. of 15 and 20 = (2 × 2 × 3 × 5) = 60  = 180 + 76 + 6960                                   60 ÷ 1 = 60, 60 × 3 = 180, 60 ÷ 15 = 4, 4 × 19 = 76 and 60 ÷ 20 =3, 3 × 23 = 69    =  325656012  = 6512 = 5512
234+556

Page No 96:

Question 12:

Find the sum:
313+414+616313+414+616

ANSWER:

We have:


                                                                                                                                                       
     313 + 414 + 616  = 103 + 174 + 376                              L.C.M. of 3, 4 and 6 = (2 × 2 × 3) = 12  = (40 + 51 + 74)12                                   {[12 ÷ 3 = 4, 4 × 10 = 40], [12 ÷ 4 = 3, 3 × 17 = 51] and [12 ÷ 6 =2, 2 × 37 = 74]}    =  16555124  = 554 = 1334     313 + 414 + 616  = 103 + 174 + 376                              L.C.M. of 3, 4 and 6 = (2 × 2 × 3) = 12  = 40 + 51 + 7412                                   12 ÷ 3 = 4, 4 × 10 = 40, 12 ÷ 4 = 3, 3 × 17 = 51 and 12 ÷ 6 =2, 2 × 37 = 74    =  16555124  = 554 = 1334
234+556

Page No 96:

Question 13:

Find the sum:
23+316+429+251823+316+429+2518

ANSWER:

We have:

                                                                                                                                                       
    23 + 316 + 429 + 2518  = 23 + 196 + 389 + 4118                              L.C.M. of 3, 6 and 9 = (2 × 3 × 3) = 18  = (12 + 57 + 76 + 41)18                                   {[18 ÷ 3 = 6, 6 × 2 = 12], [18 ÷ 6 = 3, 3 × 19 = 57], [18 ÷ 9 =2, 2 × 38 = 76] and [18 ÷ 18 = 1, 1 × 41= 41]}    =  18631183  = 313 = 1013    23 + 316 + 429 + 2518  = 23 + 196 + 389 + 4118                              L.C.M. of 3, 6 and 9 = (2 × 3 × 3) = 18  = 12 + 57 + 76 + 4118                                   18 ÷ 3 = 6, 6 × 2 = 12, 18 ÷ 6 = 3, 3 × 19 = 57, 18 ÷ 9 =2, 2 × 38 = 76 and 18 ÷ 18 = 1, 1 × 41= 41    =  18631183  = 313 = 1013
234+556

Page No 96:

Question 14:

Find the sum:
213+114+256+3712213+114+256+3712

ANSWER:

We have:
                                                                                                                                                       

    213 + 114 + 256 + 3712  = 73 + 54 + 176 + 4312                              L.C.M. of 3, 4, 6 and 12 = (2 × 2 × 3) = 12  = (28 + 15 + 34 + 43)12                                   {[12 ÷ 3 = 4, 4 × 7 = 28], [12 ÷ 4 = 3, 3 × 5 = 15], [12 ÷ 6 =2, 2 × 17 = 34] and [12 ÷ 12 = 1, 1 × 43 = 43] }    =  12010121  =  10    213 + 114 + 256 + 3712  = 73 + 54 + 176 + 4312                              L.C.M. of 3, 4, 6 and 12 = (2 × 2 × 3) = 12  = 28 + 15 + 34 + 4312                                   12 ÷ 3 = 4, 4 × 7 = 28, 12 ÷ 4 = 3, 3 × 5 = 15, 12 ÷ 6 =2, 2 × 17 = 34 and 12 ÷ 12 = 1, 1 × 43 = 43     =  12010121  =  10
234+556

Page No 96:

Question 15:

Find the sum:
2+34+158+37162+34+158+3716

ANSWER:

We have:
                                                                                                                                     

    2  + 34 + 158 + 3716  = 21 + 34 + 138 + 5516                              L.C.M. of 4, 8, and 16  =  (2 × 2 × 2 × 2) = 16  = (32 + 12 + 26 + 55)16                                   {[16 ÷ 1 = 16, 16 × 2 = 32], [16 ÷ 4 = 4, 4 × 3 = 12], [16 ÷ 8 =2, 2 × 13 = 26] and [16 ÷ 16 = 1, 1 × 55= 55]}    =  12516 =  71316    2  + 34 + 158 + 3716  = 21 + 34 + 138 + 5516                              L.C.M. of 4, 8, and 16  =  (2 × 2 × 2 × 2) = 16  = 32 + 12 + 26 + 5516                                   16 ÷ 1 = 16, 16 × 2 = 32, 16 ÷ 4 = 4, 4 × 3 = 12, 16 ÷ 8 =2, 2 × 13 = 26 and 16 ÷ 16 = 1, 1 × 55= 55    =  12516 =  71316
234+556

Page No 96:

Question 16:

Rohit bought a pencil for Rs 325325 and an eraser for Rs 27102710. What is the total cost of both the articles?

ANSWER:

Total cost of both articles = Cost of pencil + Cost of eraser 
Thus, we have:
   Rs 325 + Rs 2710 = 175 + 2710                               =  (34 + 27)10           (L.C.M. of 5 and 10 = (5 × 2) = 10)                                 = 6110 = Rs 6110   Rs 325 + Rs 2710 = 175 + 2710                               =  34 + 2710           (L.C.M. of 5 and 10 = (5 × 2) = 10)                                 = 6110 = Rs 6110
Hence, the total cost of both the articles is Rs 6110Rs 6110.

Page No 96:

Question 17:

Sohini bought 412m412m of cloth for her kurta and 223m223m of cloth for her pyjamas. Ho much cloth did she purchase in all?

ANSWER:

Total cloth purchased by Sohini = Cloth for kurta + Cloth for pyjamas
Thus, we have:
                                         (412 +  223 ) m = (92 + 83) m                (L.C.M. of 2 and 3 = (2 × 3) = 6)= ((27 + 16)6)  m                                 {[6 ÷ 2 = 3, 3 × 9 = 27] and [6 ÷ 3 = 2, 2 × 8 = 16]} = (436) m =  716 m 412 +  223  m = 92 + 83 m                (L.C.M. of 2 and 3 = (2 × 3) = 6)= 27 + 166  m                                 6 ÷ 2 = 3, 3 × 9 = 27 and 6 ÷ 3 = 2, 2 × 8 = 16 = 436 m =  716 m
∴∴ Total length of cloth purchased =  716 m 716 m

Page No 96:

Question 18:

While coming back home from his school, Kishan covered 434434 km by rickshaw and 112112 km on foot. What is the distance of his house from the school?

ANSWER:

Distance from Kishan’s house to school = Distance covered by him by rickshaw + Distance covered by him on foot
Thus, we have:
    (434 +  112 ) km =  (194 + 32) km                 = ((19  + 6)4)  km            (L.C.M .of 2 and 4 = (2 ×2) = 4)= (254) km =  614km 434 +  112  km =  194 + 32 km                 = 19  + 64  km            (L.C.M .of 2 and 4 = (2 ×2) = 4)= 254 km =  614km


Hence, the distance from Kishan’s house to school is  614 km 614 km.

Page No 96:

Question 19:

The weight of an empty gas cylinder is 16451645 kg and it contains 14231423 kg of gas. What is the weight of the cylinder filled with gas?

ANSWER:

Weight of the cylinder filled with gas = Weight of the empty cylinder + Weight of the gas inside the cylinder
Thus, we have:
   (1645 +  1423 ) kg =  (845 + 443) kg                (L.C.M. of 5 and 3 = (3 × 5) = 15)= ((252 + 220)15)  kg                                       = (47215) kg = 31715 kg 1645 +  1423  kg =  845 + 443 kg                (L.C.M. of 5 and 3 = (3 × 5) = 15)= 252 + 22015  kg                                       = 47215 kg = 31715 kg
Hence, the weight of the cylinder filled with gas is 31715 kg31715 kg.

Page No 99:

Exercise 5F

Question 1:

Find the difference:
58−1858-18

ANSWER:

Difference of like fractions = Difference of numerator ÷÷ Common denominator
58 − 18 = (5 − 1)8 = 4182 = 1258 – 18 = 5 – 18 = 4182 = 12

Page No 99:

Question 2:

Find the difference:
712−512712-512

ANSWER:

Difference of like fractions = Difference of numerator ÷÷ Common denominator
712 − 512 = (7 − 5)12 = 21126 = 16712 – 512 = 7 – 512 = 21126 = 16

Page No 99:

Question 3:

Find the difference:
437−247437-247

ANSWER:

Difference of like fractions = Difference of numerator ÷÷ Common denominator
437 − 247 = 317 − 187                    = (31 − 18)7                      = 137  437 – 247 = 317 – 187                    = 31 – 187                      = 137  

Page No 99:

Question 4:

Find the difference:
56−4956-49

ANSWER:


56 − 4956 – 49

 3  6, 9 2  2, 3 3  1, 3     1, 1 3  6, 9 2  2, 3 3  1, 3     1, 1
L.C.M. of 6 and 9 = (3 ×× 2 ×× 3) = 18
Now, we have:
56 = 5 × 36 × 3 = 1518; 49 = 4 × 29 × 2 = 818∴ 56 − 49 = 1518 − 818 = (15 − 8)18 = 71856 = 5 × 36 × 3 = 1518; 49 = 4 × 29 × 2 = 818∴ 56 – 49 = 1518 – 818 = 15 – 818 = 718

Page No 99:

Question 5:

Find the difference:
12−3812-38

ANSWER:

12 − 3812 – 38

L.C.M. of 2 and 8 = (2 ×× 2 ×× 2) = 8
Now, we have:
12 = 1 × 42 × 4 = 48 ∴ 12 − 38 = 48 − 38 = (4 − 3)8 = 1812 = 1 × 42 × 4 = 48 ∴ 12 – 38 = 48 – 38 = 4 – 38 = 18

Page No 99:

Question 6:

Find the difference:
58−71258-712

ANSWER:

58 − 71258 – 712

  2 8, 12  2 4, 6  2 2, 3   3 1, 3     1, 1  2 8, 12  2 4, 6  2 2, 3   3 1, 3     1, 1
L.C.M. of 8 and 12 = (2 ×× 2×× 2××3) = 24
Now, we have:
58 = 5 × 38 × 3 = 1524; 712 = 7 × 212 × 2 = 1424∴ 58 − 712 = 1524 − 1424 = (15 − 14)24 = 12458 = 5 × 38 × 3 = 1524; 712 = 7 × 212 × 2 = 1424∴ 58 – 712 = 1524 – 1424 = 15 – 1424 = 124

Page No 99:

Question 7:

Find the difference:
279−1815279-1815

ANSWER:

279 − 1815 = 259 − 2315 3  9, 15  3 3, 5  51, 5     1, 1L.C.M. of 9 and 15 =(3 × 3 × 5) = 45 ∴ 259 − 2315 =  (125 − 69)45 = 56 45  = 11145                                          {[45 ÷ 9 = 5, 5 × 25 = 125] and [45 ÷ 15 = 3, 3 × 23 = 69]}279 – 1815 = 259 – 2315 3  9, 15  3 3, 5  51, 5     1, 1L.C.M. of 9 and 15 =(3 × 3 × 5) = 45 ∴ 259 – 2315 =  125 – 6945 = 56 45  = 11145                                          45 ÷ 9 = 5, 5 × 25 = 125 and 45 ÷ 15 = 3, 3 × 23 = 69

Page No 99:

Question 8:

Find the difference:
358−2512358-2512

ANSWER:

358 − 2512 = 298 − 2912    2   8, 12    2  4, 6   2  2, 3    3  1, 3        1, 1  L.C.M. of 8 and 12 =(2 × 2 × 2 × 3) = 24 ∴ 298 − 2912 =  (87 − 58)24 = 29 24   = 1524                                          {[24 ÷ 8 = 3, 3 × 29 = 87] and  [24 ÷ 12 = 2, 2 × 29 = 58]}358 – 2512 = 298 – 2912    2   8, 12    2  4, 6   2  2, 3    3  1, 3        1, 1  L.C.M. of 8 and 12 =(2 × 2 × 2 × 3) = 24 ∴ 298 – 2912 =  87 – 5824 = 29 24   = 1524                                          24 ÷ 8 = 3, 3 × 29 = 87 and  24 ÷ 12 = 2, 2 × 29 = 58

Page No 99:

Question 9:

Find the difference:
2310−17152310-1715

ANSWER:

2310 − 1715 = 2310 − 2215      5 10, 15  2 2, 3  3 1, 3                                    1, 1  L.C.M. of 10 and 15 = (2 × 3 × 5) = 30= (69 − 44)30                               {[30 ÷ 10 = 3, 3 × 23 = 69] and [30 ÷ 15 = 2, 2 × 22 = 44]} = 255306  = 562310 – 1715 = 2310 – 2215      5 10, 15  2 2, 3  3 1, 3                                    1, 1  L.C.M. of 10 and 15 = (2 × 3 × 5) = 30= 69 – 4430                               30 ÷ 10 = 3, 3 × 23 = 69 and 30 ÷ 15 = 2, 2 × 22 = 44 = 255306  = 56

Page No 99:

Question 10:

Find the difference:
623−334623-334

ANSWER:

623 − 334  = 203 − 154                                   L.C.M. of 3 and 4 = (2 × 2 × 3) = 12                            = (80 − 45)12                               {[12 ÷ 3 = 4, 4 × 20 = 80] and [12 ÷ 4 = 3, 3 × 15 = 45]}= 3512 = 21112623 – 334  = 203 – 154                                   L.C.M. of 3 and 4 = (2 × 2 × 3) = 12                            = 80 – 4512                               12 ÷ 3 = 4, 4 × 20 = 80 and 12 ÷ 4 = 3, 3 × 15 = 45= 3512 = 21112 

Page No 99:

Question 11:

Find the difference:
7 − 5237 – 523

ANSWER:

7 − 523  = 71 − 173                                   L.C.M. of 1 and 3 = 3                       = (21 − 17)3                               {[3 ÷ 1 = 3, 3 × 7 = 21] and [3 ÷ 3 = 1, 1 × 17 = 17]}= 43 = 1137 – 523  = 71 – 173                                   L.C.M. of 1 and 3 = 3                       = 21 – 173                               3 ÷ 1 = 3, 3 × 7 = 21 and 3 ÷ 3 = 1, 1 × 17 = 17= 43 = 113

Page No 99:

Question 12:

Find the difference:
10 − 63810 – 638

ANSWER:

10 − 638  = 101 − 518                                   L.C.M. of 1 and 8 = 8                       = (80 − 51)8                               {[8 ÷ 1 = 8, 8 × 10 = 80] and [8 ÷ 8 = 1, 1 × 51 = 51]}= 298 = 35810 – 638  = 101 – 518                                   L.C.M. of 1 and 8 = 8                       = 80 – 518                               8 ÷ 1 = 8, 8 × 10 = 80 and 8 ÷ 8 = 1, 1 × 51 = 51= 298 = 358

Page No 99:

Question 13:

Simplify:
56−49+2356-49+23

ANSWER:

We have:

  56 − 49  + 23                                L.C.M. of 3, 6 and 9 =(2 × 3 × 3)  = 18                       = (15 − 8 + 12)18             {[18 ÷ 6 = 3, 3 × 5 = 15], [18 ÷ 9 = 2, 2 × 4 = 8] and [18 ÷ 3 = 6, 6 × 2 = 12]} = (27 − 8)18 =1918 =  1118  56 – 49  + 23                                L.C.M. of 3, 6 and 9 =2 × 3 × 3  = 18                       = 15 – 8 + 1218             18 ÷ 6 = 3, 3 × 5 = 15, 18 ÷ 9 = 2, 2 × 4 = 8 and 18 ÷ 3 = 6, 6 × 2 = 12 = 27 – 818 =1918 =  1118
3 3, 6, 93 1, 2, 32 1, 2, 1   1, 1, 13 3, 6, 93 1, 2, 32 1, 2, 1   1, 1, 1

Page No 99:

Question 14:

Simplify:
58+34−71258+34-712

ANSWER:

We have:                                                                                                                                       
  58 + 34 − 712        2 4, 8, 12  2 2, 4, 6  2 1, 2, 3   3 1, 1, 3        1,1, 1                                L.C.M. of 4, 8 and 12  =  (2 × 2 × 2 × 3) = 24= (15 + 18 −14)24                                   {[24 ÷ 8 = 3, 3 × 5 = 15], [24 ÷ 4 = 6, 6 × 3 = 18] and [24 ÷ 12 =2, 2 × 7 = 14]}  =  (33 − 14)24 = 1924   58 + 34 – 712        2 4, 8, 12  2 2, 4, 6  2 1, 2, 3   3 1, 1, 3        1,1, 1                                L.C.M. of 4, 8 and 12  =  (2 × 2 × 2 × 3) = 24= 15 + 18 -1424                                   24 ÷ 8 = 3, 3 × 5 = 15, 24 ÷ 4 = 6, 6 × 3 = 18 and 24 ÷ 12 =2, 2 × 7 = 14  =  33 – 1424 = 1924             
234+556

Page No 99:

Question 15:

Simplify:
2+1115−592+1115-59

ANSWER:

We have:                                                                                                                                            21 + 1115 − 59            3  1, 15, 9   3  1, 5, 3   5  1, 5, 1       1, 1, 1                           L.C.M. of 15  and 9 = (3 × 3 × 5) = 45  = (90 + 33 −25)45                                   {[45 ÷ 1 = 45, 45 × 2 = 90], [45 ÷ 15 = 3, 3 × 11 = 33] and [45 ÷ 9 =5, 5 × 5 = 25]}    =  (90 + 8)45 = 9845  = 2845  21 + 1115 – 59            3  1, 15, 9   3  1, 5, 3   5  1, 5, 1       1, 1, 1                           L.C.M. of 15  and 9 = (3 × 3 × 5) = 45  = 90 + 33 -2545                                   45 ÷ 1 = 45, 45 × 2 = 90, 45 ÷ 15 = 3, 3 × 11 = 33 and 45 ÷ 9 =5, 5 × 5 = 25    =  90 + 845 = 9845  = 2845
234+556

Page No 99:

Question 16:

Simplify:
534−4512+316534-4512+316

ANSWER:

We have:                                                                                                                                     
  534 − 4512 + 3 16   =  234 − 5312 + 196                     L.C.M. of 4, 12  and 6 = (2 × 2 × 3) = 12 2 4, 12, 6 2 2, 6, 3  3 1, 2, 3   2 1, 2, 1      1, 1, 1   = (69 − 53 + 38)12                                                                   {[12 ÷ 4 =3, 3 × 23 = 69], [12 ÷ 12 =1, 1 × 53 = 53] and [12 ÷ 6 =2, 2 × 19 = 38]}    =  (107 − 53)12 = 5412  =92 = 412  534 – 4512 + 3 16   =  234 – 5312 + 196                     L.C.M. of 4, 12  and 6 = (2 × 2 × 3) = 12 2 4, 12, 6 2 2, 6, 3  3 1, 2, 3   2 1, 2, 1      1, 1, 1   = 69 – 53 + 3812                                                                   12 ÷ 4 =3, 3 × 23 = 69, 12 ÷ 12 =1, 1 × 53 = 53 and 12 ÷ 6 =2, 2 × 19 = 38    =  107 – 5312 = 5412  =92 = 412
234+556

Page No 99:

Question 17:

Simplify:
2+5710−314152+5710-31415

ANSWER:

We have:                                                                                                                                           2 + 5710 −3 1415   =  21 + 5710 − 5915    5 1, 10, 15  2 1, 2, 3 3  1, 1, 3        1, 1, 1             L.C.M. of 10  and 15 = (2 × 5 × 3) = 30  = (60 + 171 −118)30                                                         {[30 ÷ 1 =30, 30 × 2 = 60], [30 ÷ 10 =3, 3 × 57 = 171] and [30 ÷ 15 =2, 2 × 59 = 118]}    =  (231 −118)30 = 11330 = 32330  2 + 5710 -3 1415   =  21 + 5710 – 5915    5 1, 10, 15  2 1, 2, 3 3  1, 1, 3        1, 1, 1             L.C.M. of 10  and 15 = (2 × 5 × 3) = 30  = 60 + 171 -11830                                                         30 ÷ 1 =30, 30 × 2 = 60, 30 ÷ 10 =3, 3 × 57 = 171 and 30 ÷ 15 =2, 2 × 59 = 118    =  231 -11830 = 11330 = 32330

Page No 99:

Question 18:

Simplify:
8−312−2148-312-214

ANSWER:

We have:                                                                                                                                    
  8 − 312 −214   =  81 − 72 − 94   2 1, 2, 4  2 1, 1, 2     1, 1, 1                     L.C.M. of 1, 2 and 4 = (2 × 2) = 4  = (32 − 14 − 9)4                                                         {[4 ÷ 1 =4, 4 × 8 = 32], [4 ÷ 2 =2, 2 × 7 = 14] and [4 ÷ 4 =1, 1 × 9 = 9]}    =  (32 − 23)4 = 94 = 214  8 – 312 -214   =  81 – 72 – 94   2 1, 2, 4  2 1, 1, 2     1, 1, 1                     L.C.M. of 1, 2 and 4 = (2 × 2) = 4  = 32 – 14 – 94                                                         4 ÷ 1 =4, 4 × 8 = 32, 4 ÷ 2 =2, 2 × 7 = 14 and 4 ÷ 4 =1, 1 × 9 = 9    =  32 – 234 = 94 = 214

