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Chapter 15 Linear Inequations Exercise Ex. 15.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 15 Linear Inequations Exercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the system of equation Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solve the system of equation 

Solution 21

Chapter 15 Linear Inequations Exercise Ex. 15.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 15 Linear Inequations Exercise Ex. 15.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 15 Linear Inequations Exercise Ex. 15.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 15 Linear Inequations Exercise Ex. 15.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Show that the following system of linear equations has no solution:

X + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.Solution 7

Question 8

Show that the solution set of the following system of linear inequalities is an unbounded region 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.Solution 8

Chapter 14 Quadratic Equations Exercise Ex. 14.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Chapter 14 Quadratic Equations Exercise Ex. 14.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

Exercise 3A

Page No 45:

Question 1:

Write the next three whole numbers after 30999.

ANSWER:

The next three whole numbers after 30999 are 31000, 31001, and 31002.

Page No 45:

Question 2:

Write the three whole numbers occurring just before 10001.

ANSWER:

Three whole numbers occurring just before 10001 are as follows:

10001 − 1 = 10000
10000 − 1 = 9999
9999 − 1 = 9998

∴ The three whole numbers just before 10001 are 10000, 9999 and 9998.

Page No 45:

Question 3:

How many whole numbers are there between 1032 and 1209?

ANSWER:

Number of whole numbers between 1032 and 1209 = (1209 − 1032) − 1
                                                                                      = 177 − 1
                                                                                        = 176

Page No 45:

Question 4:

Which is the smallest whole number?

ANSWER:

0 (zero) is the smallest whole number.

All the natural numbers along with 0 are called whole numbers.

Page No 45:

Question 5:

Write the successor of:
(i) 2540801
(ii) 9999
(iii) 50904
(iv) 61639
(v) 687890
(vi) 5386700
(vii) 6475999
(viii) 9999999

ANSWER:

(i) Successor of 2540801 = 2540801 + 1 = 2540802
(ii) Successor of 9999 = 9999 + 1 = 10000
(iii) Successor of 50904 = 50904 + 1 = 50905 
(iv) Successor of 61639 = 61639 + 1 = 61640
(v) Successor of 687890 = 687890 + 1 = 687891
(vi) Successor of 5386700 = 5386700 + 1 = 5386701
(vii) Successor of 6475999 = 6475999 + 1 = 6476000
(viii) Successor of 9999999 = 9999999 + 1 = 10000000

Page No 46:

Question 6:

Write the predecessor of:
(i) 97
(ii) 10000
(iii) 36900
(iv) 7684320
(v) 1566391
(vi) 2456800
(vii) 100000
(viii) 1000000

ANSWER:

(i) Predecessor of 97 = 97 − 1 = 96
(ii) Predecessor of 10000 = 10000 − 1 = 9999
(iii) Predecessor of 36900 = 36900 − 1 = 36899
(iv) Predecessor of 7684320 = 7684320 − 1 = 7684319
(v) Predecessor of 1566391 = 1566391 − 1 = 1566390
(vi) Predecessor of 2456800 = 2456800 − 1 = 2456799
(vii) Predecessor of 100000 = 100000 − 1 = 99999
(viii) Predecessor of 1000000 = 1000000 − 1 = 999999

Page No 46:

Question 7:

Write down three consecutive whole numbers just preceding 7510001.

ANSWER:

The three consecutive whole numbers just preceding 7510001 are as follows:

7510001 − 1 = 7510000
7510000 − 1 = 7509999
7509999 − 1 = 7509998

∴ The three consecutive numbers just preceding 7510001 are 7510000, 7509999 and 7509998.

Page No 46:

Question 8:

Write (T) for true and (F) for false against each of the following statements:
(i) Zero is the smallest natural number.
(ii) Zero is the smallest whole number.
(iii) Every whole number is a natural number.
(iv) Every natural number is a whole number.
(v) 1 is the smallest whole number.
(vi) The natural number 1 has no predecessor.
(vii) The whole number  1 has no predecessor.
(viii) The whole number 0 has no predecessor.
(ix) The predecessor of a two-digit number is never a single-digit number.
(x) The successor of a two-digit number is always a two-digit number.
(xi) 500 is the predecessor of 499.
(xii) 7000 is the successor of 6999.

ANSWER:

(i) False. 0 is not a natural number.1 is the smallest natural number.
(ii) True.
(iii) False. 0 is a whole number but not a natural number.
(iv) True. Natural numbers include 1,2,3 …, which are whole numbers.
(v) False. 0 is the smallest whole number.
(vi) True. The predecessor of 1 is 1 − 1 = 0, which is not a natural number.
(vii) False. The predecessor of 1 is 1 − 1 = 0, which is a whole number.
(viii) True. The predecessor of 0 is 0 − 1 = −1, which is not a whole number.
 (ix) False. The predecessor of a two-digit number can be a single digit number. For example, the predecessor of 10 is 10 − 1, i.e., 9.
(x) False. The successor of a two-digit number is not always a two-digit number. For example, the successor of 99 is 99 + 1, i.e., 100.
(xi) False. The predecessor of 499 is 499 − 1, i.e., 498.
(xii) True. The successor of 6999 is 6999 + 1, i.e., 7000.

Page No 48:

Exercise 3B

Question 1:

Fill in the blanks to make each of the following a true statement:
(i) 458 + 639 = 639 + ……
(ii) 864 + 2006 = 2006 + ……
(iii) 1946 + …… = 984 + 1946
(iv) 8063 + 0 = ……
(v) 53501 + (574 + 799) = 574 + (53501 + ……)

ANSWER:

(i) 458 + 639 = 639 + 458
(ii) 864 + 2006 = 2006 + 864
​(iii) 1946 + 984 = 984 + 1946
(iv) 8063 + 0 = 8063
(v) 53501 + (574 + 799) = 574 + (53501 + 799)

Page No 48:

Question 2:

Add the following numbers and check by revershing the order of the addends:
(i) 16509 + 114
(ii) 2359 + 548
(iii) 19753 + 2867

ANSWER:

(i) 16509 + 114 = 16623
 By reversing the order of the addends, we get:
  114 + 16509 = 16623  
∴ 16509 + 114 = 114 + 16509

(ii) 2359 + 548 = 2907 
 By reversing the order of the addends, we get:
     548 + 2359 = 2907
∴ 2359 + 548 = 548 + 2359

(iii) 19753 + 2867 = 22620
   By reversing the order of the addends, we get:
     2867 + 19753 = 22620
∴ 19753 + 2867 = 2867 + 19753
   

Page No 48:

Question 3:

Find the sum: (1546 + 498) + 3589.
Also, find the sum: 1546 + (498 + 3589).
Are the two sums equal?
State the property satisfied.

ANSWER:

We have:

(1546 + 498) + 3589 = 2044 + 3589 = 5633

Also, 1546 + (498 + 3589) = 1546 + 4087 = 5633

Yes, the two sums are equal.

The associative property of addition is satisfied.

Page No 48:

Question 4:

Determine each of the sums given below using suitable rearrangement.
(i) 953 + 707 + 647
(ii) 1983 + 647 + 217 + 353
(iii) 15409 + 278 + 691 + 422
(iv) 3259 + 10001 + 2641 + 9999
(v) 1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48

ANSWER:

(i) 953 + 707 + 647
953 + (707 + 647)                                   (Using associative property of addition)
= 953 + 1354  
= 2307   
 
(ii) 1983 + 647 + 217 + 353
(1983 + 647)  + (217 +353)                    (Using associative property of addition)
= 2630 + 570
=  3200


(iii) 15409 + 278 + 691 + 422
(15409 + 278) + (691 + 422)                     (Using associative property of addition)
= 15687 + 1113
= 16800

(iv) 3259 + 10001 + 2641 + 9999
(3259 + 10001) + (2641 +  9999)             (Using associative property of addition)
= 13260 + 12640
= 25900

(v)1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(1 + 2 + 3 + 4) + (96 + 97 + 98 + 99)       (Using associative property of addition)
= (10) + (390)
=  400

(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
(2 + 3 + 4 + 5) + (45 + 46 + 47 + 48)                 (Using associative property of addition)
= 14 + 186  
= 200  

Page No 48:

Question 5:

Find the sum by short method:
(i) 6784 + 9999
(ii) 10578 + 99999

ANSWER:

(i)  6784 + 9999
=  6784 + (10000 − 1)
=  (6784 + 10000) − 1                              (Using associative property of addition)
= 16784 − 1
= 16783

(ii) 10578 + 99999
= 10578 + (100000 − 1)
= (10578 + 100000) − 1                         (Using associative property of addition)
= 110578 − 1
= 110577

Page No 48:

Question 6:

For any whole numbers a, b, c, is it true that (a + b) + c = a + (c + b)? Give reasons.