Page No 99:

Question 19:

Simplify:
856−338+2712856-338+2712

ANSWER:

We have:                                                                                                                                    
  856 − 338 + 2712   =  536 − 278 + 3112      2  6, 8, 12 2 3, 4, 6 3 3, 2, 3  2 1, 2, 1        1, 1, 1           L.C.M. of 6, 8 and 12 = (2 × 2 × 2 × 3 ) = 24  = (212 − 81 + 62)24                                                         {[24 ÷ 6 =4, 4 × 53 = 212], [24 ÷ 8 =3, 3 × 27 = 81] and [24 ÷ 12 =2, 2 × 31 = 62]}    =  (274 − 81)24 = 19324 = 8124  856 – 338 + 2712   =  536 – 278 + 3112      2  6, 8, 12 2 3, 4, 6 3 3, 2, 3  2 1, 2, 1        1, 1, 1           L.C.M. of 6, 8 and 12 = (2 × 2 × 2 × 3 ) = 24  = 212 – 81 + 6224                                                         24 ÷ 6 =4, 4 × 53 = 212, 24 ÷ 8 =3, 3 × 27 = 81 and 24 ÷ 12 =2, 2 × 31 = 62    =  274 – 8124 = 19324 = 8124

Page No 99:

Question 20:

Simplify:
616−515+313616-515+313

ANSWER:

We have:                                                                                                                                     
  616 − 515 + 313      =  376 − 265 + 103        2  6, 5, 33  3, 5, 3   5 1, 5, 1       1, 1, 1      L.C.M. of 6, 5 and 3 = (2 × 5 × 3) = 30  = (185 − 156 + 100)30                         {[30 ÷ 6 =5, 5 × 37 = 185], [30 ÷ 5 =6, 6 × 26 = 156], and [30 ÷ 3 =10, 10 × 10 = 100]}  =  (285 − 156)30 = 129433010   =  4310  616 – 515 + 313      =  376 – 265 + 103        2  6, 5, 33  3, 5, 3   5 1, 5, 1       1, 1, 1      L.C.M. of 6, 5 and 3 = (2 × 5 × 3) = 30  = 185 – 156 + 10030                         30 ÷ 6 =5, 5 × 37 = 185, 30 ÷ 5 =6, 6 × 26 = 156, and 30 ÷ 3 =10, 10 × 10 = 100  =  285 – 15630 = 129433010   =  4310

Page No 99:

Question 21:

Simplify:
3+115+23−7153+115+23-715

ANSWER:

We have:                                                                                                                                        
  3 + 115 + 23 −715      =  31 + 65 + 23 − 715          5  5, 3, 15  3 1, 3, 3     1, 1, 1              L.C.M. of 1, 5, 3 and 15 = (5 × 3 ) =15  = (45 + 18 + 10 − 7)15                         {[15 ÷ 1 =15, 15 × 3 = 45], [15 ÷ 5 =3, 3 × 6 = 18], [15 ÷ 3 = 5, 5 × 2 = 10] and [15 ÷ 15 = 1, 1 × 7 = 7]}  =  (73 − 7)15 = 6622155   =225   = 425  3 + 115 + 23 -715      =  31 + 65 + 23 – 715          5  5, 3, 15  3 1, 3, 3     1, 1, 1              L.C.M. of 1, 5, 3 and 15 = (5 × 3 ) =15  = 45 + 18 + 10 – 715                         15 ÷ 1 =15, 15 × 3 = 45, 15 ÷ 5 =3, 3 × 6 = 18, 15 ÷ 3 = 5, 5 × 2 = 10 and 15 ÷ 15 = 1, 1 × 7 = 7  =  73 – 715 = 6622155   =225   = 425

Page No 99:

Question 22:

What should be added to 923923 to get 19?

ANSWER:

Let x be added to 923923 to get 19.

 ∴ 923 + x = 19Thus, we have:  x = 19 − 923      = 191 − 293                           L.C.M. of 1 and 3 is 3.       =(57 − 29)3                                 {[3 ÷ 1 = 3, 3 × 19 = 57] and [3 ÷ 3 = 1, 1 × 29 = 29]}       = 283 = 913∴ 923 + x = 19Thus, we have:  x = 19 – 923      = 191 – 293                           L.C.M. of 1 and 3 is 3.       =57 – 293                                 3 ÷ 1 = 3, 3 × 19 = 57 and 3 ÷ 3 = 1, 1 × 29 = 29       = 283 = 913923

Page No 99:

Question 23:

What should be added to 67156715 to get 815815?

ANSWER:

Let x be added to 67156715 to get 815815.
 ∴ 6715 + x = 815Therefore, we have: x  =  815 − 6715      = 415 − 9715                           L.C.M. of 5 and 15 = (5 × 3) = 15      =(123 − 97)15                                 {[15 ÷ 5 = 3, 3 × 41 = 123] and [15 ÷ 15 = 1, 1 × 97 = 97]}      = 2615 = 11115∴ 6715 + x = 815Therefore, we have: x  =  815 – 6715      = 415 – 9715                           L.C.M. of 5 and 15 = 5 × 3 = 15      =123 – 9715                                 15 ÷ 5 = 3, 3 × 41 = 123 and 15 ÷ 15 = 1, 1 × 97 = 97      = 2615 = 11115

Page No 99:

Question 24:

Subtract the sum of 359359 and 313313 from the sum of 556556 and 419419.

ANSWER:

    (556 + 419) − (359 + 313) =(356 + 379)  −(329 + 103)      2 6, 9, 3  3 3, 9, 3  3 1, 3, 1       1, 1, 1                    L.C.M. of 3, 6, 9 = (2 × 3 × 3) = 18 = [105 + 74] − [64 + 60]18                        {[18 ÷ 6 = 3, 3 × 35 = 105] and [18 ÷ 9 = 2, 2 × 37 = 74]}                                                                               {[18 ÷ 9 = 2, 2 × 32 = 64] and [18 ÷ 3 = 6, 6 × 10 = 60]} = [179] − [124]18 = 5518 = 3118    556 + 419 – 359 + 313 =356 + 379  -329 + 103      2 6, 9, 3  3 3, 9, 3  3 1, 3, 1       1, 1, 1                    L.C.M. of 3, 6, 9 = 2 × 3 × 3 = 18 = 105 + 74 – 64 + 6018                        18 ÷ 6 = 3, 3 × 35 = 105 and 18 ÷ 9 = 2, 2 × 37 = 74                                                                               18 ÷ 9 = 2, 2 × 32 = 64 and 18 ÷ 3 = 6, 6 × 10 = 60 = 179 – 12418 = 5518 = 3118

Page No 99:

Question 25:

Of 3434 and 5757, which is greater and by how much?

ANSWER:

Let us compare 34 and 5734 and 57.
3 ×× 7 = 21 and 4 ×× 5 = 20
Clearly, 21 > 20
∴ 34 > 5734 > 57
Required difference:
 = 34 − 57                              L.C.M. of 4 and 7 = (2 × 2 × 7) = 28= 21 − 2028                              {[28 ÷ 4 = 7, 7 × 3 = 21] and [28 ÷ 7 = 4, 4 × 5 = 20]}= 128= 34 – 57                              L.C.M. of 4 and 7 = 2 × 2 × 7 = 28= 21 – 2028                              28 ÷ 4 = 7, 7 × 3 = 21 and 28 ÷ 7 = 4, 4 × 5 = 20= 128
Hence, 34 is greater than 57 by 12834 is greater than 57 by 128.

Page No 99:

Question 26:

Mrs Soni bought 712712 litres of milk. Out of this milk, 534534 litres was consumed. How much milk is left with her?

ANSWER:

Amount of milk left with Mrs. Soni = Total amount of milk bought by her −- Amount of milk consumed
∴∴ Amount of milk left with Mrs. Soni =  712 − 534 = 152 − 234                           L.C.M. of 2 and 4 = (2 × 2) = 4= (30 − 23)4                               {[4 ÷ 2 = 2, 2 × 15 = 30] and [4 ÷ 4 = 1, 1 × 23 = 23]} = 74 = 134 litres=  712 – 534 = 152 – 234                           L.C.M. of 2 and 4 = 2 × 2 = 4= 30 – 234                               4 ÷ 2 = 2, 2 × 15 = 30 and 4 ÷ 4 = 1, 1 × 23 = 23 = 74 = 134 litres

∴∴ Milk left with Mrs. Soni = 134 litres134 litres

Page No 99:

Question 27:

A film show lasted for 313313 hours. Out of his time, 134134 hours was spent on advertisements. What was the actual duration of the film?

ANSWER:

Actual duration of the film = Total duration of the show −- Time spent on advertisements
                                          =(313 − 134) hours  =(103 − 74) hours                         L.C.M. of 3 and 4 = (2 × 2 × 3) = 12  =(40 − 2112) hours                               {[12 ÷ 3 = 4, 4 × 10 = 40] and [12 ÷ 4 = 3, 3 × 7 = 21]} = (1912) hours = 1712 hours =313 – 134 hours  =103 – 74 hours                         L.C.M. of 3 and 4 = 2 × 2 × 3 = 12  =40 – 2112 hours                               12 ÷ 3 = 4, 4 × 10 = 40 and 12 ÷ 4 = 3, 3 × 7 = 21 = 1912 hours = 1712 hours
Thus, the actual duration of the film was 1712 hours1712 hours.

Page No 99:

Question 28:

In one day, a rickshaw puller earned Rs 1371213712. Out of this money, he spent Rs 56345634 on food. How much money is left with him?

ANSWER:

Money left with the rickshaw puller = Money earned by him in a day −- Money spent by him on food
  = Rs (13712 − 5634)                     L.C.M. of 2 and 4=(2 × 2) = 4 = Rs (2752 − 2274)                          {[4 ÷ 2 = 2, 2 × 275 = 550] and [4 ÷ 4 = 1, 1 × 227 = 227]}= Rs (550 − 2274) = Rs (3234) = Rs 8034  = Rs 13712 – 5634                     L.C.M. of 2 and 4=2 × 2 = 4 = Rs 2752 – 2274                          4 ÷ 2 = 2, 2 × 275 = 550 and 4 ÷ 4 = 1, 1 × 227 = 227= Rs 550 – 2274 = Rs 3234 = Rs 8034 
Hence, Rs 80348034 is left with the rickshaw puller.

Page No 99:

Question 29:

A piece of wire, 234234 metres long, broke into two pieces. One piece is 5858 metre long. How long is the other piece?

ANSWER:

The length of the other piece = (Length of the wire −- Length of one piece)
   = (234 − 58) m =(114 − 58) m                        L.C.M. of 4 and 8 =(2 × 2 × 2) = 8 = (22 − 58) m                  {[8 ÷ 4 = 2, 2 × 11= 22] and [8 ÷ 8 = 1, 1 × 5 = 5]}=(178) m = 218 m = 234 – 58 m =114 – 58 m                        L.C.M. of 4 and 8 =2 × 2 × 2 = 8 = 22 – 58 m                  8 ÷ 4 = 2, 2 × 11= 22 and 8 ÷ 8 = 1, 1 × 5 = 5=178 m = 218 m
Hence, the other piece is 218 m218 m long.

Page No 99:

Exercise 5G

Question 1:

A fraction equivalent to 3535 is
(a) 3+25+23+25+2
(b) 3−25−23-25-2
(c) 3×25×23×25×2
(d) none of these

ANSWER:

(c) 3 × 25 × 23 × 25 × 2

Page No 99:

Question 2:

A fraction equivalent to 812812 is
(a) 8+412+48+412+4
(b) 8−412−48-412-4
(c) 8÷412÷48÷412÷4
(d) none of these

ANSWER:

(c) 8 ÷ 412 ÷ 48 ÷ 412 ÷ 4

Page No 100:

Question 3:

A fraction equivalent to 24362436 is
(a) 3434
(b) 2323
(c) 8989
(d) none of these

ANSWER:

 (b) 23    Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.Common factors of 24 and 36 are 1, 2, 3, 4, 6, 12.H.C.F. =12Dividing both the numerator and the denominator by 12:                                       2436 = 24 ÷ 1236 ÷ 12 = 23(b) 23    Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.Common factors of 24 and 36 are 1, 2, 3, 4, 6, 12.H.C.F. =12Dividing both the numerator and the denominator by 12:                                       2436 = 24 ÷ 1236 ÷ 12 = 23

Page No 100:

Question 4:

If 3434 is equivalent to x20x20 then the value of x is
(a) 15
(b) 18
(c)12
(d) none of these

ANSWER:

(a) 15                              

  Explanation: 
(34 = x20)                                    We have: 20 = 4 × 5So, we have to multiply the numerator by 5. ∴ x = 3 × 5 = 1534 = x20                                    We have: 20 = 4 × 5So, we have to multiply the numerator by 5. ∴ x = 3 × 5 = 15

Page No 100:

Question 5:

If 45604560 is equivalent to 3x3x then the value of x is
(a) 4
(b) 5
(c) 6
(d) none of these

ANSWER:

(a) 4

  Explanation: 
(4560 = 3x)                                  Now, 3 = 45 ÷15So, we have to divide the denominator by 15. ∴ x = 60 ÷ 15 = 44560 = 3x                                  Now, 3 = 45 ÷15So, we have to divide the denominator by 15. ∴ x = 60 ÷ 15 = 4

Page No 100:

Question 6:

Which of the following are like fractions?
(a) 25, 27, 29, 21125, 27, 29, 211
(b) 23, 34, 45, 5623, 34, 45, 56
(c) 18, 38, 58, 7818, 38, 58, 78
(d) none of these

ANSWER:

(c) 18, 38, 58, 7818, 38, 58, 78       
     (Fractions having the same denominator are called like fractions.)

Page No 100:

Question 7:

Which of the following is a proper fraction?
(a) 5353
(b) 5
(c) 125125
(d) none of these

ANSWER:

(d) none of these
In a proper fraction, the numerator is less than the denominator.

Page No 100:

Question 8:

Which of the following is a proper fractions?
(a) 7878
(b) 178178
(c) 8787
(d) none of these

ANSWER:

(a) 7878
In a proper fraction, the numerator is less than the denominator.

Page No 100:

Question 9:

Which of the following statements is correct?
(a) 34<3534<35
(b) 34>3534>35
(c) 3434 and 3535 cannot be compared

ANSWER:

(b) 34 > 3534 > 35
Between the two fractions with the same numerator, the one with the smaller denominator is the greater.

Page No 100:

Question 10:

The smallest of the fractions 35, 23, 56, 71035, 23, 56, 710 is
(a) 2323
(b) 710710
(c) 3535
(d) 5656

ANSWER:

(c) 3535

    2  5, 3, 6, 10    5 5, 3, 3, 5    3 1, 3, 3, 1        1, 1, 1, 1     2  5, 3, 6, 10    5 5, 3, 3, 5    3 1, 3, 3, 1        1, 1, 1, 1 

L.C.M. of 5, 3, 6 and 10 = (2 ×× 3 ×× 5) = 30
Thus, we have:
35 = 3 × 65 × 6 = 1830 23 =2 × 103 × 10 = 2030 56 =5 × 56 × 5  = 2530  710 =7 × 310 × 3 = 2130∴ The smallest fraction = 1830  = 3535 = 3 × 65 × 6 = 1830 23 =2 × 103 × 10 = 2030 56 =5 × 56 × 5  = 2530  710 =7 × 310 × 3 = 2130∴ The smallest fraction = 1830  = 35

Page No 100:

Question 11:

The largest of the fractions 45, 47, 49, 41145, 47, 49, 411 is
(a) 411411
(b) 4545
(c) 4747
(d) 4949

ANSWER:

( b ) 4545
Among the given fractions with the same numerator, the one with the smallest denominator is the greatest. 

Page No 100:

Question 12:

The smallest of the fractions 611, 711, 811, 911611, 711, 811, 911 is
(a) 611611
(b) 711711
(c) 811811
(d) 911911

ANSWER:

(a) 611611
 Among like fractions, the fraction with the smallest numerator is the smallest.

Page No 100:

Question 13:

The smallest of the fractions 34, 56, 712, 2334, 56, 712, 23 is
(a) 2323
(b) 3434
(c) 5656
(d) 712712

ANSWER:

(d) 712712

Explanation: 
    2  4, 6, 12, 3     2 2, 3, 6, 3     3 1, 3, 3, 3        1, 1, 1, 1     2  4, 6, 12, 3     2 2, 3, 6, 3     3 1, 3, 3, 3        1, 1, 1, 1 

​​L.C.M. of 4, 6, 12 and 3 = (2 ×× 2 ×× 3) = 12
Thus, we have:
34 = 3 × 34 × 3 = 912 56 =5 × 26 × 2 = 1012  23 =2 × 43 × 4 = 812 712Clearly, 712   is the smallest fraction.34 = 3 × 34 × 3 = 912 56 =5 × 26 × 2 = 1012  23 =2 × 43 × 4 = 812 712Clearly, 712   is the smallest fraction.

Page No 100:

Question 14:

435=?435=?
(a) 175175
(b) 235235
(c) 173173
(d) none of these

ANSWER:

(b) 235235

Page No 100:

Question 15:

347=?347=?
(a) 347347
(b) 734734
(c) 467467
(d) none of these

ANSWER:

(c) 467467
On dividing 34 by 7:
Quotient = 4
Remainder = 6
347 = 4 +67 = 467347 = 4 +67 = 467

Page No 101:

Question 16:

58+18=?58+18=?
(a) 3838
(b) 3434
(c) 6
(d) none of these

ANSWER:

(b) 3434

Explanation:Addition of like fractions = Sum of the numerators / Common denominator
= 58 + 18 = (5 + 1)8 = 6384 = 3458 + 18 = (5 + 1)8 = 6384 = 34

Page No 101:

Question 17:

58−18=?58-18=?
(a) 1414
(b) 1212
(c) 116116
(d) none of these

ANSWER:

(b) 1212
Explanation: 
58 − 18 = (5 − 1)8 = 4182 = 1258 – 18 = 5 – 18 = 4182 = 12

Page No 101:

Question 18:

334−214=?334-214=?
(a) 112112
(b) 114114
(c) 1414
(d) none of these

ANSWER:

(a) 112Explanation:334 − 214⇒154 − 94⇒(15 − 9)4⇒64 = 32 = 112(a) 112Explanation:334 – 214⇒154 – 94⇒(15 – 9)4⇒64 = 32 = 112

Page No 101:

Question 19:

56+23−49=?56+23-49=?
(a) 113113
(b) 116116
(c) 119119
(d) 11181118

ANSWER:

(d) 11181118

Explanation: 
    3  3, 6, 9    2 1, 2, 3    3 1, 1, 3        1, 1, 1     3  3, 6, 9    2 1, 2, 3    3 1, 1, 3        1, 1, 1 

    56 + 23 − 49                        ( L.C.M. of 3, 6 and 9 = (2 × 3 × 3) = 18) = (15 + 12 −8)18                          {[18 ÷ 6 = 3,  3 × 5 = 15], [18 ÷ 3 = 6, 6 × 2 = 12] and [18 ÷ 9 = 2, 2 × 4 = 8]} = (27 − 8)18 = 1918 = 1118    56 + 23 – 49                        ( L.C.M. of 3, 6 and 9 = 2 × 3 × 3 = 18) = 15 + 12 -818                          18 ÷ 6 = 3,  3 × 5 = 15, 18 ÷ 3 = 6, 6 × 2 = 12 and 18 ÷ 9 = 2, 2 × 4 = 8 = 27 – 818 = 1918 = 1118

Page No 101:

Question 20:

Which is greater: 313 or 3310313 or 3310?
(a) 313313
(b) 33103310
(c) both are equal

ANSWER:

(a) 313313

Explanation:
Let us compare  313 and 3310 or 103 and 3310 313 and 3310 or 103 and 3310 .
10 ⨯ 10 = 100 and 3 ​⨯ 33 = 99
Clearly, 100 > 99
∴ 103>3310 or 313 >3310103>3310 or 313 >3310

Page No 103:

Exercise 5H

Question 1:

Define a fraction. Give five examples of fractions.

ANSWER:

A fraction is defined as a number representing a part of a whole, where the whole may be a single object or a group of objects.

Examples: 57 , 85 , 23 , 43 , 4957 , 85 , 23 , 43 , 49

Page No 103:

Question 2:

What fraction of an hour is 35 minutes?

ANSWER:

An hour has 60 minutes.
∴∴ Fraction for 35 minutes = 3576012  = 7123576012  = 712
Hence, 712712 part of an hour is equal to 35 minutes.

Page No 103:

Question 3:

Find the equivalent fraction of 5/8 with denominator 56.

ANSWER:

56 = 8 ⨯ 7
So, we need to multiply the numerator by 7.
∴∴ 58 = 5 × 78 × 7 = 355658 = 5 × 78 × 7 = 3556
Hence, the required fraction is 35563556.

Page No 103:

Question 4:

Represent 235235 on the number line.

ANSWER:

Let OA = AB = BC = 1 unit
∴∴ OB = 2 units and OC = 3 units
Divide BC into 5 equal parts and take 3 parts out to reach point P.
Clearly, point P represents the number 235235.

Page No 103:

Question 5:

Find the sum 245+1310+3115245+1310+3115.