ANSWER:

For any whole numbers a, b and c, we have:
(a + b) + c = a + ( b + c​) 

Let a = 2, b = 3 and c = 4 [we can take any values for a, b and c]

LHS = (a + b​) + c
= (2 + 3) + 4
= 5 + 4
= 9


RHS = a + (c + b)
= a + (b + c)       [∵ Whole numbers follow the commutative law]                    
= 2 + (3 + 4)
= 2 + 7
= 9

∴ This shows that associativity (in addition) is one of the properties of whole numbers.

Page No 48:

Question 7:

Complete each one of the following magic squares by supplying the missing numbers:
(i)

	9	2
	5
8

(ii)

16	2
	10
		4

(iii) 

2	15	16
9	12
		7	10
14			17

(iv) 

	18	17	4
		14	11
	9	10
19			16

ANSWER:

  In a magic square, the  sum of each row is equal to the sum of each column and the sum of each main diagonal.  By using this concept, we have:
(i)

4	9	2
3	5	7
8	1	6

(ii)

16	2	12
6	10	14
8	18	4

(iii)

2	15	16	5
9	12	11	6
13	8	7	10
14	3	4	17

(iv)

7	18	17	4
8	13	14	11
12	9	10	15
19	6	5	16

Page No 48:

Question 8:

Write (T) for true and (F) for false for each of the following statements:
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two even numbers is an even number.
(iii) The sum of an even number and an odd number is an odd number.

ANSWER:

(i)  F (false). The sum of two odd numbers may not be an odd number. Example: 3 + 5 = 8, which is an even number.
  
(ii) T (true). The sum of two even numbers is an even number. Example: 2 + 4 = 6, which is an even number.
  
(iii) T (true). The sum of an even and an odd number is an odd number. Example: 5 + 4 = 9, which is an odd number.

Page No 49

Exercise 3C

Question 1:

Perform the following subtractions. Check your results by the corresponding additions.
(i) 6237 − 694
(ii) 21205 − 10899
(iii) 100000 − 78987
(iv) 1010101 − 656565

ANSWER:

(i) Subtraction: 6237 − 694 = 5543
   Addition: 5543 + 694 = 6237


(ii) Subtraction: 21205 − 10899 = 10306
  Addition: 10306 + 10899 = 21205


(iii) Subtraction: 100000 − 78987 = 21013
  Addition: 21013 + 78987 = 100000 


(iv) Subtraction: 1010101 − 656565 = 353536
 Addition: 353536 + 656565 = 1010101 

Page No 49:

Question 2:

Replace each * by the correct digit in each of the following:
(i) 
(ii) 
(iii) 
(iv) 

ANSWER:

(i)   917 − *5* =  5*8

  
  ⇒ 917 − 359 =  558

(ii) 6172 − **69 =  29**
    

⇒ 6172 − 3269 = 2903

(iii) 5001003 − **6987 =  484****
  

 ⇒ 5001003 − 155987 = 4845016

(iv)  1000000 − ****1 = *7042*
    

  ⇒ 1000000 − 29571 = 970429

Page No 49:

Question 3:

Find the difference:
(i) 463 − 9
(ii) 5632 − 99
(iii) 8640 − 999
(iv) 13006 − 9999

ANSWER:

(i) 463 − 9
= 463 − 10 + 1
= 464 − 10
= 454

(ii) 5632 − 99
= 5632 − 100 + 1
= 5633 − 100
=  5533

(iii) 8640 − 999
= 8640 − 1000 + 1
= 8641 − 1000
= 7641

(iv) 13006 − 9999
= 13006 − 10000 + 1
= 13007 − 10000
= 3007

Page No 50:

Question 4:

Find the difference between the smallest number of 7 digits and the largest number of 4 digits.