ANSWER:

We have:
   245 + 1310 + 3115 = 145 + 1310 + 4615         5  5, 10, 15  2 1, 2, 3  3 1, 1, 3      1, 1, 1                   L.C.M. of 5, 10 and 15 = (5 × 2 × 3) = 30 = 84 + 39 + 9230                   {[30 ÷ 5 = 6, 6 × 14 = 84], [30 ÷ 10 = 3, 3 × 13 = 39] and [30 ÷ 15 = 2, 2 × 46 = 92]} = 21543306 = 436 = 716   245 + 1310 + 3115 = 145 + 1310 + 4615         5  5, 10, 15  2 1, 2, 3  3 1, 1, 3      1, 1, 1                   L.C.M. of 5, 10 and 15 = 5 × 2 × 3 = 30 = 84 + 39 + 9230                   30 ÷ 5 = 6, 6 × 14 = 84, 30 ÷ 10 = 3, 3 × 13 = 39 and 30 ÷ 15 = 2, 2 × 46 = 92 = 21543306 = 436 = 716

Page No 103:

Question 6:

The cost of a pen is Rs 16231623 and that of a pencil is Rs 416416.
Which costs more and by how much?

ANSWER:

Cost of a pen = Rs 1623 = Rs 503 = Rs 50 × 23 × 2 = Rs 1006Rs 1623 = Rs 503 = Rs 50 × 23 × 2 = Rs 1006

Cost of a pencil = Rs 416 = Rs 256 Rs 416 = Rs 256 
 1006 > 256 ∴ Rs 1623 >Rs  4161006 > 256 ∴ Rs 1623 >Rs  416
So, the cost of a pen is more than the cost of a pencil.
Difference between their costs:
  = Rs (503 − 256) = Rs (100 − 256) = Rs (752562) = Rs (252) = Rs 1212 = Rs 503 – 256 = Rs 100 – 256 = Rs 752562 = Rs 252 = Rs 1212
Hence, the cost of a pen is Rs 12121212 more than the cost of a pencil.

Page No 103:

Question 7:

Of 3434 and 5757, which is greater and by how much?

ANSWER:

Let us compare 34 and 5734 and 57.
By cross multiplying:
3 ⨯ 7 = 21 and ​4 ⨯ 5 = 20
Clearly, 21 > 20
∴​34>5734>57
 Their difference:
   34 − 57                 L.C.M. of 4 and 7 = (2 × 2 × 7) = 28= 21 − 2028                {[28 ÷ 4 = 7, 7 × 3 = 21] and  [28 ÷ 7 = 4,  4 × 5 = 20]}= 128   34 – 57                 L.C.M. of 4 and 7 = 2 × 2 × 7 = 28= 21 – 2028                28 ÷ 4 = 7, 7 × 3 = 21 and  28 ÷ 7 = 4,  4 × 5 = 20= 128 
Hence, 34 is greater than 57 by 128.34 is greater than 57 by 128.

Page No 103:

Question 8:

Convert the fractions 12, 23, 4912, 23, 49 and 5656 into like fractions.

ANSWER:

 The given fractions are 12, 23, 49, 56. The given fractions are 12, 23, 49, 56.
L.C.M. of 2, 3, 9 and 6 = (2 ⨯ 3 ​⨯ 3) = 18
Now, we have:
12 = 1 × 92 × 9 = 918 23 = 2 × 63 × 6 = 121849 = 4 × 29 × 2 = 818 56 = 5 × 36 × 3 = 1518Hence, 918, 1218, 818 and 1518 are like fractions.12 = 1 × 92 × 9 = 918 23 = 2 × 63 × 6 = 121849 = 4 × 29 × 2 = 818 56 = 5 × 36 × 3 = 1518Hence, 918, 1218, 818 and 1518 are like fractions.

Page No 103:

Question 9:

Find the equivalent fraction of 3535 having denominator 30.

ANSWER:

Let 35 = 30Let 35 = 30

30 = 5 ​⨯ 6 
So, we have to multiply the numerator by 6 to get the equivalent fraction having denominator 30.

∴ 35 = 3 × 65 × 6 = 183035 = 3 × 65 × 6 = 1830

Thus, 1830 is the equivalent fraction of 35.1830 is the equivalent fraction of 35.

Page No 103:

Question 10:

Reduce 84988498 to the simplest form.

ANSWER:

The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.
The factors of 98 are 1, 2, 7, 14, 49, 98.
The common factors of 84 and 98 are 1, 2, 7, 14.
The H.C.F. of 84 and 98 is 14.
Dividing both the numerator and the denominator by the H.C.F.:
8498 = 84 ÷ 1498 ÷ 14 = 678498 = 84 ÷ 1498 ÷ 14 = 67

Page No 103:

Question 11:

24112411 is an example of
(a) a proper fraction
(b) an improper fraction
(c) a mixed fraction
(d) none of these

ANSWER:

(b) an improper fraction

In an improper fraction, the numerator is greater than the denominator.

Page No 103:

Question 12:

3838 is an example of
(a) a proper fraction
(b) an improper fraction
(c) a mixed fraction
(d) none of these

ANSWER:

(a) proper fraction

In a proper fraction, the numerator is less than the denominator.

Page No 103:

Question 13:

3838 and 512512 on comparison give
(a) 38>51238>512
(b) 38<51238<512
(c) 38=51238=512
(d) none of these

ANSWER:

(b) 38<51238<512

Considering 38 and 51238 and 512:

On cross multiplying, we get:3 × 12 = 36 and 8 × 5 = 40Clearly, 36 < 40∴ 38 < 512On cross multiplying, we get:3 × 12 = 36 and 8 × 5 = 40Clearly, 36 < 40∴ 38 < 512

Page No 103:

Question 14:

The largest of the fractions 23, 59, 1223, 59, 12 and 712712 is
(a) 2323
(b) 5959
(c) 712712
(d) 1212

ANSWER:

(a) 2323

Explanation:
L.C.M. of 3, 9, 2 and 12 = ( 2 ⨯ 2 ⨯ 3 ​⨯ 3) = 36
Now, we have:
23 = 2 × 123 × 12  = 2436 59 = 5 × 49 × 4 = 203612 = 1 × 182 × 18 = 1836 712 = 7 × 312 × 3 = 2136Hence, 2436 = 23 is the largest fraction.23 = 2 × 123 × 12  = 2436 59 = 5 × 49 × 4 = 203612 = 1 × 182 × 18 = 1836 712 = 7 × 312 × 3 = 2136Hence, 2436 = 23 is the largest fraction.

Page No 103:

Question 15:

334−112=?334-112=?
(a) 212212
(b) 214214
(c) 112112
(d) 114114

ANSWER:

(b) 214214
Explanation:

334 − 112 = 154 − 32             (L.C.M. of 2 and 4 = (2 × 2) = 4)                       = 15 − 64                        = 94 = 214334 – 112 = 154 – 32             (L.C.M. of 2 and 4 = 2 × 2 = 4)                       = 15 – 64                        = 94 = 214

Page No 103:

Question 16:

Which of the following are like fractions?
(a) 23, 34, 45, 5623, 34, 45, 56
(b) 25, 27, 29, 21125, 27, 29, 211
(c) 18, 38, 58, 7818, 38, 58, 78
(d) none of these

ANSWER:

(c) 18, 38, 58, 7818, 38, 58, 78
Like fractions have same the denominator.

Page No 104:

Question 17:

?−821=821?-821=821
(a) 0
(b) 1
(c) 218218
(d) 16211621

ANSWER:

(d) 16211621

?  − 821 = 821? = 821 + 821 = 1621?  – 821 = 821? = 821 + 821 = 1621 

Page No 104:

Question 18:

Fill in the blanks:
(i) 923+……=19923+……=19
(ii) 616−?=2930616-?=2930
(iii) 7 − 523=……7 – 523=……
(iv) 72907290 reduced to simples form is ……
(v) 4254=7    4254=7    

ANSWER:

(i) 923 +……=19……=19 − 923 ……= 191 − 293  ……=57 − 293 ……= 283 …..= 913(i) 923 +……=19……=19 – 923 ……= 191 – 293  ……=57 – 293 ……= 283 …..= 913                       


(ii)                    
616− ? =2930 ? = 616 − 2930 ? = 376  − 2930 ? = 185 − 2930 ? = 15626305 ? = 515616- ? =2930 ? = 616 – 2930 ? = 376  – 2930 ? = 185 – 2930 ? = 15626305 ? = 515
 

(iii)                   
  7 − 523 = 71 − 173 = 21 −173 = 43 = 1137 – 523 = 71 – 173 = 21 -173 = 43 = 113

(iv)
 
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.The common factors of 72 and 90 are 1, 2, 3, 6, 9, 18.H.C.F. of 72 and 90 is 18. 72 ÷ 1890 ÷ 18 = 45The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.The common factors of 72 and 90 are 1, 2, 3, 6, 9, 18.H.C.F. of 72 and 90 is 18. 72 ÷ 1890 ÷ 18 = 45

 (v) 

4254 =79⇒42 ÷ 654 ÷ 6 = 794254 =79⇒42 ÷ 654 ÷ 6 = 79

Page No 104:

Question 19:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(a) 313>3310313>3310.
(b) 8−156=7168-156=716.
(c) 12, 1312, 13and 1414 are like fractions.
(d) 3535 lies between 3 and 5.
(e) Among 12, 13, 34, 4312, 13, 34, 43 the largest fractions is 4343.

ANSWER:

(a) T                     
(b) F                   (81 − 116 = 48 −116 = 376 = 616)81 – 116 = 48 -116 = 376 = 616
(c) F                    (Because like fractions have the same denominator.)
​(d) F                    (It lies between 0 and 1 as all proper fractions are less than 1.)
(e) T                    (Because it is an improper fraction, while the others are proper fractions.)

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RS Agarwal Solution | Class 6th | Chapter-6 | Simplification | Edugrown

Exercise 6A

Page No 107:

Question 1:

Simplify:
21 – 12 ÷ 3 × 2

ANSWER:

Given expression:
= 21 – 12 ÷ 3 ⨯ 2
= 21 – 4 ​⨯ 2               [Performing division]
= 21 –  8                     [Performing multiplication]
= 13                           [Performing subtraction]

Page No 107:

Question 2:

Simplify:
16 + 8 ÷ 4 – 2 × 3

ANSWER:

Given expression:
= 16 + 8 ÷ 4 – 2 ⨯ 3
= 16 + 2 – 2 ​⨯ 3               [Performing division]
= 16 + 2 – 6                     [Performing multiplication]
= 18 – 6                           [Performing addition]
= 12                               ​ [Performing subtraction]

Page No 107:

Question 3:

Simplify:
13 – (12 – 6 ÷ 3)

ANSWER:

Given expression:
= 13 – (12 – 6 ÷ 3 ) 
= 13 – (12 – 2)       [Performing division]
= 13 – 10 = 3        [Performing subtraction]

Page No 107:

Question 4:

Simplify:
19 – [4 + {16 – (12 – 2)}]

ANSWER:

Given expression:
= 19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]                        [Removing parentheses] 
= 19 – [4 + 6]                                    [Removing braces] 
= 19 – 10                                           [Removing square brackets] 
= 9

Page No 107:

Question 5:

Simplify:
36 – [18 – {14 – (15 – 4 ÷2 ×2)}]

ANSWER:

Given expression:
= 36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]
= 36 – [18 – {14 – (15 – 2 × 2)}]                   [Performing division]  
= 36 – [18 – {14 – (15 – 4)}]                         [Performing multiplication]   
= 36 – [18 – {14 – 11}]                                [Removing parentheses]  
= 36 – [18 – 3]                                            [Removing braces] 
​= 36 – 15                                                   [Removing square brackets] 
= 21

Page No 107:

Question 6:

Simplify:
27 – [18 – {16 – (5 – ¯4 – 1)}]

ANSWER:

Given expression:
= 27 – [18 – {16 – (5 – 4-1)}]
= 27 – [18 – {16 – (5 – 3)}]                       [Removing bar]  
= 27 – [18 – {16 – 2}]                              [Removing parentheses]  
= 27 – [18 – 14]                                       [Removing braces]                             
= 27 – 4                                                 [Removing square brackets]                             
​= 23                                                            

Page No 107:

Question 7:

Simplify:
445÷35 of 5+45×310-15

ANSWER:

Given expression:
  = 445 ÷35 of 5 + 45×310 – 15= 445 ÷35 × 51 + 45×310 – 15                         (Removing ‘of ‘)= 245 ÷31 + 45×310 – 15                                 =245×13 + 45×310 – 15                                        (Removing ‘÷’)= 85 + 45×310 – 15                                                 (Removing ‘×’)=85 +625 – 15                                                           (Removing ‘×’)= 40 + 6 – 525 = 41 25 = 11625

Page No 107:

Question 8:

Simplify:
(23+49) of 35 ÷ 123×114-13

ANSWER:

Given expression:
 = (23 + 49) of 35 ÷123× 114 – 13 = (6 +49) of 35 ÷123× 114 – 13                      (Removing parentheses)= 109×35÷123× 114 – 13                                (Removing ‘of’)= 23÷53× 114 – 13                                    =23×35 ×54 – 13                                                 (Removing ‘÷’ )= 25×54 – 13                                     = 12 – 13                                                                  (Removing ‘×’ )= (3-2)6 = 16

Page No 107:

Question 9:

Simplify:
713÷23 of 215+138÷234-112

ANSWER:

The given expression
 = 713÷23 of 215 + 138 ÷234 – 112 = 223 ÷23 of 115 + 118 ÷ 114 – 32  = 223 ÷23 ×115 + 118 ÷ 114 – 32                             (Removing ‘of’) =223 ÷2215 + 118 ÷ 114 – 32        = 223×1522 +118 ÷ 114 – 32                    (Removing ‘÷’) =  5 + 118 × 411 – 32                                        (Removing ‘÷’)= 5 + 12 – 32                               ( On simplifying)= 10 +1 -32 = 82  = 4

Page No 107:

Question 10:

Simplify:
517-{3310÷(245-710)}

ANSWER:

Given expression:
= 517 – {3310÷(245 – 710)}= 367 – {3310÷(145 – 710)}= 367 – {3310÷(28-710)}                            = 367 – {3310÷2110}                                         (Removing parentheses)= 367 – {3310×1021}                                          (Removing ‘÷’)= 367 – 117                                                          (Removing braces)= 36 – 117 = 257 = 347                                  (Simplifying)

Page No 107:

Question 11:

Simplify:
934÷[216+{413-(112+134)}]

ANSWER:

Given expression:
= 934 ÷ [216 + {413 – (112 + 134)}] = 394 ÷ [136 + {133 – (32 + 74)}] = 394 ÷ [136 + {133 – (6+74)}]                            = 394 ÷ [136 + {133 – 134}]                                         (Removing parentheses)= 394 ÷ [136 + {52 -3912}]                                          = 394 ÷ [136 + 1312]                                                          (Removing braces)= 394 ÷ [26 +1312]     = 394 ÷3912                                                                            (Removing square brackets)= 394 × 1239 = 3                                                                    (Removing ‘÷’)

Page No 107:

Question 12:

Simplify:
4110-[212-{56-(25+310-415)}]

ANSWER:

Given expression:
 = 4110- [212 – {56 – (25 + 310-415)}]  = 4110 – [52 – {56 – (25 +  310-415)}]  = 4110 – [52 – {56 – (12 +9 – 8 30)}]                         = 4110 – [52 – {56 – 1330}]                                        (Removing parentheses) = 4110 – [52 – {25 -1330}]                                         = 4110 – [52 – 1230]                                                       (Removing braces)= 4110 – [75 -1230]     = 4110 -6330                                                                       (Removing square brackets)= 123 – 6330 = 6030 = 2                                                                   

Page No 107:

Question 13:

Simplify:
156+[223-{334 (345÷912)}]

ANSWER:

Given expression:
= 156+ [223 – {334 (345÷ 912)}] = 116 + [83 – {154 (195 ÷ 192)}]  = 116 + [83 – {154 (195 × 219)}]                          = 116 + [83 – {154 × 25}]                                        (Removing parentheses)= 116 + [83 – 32]                                                         (Removing braces)                           = 116 + [16 – 96]                                                        = 116 + 76                                                                        (Removing square brackets)= 186 = 3                                                                   

Page No 107:

Question 14:

Simplify:
445÷{215-12(114-¯14-15)}

ANSWER:

Given expression:
= 445 ÷ {215 – 12(114 – 14 – 15)}= 245 ÷ {115 – 12(54 – 14 – 15)} = 245 ÷ {115 – 12(54 – 120)}                             (Removing bar)                  = 245 ÷ {115 – 12(25 – 120)}                                            = 245 ÷ {115 – 12 × 2420}                                     (Removing parentheses)=245 ÷ {115 – 1220}                                                 (Removing ‘×’)                           = 245 ÷ {44 – 1220}                                                                                                   = 245÷3220                                                                    (Removing braces)   = 245 × 2032                                                                  (Removing ‘÷’)= 34 × 4 = 3                                                                   

Page No 107:

Question 15:

Simplify:
712-[214÷{114-12(32-¯13-16)}]

ANSWER:

Given expression:
= 712-[214÷{ 114- 12(32 – 13 – 16)}]= 152-[94÷{ 54- 12(32 – 13 – 16)}]= 152-[94÷{ 54- 12(32 – 16)}]                                    (Removing bar)                  = 152-[94÷{ 54- 12(9 -16)}]                                         = 152-[94÷{ 54- 12×43}]                                              (Removing parentheses)=  152-[94÷{ 54- 23}]                                                       (Removing ‘×’)                           = 152-[94÷{  15 – 812}]                                                      (Removing braces)                                            =152-[94÷712]                                                                       = 152-[94×127]                                                                     (Removing ‘÷’)= 152 – 277                                                                               (Removing square brackets)  = 105 – 5414 =5114 = 3914

Page No 108:

Exercise 6B

Question 1:

8 + 4 ÷ 2 × 5 = ?
(a) 30
(b) 50
(c) 18
(d) none of these

ANSWER:

(c ) 18

Explanation:
= 8 + 4 ÷ 2 × 5
= 8 + 2 × 5
= 8 + 10 = 18 

Page No 108:

Question 2:

54 ÷ 3 of 6 + 9 = ?
(a) 117
(b) 12
(c) 65
(d) none of these

ANSWER:

 ( b ) 12

Explanation:
= 54 ÷ 3 of 6 + 9
= 54 ÷ (3  x 6) + 9
= 54 ÷ 18 + 9
= 3 + 9 = 12

Page No 108:

Question 3:

13 − (12 − 6 ÷ 3) = ?
(a) 11
(b) 3
(c) 73
(d) none of these

ANSWER:

(b ) 3

Explanation:
= 13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – 10 = 3

Page No 108:

Question 4:

1001 ÷ 11 of 13 = ?
(a) 7
(b) 1183
(c) 847
(d) none of these

ANSWER:

(a ) 7

Explanation:
= 1001 ÷  11 of 13 
= 1001 ÷ ( 11 x 13)
= 1001 ÷ 143 = 7

Page No 108:

Question 5:

133 + 28 ÷ 7 − 8 × 2 = ?
(a) 7
(b) 121
(c) 30
(d) none of these

ANSWER:

(b) 121

Explanation:
Given expression:
= 133 + 28 ÷ 7 – 8 ⨯ 2
= 133 + 4 – 8 ​⨯2              [Performing division]
= 133 + 4 – 16                 [Performing multiplication]
= 137 – 16                       [Performing addition]
= 121                              [Performing subtraction]

Page No 108:

Question 6:

3640 − 14 ÷ 7 × 2 = ?
(a) 3636
(b) 1036
(c) 1819
(d) none of these

ANSWER:

(a) 3636

Explanation:
Given expression:
= 3640 – 14 ÷ 7 ⨯ 2
​= 3640 – 2  ​⨯2               [Performing division]
= 3640 – 4                    [Performing multiplication]
= 3636                          [Performing subtraction]

Page No 108:

Question 7:

100 × 10 − 100 + 2000 ÷ 100 = ?
(a) 29
(b) 920
(c) none of these

ANSWER:

(b) 920

Explanation:
Given expression:
= 100 ⨯ 10 – 100 + 2000 ÷ 100
​= 100 ⨯ 10 – 100 + 20                  [Performing division]
= 1000 – 100 + 20                        [Performing multiplication]
= 1020 -100                                 [Performing addition]
= 920                                   ​        [Performing subtraction]

Page No 108:

Question 8:

27 – [18 – {16 – (5 – ¯4 – 1)] = ?
(a) 25
(b) 23
(c) none of these

ANSWER:

(b) 23
Explanation:

Given expression:
= 27 – [18 – {16 – (5 – 4-1)}]
=  27 – [18 – {16 – (5 – 3)}]                          (Removing bar)
= 27 – [18 – {16 – 2}]                                   (Removing parentheses)
= 27 – [18 – 14]                                            (Removing braces)
= 27 – 4                                                        (Removing square brackets)
= 23

Page No 108:

Question 9:

32 – [48 ÷ {36 – (27 – ¯16 – 9)}] = ?
(a) 29
(b) 52017
(c) none of these

ANSWER:

(a) 29
Explanation:
Given expression:
= 32 – [48 ÷ {36 – ( 27 – 16-9)}]
=  32 – [48 ÷ {36 – ( 27 – 7)}]                    (Removing bar)
= 32 – [48 ÷ {36 -20}]                               (Removing parentheses)
= 32 – [48 ÷ 16]                                          (Removing braces)
= 32 – 3                                                      (Removing square brackets)
= 29

Page No 108:

Question 10:

8 − [28 ÷ {34 − (36 − 18 ÷ 9 × 8)}] = ?
(a) 6
(b) 649
(c) none of these

ANSWER:

(a ) 6

Explanation: 
Given expression:
= 8 – [28 ÷ {34 – (36 – 18 ÷ 9 × 8)}]                        [Performing division]
= 8 – [28 ÷ {34 – (36 – 2 ×  8)}]                               [Performing multiplication]
= 8 – [28 ÷ {34 – (36 – 16)}]
= 8 – [28 ÷ {34 – 20}]                                              [Removing parentheses]
= 8 – [28 ÷ 14]                                                         [Removing braces]
= 8 – 2 = 6                                                               [Removing square brackets]

Page No 109:

Exercise 6C

Question 1:

Simplify:
16 – [11 + {8 – 6 – ¯4 – 2}]

ANSWER:

We have:
    16 – [11 + {8 – (6 -4-2 )}]
= ​16 – [11 + {8 – (6 – 2)}]         (Removing bar)
= ​16 – [11 + {8 – 4}]                 
= ​16 – [11 + 4]                           (Removing barces)
= 16 – 15                                          (Removing square brackets)
= 1

Page No 109:

Question 2:

Simplify:
16 + 8 ÷ 4 – 2 × 3

ANSWER:

We have:
  16 + 8 ÷ 4 – 2 × 3
= 16 + 2 – 2 × 3             (Division)
= 16 + 2 – 6                   (Multiplication)
= 18 – 6                           (Addition)
= 12

Page No 109:

Question 3:

Simplify:
36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]

ANSWER:

Given expression:
   36 – [18 – {14 – (15 – 4 ÷ 2 × 2)}]                                     [Performing division]
= 36 – [18 – {14 – (15 – 2 × 2)}]                                           [Performing multiplication]
= 36 – [18 – {14 – (15 – 4)}]      
= 36 – [18 – {14 – 11}]                                                        [Removing parentheses]
= 36 – [18 – 3]                                                                    [Removing braces]
= 36 – 15                                                                           [Removing square brackets]
= 21

Page No 109:

Question 4:

15 + 5 ÷ 5 × 2 = ?
(a) 8
(b) 2
(c) 17
(d) none of these

ANSWER:

(c) 17
We have:
15 + 5 ÷ 5 x 2 
= 15 +1 x 2              (Performing division)
= 15 + 2                  (Performing multiplication)
= 17

Page No 109:

Question 5:

16 – 2 ÷ 7 + 6 × 2 = ?
(a) 16
(b) 14
(c) 2757
(d) none of these

ANSWER:

(c) 2757
We have:
16 – 2 ÷ 7 +  6 × 2  
= 16 – (27) + 6 × 2                                                                [Performing division]
= 16 – (27) + 12                                                                    [Performing multiplication]
= 28 – (27)                                                                            [Performing addition]
=  281 -27  = (196 -2)7  = 1947 = 2757                                   [Performing subtraction]

Page No 109:

Question 6:

54 ÷ 3 of 6 + 9 = ?
(a) 117
(b) 12
(c) 65
(d) none of these

ANSWER:

(b)12We have: 
    54 ÷ 3 of 6 + 9
= 54 ÷ 18 + 9
= 3 + 9
= 12

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RS Agarwal Solution | Class 6th | Chapter-1 | Number System | Edugrown

Exercise 1A

Question 1:

Write the numeral for each of the following numbers:
(i) Nine thousand eighteen
(ii) Fifty-four thousand seventy-three
(iii) Three lakh two thousand five hundred six
(iv) Twenty lakh ten thousand eight
(v) Six crore five lakh fifty-seven
(vi) Two crore two lakh two thousand two hundred two
(vii) Twelve crore twelve lakh twelve thousand twelve
(viii) Fifteen crore fifty lakh twenty thousand sixty-eight

ANSWER:

(i) Nine thousand eighteen = 9018
(ii) Fifty-four thousand seventy-three = 54073
(iii) Three lakh two thousand five hundred six = 302506
(iv) Twenty lakh ten thousand eight = 2010008
(v) Six crore five lakh fifty-seven = 60500057
(vi) Two crore two lakh two thousand two hundred two = 20202202
(vii) Twelve crore twelve lakh twelve thousand twelve = 121212012
(viii) Fifteen crore fifty lakh twenty thousand sixty-eight = 155020068

Page No 5:

Question 2:

Write each of the following numbers in words:
(i) 63,005
(ii) 7,07,075
(iii) 34,20,019
(iv) 3,05,09,012
(v) 5,10,03,604
(vi) 6,18,05,008
(vii) 19,09,09,900
(viii) 6,15,30,807
(ix) 6,60,60,060

ANSWER:

(i) 63,005 = Sixty-three thousand five
(ii) 7,07,075 =  Seven lakh seven thousand seventy-five
(iii) 34,20,019 = Thirty-four lakh twenty thousand nineteen
(iv) 3,05,09,012 = Three crore five lakh nine thousand twelve
(v) 5,10,03,604 = Five crore ten lakh three thousand six hundred four
(vi) 6,18,05,008 = Six crore eighteen lakh five thousand eight
(vii) 19,09,09,900 = Nineteen crore nine lakh nine thousand nine hundred
(viii) 6,15,30,807 = Six crore fifteen lakh thirty thousand eight hundred seven
(ix) 6,60,60,060 = Six crore sixty lakh sixty thousand sixty

Page No 5:

Question 3:

Write each of the following numbers in expanded form:
(i) 15,768
(ii) 3,08,927
(iii) 24,05,609
(iv) 5,36,18,493
(v) 6,06,06,006
(iv) 9,10,10,510

ANSWER:

(i) 15,768 = (1 x 10000) + (5 x 1000) + (7 x 100) + (6 x 10) + (8 x 1)

(ii) 3,08,927 = (3 x 100000) + (8 x 1000) + (9 x 100) + (2 x 10) + (7 x 1)

(iii) 24,05,609 = (2 x 1000000) + (4 x 100000) + (5 x 1000) + (6 x 100) + (9 x 1)

(iv) 5,36,18,493 = (5 x 10000000) + (3 x 1000000) + (6 x 100000) + (1 x 10000) + (8 x 1000) + (4 x 100) + (9 x 10) + (3 x 1)

(v) 6,06,06,006 = (6 x 10000000) + (6 x 100000) + (6 x 1000) + (6 x 1)

(iv) 9,10,10,510 = (9 x 10000000) + (1 x 1000000) + (1 x 10000) + (5 x 100) + (1 x 10)

Page No 6:

Question 4:

Write the corresponding numeral for each of the following:
(i) 6 × 10000 + 2 × 1000 + 5 × 100 + 8 × 10 + 1
(ii) 5 × 100000 + 8 × 10000 + 1 × 1000 + 6 × 100 + 2 × 10 + 3 × 1
(iii) 2 × 10000000 + 5 × 100000 + 7 × 1000 + 9 × 100 + 5 × 1
(iv) 3 × 1000000 + 4 × 100000 + 6 × 1000 + 5 × 100 + 7 × 1

ANSWER:

(i) 6 × 10000 + 2 × 1000 + 5 × 100 + 8 × 10 + 4 x 1 = 62,584

(ii) 5 × 100000 + 8 × 10000 + 1 × 1000 + 6 × 100 + 2 × 10 + 3 × 1 = 5,81,623

(iii) 2 × 10000000 + 5 × 100000 + 7 × 1000 + 9 × 100 + 5 × 1 = 2,05,07,905

(iv) 3 × 1000000 + 4 × 100000 + 6 × 1000 + 5 × 100 + 7 × 1 = 34,06,507

Page No 6:

Question 5:

Find the difference between the place values of the two nines in 79520986.

ANSWER:

The place value of 9 at ten lakhs place = 90 lakhs = 9000000
The place value of 9 at hundreds place = 9 hundreds = 900
∴∴ Required difference = (9000000 ‒ 900) = 8999100

Page No 6:

Question 6:

Find the difference between the place value and the face value of 7 in 27650934.

ANSWER:

The place value of 7 in 27650934 = 70 lakhs = 70,00,000
The face value of 7 in 27650934 = 7
∴∴ Required difference = (7000000 ‒ 7) = 69,99,993

Page No 6:

Question 7:

How many 6-digit numbers are there in all?

ANSWER:

The largest 6-digit number = 999999
The smallest 6-digit number = 100000
∴∴ Total number of 6-digit numbers = (999999 ‒ 100000) + 1
                                                 = 899999 + 1
                                                 = 900000
                                                 = 9 lakhs

Page No 6:

Question 8:

How many 7-digit numbers are there in all?

ANSWER:

The largest 7-digit number = 9999999
The smallest 7-digit number = 1000000
∴ Total number of 7-digit numbers = (9999999 – 1000000) + 1
                                                     = 8999999 + 1
                                                     = 9000000
                                                     = Ninety lakhs

Page No 6:

Question 9:

How many thousands make a lakh?

ANSWER:

One lakh (1,00,000) is equal to one hundred thousand (100 ×× 1000).
Thus, one hundred thousands make a lakh.

Page No 6:

Question 10:

How many thousands make a crore?

ANSWER:

One crore (1,00,00,000) is equal to one hundred lakh (10,000 ×× 1,000).
Thus, 10,000 thousands make a crore.

Page No 6:

Question 11:

Find the difference between the number 738 and that obtained on reversing its digits.

ANSWER:

The given number is 738.
On reversing the digits of this number, we get 837.
∴ Required difference = 837 ‒ 738 = 99

Page No 6:

Question 12:

What comes just after 9547999?

ANSWER:

The number just after 9547999 is 9547999 + 1 = 9548000.

Page No 6:

Question 13:

What comes just before 9900000?

ANSWER:

The number just before 9900000 is 9900000 ‒ 1 = 9899999.

Page No 6:

Question 14:

What comes just before 10000000?

ANSWER:

The number just before 10000000 is 10000000 ‒ 1 = 9999999.

Page No 6:

Question 15:

Write all 3-digit numbers using 2, 3, 4, taking each digit only once.

ANSWER:

The 3-digit numbers formed by 2, 3 and 4 by taking each digit only once are 234, 324, 243, 342, 423 and 432.

Page No 6:

Question 16:

Write the smallest number of different digits formed by using the digits 3, 1, 0, 5 and 7.

ANSWER:

The smallest number formed by using each of the given digits (i.e, 3,1,0,5 and 7) only once is 10357.

Page No 6:

Question 17:

Write the largest number of different digits formed by using the digits 2, 4, 0, 3, 6 and 9.

ANSWER:

The largest number formed by using each of the given digits only once is 964320.

Page No 6:

Question 18:

Rewrite each of the following numerals with proper commas, using the international place-value chart. Also, write the number name of each in the international system.
(i) 735821
(ii) 6057894
(iii) 56943821
(iv) 37502093
(v) 89350064
(vi) 90703006

ANSWER:

Representation of the numbers on the international place-value chart:
 

PeriodsMillionsThousandsOnes
PlaceHundred
millions		Ten millionsMillionsHundred thousandsTen
thousands		ThousandsHundredsTensOnes
HMTMMH ThT ThThHTO
(i)   735821
(ii)  6057894
(iii) 56943821
(iv) 37502093
(v) 89350064
(vi) 90703006
  CroreTen lakhsLakhsTen ThousandThousandHundredTensOnes

The number names of the given numbers in the international system:

(i) 735,821 = Seven hundred thirty-five thousand eight hundred twenty-one
(ii) 6,057,894 = Six million fifty-seven thousand eight hundred ninety-four
(iii) 56,943,821 = Fifty-six million nine hundred forty-three thousand eight hundred twenty-one
(iv) 37,502,093 = Thirty-seven million five hundred two thousand ninety-three
(v) 89,350,064 = Eighty-nine millions three hundred fifty thousand sixty-four
(vi) 90,703,006 = Ninety million seven hundred three thousand and six

Page No 6:

Question 19:

Write each of the following in figures in the international place-value chart:
(i) Thirty million one hundred five thousand sixty-three
(ii) Fifty-two million two hundred five thousand six
(iii) Five million five thousand five

ANSWER:

PeriodsMillionsThousandsOnes
PlaceHundred millionsTen millionsMillionsHundred thousandsTen thousandsThousandsHundredsTensOnes
HMTMMH ThT ThThHTO
(i) 30105063
(ii) 52205006
(iii)   5005005

Page No 8:

Question 1:

Fill in each of the following boxes with the correct symbol > or <:
1003467          9879651003467          987965

ANSWER:

1003467 >> 987965

We know that a 7-digit number is always greater than a 6-digit number. Since 1003467 is a 7-digit number and 987965 is a 6-digit number, 1003467 is greater than 987965.

Page No 8:

Question 2:

Fill in each of the following boxes with the correct symbol > or <:
3572014          102354013572014          10235401

ANSWER:

3572014 << 10235401

We know that a 7-digit number is always less than an 8-digit number. Since 3572014 is a 7-digit number and 10235401 is an 8-digit number, 3572014 is less than 10235401.

Page No 8:

Question 3:

Fill in each of the following boxes with the correct symbol > or <:
3254790          32601523254790          3260152

ANSWER:

Both the numbers have the digit 3 at the ten lakhs places.
Also, both the numbers have the digit 2 at the lakhs places.
However, the digits at the ten thousands place in 3254790 and 3260152 are 5 and 6, respectively.
Clearly, 5 < 6
∴ 3254790 < 3260152

Page No 8:

Question 4:

Fill in each of the following boxes with the correct symbol > or <:
10357690          1124356710357690          11243567

ANSWER:

Both have the digit 1 at the crores places.
However, the digits at the ten lakhs places in 10357690 and 11243567 are 0 and 1, respectively.
Clearly, 0 < 1
∴ 10357690 < 11243567

Page No 8:

Question 5:

Fill in each of the following boxes with the correct symbol > or <:
27596381          796541227596381          7965412

ANSWER:

27596381 > 7965412

We know that an 8-digit number is always greater than a 7-digit number. Since 7965412 is a 7-digit number and  27596381 is an 8-digit number, 27596381 is greater than 7965412.

Page No 8:

Question 6:

Fill in each of the following boxes with the correct symbol > or <:
47893501          4789402147893501          47894021

ANSWER:

Both the numbers have the same digits, namely 4, 7, 8 and 9, at the crores, ten lakhs, lakhs and ten thousands places, respectively.
However, the digits at the thousands place in 47893501 and 47894021 are 3 and 4, respectively.
Clearly, 3 < 4
∴ 47893501 < 47894021

Page No 8:

Question 7:

Arrange the following numbers in descending order:
63521047, 7354206, 63514759, 7355014, 102345680

ANSWER:

102345680 is a 9-digit number.

63521047 and 63514759 are both 8-digit numbers.
Both the numbers have the same digits, namely 6, 3 and 5, at the crores, ten lakhs and lakhs places, respectively.
However, the digits at the ten thousands place in 63521047 and 63514759 are 2 and 1, respectively.
Clearly, 2 > 1
∴ 63521047 > 63514759

7355014 and 7354206 are both 7-digit numbers.
Both the numbers have the same digits, namely 7, 3 and 5 at the crores, ten lakhs and lakhs places, respectively.
However, the digits at the ten thousands place in 7355014 and 7354206 are 5 and 4, respectively.
Clearly, 5> 4
∴ 7355014 > 7354206

The given numbers in descending order are:
102345680 > 63521047 > 63514759 > 7355014 > 7354206

Page No 8:

Question 8:

Arrange the following numbers in descending order:
5032786, 23794206, 5032790, 23756819, 987876

ANSWER:

23794206 and 23756819 are both 8-digit numbers.
Both the numbers have the same digits, namely 2, 3 and 7 at the crores, ten lakhs and lakhs places, respectively.
However, the digits at the ten thousands place in 23794206 and 23756819 are 9 and 5, respectively.
Clearly, 9 > 5
∴ 23794206  > 23756819

5032790 and 5032786 are both 7-digit numbers.
Both the numbers have the same digits, namely 5, 0, 3, 2 and 7, at the ten lakhs, lakhs, ten thousands, thousands and hundreds places, respectively.
However, the digits at the tens place in 5032790 and 5032786 are 9 and 8, respectively.
Clearly,   9 > 8
∴ 5032790 > 5032786

987876 is a 6-digit number.

The given numbers in descending order are:
23794206  > 23756819 > 5032790 > 5032786 > 987876

Page No 8:

Question 9:

Arrange the following numbers in descending order:
190909, 1808088, 16060666, 16007777, 181888, 1808090

ANSWER:

16060666 and 16007777 are both 8-digit numbers.
Both the numbers have the same digits, namely 1, 6 and 0, at the crores, ten lakhs and lakhs places, respectively.
However, the digits at the ten thousands place in 16060666 and 16007777 are 6 and 0, respectively.
Clearly, 6 > 0
∴ 16060666 > 16007777

1808090 and 1808088 are both 7-digit numbers.
Both the numbers have the same digits , namely 1, 8, 0, 8 and 0, at the ten lakhs, lakhs, ten thousands, thousands and hundreds places, respectively.
However, the digits at the tens place in 1808090 and 1808088 are 9 and 8, respectively.
Clearly, 9 > 8
∴ 1808090 > 1808088

190909 and 181888 are both 6-digit numbers.
Both the numbers have the same digit, 1, at the lakhs place.
However, the digits at the ten thousands place in 190909 and 181888 are 9 and 8, respectively.
Clearly, 9 > 8
∴ 190909 > 181888

Thus, the given numbers in descending order are:
16060666 > 16007777 > 1808090 > 1808088 >190909 > 181888

Page No 8:

Question 10:

Arrange the following numbers in descending order:
199988, 1704382, 200175, 1702497, 201200, 1712040

ANSWER:

1712040, 1704382 and 1702497 are all 7-digit numbers.
The three numbers have the same digits, namely 1 and 7, at the ten lakhs and lakhs places, respectively. 
However, the digits at the ten thousands place in 1712040, 1704382 and 1702497 are 1, 0 and 0.
∴ 1712040  is the largest.
Of the other two numbers, the respective digits at the thousands place are 4 and 2.
Clearly, 4 > 2
∴ 1704382 > 1702497

201200, 200175 and 199988 are all 6-digit numbers.
At the lakhs place, we have 2 > 1.
So, 199988 is the smallest of the three numbers.

The other two numbers have the same digits, namely 2 and 0, at the lakhs and ten thousands places, respectively.
However, the digits at the thousands place in 201200 and 200175 are 1 and 0, respectively.
Clearly, 1 > 0
∴ 201200 > 200175

The given numbers in descending order are:
1712040 > 1704382 > 1702497 > 201200 > 200175 > 199988

Page No 8:

Question 11:

Arrange the following numbers in ascending order:
9873426, 24615019, 990357, 9874012, 24620010

ANSWER:

990357 is 6 digit number.

9873426 and 9874012 are both 7-digit numbers.
Both the numbers have the same digits, namely 9, 8 and 7, at the ten lakhs, lakhs and ten thousands places, respectively.
However, the digits at the thousands place in 9873426 and 9874012 are 3 and 4, respectively.
Clearly, 4 < 7
∴ 9873426 <  9874012

24615019 and  24620010 are both 8-digit numbers.

Both the numbers have the same digits, namely 2, 4 and 6, at the crores, ten lakhs and lakhs places, respectively.
However, the digits at the ten thousands place in 24615019 and 24620010 are 2 and 1, respectively.
Clearly, 1 < 2
∴ 24615019 < 24620010

The given numbers in ascending order are:
990357 < 9873426 <  9874012 < 24615019 < 24620010

Page No 8:

Question 12:

Arrange the following numbers in ascending order:
56943201, 5694437, 56944000, 5695440, 56943300

ANSWER:

5694437 and 5695440 are both 7-digit numbers.
Both have the same digit, i.e., 5 at the ten lakhs place.
Both have the same digit, i.e., 6 at the lakhs place.
Both have the same digit, i.e., 9 at the ten thousands place.
However, the digits at the thousand place in 5694437 and 5695440 are 4 and 5, respectively.
Clearly, 4 < 5
∴ 5694437 < 5695440

56943201, 56943300 and 56944000 are all 8-digit numbers.
They have the same digit, i.e., 5 at the crores place.
They have the same digit, i.e., 6 at the ten lakhs place.
They have the same digit, i.e., 9 at the lakhs place.
They have the same digit, i.e., 4 at the ten thousands place.
However, at the thousands place, one number has 4 while the others have 3 .
∴ 56944000 is the largest.

The other two numbers have 3 and 2 at their hundreds places.
Clearly, 2 <3
∴ 56943201 < 56943300

The given numbers in ascending order are:
5694437 < 5695440 < 56943201 < 56943300 < 56944000

Page No 8:

Question 13:

Arrange the following numbers in ascending order:
700087, 8014257, 8015032, 10012458, 8014306

ANSWER:

700087 is 6-digit number.