ANSWER:

Smallest seven-digit number = 1000000
Largest four-digit number = 9999              
∴ Their difference = 1000000 − 9999
                               =1000000 − 10000 + 1
                                =1000001 − 10000
                                 =990001

Page No 50:

Question 5:

Ravi opened his account in a bank by depositing Rs 136000. Next day he withdrew Rs 73129 from it. How much money was left in his account?

ANSWER:

Money deposited by Ravi = Rs 1,36,000
Money withdrawn by Ravi= Rs 73,129
Money left in his account  =  money deposited − money withdrawn
                                           = Rs (136000 − 73129)
                                             = Rs 62871

∴ Rs 62,871 is left in Ravi’s account.

Page No 50:

Question 6:

Mrs Saxena withdrew Rs 100000 from her bank account. She purchased a TV set for Rs 38750, a refrigerator for Rs 23890 and jewellery worth Rs 35560. How much money was left with her?

ANSWER:

Money withdrawn by Mrs Saxena  = Rs 1,00,000
Cost of the TV set = Rs 38,750
Cost of the refrigerator = Rs 23,890
Cost of the jewellery = Rs 35,560
Total money spent = Rs (38750 + 23890 + 35560) = Rs 98200

Now, money left = money withdrawn − money spent
                            = Rs (100000 − 98200)
                            = Rs 1800

∴ Rs 1,800 is left with Mrs Saxena.

Page No 50:

Question 7:

The population of a town was 110500. In one year it increased by 3608 due to new births. However, 8973 persons died or left the town during the year. What was the population at the end of the year?

ANSWER:

Population of the town = 110500
Increased population = 110500 + 3608 = 114108
Number of persons who died or left the town = 8973
Population at the end of the year = 114108 − 8973 = 105135

∴ The population at the end of the year will be 105135.

Page No 50:

Question 8:

Find the whole number n when:
(i) n + 4 = 9
(ii) n + 35 = 101
(iii) n − 18 = 39
(iv) n − 20568 = 21403

ANSWER:

(i) n + 4 = 9
⇒ n = 9 − 4 = 5

(ii) n + 35 = 101
⇒ n = 101 − 35 = 66

(iii) n – 18 = 39
 ⇒ n =  18 + 39 = 57

(iv) n − 20568 = 21403
  ⇒ n  = 21403 + 20568 = 41971

Page No 53:

Exercise 3D

Question 1:

Fill in the blanks to make each of the following a true statement:
(i) 246 × 1 = ……
(ii) 1369 × 0 = …….
(iii) 593 × 188 = 188 × …….
(iv) 286 × 753 = …… × 286
(v) 38 × (91 × 37) = …… × (38 × 37)
(vi) 13 × 100 × …… = 1300000
(vii) 59 × 66 + 59 × 34 = 59 × (…… + ……)
(viii) 68 × 95 = 68 × 100 − 68 × …….

ANSWER:

(i) 246 × 1 = 246
(ii) 1369 × 0 = 0
(iii) 593 × 188  = 188 × 593
(iv) 286 × 753 = 753 × 286
(v) 38 × (91 × 37) = 91 × (38 × 37)
(vi) 13 × 100 × 1000 = 1300000
(vii) 59 × 66 + 59  ×  34 = 59 × ( 66 + 34)
(viii) 68 × 95 = 68 × 100 − 68 × 5 

Page No 53:

Question 2:

State the property used in each of the following statements:
(i) 19 × 17 = 17 × 19
(ii) (16 × 32) is a whole number
(iii) (29 × 36) × 18 = 29 × (36 × 18)
(iv) 1480 × 1 = 1480
(v) 1732 × 0 = 0
(vi) 72 × 98 + 72 × 2 = 72 × (98 + 2)
(vii) 63 × 126 − 63 × 26 = 63 × (126 − 26)

ANSWER:

(i) Commutative law in multiplication
(ii) Closure property
(iii) Associativity of multiplication
(iv) Multiplicative identity 
(v) Property of zero
(vi) Distributive law of multiplication over addition
(vii) Distributive law of multiplication over subtraction