8014257, 8014306 and 8015032 are all 7-digit numbers.
They have the same digits, namely 8, 0 and 1, at the ten lakhs, lakhs and ten thousands places, respectively.
But, at the thousands place, one number has 5 while the other two numbers have 4.
Here, 801503 is the largest.
The other two numbers have 2 and 3 at their hundreds places.
Clearly, 2 < 3
∴ 8014306  < 8015032

10012458 is an 8-digit number.

The given numbers in ascending order are:
700087 <  8014257 <  8014306  < 8015032 < 10012458

Page No 8:

Question 14:

Arrange the following numbers in ascending order:
1020304, 893245, 980134, 1021403, 893425, 1020216

ANSWER:

893245, 893425 and 980134 are all 6-digit numbers.
Among the three, 980134 is the largest.
The other two numbers have the same digits, namely 8, 9 and 3, at the lakhs, ten thousands and thousands places, respectively.
However, the digits at the hundreds place in 893245 and 893425 are 2 and 4, respectively.
Clearly, 2 < 4
∴ 893245 < 893425

1020216, 1020304 and 1021403 are all 7-digit numbers.
They have the same digits, namely 1, 0 and 2, at the ten lakhs, lakhs and ten thousands places, respectively.
At the thousands place, 1021403 has 1.
The other two numbers have the digits 2 and 3 at their hundreds places.
Clearly, 2 < 3
∴ 1020216 < 1020304

The given numbers in ascending order are:
893245 < 893425 <  980134 < 1020216 < 1020304 < 1021403

Exercise 1B

Question 1:

The number of persons who visited the holy shrine of Mata Vaishno Devi during last two consecutive years was 13789509 and 12976498 respectively. How many persons visited the shrine during these two years?

ANSWER:

Number of persons who visited the holy shrine in the first year = 13789509
Number of persons who visited the holy shrine in the second year = 12976498
∴ Number of persons who visited the holy shrine during these two years = 13789509 + 12976498 = 26766007

Page No 11:

Question 2:

Last year, three sugar factories in a town produced 24809565 bags, 18738576 bags and 9564568 bags of sugar respectively. How many bags were produced by all the three factories during last year?

ANSWER:

Bags of sugar produced by the first factory in last year = 24809565
Bags of sugar produced by the second factory in last year = 18738576
Bags of sugar produced by the third sugar factory in last year = 9564568
∴ Total number of bags of sugar were produced by the three factories during last year = 24809565 + 18738576 + 9564568
                                                                                                                                   = 53112709

Page No 11:

Question 3:

A number exceeds 37684955 by 3615045. What is that number?

ANSWER:

New number = Sum of 37684955 and 3615045
                      = 37684955 + 3615045
                      = 41300000

Page No 11:

Question 4:

There were three candidates in an election. They received 687905 votes, 495086 votes and 93756 votes respectively. The number of invalid votes was 13849. If 25467 persons did not vote, find how many votes were registered.

ANSWER:

Total number of votes received by the three candidates = 687905 + 495086 + 93756 = 1276747
Number of invalid votes = 13849
Number of persons who did not vote = 25467
∴ Total number of registered voters = 1276747 + 13849 + 25467
                                                        = 1316063

Page No 11:

Question 5:

A survey conducted on an Indian state shows that 1623546 people have only primary education; 9768678 people have secondary education; 6837954 people have higher education and 2684536 people are illiterate. If the number of children below the age of school admission is 698781, find the total population of the state.

ANSWER:

People who had only primary education = 1623546
People who had secondary education = 9768678
People who had higher education = 6837954
Illiterate people in the state = 2684536
Children below the age of school admission = 698781
∴ Total population of the state = 1623546 + 9768678 + 6837954 + 2684536 + 698781
                                                 = 21613495

Page No 11:

Question 6:

In a particular year a company produced 8765435 bicycles. Next year, the number of bicycles produced was 1378689 more than those produced in the preceding year.
How many bicycles were produced during the second year?
How many bicycles were produced during these two years?

ANSWER:

Bicycles produced by the company in the first year = 8765435
Bicycles produced by the company in the second year = 8765435 + 1378689
                                                                                        = 10144124

∴ Total number of bicycles produced during these two years = 8765435 + 10144124
                                                                                                       = 18909559

Page No 11:

Question 7:

The sale receipt of a company during a year was Rs 20956480. Next year, it increased by Rs 6709570. What was the total sale receipt of the company during these two years?

ANSWER:

Sale receipts of a company during the first year = Rs 20956480
Sale receipts of the company during the second year = Rs 20956480 + Rs 6709570
                                                                             = Rs 27666050

∴ Total number of sale receipts of the company during these two years = Rs 20956480 + Rs 27666050
                                                                                                                       = Rs 48622530

Page No 11:

Question 8:

The total population of a city is 28756304. If the number of males is 16987059, find the number of females in the city.

ANSWER:

Total population of the city = 28756304
Number of males in the city = 16987059
∴ Number of females in the city =  28756304 ‒ 16987059
                                                         = 11769245

Page No 12:

Question 9:

By how much is 13246510 larger than 4658642?

ANSWER:

Required number = 13246510 ‒ 4658642 = 8587868
∴ 13246510 is larger than 4658642 by 8587868.

Page No 12:

Question 10:

By how much is 5643879 smaller than one crore?

ANSWER:

Required number = 1 crore ‒ 564387
                             = 10000000 ‒ 5643879
                             = 4356121

∴ 5643879 is smaller than one crore by 4356121.

Page No 12:

Question 11:

What number must be subtracted from 11010101 to get 2635967?

ANSWER:

11010101 ‒ required number = 2635967

Thus, required number = 11010101 ‒ 2635967
                                        = 8374134

∴ The number 8374134 must be subtracted from 11010101 to get 2635967.

Page No 12:

Question 12:

The sum of two numbers is 10750308. If one of them is 8967519, what is the other number?

ANSWER:

Sum of the two numbers = 10750308
One of the number = 8967519

∴ The other number = 10750308 ‒ 8967519
                                    = 1782789

Page No 12:

Question 13:

A man had Rs 20000000 with him. He spent Rs 13607085 on buying a school building. How much money is left with him?

ANSWER:

Initial amount with the man = Rs 20000000
Amount spent on buying a school building = Rs 13607085

∴ Amount left with the man = Rs 20000000 ‒ Rs 13607085
                                               = Rs 6392915

Page No 12:

Question 14:

A society needed Rs 18536000 to buy a property. It collected Rs 7253840 as membership fee, took a loan of Rs 5675450 from a bank and collected Rs 2937680 as donation. How much is the society still short of?

ANSWER:

Money need by the society to buy the property = Rs 18536000
Amount collected as membership fee = Rs 7253840
Amount taken on loan from the bank = Rs 5675450
Amount collected as donation = Rs 2937680

∴ Amount of money short = Rs 18536000 ‒ (Rs 7253840 + Rs 5675450 + Rs 2937680)
                                             = Rs 18536000 ‒  Rs 15866970
                                              = Rs 2669030

Page No 12:

Question 15:

A man had Rs 10672540 with him. He gave Rs 4836980 to his wife, Rs 3964790 to his son and the rest to his daughter. How much money was received by the daughter?

ANSWER:

Initial amount with the man = Rs 10672540
Amount given to his wife = Rs 4836980
Amount given to his son = Rs 3964790

∴ Amount received by his daughter = Rs 10672540 ‒ (Rs 4836980 + Rs 3964790)
                                                            = Rs 10672540 ‒ Rs 8801770
                                                             = Rs 1870770

Page No 12:

Question 16:

The cost of a chair is Rs 1485. How much will 469 such chairs cost?

ANSWER:

Cost of one chair = Rs 1485
Cost of 469 chairs = Rs 1485 ×× 469
                              = Rs 696465

∴ Cost of 469 chairs is Rs 696465.

Page No 12:

Question 17:

How much money was collected from 1786 students of a school for a charity show if each student contributed Rs 625?

ANSWER:

Contribution from one student for the charity program = Rs 625
Contribution from 1786 students = Rs 625 x 1786 = Rs 1116250

∴ Rs 1116250 was collected from 1786 students for the charity program.

Page No 12:

Question 18:

A factory produces 6985 screws per day. How many screws will it produce in 358 days?

ANSWER:

Number of screws produced by the factory in one day = 6985
Number of screws produced in 358 days = 6985 x 358
                                                                    = 2500630

∴ The factory will produce 2500630 screws in 358 days.

Page No 12:

Question 19:

Mr Bhaskar saves  Rs 8756 every month. How much money will he save in 13 years?

ANSWER:

We know that
1 year = 12 months
13 years = 13 x 12 = 156 months

Now, we have:
Amount saved by Mr Bhaskar in one month = Rs 8756
Amount saved in 156 months = Rs 8756 ×× 156 = Rs 1365936

∴ Mr Bhaskar will save Rs 1365936 in 13 years.

Page No 12:

Question 20:

A scooter costs Rs 36725. How much will 487 such scooters cost?

ANSWER:

Cost of one scooter = Rs 36725
Cost of 487 scooter = Rs 36725 ×× 487
                                = Rs 17885075

∴ The cost of 487 scooters will be Rs 17885075.

Page No 12:

Question 21:

An aeroplane covers 1485 km in 1 hour. How much distance will it cover in 72 hours?

ANSWER:

Distance covered by the aeroplane in one hour = 1485 km
Distance covered in 72 hours = 1485 km ×× 72 = 106920 km

∴ The distance covered by the aeroplane in 72 hours will be 106920 km.

Page No 12:

Question 22:

The product of two numbers is 13421408. If one of the numbers is 364, find the other.

ANSWER:

Product of two numbers = 13421408
One of the number = 364

∴ The other number = 13421408 ÷ 364                                   
                                   = 36872

Page No 12:

Question 23:

If 36 flats cost Rs 68251500, what is the cost of each such flat?

ANSWER:

Cost of 36 flats = Rs 68251500
Cost of one flat = Rs 68251500 ÷ 36
                             = Rs 1895875

∴ Each flat costs Rs 1895875.

Page No 12:

Question 24:

The mass of a cylinder filled with gas is 30 kg 250 g and the mass of the empty cylinder is 14 kg 480 g. How much is the mass of the gas contained in it?

ANSWER:

We know that 1 kg = 1000 g
Now, mass of the gas-filled cylinder = 30 kg 250 g = 30.25 kg
Mass of an empty cylinder = 14 kg 480 g = 14.48 kg

∴ Mass of the gas contained in the cylinder = 30.25 kg ‒ 14.48 kg
                                                                             = 15.77 kg = 15 kg 770 g

Page No 12:

Question 25:

From a cloth 5 m long, a piece of length 2 m 85 cm is cut off. What is the length of the remaining piece?

ANSWER:

We know that 1 m = 100 cm
Length of the cloth = 5 m
Length of the piece cut off from the cloth = 2 m 85 cm

∴ Length of the remaining piece of cloth = 5 m ‒ 2.85 m
                                                                        = 2.15 m = 2 m 15 cm
                                                               

Page No 12:

Question 26:

In order to make a shirt, a length of 2 m 75 cm of cloth is needed. How much length of the cloth will be required for 16 such shirts?

ANSWER:

We know that 1 m = 100 cm
Now, length of the cloth required to make one shirt = 2 m 75 cm
Length of the cloth required to make 16 such shirts = 2 m 75 cm ×× 16
                                                                             = 2.75 m ×× 16
                                                                             = 44 m

∴ The length of the cloth required to make 16 shirts will be 44 m.

Page No 12:

Question 27:

For making 8 trousers of the same size, 14 m 80 cm of cloth is needed. How much cloth will be required for each such trouser?

ANSWER:

We know that 1 m = 100 cm
Cloth needed for making 8 trousers = 14 m 80 cm
Cloth needed for making 1 trousers = 14 m 80 cm ÷ 8
                                                           = 14 .8 m ÷ 8
                                                            = 1.85 m = 1 m 85 cm

∴ 1 m 85 cm of cloth will be required to make one shirt.

Page No 12:

Question 28:

The mass of a brick is 2 kg 750 g. What is the total mass of 14 such bricks?

ANSWER:

We know that 1 kg = 1000 g
Now, mass of one brick = 2 kg 750 g
∴ Mass of 14 such bricks = 2 kg 750 g ×× 14
                                              = 2.75 kg ×× 14
                                                = 38.5 kg = 38 kg 500 g

Page No 12:

Question 29:

The total mass of 8 packets, each of the same size, is 10 kg 600 g. What is the mass of each such packet?

ANSWER:

We know that 1 kg = 1000 g
Now, total mass of 8 packets of the same size = 10 kg 600 g
∴ Mass of one such packet = 10 kg 600 g ÷ 8
                                                 = 10.6 kg ÷ 8
                                                    = 1.325 kg = 1 kg 325 g

Page No 12:

Question 30:

A rope of length 10 m has been divided into 8 pieces of the same length. What is the length of each piece?

ANSWER:

Length of the rope divided into 8 equal pieces = 10 m
Length of one piece = 10 m ÷ 8
                                 = 1.25 m = 1 m 25 cm     [∵ 1 m = 100 cm]

Page No 14:

Exercise 1C

Question 1:

Round each of the following numbers to the nearest ten:
(a) 36
(b) 173
(c) 3869
(d) 16378

ANSWER:

(i) In 36, the ones digit is 6 > 5.
     ∴ The required rounded number = 40

(ii) In 173, the ones digit is 3 < 5.
     ∴ The required rounded number = 170

(iii) In 3869, the ones digit is 9 > 5.
     ∴ The required rounded number = 3870

(iv) In 16378, the ones digit is 8 > 5.
     ∴ The required rounded number = 16380

Page No 14:

Question 2:

Round each of the following numbers to the nearest hundred:
(a) 814
(b) 1254
(c) 43126
(d) 98165

ANSWER:

(i) In 814, the tens digit is 1 < 5.
     ∴ The required rounded number = 800

(ii) In 1254, the tens digit is 5 = 5
      ∴ The required rounded number = 1300

(iii) In 43126, the tens digit is 2 < 5
      ∴ The required rounded number = 43100

(iv) In 98165, the tens digit is 6 > 5
      ∴ The required rounded number = 98200

Page No 14:

Question 3:

Round each of the following numbers to the nearest thousand:
(a) 793
(b) 4826
(c) 16719
(d) 28394

ANSWER:

(i) In 793, the hundreds digit is 7 > 5
      ∴ The required rounded number = 1000

(ii) In 4826, the hundreds digit is 8 > 5
      ∴ The required rounded number = 5000

(iii) In 16719, the hundreds digit is 7 > 5
      ∴ The required rounded number = 17000

(iv) In 28394, the hundreds digit is 3 < 5
      ∴ The required rounded number = 28000

Page No 14:

Question 4:

Round each of the following numbers to the nearest ten thousand:
(a) 17514
(b) 26340
(c) 34890
(d) 272685

ANSWER:

(i) In 17514, the thousands digit is 7 > 5
      ∴ The required rounded number = 20000

(ii) In 26340, the thousands digit is 6 > 5
      ∴ The required rounded number = 30000

(iii) In 34890, the thousands digit is 4 < 5
      ∴ The required rounded number = 30000

(iv) In 272685, the thousands digit is 2 < 5
      ∴ The required rounded number = 270000  

Page No 14:

Question 5:

Estimate each sum to the nearest ten:
(57 + 34)

ANSWER:

57 estimated to the nearest ten = 60
34 estimated to the nearest ten = 30

∴ The required estimation = (60 + 30) = 90

Page No 14:

Question 6:

Estimate each sum to the nearest ten:
(43 + 78)

ANSWER:

43 estimated to the nearest ten = 40
78 estimated to the nearest ten = 80
∴ The required estimation = (40 + 80) = 120

Page No 14:

Question 7:

Estimate each sum to the nearest ten:
(14 + 69)

ANSWER:

14 estimated to the nearest ten = 10
69 estimated to the nearest ten = 70
∴ The required estimation = (10 + 70) = 80

Page No 14:

Question 8:

Estimate each sum to the nearest ten:
(86 + 19)

ANSWER:

86 estimated to the nearest ten = 90
19 estimated to the nearest ten = 20
∴ The required estimation = (90 + 20) = 110

Page No 14:

Question 9:

Estimate each sum to the nearest ten:
(95 + 58)

ANSWER:

95 estimated to the nearest ten = 100
58 estimated to the nearest ten = 60
∴ The required estimation = (100 + 60) = 160

Page No 14:

Question 10:

Estimate each sum to the nearest ten:
(77 + 63)

ANSWER:

77 estimated to the nearest ten = 80
63 estimated to the nearest ten = 60
∴ The required estimation = (80 + 60) = 140

Page No 14:

Question 11:

Estimate each sum to the nearest ten:
(356 + 275)

ANSWER:

356 estimated to the nearest ten = 360
275 estimated to the nearest ten = 280
∴ The required estimation = (360 + 280) = 640

Page No 14:

Question 12:

Estimate each sum to the nearest ten:
(463 + 182)

ANSWER:

463 estimated to the nearest ten = 460
182 estimated to the nearest ten = 180
∴ The required estimation = (460 + 180) = 640

Page No 14:

Question 13:

Estimate each sum to the nearest ten:
(538 + 276)

ANSWER:

538 estimated to the nearest ten = 540
276 estimated to the nearest ten = 280
∴ The required estimation = (540 + 280) = 820

Page No 14:

Question 14:

Estimate each sum to the nearest hundred:
(236 + 689)

ANSWER:

236 estimated to the nearest hundred = 200
689 estimated to the nearest hundred = 700
∴ The required estimation = (200 + 700) = 900

Page No 14:

Question 15:

Estimate each sum to the nearest hundred:
(458 + 324)

ANSWER:

458 estimated to the nearest hundred = 500
324 estimated to the nearest hundred = 300
∴ The required estimation = (500 + 300) = 800

Page No 14:

Question 16:

Estimate each sum to the nearest hundred:
(170 + 395)

ANSWER:

170 estimated to the nearest hundred = 200
395 estimated to the nearest hundred = 400
∴ The required estimation = (200 + 400) = 600

Page No 15:

Question 17:

Estimate each sum to the nearest hundred:
(3280 + 4395)

ANSWER:

3280 estimated to the nearest hundred = 3300
4395 estimated to the nearest hundred = 4400
∴ The required estimation = (3300 + 4400) = 7700

Page No 15:

Question 18:

Estimate each sum to the nearest hundred:
(5130 + 1410)

ANSWER:

5130 estimated to the nearest hundred = 5100
1410 estimated to the nearest hundred = 1400
∴ The required estimation = (5100 + 1400) = 6500

Page No 15:

Question 19:

Estimate each sum to the nearest hundred:
(10083 + 29380)

ANSWER:

10083 estimated to the nearest hundred = 10100
29380 estimated to the nearest hundred = 29400
∴ The required estimation = (10100 + 29400) = 39500

Page No 15:

Question 20:

Estimate each sum to the nearest thousand:
(32836 + 16466)

ANSWER:

32836 estimated to the nearest thousand = 33000
16466 estimated to the nearest thousand = 16000
∴ The required estimation = (33000 + 16000) = 49000

Page No 15:

Question 21:

Estimate each sum to the nearest thousand:
(46703 + 11375)

ANSWER:

46703 estimated to the nearest thousand = 47000
11375 estimated to the nearest thousand = 11000
∴ The required estimation = (47000 + 11000) = 58000

Page No 15:

Question 22:

Estimate each sum to the nearest thousand:
There are 54 balls in box A and 79 balls in box B. Estimate the total number of balls in both the boxes taken together.