Page No 53:

Question 3:

Find the value of each of the following using various properties:
(i) 647 × 13 + 647 × 7
(ii) 8759 × 94 + 8759 × 6
(iii) 7459 × 999 + 7459
(iv) 9870 × 561 − 9870 × 461
(v) 569 × 17 + 569 × 13 + 569 × 70
(vi) 16825 × 16825 − 16825 × 6825

ANSWER:

(i) 647 × 13 + 647 × 7
=  647 × (13 + 7)
= 647 ×  20
= 12940                                 (By using distributive property)

(ii)  8759 × 94 + 8759 × 6
= 8759 × (94 + 6)
= 8759 ×  100
= 875900                              (By using distributive property)

(iii) 7459 × 999 + 7459
= 7459× (999 + 1)
= 7459 × 1000
= 7459000                         (By using distributive property)

(iv) 9870 × 561 − 9870 × 461
= 9870 × (561 − 461)
= 9870 × 100                           
= 987000                           (By using distributive property)

(v)  569 × 17 + 569 × 13 + 569 × 70
    = 569 × (17+ 13+ 70)
   = 569  × 100
   = 56900                            (By using distributive property)

(vi) 16825 × 16825 − 16825 × 6825
= 16825 × (16825 − 6825)
=  16825 × 10000
= 168250000                        (By using distributive property)

Page No 53:

Question 4:

Determine each of the following products by suitable rearrangements:
(i) 2 × 1658 × 50
(ii) 4 × 927 × 25
(iii) 625 × 20 × 8 × 50
(iv) 574 × 625 × 16
(v) 250 × 60 × 50 × 8
(vi) 8 × 125 × 40 × 25

ANSWER:

(i) 2 × 1658 × 50
= (2 × 50) × 1658
= 100 × 1658
= 165800

(ii) 4 × 927 × 25
= (4 × 25) × 927
= 100 × 927
= 92700

(iii) 625 × 20 × 8 × 50  
= (20  × 50) ×  8 × 625
= 1000 ×  8 × 625
= 8000 × 625
= 5000000

(iv) 574 × 625 × 16
= 574 × (625 × 16)
=  574 × 10000
= 5740000

(v)  250 × 60 × 50 × 8
= (250 × 8) × (60 × 50)
=  2000  × 3000
=  6000000

(vi)  8 × 125 × 40 × 25
=  (8 × 125) × (40 × 25)
= 1000 × 1000
= 1000000

Page No 53:

Question 5:

Find each of the following products, using distributive laws:
(i) 740 × 105
(ii) 245 × 1008
(iii) 947 × 96
(iv) 996 × 367
(v) 472 × 1097
(vi) 580 × 64
(vii) 439 × 997
(viii) 1553 × 198

ANSWER:

(i)  740 × 105
= 740 × (100 + 5)
= 740 × 100 + 740 × 5                    (Using distributive law of multiplication over addition)
= 74000 + 3700
= 77700

(ii) 245 × 1008
= 245 × (1000 + 8)
= 245 × 1000 + 245 × 8                  (Using distributive law of multiplication over addition)
= 245000 + 1960
= 246960

(iii) 947 × 96
= 947 × ( 100 − 4)
=  947 × 100 − 947 × 4                    (Using distributive law of multiplication over subtraction)
= 94700 − 3788
= 90912

(iv)  996 × 367
=  367 × (1000 − 4)
=   367 × 1000 − 367 × 4             (Using distributive law of multiplication over subtraction)
= 367000 × 1468
= 365532

(v) 472 × 1097
= 472 × ( 1000 + 97)  
= 472 × 1000 + 472 × 97                 (Using distributive law of multiplication over addition)
= 472000 + 45784
 = 517784

(vi)  580 × 64
=  580 × (60 + 4)
=  580 × 60 + 580 × 4                        (Using distributive law of multiplication over addition)
= 34800 + 2320
= 37120

(vii) 439 × 997
= 439 × (1000 − 3)
= 439 × 1000 − 439 × 3                   (Using distributive law of multiplication over subtraction)
= 439000 − 1317
= 437683