ANSWER:

Number of balls in box A = 54
Number of balls in box B = 79
Estimated number of balls in box A = 50
Estimated number of balls in box B = 80
∴ Total estimated number of balls in both the boxes = (50 + 80) = 130

Page No 15:

Question 23:

Estimate each difference to the nearest ten:
(53 − 18)

ANSWER:

We have,
53 estimated to the nearest ten = 50
18 estimated to the nearest ten = 20
∴ The required estimation = (50 ‒ 20) = 30

Page No 15:

Question 24:

Estimate each difference to the nearest ten:
(97 − 38)

ANSWER:

100 estimated to the nearest ten = 100
38 estimated to the nearest ten = 40
∴ The required estimation = (100 ‒ 40) = 60

Page No 15:

Question 25:

Estimate each difference to the nearest ten:
(409 − 148)

ANSWER:

409 estimated to the nearest ten = 410
148 estimated to the nearest ten = 150
∴ The required estimation = (410 ‒ 150) = 260

Page No 15:

Question 26:

Estimate each difference to the nearest hundred:
(678 − 215)

ANSWER:

678 estimated to the nearest hundred = 700
215 estimated to the nearest hundred = 200
∴ The required estimation = (700 ‒ 200) = 500

Page No 15:

Question 27:

Estimate each difference to the nearest hundred:
(957 − 578)

ANSWER:

957 estimated to the nearest hundred = 1000
578 estimated to the nearest hundred = 600
∴ The required estimation = (1000 ‒ 600) = 400

Page No 15:

Question 28:

Estimate each difference to the nearest hundred:
(7258 − 2429)

ANSWER:

7258 estimated to the nearest hundred = 7300
2429 estimated to the nearest  hundred = 2400
∴ The required estimation = (7300 ‒ 2400) = 4900

Page No 15:

Question 29:

Estimate each difference to the nearest hundred:
(5612 − 3095)

ANSWER:

5612 estimated to the nearest hundred = 5600
3095 estimated to the nearest hundred = 3100
∴ The required estimation = (5600 ‒ 3100) = 2500

Page No 15:

Question 30:

Estimate each difference to the nearest thousand:
(35863 − 27677)

ANSWER:

35863 estimated to the nearest thousand = 36000
27677 estimated to the nearest  thousand = 28000
∴ The required estimation = (36000 ‒ 28000) = 8000

Page No 15:

Question 31:

Estimate each difference to the nearest thousand:
(47005 − 39488)

ANSWER:

47005 estimated to the nearest thousand = 47000
39488 estimated to the nearest  thousand = 39000
∴ The required estimation = (47000 ‒ 39000) = 8000

Page No 15:

Exercise 1D

Question 1:

Estimate each of the following products by rounding off each number to the nearest ten:
38 × 63

ANSWER:

38 estimated to the nearest ten = 40
63 estimated to the nearest ten = 60
∴ The required estimation = (40 ×× 60) = 2400

Page No 15:

Question 2:

Estimate each of the following products by rounding off each number to the nearest ten:
54 × 47

ANSWER:

54 estimated to the nearest ten = 50
47 estimated to the nearest ten = 50
∴ The required estimation = (50 ×× 50) = 2500

Page No 15:

Question 3:

Estimate each of the following products by rounding off each number to the nearest ten:
28 × 63

ANSWER:

28 estimated to the nearest ten = 30
63 estimated to the nearest ten = 60
∴ The required estimation = (30 ×× 60) = 1800

Page No 15:

Question 4:

Estimate each of the following products by rounding off each number to the nearest ten:
42 × 75

ANSWER:

42 estimated to the nearest ten = 40
75 estimated to the nearest ten = 80
∴ The required estimation = (40 ×× 80) = 3200

Page No 15:

Question 5:

Estimate each of the following products by rounding off each number to the nearest ten:
64 × 58

ANSWER:

64 estimated to the nearest ten = 60
58 estimated to the nearest ten = 60
∴ The required estimation = (60 ×× 60) = 3600

Page No 15:

Question 6:

Estimate each of the following products by rounding off each number to the nearest ten:
15 × 34

ANSWER:

15 estimated to the nearest ten = 20
34 estimated to the nearest ten = 30
∴ The required estimation = (20 ×× 30) = 600

Page No 16:

Question 7:

Estimate each of the following products by rounding off each number to the nearest hundred:
376 × 123

ANSWER:

376 estimated to the nearest hundred = 400
123 estimated to the nearest hundred = 100
∴ The required estimation = (400 ×× 100) = 40000

Page No 16:

Question 8:

Estimate each of the following products by rounding off each number to the nearest hundred:
264 × 147

ANSWER:

264 estimated to the nearest hundred = 300
147 estimated to the nearest hundred = 100
∴ The required estimation = (300 ×× 100) = 30000

Page No 16:

Question 9:

Estimate each of the following products by rounding off each number to the nearest hundred:
423 × 158

ANSWER:

423 estimated to the nearest hundred = 400
158 estimated to the nearest hundred = 200
∴ The required estimation = (400 ×× 200) = 80000

Page No 16:

Question 10:

Estimate each of the following products by rounding off each number to the nearest hundred:
509 × 179

ANSWER:

509 estimated to the nearest hundred = 500
179 estimated to the nearest hundred = 200
∴ The required estimation = (500 ×× 200) = 100000

Page No 16:

Question 11:

Estimate each of the following products by rounding off each number to the nearest hundred:
392 × 138

ANSWER:

392 estimated to the nearest hundred = 400
138 estimated to the nearest hundred = 100
∴ The required estimation = (400 ×× 100) = 40000

Page No 16:

Question 12:

Estimate each of the following products by rounding off each number to the nearest hundred:
271 × 339

ANSWER:

271 estimated to the nearest hundred = 300
339 estimated to the nearest hundred = 300
∴ The required estimation = (300 ×× 300) = 90000

Page No 16:

Question 13:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
183 × 154

ANSWER:

183 estimated upwards = 200
154 estimated downwards = 100
∴ The required product = (200 ×× 100) = 20000

Page No 16:

Question 14:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
267 × 146

ANSWER:

267 estimated upwards = 300
146 estimated downwards = 100
∴ The required product = (300 ×× 100) = 30000

Page No 16:

Question 15:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
359 × 76

ANSWER:

359 estimated upwards = 400
76 estimated downwards = 70
∴ The required product = (400 ×× 70) =28000

Page No 16:

Question 16:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
472 × 158

ANSWER:

472 estimated upwards = 500
158 estimated downwards = 100
∴ The required product = (500 ×× 100) = 50000

Page No 16:

Question 17:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
680 × 164

ANSWER:

680 estimated upwards = 700
164 estimated downwards = 100
∴ The required product = (700 ×× 100) = 70000

Page No 16:

Question 18:

Estimate each of the following products by rounding off the first number upwards and the second number downwards:
255 × 350

ANSWER:

255 estimated upwards = 300
350 estimated downwards = 300
∴ The required product = (300 ×× 300) = 90000

Page No 16:

Question 19:

Estimate each of the following products by rounding off the first number downwards and the second number upwards:
356 × 278

ANSWER:

356 estimated downwards = 300
278 estimated upwards = 300
∴ The required product = (300 ×× 300) = 90000

Page No 16:

Question 20:

Estimate each of the following products by rounding off the first number downwards and the second number upwards:
472 × 76

ANSWER:

472 estimated downwards = 400
76 estimated upwards = 80
∴ The required product = (400 ×× 80) = 32000

Page No 16:

Question 21:

Estimate each of the following products by rounding off the first number downwards and the second number upwards:
578 × 369

ANSWER:

578 estimated downwards = 500
369 estimated upwards = 400
∴  The required product = (500 ×× 400) = 200000

Page No 16:

Exercise 1E

Question 1:

Find the estimated quotient for each of the following:
87 ÷ 28

ANSWER:

87 ÷ 28 is approximately equal to 90 ÷ 30 = 3.

Page No 16:

Question 2:

Find the estimated quotient for each of the following:
83 ÷ 17

ANSWER:

The estimated quotient for 83 ÷ 17 is approximately equal to 80 ÷ 20 = 8 ÷ 2 = 4.

Page No 16:

Question 3:

Find the estimated quotient for each of the following:
75 ÷ 23

ANSWER:

The estimated quotient of 75 ÷ 23 is approximately equal to 80 ÷ 20 = 8 ÷ 2 = 4.

Page No 16:

Question 4:

Find the estimated quotient for each of the following:
193 ÷ 24

ANSWER:

The estimated quotient of 193 ÷ 24 is approximately equal to 200 ÷ 20 = 20 ÷ 2 = 10.

Page No 16:

Question 5:

Find the estimated quotient for each of the following:
725 ÷ 23

ANSWER:

The estimated quotient of 725 ÷ 23 is approximately equal to 700 ÷ 20 = 70 ÷ 2 = 35.

Page No 16:

Question 6:

Find the estimated quotient for each of the following:
275 ÷ 25

ANSWER:

The estimated quotient of 275 ÷ 25 is approximately equal to 300 ÷ 30 = 30 ÷ 3 = 10.

Page No 16:

Question 7:

Find the estimated quotient for each of the following:
633 ÷ 33

ANSWER:

The estimated quotient of 633 ÷ 33 is approximately equal to 600 ÷ 30 = 60 ÷ 3 = 20.

Page No 16:

Question 8:

Find the estimated quotient for each of the following:
729 ÷ 29

ANSWER:

729 ÷ 29 is approximately equal to 700 ÷ 30 or 70 ÷ 3, which is approximately equal to 23.

Page No 16:

Question 9:

Find the estimated quotient for each of the following:
858 ÷ 39

ANSWER:

858 ÷ 39 is approximately equal to 900 ÷ 40 or 90 ÷ 4, which is approximately equal to 23.

Page No 16:

Question 10:

Find the estimated quotient for each of the following:
868 ÷ 38

ANSWER:

868 ÷ 38 is approximately equal to 900 ÷ 40 or 90 ÷ 4, which is approximately equal to 23.

Page No 19:

Exercise 1F

Question 1:

Express each of the following as a Roman numeral:
(i) 2
(ii) 8
(iii) 14
(iv) 29
(v) 36
(vi) 43
(vii) 54
(viii) 61
(ix) 73
(x) 81
(xi) 91
(xii) 95
(xiii) 99
(xiv) 105
(xv) 114

ANSWER:

We may write these numbers as given below:
(i) 2 = II
(ii) 8 = (5 + 3) = VIII
(iii) 14 = (10 + 4) = XIV
(iv) 29 = ( 10 + 10 + 9 ) = XXIX
(v) 36 = (10 + 10 + 10 + 6) = XXXVI
(vi) 43 = (50 – 10) + 3 = XLIII
(vii) 54 = (50 + 4) = LIV
(viii) 61= (50 + 10 + 1) = LXI
(ix) 73 = ( 50 + 10 + 10 + 3) = LXXIII
(x) 81 = (50 + 10 + 10 + 10 + 1) = LXXXI
(xi) 91 =(100 – 10) + 1 = XCI
(xii) 95 = (100 – 10) + 5 = XCV
(xiii) 99 = (100 – 10) + 9 = XCIX
(xiv) 105 = (100 + 5) = CV
(xv) 114 = (100 + 10) + 4 = CXIV

Page No 19:

Question 2:

Express each of the following as a Roman numeral:
(i) 164
(ii) 195
(iii) 226
(iv) 341
(v) 475
(vi) 596
(vii) 611
(viii) 759

ANSWER:

We may write these numbers in Roman numerals as follows:

(i) 164 = (100 + 50 + 10 + 4) = CLXIV
(ii) 195 = 100 + (100 – 10) + 5 = CXCV
(iii) 226 = (100 + 100 + 10 + 10 + 6) = CCXXVI
(iv) 341= 100 + 100+ 100 + (50 -10) + 1 = CCCXLI
(v) 475 = (500 – 100) + 50 + 10 + 10 + 5 = CDLXXV
(vi) 596 = 500 +  (100 – 10) + 6 = DXCVI
(vii) 611= 500 + 100 + 11 = DCXI
(viii) 759 = 500 + 100 + 100 + 50 + 9 = DCCLIX

Page No 19:

Question 3:

Write each of the following as a Hindu-Arabic numeral:
(i) XXVII
(ii) XXXIV
(iii) XLV
(iv) LIV
(v) LXXIV
(vi) XCI
(vii) XCVI
(viii) CXI
(ix) CLIV
(x) CCXXIV
(xi) CCCLXV
(xii) CDXIV
(xiii) CDLXIV
(xiv) DVI
(xv) DCCLXVI

ANSWER:

We can write the given Roman numerals in Hindu-Arabic numerals as follows:

(i) XXVII = 10 + 10 + 7 = 27
(ii) XXXIV = 10 + 10 + 10 + 4 = 34
(iii) XLV = (50 − 10 ) + 5 = 45
(iv) LIV = 50 + 4 = 54
(v) LXXIV = 50 + 10 + 10 + 4 = 74
(vi) XCI = (100 − 10) + 1 = 91
(vii) XCVI = (100 − 10) + 6 = 96
(viii) CXI = 100 + 10 + 1= 111
(ix) CLIV = 100 + 50 + 4 = 154
(x) CCXXIV = 100 + 100 + 10 + 10 + 4 = 224
(xi) CCCLXV = 100 +  100 + 100 + 50 + 10 + 5 = 365
(xii) CDXIV = (500 − 100) + 10 + 4 = 414
(xiii) CDLXIV = (500 − 100) + 50 + 10 + 4 = 464
(xiv) DVI = 500 + 6= 506
(xv) DCCLXVI = 500 + 100 + 100 + 50 + 10 + 6 = 766

Page No 19:

Question 4:

Show that each of the following is meaningless. Give reason in each case.
(i) VC
(ii) IL
(iii) VVII
(iv) IXX

ANSWER:

(i) VC is wrong because V, L and D are never subtracted.
(ii) IL is wrong because I can be subtracted from V and X only.
(iii) VVII is wrong because V, L and D are never repeated.
(iv) IXX is wrong because X (ten) must be placed before IX (nine).

Page No 20:

Exercise 1G

Question 1:

Mark against the correct answer
The place value of 6 in the numeral 48632950 is
(a) 6
(b) 632950
(c) 600000
(d) 486

ANSWER:

Option c is correct.

Place value of 6 = 6 lakhs = (6 ×× 100000) = 600000

Page No 20:

Question 2:

Mark against the correct answer
The face value of 4 in the numeral 89247605 is
(a) 4
(b) 40000
(c) 47605
(d) 8924

ANSWER:

Option a is correct.

The face value of a digit remains as it is irrespective of the place it occupies in the place value chart.
Thus, the face value of 4 is always 4 irrespective of where it may be.

Page No 20:

Question 3:

Mark against the correct answer
The difference between the place value and the face value of 5 in the numeral 78653421 is
(a) 53416
(b) 4995
(c) 49995
(d) none of these

ANSWER:

Option c is correct.

Place value of 5 = 5 ×× 10000 = 50000
Face value of 5 = 5

∴ Required difference = 50000 − 5 = 49995

Page No 20:

Question 4:

Mark against the correct answer
The smallest counting number is
(a) 0
(b) 1
(c) 10
(d) none of these

ANSWER:

Option b is correct.

The smallest counting number is 1.

Page No 20:

Question 5:

Mark against the correct answer
How many 4-digit numbers are there?
(a) 8999
(b) 9000
(c) 8000
(d) none of these

ANSWER:

Option b is correct.

The largest four-digit number = 9999
The smallest four-digit number = 1000
Total number of all four-digit numbers = (9999 − 1000) + 1
                                                                 = 8999 + 1
                                                                      = 9000
                                           

Page No 20:

Question 6:

Mark against the correct answer
How many 7-digit numbers are there?
(a) 8999999
(b) 9000000
(c) 10000000
(d) none of these

ANSWER:

Option b is correct.

The largest seven-digit number = 9999999
The smallest seven-digit number = 1000000
Total number of seven-digit numbers = (9999999 − 1000000) + 1
                                                            = 8999999 + 1
                                                             = 9000000
                                        

Page No 20:

Question 7:

Mark against the correct answer
How many 8-digit numbers are there?
(a) 99999999
(b) 89999999
(c) 90000000
(d) none of these

ANSWER:

Option c is correct.

The largest eight-digit number = 99999999
The smallest eight-digit number = 10000000
Total number of eight-digit numbers = (99999999 − 10000000) + 1
                                                             = 89999999 + 1
                                                               = 90000000
                                               

Page No 20:

Question 8:

Mark against the correct answer
What comes just before 1000000?
(a) 99999
(b) 999999
(c) 9999999
(d) none of these

ANSWER:

Option b is correct.

The number just before 1000000 is 999999 (i.e., 1000000 − 1).

Page No 20:

Question 9:

Mark against the correct answer
Which of the following is not meaningful?
(a) VX
(b) XV
(c) XXV
(d) XXXV

ANSWER:

Option a is correct.

V, L and D are never subtracted. Thus, VX is wrong.

Page No 20:

Question 10:

Mark against the correct answer
Which of the following is not meaningful?
(a) CI
(b) CII
(c) IC
(d) XC

ANSWER:

Option c is correct.

I can be subtracted from V and X only. Thus, IC is wrong.

Page No 20:

Question 11:

Mark against the correct answer
Which of the following is not meaningful?
(a) XIV
(b) XVV
(c) XIII
(d) XXII

ANSWER:

Option b is correct.

V, L and D are never repeated. Thus, XVV is meaningless.

Page No 21:

Exercise 1G

Question 1:

Write each of the following numerals in words:
(i) 16, 06, 23, 708
(ii) 14, 23, 08, 915

ANSWER:

(i) Sixteen crore six lakh twenty-three thousand seven hundred eight
(ii) Fourteen crore twenty-three lakh eight thousand nine hundred fifteen

Page No 21:

Question 2:

Write each of the following numerals in words:
(i) 80, 060, 409
(ii) 234, 150, 319

ANSWER:

(i) Eighty million sixty thousand four hundred nine
(ii) Two hundred thirty-four million one hundred fifty thousand three hundred nineteen

Page No 21:

Question 3:

Arrange the following numbers in ascending order:
3903216, 19430124, 864572, 6940513, 16531079

ANSWER:

We have,
864572 is a 6-digit number.

3903216 and  6940513 are seven-digit numbers.
At the ten lakhs place, one number has 3, while the second number has 6.
Clearly, 3 < 6
∴ 3903216 <  6940513

16531079  and 19430124 are eight-digit numbers.
At the crores place, both the numbers have the same digit, namely 1.
At the ten lakhs place, one number has 6, while the second number has 9.
Clearly, 6 < 9
∴ 16531079  < 19430124

The given numbers in ascending order are:
864572 < 3903216 < 6940513 < 16531079 < 19430124

Page No 21:

Question 4:

Arrange the following numbers in descending order:
54796203, 4675238, 63240613, 5125648, 589623

ANSWER:

63240613 and 54796203 are both eight-digit numbers.
At the crores place, one number has 6, while the second number has 5.
Clearly, 5 < 6
∴ 63240613 > 54796203

5125648 and 4675238 are both seven-digit numbers.
However, at the ten lakhs place, one number has 5, while the second number has 4.
Clearly, 4 < 5
∴ 5125648 > 4675238

589623 is a six-digit number.

The given numbers in descending order are:
63240613 > 54796203 > 5125648 > 4675238 > 589623

Page No 21:

Question 5:

How many 7-digit numbers are there in all?

ANSWER:

The largest seven-digit number = 9999999
The smallest seven-digit number  = 1000000
Number of all seven-digits numbers = (9999999 − 1000000) + 1
                                                    = 899999 + 1
                                                    = 9000000

Hence, there is a total of ninety lakh 7-digit numbers.

Page No 21:

Question 6:

Write the largest and smallest numbes using each of the digits 1, 4, 6, 8, 0 only once and find their difference.

ANSWER:

The largest number using each of the digits: 1, 4, 6, 8 and 0, is 86410.
The smallest  number using each of the digits: 1, 4, 6, 8 and 0, is 10468.
∴ Required difference = 86410 − 10468
                                       = 75942

Page No 21:

Question 7:

Write the Hindu-Arabic numeral for each of the following:
(i) CCXLII
(ii) CDLXV
(iii) LXXVI
(iv) DCCXLI
(v) XCIV
(vi) CXCIX

ANSWER:

(i) CCXLII = 100 + 100 + (50 − 10) + 2 = 242
(ii) CDLXV = (500 − 100) + 50 + 10 + 5 = 465
(iii) LXXVI = 50 + 10 + 10 + 6 = 76
(iv) DCCXLI = 500 + 100 + 100 + ( 50 − 10) + 1 = 741
(v) XCIV = (100 − 10) + 4 = 94
(vi) CXCIX = 100 + (100 − 10) + 9 = 199

Page No 21:

Question 8:

Write the Roman numeral for each of the following:
(i) 84
(ii) 99
(iii) 145
(iv) 406
(v) 519

ANSWER:

(i) 84 = 50 + 30 + 4 = LXXXIV
(ii) 99 = 90 + 9 =  XCIX
(iii) 145 = 100 + (50 − 10) + 5 = CXLV
(iv) 406 = 400 + 6 = CDVI
(v) 519 = 500 +10 + 9 = DXIX

Page No 21:

Question 9:

Write the successor and predecessor of 999999 and find their difference.

ANSWER:

Successor of 999999 = 999999 + 1 = 1000000
Predecessor of 999999 = 999999 − 1 = 999998
∴ Required difference = 1000000 − 999998
                                       = 2

Page No 21:

Question 10:

Round off each of the following to the nearest thousand:
(i) 1046
(ii) 973
(iii) 5624
(iv) 4368

ANSWER:

(i) The number is 1046. Its digit at the hundreds place is 0 < 5.
     So, the given number is rounded off to the nearest thousand as 1000.

(ii) The number is 973. Its digit at the hundreds place is 9 > 5.
      So, the given number is rounded off to the nearest thousand as 1000.

(iii) The number is 5624. Its digit at the hundreds place is 6 > 5.
       So, the given number is rounded off to the nearest thousand as 6000.

(iv) The number is 4368. Its digit at the hundreds place is 3 < 5.
       So, the given number is rounded off to the nearest thousand as 4000.

Page No 21:

Question 11:

Which of the  following Roman numerals is correct?
(a) XC
(b) XD
(c) DM
(d) VL

ANSWER:

Option (a) is correct.

X can be subtracted from L and C only.
i.e., XC = ( 100 − 10 ) = 90

Page No 21:

Question 12:

1 Lakh = …… thousands.
(a) 10
(b) 100
(c) 1000
(d) none of these

ANSWER:

Option (b) is correct.

One lakh (100000) is equal to one hundred thousand (100,000).

Page No 21:

Question 13:

No Roman numeral can be repeated more than ….. times.
(a) two
(b) three
(c) four
(d) none of these

ANSWER:

Option (b) is correct.

No Roman numeral can be repeated more than three times.

Page No 21:

Question 14:

How many times does the digit 9 occur between 1 and 100?
(a) 11
(b) 15
(c) 18
(d) 20

ANSWER:

Option (d) is correct.

Between 1 and 100, the digit 9 occurs in 9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98 and 99.
∴ The digit occurs 20 times between 1 and 100.

Page No 21:

Question 15:

(7268 − 2427) estimated to the nearest hundred is
(a) 4900
(b) 4800
(c) 4841
(d) 5000

ANSWER:

Option (a) is correct.

7268 will be rounded off to the nearest hundred as 7300.
2427 will be rounded of  to the nearest hundred as 2400.
∴ 7300 − 2400 = 4900

Page No 21:

Question 16:

One million = …… .
(a) 1 lakh
(b) 10 lakh
(c) 100 lakh
(d) 1 crore

ANSWER:

Option (b) is correct.

1 million (1,000,000) = 10 lakh (10 ×× 1,00,000)

Page No 21:

Question 17:

1512 when round off to the nearest hundred is
(a) 1600
(b) 1500
(c) 1510
(d) none of these

ANSWER:

Option (b) is correct.

The number is 1512. Its digit at the tens place is 1 < 5.
So, the given number is rounded off to the nearest hundred as 1500.

Page No 21:

Question 18:

Which of the symbols are never repeated?
(a) V, X and C
(b) V, X and D
(c) V, L and D
(d) L, K and C

ANSWER:

Option (c) is correct.

In Roman numerals, V, L and D are never repeated and never subtracted.

Page No 21:

Question 19:

Write 86324805 separating periods in HIndu-Arabic system.

ANSWER:

Periods:     Crores      Lakhs           Thousands           Hundreds            Tens          Ones
Digits:            8             63                    24                        8                       0                5

Using commas, we write the given number as 8,63,24,805.

Page No 21:

Question 20:

Fill in the blanks:
(i) 1 crore = …… lakh
(ii) 1 crore = …… million
(iii) 564 when estimated to the nearest hundred is …… .
(iv) The smallest 4-digit number with four different digits is …… .

ANSWER:

(i) 1 crore =  100 lakh
(ii) 1 crore = 10 million
(iii) 564 when estimated to the nearest hundred is 600.
(iv) The smallest 4-digit number with four different digits is 1023.

Page No 22:

Question 21:

Write ‘T’ for true and ‘F’ for false
The difference in the face value and the place value of 5 in 85419 is 85414.

ANSWER:

F

Place value of 5 in 85419 = 5000
Face value of 5 in 85419 = 5
∴ Their difference = 5000 − 5 = 4995

Page No 22:

Question 22:

Write ‘T’ for true and ‘F’ for false
In Roman numerals V, L and D are never subtracted.

ANSWER:

T

In Roman numerals, V, L and D are never repeated and never subtracted.

Page No 22:

Question 23:

Write ‘T’ for true and ‘F’ for false
The successor of the greatest 5-digit number is 100000.

ANSWER:

T
Greatest five-digit number = 99999
Successor of 99999 = 99999 + 1 = 100000

Page No 22:

Question 24:

Write ‘T’ for true and ‘F’ for false
The estimated value of 46,530 to the nearest hundred is 46500.

ANSWER:

T

The number is 46,530. Its digit at the tens place is 3 < 5.
So, the number 46,530 is rounded off to the nearest hundred as 46,500.

Page No 22:

Question 25:

Write ‘T’ for true and ‘F’ for false
100 lakhs make a million.

ANSWER:

F

10 lakhs = 1 million

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RS Agarwal Solution | Class 11th | Chapter-8 |   Permutations | Edugrown

Exercise Ex. 8.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Exercise Ex. 8.2

Solution 1

Solution 2

3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

∴ 3! + 4! = 6 + 24 = 30

7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

∴ 3! + 4! ≠ 7!Solution 3

Solution 4

Solution 5

Exercise Ex. 8.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Exercise Ex. 8.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

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RS Agarwal Solution | Class 11th | Chapter-7 |   Linear Equations in Two Variable | Edugrown

Exercise Ex. 7A

Question 1

Solve each of the following systems of equations graphically:

2x + 3y = 2

x – 2y = 8Solution 1

Question 3

Solve each of the following systems of equations graphically:

2x + 3y = 8

x – 2y + 3 = 0Solution 3

Question 4

Solve each of the following systems of equations graphically:

2x – 5y + 4 = 0

2x + y – 8 = 0Solution 4

Question 5

Solve each of the following systems of equations graphically:

3x + 2y = 12, 5x – 2y = 4.Solution 5

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.Question 6

Solve each of the following systems of equations graphically:

3x + y + 1 = 0

2x – 3y + 8 = 0Solution 6

Question 7

Solve each of the following systems of equations graphically:

2x + 3y + 5 = 0, 3x – 2y – 12 = 0.Solution 7

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.Question 8

Solve each of the following systems of equations graphically:

2x – 3y + 13 = 0, 3x – 2y + 12 = 0.Solution 8

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.Question 9

Solve each of the following systems of equations graphically:

2x + 3y – 4 = 0, 3x – y + 5 = 0.Solution 9

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.Question 10

Solve each of the following systems of equations graphically:

x + 2y + 2 = 0

3x + 2y – 2 = 0Solution 10

Question 11

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 3 = 0, 2x + 3y – 4 = 0.Solution 11

Question 12

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

2x – 3y + 4 = 0, x + 2y – 5 = 0.Solution 12

Question 13

Solve the following system of linear equations graphically:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0

Find the area of the region bounded by these lines and the x-axis.Solution 13

Question 14

Solve the following system of linear equation graphically:

x – y + 1 = 0, 3x + 2y – 12 = 0

Calculate the area bounded by these lines and x-axis.Solution 14

Question 15

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – 2y + 2 = 0, 2x + y – 6 = 0.Solution 15

Question 16

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.Solution 16

Question 17

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

4x – y – 4 = 0, 3x + 2y – 14 = 0.Solution 17

Question 18

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

x – y – 5 = 0, 3x + 5y – 15 = 0.Solution 18

Question 19

Solve the following system of linear equations graphically:

2x – 5y + 4 = 0, 2x + y – 8 = 0

Find the point, where these lines meet the y-axisSolution 19

Question 20

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

5x – y – 7 = 0, x – y + 1 = 0.Solution 20

Question 21

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

2x – 3y = 12, x + 3y = 6.Solution 21

Question 22

Show graphically that each of the following given systems of equations has infinitely many solutions:

2x + 3y = 6, 4x + 6y = 12.Solution 22

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.Question 23

Show graphically that the system of equations 3x – y = 5, 6x – 2y = 10 has infinitely many solutions.Solution 23

Question 24

Show graphically that the system of equations 2x + y = 6, 6x + 3y = 18 has infinitely many solutions.Solution 24

Question 25

Show graphically that each of the following given systems of equations has infinitely many solutions:

x – 2y = 5, 3x – 6y = 15.Solution 25

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. Question 26

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

x – 2y = 6, 3x – 6y = 0Solution 26

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 27

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + 3y = 4, 4x + 6y = 12.Solution 27

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 28

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

2x + y = 6, 6x + 3y = 20.Solution 28

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.Question 29

Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.Solution 29

Exercise Ex. 7B

Question 1

Solve for x and y

x + y = 3

4x – 3y = 26Solution 1

Question 2

Solve for x and y:

Solution 2

Question 3

Solve for x and y

2x + 3y = 0

3x + 4y = 5Solution 3

Question 4

Solve for x and y

2x – 3y = 13

7x – 2y = 20Solution 4

Question 5

Solve for x and y

3x – 5y – 19 = 0

-7x + 3y + 1 = 0Solution 5

Question 6

Solve for x and y:

2x – y + 3 = 0, 3x – 7y + 10 = 0.Solution 6

Question 7

Solve for x and y:

Solution 7

Question 8

Solve for x and y

Solution 8

Question 9

Solve for x and y

Solution 9

Question 10

Solve for x and y

Solution 10

Question 11

Solve for x and y

Solution 11

Question 12

Solve for x and y

Solution 12

Question 13

Solve for x and y:

0.4x + 0.3y = 1.7, 0.7x – 0.2y = 0.8.Solution 13

Question 14

Solve for x and y:

0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74.Solution 14

Question 15

Solve for x and y

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2Solution 15

Question 16

Solve for x and y

6x + 5y = 7x + 2y + 1 = 2(x + 6y – 1)Solution 16

Question 17

Solve for x and y

Solution 17

Question 18

×Solve for x and y

Solution 18

Putting the given equations become

5u + 6y = 13—(1)

3u + 4y = 7 —-(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52—(3)

18u + 24y = 42—(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 – 25

6y = -12

y = -2

Question 19

Solve for x and y

Solution 19

The given equations are and 

Putting 

x + 6v = 6 —-(1)

3x – 8v = 5—(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24—(3)

9x – 24v = 15 —(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 – 3 = 3

solution is x = 3, y = 2Question 20

Solve for x and y

Solution 20

Putting in the given equation

2x – 3v = 9 —(1)

3x + 7v = 2 —(2)

Multiplying (1) by7 and (2) by 3

14x – 21v = 63 —(3)

9x + 21v = 6 —(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 – 3v = 9

         -3v = 9 – 6

     -3v= 3

       v = -1

the solution is x = 3, y = -1Question 21

Solve for x and y

Solution 21

Question 22

Solve for x and y

Solution 22

Putting in the equation

9u – 4v = 8 —(1)

13u + 7v = 101—(2)

Multiplying (1) by 7 and (2) by 4, we get

63u – 28v = 56—(3)

52u + 28v = 404—(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 – 4v = 8

36 – 4v = 8

-4v = 8 – 36

-4v = -28

Question 23

Solve for x and y

Solution 23

Putting in the given equation, we get

5u – 3v = 1 —(1)

Multiplying (1) by 4 and (2) by 3, we get

20u – 12v = 4—-(3)

27u + 12v = 90—(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) – 3v = 1

10 – 3v = 1

-3v = 1 – 10 -3v = -9 

 v = 3

Question 24

Solve for x and y:

Solution 24

Question 25

Solve for x and y

4x + 6y = 3xy

8x + 9y = 5xy;Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3—(3)

8v + 9u = 5—(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 —(5)

48v + 54u = 30 —(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4Question 26

Solve for x and y:

Solution 26

Question 27

Solve for x and y:

Solution 27

Question 28

Solve for x and y

Solution 28

Putting 

3u + 2v = 2—-(1)

9u – 4v = 1—-(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4—-(3)

9u – 4v = 1—-(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is Question 29

Solve for x and y

Solution 29

The given equations are

Putting 

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5Question 30

Solve for x and y

Solution 30

Putting in the equation, we get

44u + 30v = 10—-(1)

55u + 40v = 13—-(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40—(3)

165u + 120v = 39—(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 – 8 = 3

the solution is x = 8, y = 3Question 31

Solve for x and y:

Solution 31

Question 32

Solve for x and y

71x + 37y = 253

37x + 71y = 287Solution 32

The given equations are

71x + 37y = 253—(1)

37x + 71y = 287—(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

solution is x = 2, y = 3Question 33

Solve for x and y

217x + 131y = 913

131x + 217y = 827Solution 33

217x + 131y = 913—(1)

131x + 217y = 827—(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5—-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

   x – y = 1—(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

solution is x = 3, y = 2Question 34

Solve for x and y:

23x – 29y = 98, 29x – 23y = 110.Solution 34

Question 35

Solve for x and y:

Solution 35

Question 36

Solve for x and y:

Solution 36

Question 37

Solve for x and y

where Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54—(5)

125u – 60v = —(6)

Adding (5) and (6), we get

Question 38

Solve for x and y:

Solution 38

Question 39

Solve for x and y:

Solution 39

Question 40

Solve for x and y:

x + y = a + b, ax – by = a2 – b2.Solution 40

Question 41

Solve for x and y

Solution 41

Question 42

Solve for x and y:

px + py = p – q, qx – py = p + q.Solution 42

Question 43

Solve for x and y:

Solution 43

Question 44

Solve for x and y

6(ax + by) = 3a + 2b

6(bx – ay) = 3b – 2aSolution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting  in(1), we get

Hence, the solution is  Question 45

Solve for x and y:

ax – by = a2 + b2, x + y = 2a.Solution 45

Question 46

Solve for x and y

bx – ay + 2ab = 0Solution 46

Question 47

Solve for x and y

x + y = 2abSolution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by 

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = abQuestion 48

Solve for x and y:

x + y = a + b, ax – by = a2 – b2.Solution 48

Question 49

Solve for x and y:

a2x + b2y = c2, b2x + a2y = d2.Solution 49

Question 50

Solve for x and y:

Solution 50

Exercise Ex. 7C

Question 1

Solve for x and y by method of cross multiplication:

x + 2y + 1 = 0

2x – 3y – 12 = 0Solution 1

x + 2y + 1 = 0 —(1)

2x – 3y – 12 = 0 —(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solutionQuestion 2

Solve for x and y by method of cross multiplication:

3x – 2y + 3 = 0

4x + 3y – 47 = 0Solution 2

3x – 2y + 3 = 0

4x + 3y – 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9Question 3

Solve for x and y by method of cross multiplication:

6x – 5y – 16 = 0

7x – 13y + 10 = 0Solution 3

6x – 5y – 16 = 0

7x – 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4Question 4

Solve for x and y by method of cross multiplication:

3x + 2y + 25 = 0

2x + y + 10 = 0Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20Question 5

Solve for x and y by method of cross multiplication:

2x +5y = 1

2x + 3y = 3Solution 5

2x + 5y – 1 = 0 —(1)

2x + 3y – 3 = 0—(2)

By cross multiplication we have

the solution is x = 3, y = -1Question 6

Solve for x and y by method of cross multiplication:

2x + y – 35 = 0

3x + 4y – 65 = 0Solution 6

2x + y – 35 = 0

3x + 4y – 65 = 0

By cross multiplication, we have

Question 7

Solve each of the following systems of equations by using the method of cross multiplication:

7x – 2y = 3, 22x – 3y = 16.Solution 7

Question 8

Solve for x and y by method of cross multiplication:

Solution 8

Question 9

Solve for x and y by method of cross multiplication:

Solution 9

Taking 

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have

the solution is Question 10

Solve for x and y by method of cross multiplication:

Solution 10

Let in the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

Question 11

Solve for x and y by method of cross multiplication:

Solution 11

Question 12

Solve for x and y by method of cross multiplication:

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0Solution 12

2ax + 3by – (a + 2b) = 0

3ax+ 2by – (2a + b) = 0

By cross multiplication, we have

Question 13

Solve each of the following systems of equations by using the method of cross multiplication:

Solution 13

Exercise Ex. 7D

Question 1

Show that the following system of equations has a unique solution:

3x + 5y = 12, 5x + 3y = 4

Also, find the solution of the given system of equations.Solution 1

Question 2

Show that each of the following systems of equations has a unique solution and solve it:

2x – 3y = 17, 4x + y = 13.Solution 2

Question 3

Show that the following system of equations has a unique solution:

Also, find the solution of the given system of equationsSolution 3

Question 4

Find the value of k for which each of the following systems of equations has a unique solution:

2x + 3y – 5 = 0, kx – 6y – 8 = 0.Solution 4

Question 5

Find the value of k for which each of the following systems of equations has a unique solution:

x – ky = 2, 3x + 2y + 5 = 0.Solution 5

Question 6

Find the value of k for which each of the following systems of equations has a unique solution:

5x – 7y – 5 = 0, 2x + ky – 1 = 0.Solution 6

Question 7

Find the value of k for which each of the following systems of equations has a unique solution:

4x + ky + 8 = 0, x + y + 1 = 0.Solution 7

Question 8

Find the value of k for which each of the following systems of equations has a unique solution:

4x – 5y = k , 2x – 3y = 12Solution 8

4x – 5y – k = 0, 2x – 3y – 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solutionQuestion 9

Find the value of k for which each of the following systems of equations has a unique solution:

kx + 3y = (k – 3),12x + ky = kSolution 9

kx + 3y – (k – 3) = 0

12x + ky – k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solutionQuestion 10

Show that the system of equations

2x – 3y = 5, 6x – 9y = 15

has an infinite number of solutions.Solution 10

2x – 3y – 5 = 0, 6x – 9y – 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutionsQuestion 11

Show that the system of equations   Solution 11

Question 12

For what value of k, the system of equations

kx + 2y = 5, 3x – 4y = 10

has (i) a unique solution (ii) no solution?Solution 12

kx + 2y – 5 = 0

3x – 4y – 10 = 0

These equations are of the form

This happens when 

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have 

Hence, the given system of equations has no solution if Question 13

For what value of k, the system of equations

x + 2y = 5, 3x + ky + 15 = 0

has (i) a unique solution (ii) no solution?Solution 13

x + 2y – 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.Question 14

For what value of k, the system of equations

x + 2y = 3, 5x + ky + 7 = 0

has (i) a unique solution (ii) no solution?

Is there any value of k for which the given system has an infinite number of solutions?Solution 14

x + 2y – 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if 

For infinite number of solutions we must have

This is never possible since 

There is no value of k for which system of equations has infinitely many solutionsQuestion 15

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3kSolution 15

2x + 3y – 7 = 0

(k – 1)x + (k + 2)y – 3k = 0

These are of the form

This hold only when

Now the following cases arises

Case : I

Case: II

Case III

For k = 7, there are infinitely many solutions of the given system of equationsQuestion 16

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

2x + (k – 2)y =k

6x + (2k – 1)y = (2k + 5)Solution 16

2x + (k – 2)y – k = 0

6x + (2k – 1)y – (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

Case (2)

Case (3)

Thus, for k = 5 there are infinitely many solutionsQuestion 17

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

kx + 3y = (2k +1)

2(k + 1)x + 9y = (7k + 1)Solution 17

kx + 3y – (2k +1) = 0

2(k + 1)x + 9y – (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case – (1)

Case (2)

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutionsQuestion 18

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

5x + 2y = 2k

2(k + 1)x + ky = (3k + 4)Solution 18

5x + 2y – 2k = 0

2(k +1)x + ky – (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.Question 19

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 1)x – y = 5

(k + 1)x + (1 – k)y = (3k + 1)Solution 19

(k – 1)x – y – 5 = 0

(k + 1)x + (1 – k)y – (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

k = 3 is common value for which the number of solutions is infinitely manyQuestion 20

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

(k – 3)x + 3y = k, kx + ky = 12.Solution 20

Question 21

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(a – 1)x + 3y = 2

6x + (1 – 2b)y = 6Solution 21

(a – 1)x + 3y – 2 = 0

6x + (1 – 2b)y – 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4Question 22

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5

3x + (b – 1)y = 2Solution 22

(2a – 1)x + 3y – 5 = 0

3x + (b – 1)y – 2 = 0

These equations are of the form

These holds only when

Question 23

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x – 3y = 7

(a + b)x + (a + b – 3)y = (4a + b)Solution 23

2x – 3y – 7 = 0

(a + b)x + (a + b – 3)y – (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1Question 24

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y=7, (a + b + 1)x +(a + 2b + 2)y = 4(a + b)+ 1.Solution 24

Question 25

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, (a + b)x + (2a – b)y = 21.Solution 25

Question 26

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.Solution 26

Question 27

Find the value of k for which each of the following systems of equations no solution:

8x + 5y = 9, kx + 10y = 15.Solution 27

Question 28

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = 3, 12x + ky = 6.Solution 28

Question 29

Find the value of k for which each of the following systems of equations no solution:

Solution 29

Question 30

Find the value of k for which each of the following systems of equations no solution:

kx + 3y = k – 3, 12x + ky = k.Solution 30

Question 31

Find the value of k for which the system of equations

5x – 3y = 0;2x + ky = 0

has a nonzero solution.Solution 31

We have 5x – 3y = 0 —(1)

2x + ky = 0—(2)

Comparing the equation with

These equations have a non – zero solution if 

Exercise Ex. 7E

Question 1

5 chairs and 4 tables together cost Rs.5600, while 4 chairs and 3 tables together cost Rs.4340. Find the cost of a chair and that of a table.Solution 1

Question 2

23 spoons and 17 forks together cost Rs.1770, while 17 spoons and 23 forks together cost Rs.1830. Find the cost of a spoon and that of a fork.Solution 2

Question 3

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all Rs.19.50, how many coins of each kind does she have?Solution 3

Question 4

The sum of two numbers is 137 and their difference is 43. Find the numbers.Solution 4

Let the two numbers be x and y respectively.

Given:

x + y = 137 —(1)

x – y = 43 —(2)

Adding (1) and (2), we get

2x = 180

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

  = 47

Hence, the two numbers are 90 and 47.Question 5

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 —(1)

4x – 7y = 2 —(2)

Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644 —(3)

12x – 21y = 6 —(4)

Adding (3) and (4), we get

Putting x = 25 in (1), we get

2 × 25 + 3y = 92

50 + 3y = 92

 3y = 92 – 50

y = 14

Hence, the first number is 25 and second is 14Question 6

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 —(1)

4x – y = 138 —(2)

Adding (1) and (2), we get

Putting x = 40 in (1), we get

3 × 40 + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.Question 7

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the number.Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x – 45 = y

2x – y = 45—(1)

and

2y – x = 21

 -x + 2y = 21—(2)

Multiplying (1) by 2 and (2) by 1

4x – 2y = 90—(3)

-x + 2y = 21 —(4)

Adding (3) and (4), we get

3x = 111

Putting x = 37 in (1), we get

2 × 37 – y = 45

 74 – y = 45

y = 29

Hence, the greater and the smaller numbers are 37 and 29.Question 8

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 —(1)

And

5y = x × 3 + 5

-3x + 5y = 5 —(2)

Adding (1) and (2), we get

y = 13

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.Question 9

If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.Solution 9

Let the required numbers be x and y respectively.