(viii) 1553 × 198
= 1553 × (200 − 2)
= 1553 × 200 − 1553 × 2                 (Using distributive law of multiplication over subtraction)
= 310600 − 3106
= 307494

Page No 53:

Question 6:

Find each of the following products, using distributive laws:
(i) 3576 × 9
(ii) 847 × 99
(iii) 2437 × 999

ANSWER:

Distributive property of multiplication over addition states that a (b + c) = ab + ac
Distributive property of multiplication over subtraction  states that a (b − c) = ab – ac
(i) 3576  ​×  9
= 3576 × (10 − 1)
= 3576 ​× 10 − 3576 × 1
= 35760 − 3576
= 32184

(ii) 847 ×  99
= 847 × (100 − 1)
= 847 × 100 − 847 × 1
=  84700 − 847
= 83853

(iii) 2437 × 999
= 2437 × (1000 − 1)
= 2437 × 1000 − 2437 × 1
=  2437000 − 2437
= 2434563

Page No 54:

Question 7:

Find the products:
(i) 

(ii) 

(iii) 

(iv) 

ANSWER:

(i)

458 × 67 = 30686

(ii)

3709 × 89 = 330101

(iii)


4617 × 234 = 1080378

(iv)

15208 × 542 = 8242736

Page No 54:

Question 8:

Find the product of the largest 3-digit number and the largest 5-digit number.

ANSWER:

Largest three-digit number = 999
Largest five-digit number = 99999
∴ Product of the two numbers = 999 × 99999
                                                    = 999 × (100000 − 1)                  (Using distributive law)
                                                     = 99900000 − 999
                                                      = 99899001

Page No 54:

Question 9:

A car moves at a uniform speed of 75 km per hour. How much distance will it cover in 98 hours?

ANSWER:

Uniform speed of a car = 75 km/h

Distance = speed × time
               = 75 × 98
               =75 × (100 − 2)                     (Using distributive law)
               =75 × 100 − 75 × 2
               =7500 − 150
              = 7350 km

∴ The distance covered in 98 h is 7350 km.

Page No 54:

Question 10:

A dealer purchased 139 VCRs. If the cost of each set is Rs 24350, find the cost of all the sets together.

ANSWER:

Cost of 1 VCR set = Rs 24350 
Cost of 139 VCR sets = 139 × 24350
                                  =24350 × (140 − 1)                 (Using distributive property)
                                  =24350 × 140 − 24350
                                  =3409000 − 24350
                                   = Rs. 3384650 

∴ The cost of all the VCR sets is Rs 33,84,650.

Page No 54:

Question 11:

A housing society constructed 197 houses. If the cost of construction for each house is Rs 450000, what is the total cost for all the houses?

ANSWER:

Cost of construction of 1 house = Rs 450000
Cost of construction of 197 such houses = 197 × 450000
                                                                = 450000 × (200 − 3)
                                                                = 450000 × 200 − 450000 × 3               [Using distributive property of multiplication over subtraction]
                                                                = 90000000 − 1350000
                                                                = 88650000 

∴ The total cost of construction of 197 houses is Rs 8,86,50,000.

Page No 54:

Question 12:

50 chairs and 30 blackboards were purchased for a school. If each chair costs Rs 1065 and each blackboard costs Rs 1645, find the total amount of the bill.

ANSWER:

Cost of a chair = Rs 1065
Cost of a blackboard = Rs 1645
Cost of 50 chairs = 50 × 1065 = Rs 53250
Cost of 30 blackboards = 30 × 1645 = Rs 49350
∴ Total amount of the bill = cost of 50 chairs + cost of 30 blackboards 
                                           = Rs (53250 + 49350)
                                            = Rs 1,02,600

Page No 54:

Question 13:

There are six sections of Class VI in a school and there are 45 students in each section. If the monthly charges from each student be Rs 1650, find the total monthly collection from Class VI.

ANSWER:

Number of student in 1 section = 45
Number of students in 6 sections = 45 × 6 = 270
Monthly charges from 1 student = Rs 1650
∴ Total monthly collection from class VI = Rs 1650 × 270 = Rs 4,45,500

Page No 54:

Question 14:

The product of two whole numbers is zero. What do you conclude?

ANSWER:

If the product of two whole numbers is zero, then one of them is definitely zero.
Example: 0 × 2 = 0 and 0 × 15 = 0

If the product of whole numbers is zero, then both of them may be zero.
i.e., 0 × 0 = 0

Now, 2 × 5 = 10. Here, the product will be non-zero because the numbers to be multiplied are not equal to zero.

Page No 54:

Question 15:

Fill in the blanks:
(i) Sum of two odd numbers is an …… number.
(ii) Product of two odd numbers is an …… number.
(iii) a ≠ 0a ≠ 0 and a × a = a ⇒⇒ a = ?

ANSWER:

(i) Sum of two odd numbers is an even number. Example: 3 + 5 = 8, which is an even number.
(ii) Product of two odd numbers is an odd number. Example: 5 × 7 = 35, which is an odd number.
(iii)  a ≠ 0 anda × a = a 
     Given: a × a = a
                ⇒ a = aa=1aa=1, a ≠ 0             

Page No 56:

Exercise 3E

Question 1:

Divide and check your answer by the corresponding multiplication in each of the following:
(i) ​1936 ​÷ 36
(ii) 19881 ​÷ 47
(iii) 257796 ​÷ 341
(iv) 612846 ​÷ 582
(v) 34419 ​÷ 149
(vi) 39039 ​÷ 1001

ANSWER:

(i)


Dividend = 1936, Divisor = 36 , Quotient = 53 , Remainder = 28
Check: Divisor × Quotient + Remainder  =  36 × 53 + 28
                                                              = 1936
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.
(ii) 19881 ​÷ 47
    
 Dividend = 19881, Divisor = 47 , Quotient = 423, Remainder = 0
Check: Divisor ×Quotient + Remainder= 47 × 423 + 0
                                                           = 19881
                                                           =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iii)

Dividend = 257796 , Divisor = 341 , Quotient = 756 , Remainder = 0
Check : Divisor × Quotient + Remainder = 341 × 756 + 0
                                                              = 257796
                                                             = Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iv) 612846 ​÷ 582 
   

Dividend = 612846 , Divisor = 582, Quotient = 1053 , Remainder = 0
Check :  Divisor × Quotient + Remainder= 582 × 1053 + 0
                                                              = 612846
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(v) 34419 ​÷ 149
    
Dividend = 34419, Divisor = 149 , Quotient = 231, Remainder = 0
Check : Divisor × Quotient + Remainder  = 149 × 231 + 0
                                                               = 34419
                                                               =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.
 
(vi) 39039 ​÷ 1001

  
Dividend = 39039 , Divisor = 1001 , Quotient = 39 , Remainder = 0
Check : Divisor × Quotient + Remainder = 1001 × 39 + 0
                                                              = 39039
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

Page No 56:

Question 2:

Divide, and find out the quotient and remainder. Check your answer.
(i) 6971 ​÷ 47
(ii) 4178 ​÷ 35
(iii) 36195 ​÷ 153
(iv) 93575 ​÷ 400
(v) 23025 ​÷ 1000
(vi) 16135 ​÷ 875

ANSWER:

(i)  6971 ​÷ 47
     
Quotient = 148 and Remainder = 15
Check: Divisor × Quotient + Remainder = 47 × 148 + 15
                                                                  = 6971
                                                                 = Dividend
∴ Dividend = Divisor × Quotient + Remainder 
Verified.

(ii)  4178 ​÷ 35

       
Dividend = 119 and Remainder = 13
Check: Divisor × Quotient + remainder = 35 ×  119 + 13 
                                                                    = 4178
                                                                   = Dividend

∴ Dividend= Divisor × Quotient + Remainder
Verified.
(iii) 36195 ​÷ 153
     
Quotient = 236 and Remainder = 87
Check: Divisor × Quotient + Remainder =  153 × 236 + 87
                                                            = 36195
                                                            = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(iv) 93575 ​÷ 400
      

Quotient = 233 and Remainder = 375
Check: Divisor × Quotient + Remainder =  400 ×  233 + 375
                                                              = 93575
                                                              = Dividend
∴ Dividend= Divisor × Quotient + Remainder
Verified.