Then,

Therefore,

2x – y =-2—(1)

11x – 5y = 24 —(2)

Multiplying (1) by 5 and (2) by 1

10x – 5y = -10—(3)

11x – 5y = 24—(4)

Subtracting (3) and (4) we get

x = 34

putting x = 34 in (1), we get

 2 × 34 – y = -2

 68 – y = -2

-y = -2 – 68

y = 70

Hence, the required numbers are 34 and 70.Question 10

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.Solution 10

Let the numbers be x and y respectively.

According to the question:

x – y = 14 —(1)

From (1), we get

x = 14 + y —(3)

putting x = 14 + y in (2), we get

Putting y = 9 in (1), we get

x – 9 = 14

  x = 14 + 9 = 23

Hence the required numbers are 23 and 9Question 11

The sum of the digits of a two digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.Solution 11

Let the ten’s digit be x and units digit be y respectively.

Then,

x + y = 12—(1)

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x – (10x + y) = 18

10y + x – 10x – y = 18

9y – 9x = 18

y – x = 2 —-(2)

Adding (1) and (2), we get

Putting y = 7 in (1), we get

x + 7 = 12

x = 5

Number= 10x + y

                  = 10 × 5 + 7

                  = 50 + 7

                  = 57

Hence, the number is 57.
Question 12

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.Solution 12

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

 10x + y = 7(x + y)

     10x + y = 7x + 7y

       3x – 6y = 0—(1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

      10x – x + y – 10y = 27

        9x – 9y = 27

         (x – y) = 27

           x – y = 3—(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0—(3)

6x – 6y = 18 —(4)

Subtracting (3) from (4), we get

Putting x = 6 in(1), we get

3 × 6 – 6y = 0

18 – 6y = 0

Number = 10x + y

            = 10 × 6 + 3

            = 60 + 3

            = 63

Hence the number is 63.
Question 13

The sum of the digits of a two-digit number is 15. The number of obtained by interchanging the digits exceeds the given number by 9. Find the number.Solution 13

Let the ten’s digit and unit’s digits of required number be x and y respectively.

Then,

x + y = 15—(1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x – (10x + y) = 9

10y + x – 10x – y = 9

       9y – 9x = 9

Add (1) and (2), we get

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

                         = 10 × 7 + 8

                         = 70 + 8

                         = 78

Hence the required number is 78.
Question 14

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.Solution 14

Let the ten’s and unit’s of required number be x and y respectively.

Then,required number =10x + y

According to the given question:

 10x + y = 4(x + y) + 3

 10x + y = 4x + 4y + 3

6x – 3y = 3

  2x – y = 1 —(1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

 x – y = -2—(2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

                         = 10 × 3 + 5

                         = 30 + 5

                         = 35

Hence, required number is 35.
Question 15

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.Solution 15

Let the ten’s digit and unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder

According to the given questiion:

10x + y = 6 × (x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 —(1)

Number obtained by reversing the digits is 10y + x

 10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) =9

(x – y) = 1—(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 —(3)

5x – 5y = 5 —(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1), we get

x =5 and y= 4

Hence, required number is 54.Question 16

A two – digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchanged their places. Find the number.Solution 16

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y

Also,

(10x + y) + 18 = 10y + x

9x – 9y = -18

9(y – x) = 18—(1)

y – x = 2

Now, 

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5 × 10 + 7

                                        = 57Question 17

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.Solution 17

Let the ten’s and unit’s digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

(10x + y) – 63 = (10y + x)

9x – 9y = 63

x – y = 7—(1)

Now,

Adding (1) and (2), we get

Putting x = 9 in (1), we get

9 – y = 7

y = 9 – 7

y =2

x = 9, y = 2

Hence, the required number = 9 × 10 + 2

                                        = 92.Question 18

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.Solution 18

Question 19

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes Find the fraction.Solution 19

Let the numerator and denominator of fraction be x and y respectively.

According to the question:

x + y = 8—(1)

And

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24—(3)

4x – 3y = -3 —(4)

Add (3) and (4), we get

Putting x = 3 in (1), we get

3 + y= 8

y = 8 – 3

y = 5

x = 3, y = 5

Hence, the fraction is Question 20

If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fractionSolution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

2 × 3 – 4

-y = -4 -6

y = 10

x = 3 and y = 10

Hence the fraction is Question 21

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes Find the fraction.Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is .

According to the given question:

y = x + 11

y- x = 11—(1)

and

-3y + 4x = -8 —(2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44—(3)

-3y + 4x = -8—(4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11 

  x = 25

x = 25, y = 36

Hence the fraction is Question 22

Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is 

Subtracting (1) from (2), we get

x = 15

Putting x = 15 in (1), we get

2 × 15 – y = 4

 30 – y = 4

 y = 26

x = 15 and y = 26

Hence the given fraction is Question 23

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.Solution 23

Question 24

The sum of two numbers is 16 and the sum of their reciprocals is Find the numbers.Solution 24

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16—(1)

And

—(2)

From (2),

xy = 48

We know,

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1),

y = 16 – x

   = 16 – 12

   = 4

The required numbers are 12 and 4Question 25

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.Solution 25

Let the number of student in class room A and B be x and y respectively.

When 10 students are transferred from A to B:

x – 10 = y + 10

 x – y = 20—(1)

When 20 students are transferred from B to A:

 2(y – 20) = x + 20

 2y – 40 = x + 20

-x + 2y = 60—(2)

Adding (1) and (2), we get

y = 80

Putting y = 80 in (1), we get

x – 80 = 20

 x = 100

Hence, number of students of A and B are 100 and 80 respectively.Question 26

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays Rs.1330, and travelling 90 km, he pays Rs.1490. Find the fixed charges and rate per km.Solution 26

Question 27

A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs.4500, whereas a student B who takes food for 30 days, has to pay Rs.5200. Find the fixed charges per month and the cost of the food per day.Solution 27

Question 28

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received Rs. 1350 as an annual interest. Had he interchanged the amounts invested, he would have received Rs.45 less as interest. What amounts did he invest at different rates?Solution 28

Question 29

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves Rs.9000 per month, find the monthly income of eachSolution 29

Question 30

A man sold a chair and a table together for Rs.1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for Rs.1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.Solution 30

Question 31

Points A and B are 70km apart on a highway. A car starts from A and another starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

Distance covered by the car P in 7 hours = 7x km

Distance covered by the car Q in 7 hours = 7y km

Let the cars meet at point M.

AM = 7x km and BM = 7y km

AM – BM = AB

 7x – 7y = 70

7(x – y) = 70

x – y = 10 —-(1)

Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km


In this case let the cars meet at a point N.

AN = x km and BN = y km

AN + BN = AB

 x + y = 70—(2)

Adding (1) and (2), we get

2x = 80

 x = 40

Putting x = 40 in (1), we get

40 – y = 10

 y = (40 – 10) = 30

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.Question 32

A train covered a certain distance at a uniform speed. If the train had been 5kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3hours less than the scheduled time. Find the length of the journey.Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

Case I:

Speed = (x + 5)km/h and time taken = (y – 3)hour

Distance covered = (x + 5)(y – 3)km

(x + 5) (y – 3) = xy

xy + 5y -3x -15 = xy

5y – 3x = 15 —(1)

Case II:

Speed (x – 4)km/hr and time taken = (y + 3)hours

Distance covered = (x – 4)(y + 3) km

(x – 4)(y + 3) = xy

xy -4y + 3x -12 = xy

 3x – 4y = 12 —(2)

Multiplying (1) by 4 and (2) by 5, we get

20y – 12x = 60 —(3)

-20y + 15x = 60 —(4)

Adding (3) and (4), we get

3x = 120

or x = 40

Putting x = 40 in (1), we get

5y – 3 × 40 = 15 

  5y = 135  

  y = 27

Hence, length of the journey is (40 × 27) km = 1080 kmQuestion 33

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.Solution 33

Question 34

Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.Solution 34

Question 35

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.Solution 35

Question 36

A boat goes 12km upstream and 40km downstream in 8hours. It can go 16km upstream and 32 downstream in the same times. Find the speed of the boat in still water and the speed of the stream.Solution 36

Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.

Then,

Speed upstream = (x – y)km/hr

Speed downstream = (x + y) km/hr

Time taken to cover 12 km upstream = 

Time taken to cover 40 km downstream = 

Total time taken = 8hrs

Again, time taken to cover 16 km upstream = 

Time taken to taken to cover 32 km downstream = 

Total time taken = 8hrs

Putting 

12u + 40v = 8 

   3u + 10v = 2 —(1)

and

16u + 32v = 8 

   2u + 4v = 1—(2)

Multiplying (1) by 4 and (2) by 10, we get

12u + 40v = 8—(3)

20u + 40v = 10 —(4)

Subtracting (3) from (4), we get

Putting in (3), we get

On adding (5) and (6), we get

2x = 12 

  x = 6

Putting x = 6 in (6) we get

6 + y = 8

y = 8 – 6 = 2

x = 6, y = 2

Hence, the speed of the boat in still water = 6 km/hr and speedof the stream = 2km/hrQuestion 37

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.Solution 37

Let man’s 1 day’s work be and 1 boy’s day’s work be 

Also let  and 

Multiplying (1) by 6 and (2) by 5 we get

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.Question 38

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

 x – y = 3 —-(1)

Also,

(x + 3)(y – 2) = xy

 3y – 2x = 6 —-(2)

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 —(3)

3y – 2x = 6 —(4)

Adding (3) and (4), we get

y = 12

Putting y = 12 in (1), we get

x – 12 = 3

 x= 15

x = 15, y = 12

Hence length = 15 metres and breadth = 12 metresQuestion 39

The area of a rectangle gets reduced by 8 , when its length is reduced by 5m and its breadth is increased by 3m. If we increase the length by 3m and breadth by 2m, the area is increased by 74 . Find the length and breadth of the rectangle.Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy – (x – 5)(y + 3) = 8

xy – [xy – 5y + 3x -15] = 8

xy – xy + 5y – 3x + 15 = 8 

 3x – 5y = 7 —(1)

And

(x + 3)(y + 2) – xy = 74

 xy + 3y +2x + 6 – xy = 74

2x + 3y = 68—(2)

Multiplying (1) by 3 and (2) by 5, we get

9x – 15y = 21—(3)

10x + 15y = 340—(4)

Adding (3) and (4), we get

Putting x = 19 in (3) we get

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10mQuestion 40

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m. If the length. is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.Solution 40

Question 41

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs Rs.4150 while one full and one half reserved first class tickets cost Rs.6255. What is the basic first class full fare and what is the reservation charge?Solution 41

Question 42

Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.Solution 42

Question 43

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of man and his son.Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)


x – 2 = 5y – 10

  x – 5y = -8 —(1)

Two years later:

(x + 2) = 3(y + 2) + 8


x + 2 = 3y + 6 + 8

 x – 3y = 12 —(2)

Subtracting (2) from (1), we get

-2y = -20

  y = 10

Putting y = 10 in (1), we get

x – 5 10 = -8

x – 50 = -8

 x = 42

Hence the present ages of the man and the son are 42 years and 10 respectively.Question 44

If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95years. Find the age of the mother and that of the son.Solution 44

Let the present ages of  the mother and her son be x and y respectively.

According to the given question:

x + 2y = 70—(1)

and

2x + y = 95—(2)

Multiplying (1) by 1 and (2) by 2, we get

x + 2y = 70 —(3)

4x + 2y = 190—(4)

Subtracting (3) from (4), we get

Putting x = 40 in (1), we get

40 + 2y = 70

2y = 30

 y = 15

x = 40, y = 15

Hence, the ages of the mother and the son are 40 years and 15 years respectively.Question 45

The present age of a woman is 3 years more than three times the age of her daughter; three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3—(1)

Three years later:

 (x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

 x – 2y = 13—(2)

Subtracting (2) from (1), we get

 y = 10

Putting y = 10 in (1), we get

x – 3 × 10 = 3

 x = 33

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.Question 46

On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains Rs.7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains Rs.13. Find the actual price of each of the tea set and the lemon set.Solution 46

Question 47

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid Rs.27 for a book kept for 7 days, while Tanvy paid Rs. 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.Solution 47

Question 48

A chemist has one solution containing 50% acid and a second one containing 25% arid. How much of each should be used to make 10 litres of a 40% arid solution?Solution 48

Question 49

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given: Pure gold is 24-carat)Solution 49

Question 50

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of adds to be mixed to form the mixture.Solution 50

Question 51

The larger of the two supplementary angles exceeds the smaller by 18o.  Find them.Solution 51

Question 52

Solution 52

Question 58

Solution 58

Exercise Ex. 7F

Question 1

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.Solution 1

Question 2

Find the value of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.Solution 2

Question 3

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.Solution 3

Question 4

For what value of k will the following pair of linear equations have no ‘solution?

2x + 3y = 9, 6x + (k – 2)y = (3k – 2).Solution 4

Question 5

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.Solution 5

Question 6

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.Solution 6

Question 7

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.Solution 7

Question 8

The cost of 5 pens and 8 pencils is Rs.120, while the cost of 8 pens and 5 pencils is Rs.153. Find the cost of 1 pen and that of I pencil.Solution 8

Question 9

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.Solution 9

Question 10

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.Solution 10

Question 11

A man purchased 47 stamps of 20 p and 25 p for Rs.10. Find the number of each type of stamps.Solution 11

Question 12

A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x +y).Solution 15

Question 16

Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.Solution 16

Question 17

Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.Solution 17

Question 18

Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.Solution 18

Question 19

Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.Solution 19

Question 20

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.Solution 20

Question 21

Solution 21

Exercise MCQ

Question 1

If 2x + 3y = 12 and 3x – 2y = 5 then

(a) x = 2, y = 3

(b) x = 2, y = -3

(c) x = 3, y = 2

(d) x = 3, y = -2Solution 1

Question 2

(a) x = 4, y = 2

(b) x = 5, y = 3

(c) x = 6, y = 4

(d) x = 7, y = 5Solution 2

Question 3

(a) x = 2, y = 3

(b) x = -2, y = 3

(c) x = 2, y = -3

(d) x = -2, y = -3Solution 3

Question 4

Solution 4

Question 5

(a) x = 1, y = 1

(b) x = -1, y = -1

(c) x = 1, y = 2

(d) x = 2, y = 1Solution 5

Question 6

Solution 6

Question 7

If 4x+6y=3xy and 8x+9y=5xy then

(a) x=2, y=3

(b) x=1, y=2

(c) x=3, y=4

(d) x=1, y=-1Solution 7

Question 8

If 29x+37y=103 and 37x+29y=95 then

(a) x=1, y=2

(b) x=2, y=1

(c) x=3, y=2

(d) x=2, y=3Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The system kx – y = 2 and 6x – 2y = 3  has a unique solution only when

(a) k = 0

(b) k ≠ 0

(c) k = 3

(d) k ≠ 3Solution 11

Question 12

The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

(a) k = -6

(b) k ≠ -6

(c) k = 0

(d) k ≠ 0Solution 12

Question 13

The system x+2y=3 and 5x+ky+7=0 has no solution, when

(a) k=10

(b) k≠10

(c) 

(d) K=-21Solution 13

Question 14

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is

Solution 14

Question 15

For what value of k do the equations kx – 2y = 3 and

3x + y = 5 represent two lines intersecting at a unique point?

(a) k=3

(b) k=-3

(c) k=6

(d) all real values except -6Solution 15

Question 16

The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 16

We space know space that comma the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0 has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2. So comma space the space pair space of space equations space has space no space solution.

Question 17

The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solutionSolution 17

We space know space that comma the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0 has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2. So comma space the space pair space of space equations space has space no space solution.

Question 18

If a pair of linear equations is consistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 18

Question 19

If a pair of linear equations is inconsistent then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincidentSolution 19

Question 20

In a ΔABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠B = ?

(a) 20° 

(b) 40° 

(c) 60° 

(d) 80° Solution 20

Question 21

In a cyclic quadrilateral ABCD, it is being given that

∠A = (x + y + 10) °, ∠B = (y + 20) °,

∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?

(a) 70° 

(b) 80° 

(c) 100° 

(d) 110° Solution 21

Question 22

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96

(b) 69

(c) 87

(d) 78Solution 22

Question 23

Solution 23

Question 24

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(c) 47 years

(d) 40 yearsSolution 24

Question 25

The graphs of the equations 6x – 2y + 9 = 0 and

3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 25

Question 26

The graphs of the equations 2x+3y-2=0 and x-2y-8=0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 26

Question 27

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each otherSolution 27

Exercise FA

Question 1

The graphic representation of the equations x+2y=3 and 2x+4y+7=0 gives a pair of

(a) parallel lines

(b) intersecting lines

(c) coincident lines

(d) none of theseSolution 1

Question 2

If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a+b have an infinite number of solutions then

(a) a= 5, b = 1

(b) a = -5, b = 1

(c) a = 5, b = -1

(d) a = -5, b = -1Solution 2

Question 3

The pair of equations 2x+y=5, 3x+2y=8 has

(a) a unique solution

(b) two solutions

(c) no solution

(d) infinitely many solutionsSolution 3

Question 4

If x = -y and y > 0, which of the following is wrong?

(a) x2y > 0

(b) x + y = 0

(c) xy < 0

(d)  Solution 4

Question 5

Solution 5

Question 6

For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?Solution 6

Question 7

Solution 7

Question 8

Solve the system of equations x – 2y = 0, 3x + 4y = 20.Solution 8

Question 9

Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.Solution 9

Question 10

The difference between two numbers is 26 and one number is three times the other. Find the numbers.Solution 10

Question 11

Solve : 23x+29y=98, 29x+23y=110.Solution 11

Question 12

Solve : 6x+3y=7xy and 3x+9y = 11xy.Solution 12

Question 13

Find the value of k for which the system of equations 3x+y=1 and kx+2y=5 has (i) a unique solution, (ii) no solution.Solution 13

Question 14

In a ΔABC, ∠C =3∠B =2(∠A+∠B). Find the measure of each one of the ∠A, ∠B and ∠C. Solution 14

Question 15

5 pencils and 7 pens together cost Rs. 195 while 7 pencils and 5 pens together cost Rs. 153. Find the cost of each one of the pencil and the pen.Solution 15

Question 16

Solve the following system of equations graphically :

2x-3y=1, 4x-3y+1=0.Solution 16

Since the intersection of the lines is the point with coordinates (-1, -1), x = -1 and y = -1.Question 17

Find the angles of a cyclic quadrilateral ABCD in which ∠A =(4x+20)°, ∠B=(3x-5)°, ∠C=(4y)° and ∠D=(7y+5)° Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

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RS Agarwal Solution | Class 11th | Chapter-2 |   Relations | Edugrown

Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Given colon space straight f open parentheses straight x close parentheses equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction Let space straight y equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction rightwards double arrow straight y plus yx squared equals straight x squared rightwards double arrow straight y equals straight x squared minus yx squared rightwards double arrow straight y equals straight x squared open parentheses 1 minus straight y close parentheses rightwards double arrow fraction numerator straight y over denominator 1 minus straight y end fraction equals straight x squared rightwards double arrow straight x equals square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root straight x space is space defined space only space if space fraction numerator straight y over denominator 1 minus straight y end fraction greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 minus straight y greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 greater or equal than straight y straight y not equal to 1 Therefore comma space straight y greater or equal than 0 space and space 1 greater than straight y rightwards double arrow 0 less or equal than straight y less than 1 rightwards double arrow straight y element of left square bracket 0 comma space 1 right parenthesis rightwards double arrow fraction numerator straight x squared over denominator 1 plus straight x squared end fraction element of left square bracket 0 comma space 1 right parenthesis Hence comma space the space range space of space straight f space is space left square bracket 0 comma space 1 right parenthesis.

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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RS Agarwal Solution | Class 11th | Chapter-3 |   Functions | Edugrown

Exercise Ex. 2.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Ex. 2.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Exercise Ex. 2.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Given colon space straight f open parentheses straight x close parentheses equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction Let space straight y equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction rightwards double arrow straight y plus yx squared equals straight x squared rightwards double arrow straight y equals straight x squared minus yx squared rightwards double arrow straight y equals straight x squared open parentheses 1 minus straight y close parentheses rightwards double arrow fraction numerator straight y over denominator 1 minus straight y end fraction equals straight x squared rightwards double arrow straight x equals square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root straight x space is space defined space only space if space fraction numerator straight y over denominator 1 minus straight y end fraction greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 minus straight y greater or equal than 0 rightwards double arrow straight y greater or equal than 0 space and space 1 greater or equal than straight y straight y not equal to 1 Therefore comma space straight y greater or equal than 0 space and space 1 greater than straight y rightwards double arrow 0 less or equal than straight y less than 1 rightwards double arrow straight y element of left square bracket 0 comma space 1 right parenthesis rightwards double arrow fraction numerator straight x squared over denominator 1 plus straight x squared end fraction element of left square bracket 0 comma space 1 right parenthesis Hence comma space the space range space of space straight f space is space left square bracket 0 comma space 1 right parenthesis.

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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RS Agarwal Solution | Class 11th | Chapter-5|  Complex Numbers and Quadratic Equations | Edugrown

Exercise Ex. 5.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Exercise Ex. 5.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Exercise Ex. 5.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

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