(v)  23025 ​÷ 1000 
        
Quotient = 23 and remainder = 25
Check: Divisor × Quotient + Remainder =1000  × 23 + 25
                                                                = 23025
                                                                = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(vi) 16135 ​÷ 875
        
Quotient = 18 and Remainder = 385
Check: Divisor × Quotient + Remainder =875  ×  18 + 385
                                                                 = 16135
                                                                = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.

Page No 56:

Question 3:

Find the value of
(i) 65007 ​÷ 1
(ii) 0 ​÷ 879
(iii) 981 + 5720 ​÷ 10
(iv) 1507 − (625 ÷ 25)
(v) 32277 ÷ (648 − 39)
(vi) (1573 ÷ 1573) − (1573 ÷ 1573)

ANSWER:

(i) 65007 ​÷ 1 = 65007

(ii) 0 ​÷ 879  = 0

(iii) 981 + 5720 ​÷ 10
= 981 + (5720 ​÷ 10)                               (Following DMAS property)
= 981 + 572
= 1553

(iv) 1507 − (625 ÷ 25)                             (Following BODMAS property)
= ​​1507 − 25
= 1482

(v) 32277 ÷ (648 − 39)                                 (Following BODMAS property)
= ​32277 ÷ (609)
=  53

(vi)  (1573 ÷ 1573) − (1573 ÷ 1573)            (Following BODMAS property)
= 1 − 1
= 0 

Page No 56:

Question 4:

Find a whole number n such that n ÷ n = n.

ANSWER:

Given:  n ÷ n = n
⇒  nnnn= n 
​⇒  n = n^​2

i.e., the whole number n is equal to n^​2.

∴ The given whole number must be 1.                  

Page No 56:

Question 5:

The product of two numbers is 504347. If one of the numbers is 317, find the other.

ANSWER:

Let x and y be the two numbers.

Product of the two numbers = x × y = 504347

If x = 317, we have:

317 × y = 504347
⇒ y = 504347 ÷ 317
      


 y=  1591

∴ The other number is 1591.

Page No 56:

Question 6:

On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.

ANSWER:

Dividend = 59761, quotient = 189, remainder = 37 and divisor = ?
  
Dividend = divisor × quotient + remainder
⇒ 59761 = divisor × 189 + 37
⇒ 59761 − 37 = divisor × 189
⇒ 59724 = divisor × 189
⇒ Divisor = 59724 ​÷ 189                  
              
                                                  
Hence, divisor =316                                    
                                                            

Page No 56:

Question 7:

On dividing 55390 by 299, the remainder is 75. Find the quotient using the division algorethm.

ANSWER:

Here, Dividend = 55390, Divisor = 299 and Remainder = 75
We have to find the quotient.
Now, Dividend = Divisor × Quotient + Remainder 
⇒ 55390 = 299 × Quotient + 75
⇒ 55390 − 75 = 299 × Quotient
⇒ 55315 = 299 × Quotient 
⇒ Quotient = 55315 ​÷ 299                                 
                                                         
                                                                           
Hence, quotient =185       

Page No 56:

Question 8:

What least number must be subtracted from 13601 to get a number exactly divisible by 87?

ANSWER:

First, we will divide 13601 by 87.



Remainder = 29
So, 29 must be subtracted from 13601 to get a number exactly divisible by 87.
i.e., 13601 − 29 = 13572

Now, we have:

  
∴ 29 must be subtracted from 13601 to make it divisible by 87.

Page No 56:

Question 9:

What least number must be added to 1056 to get a number exactly divisible by 23?

ANSWER:

First, we will divide 1056 by 23.
        

Required number = 23 − 21 = 2
So, 2 must be added to 1056 to make it exactly divisible by 23.
i.e., 1056 + 2 = 1058

Now, we have:



∴ 1058 is exactly divisible by 23.
                         

Page No 56:

Question 10:

Find the largest 4-digit number divisible by 16.

ANSWER:

We have to find the largest four digit number divisible by 16 .
The largest four-digit number = 9999
Therefore, dividend =9999
Divisor =16


Here, we get remainder =15
Therefore, 15 must be subtracted from 9999 to get the largest four digit number that is divisible by 16.
i.e., 9999 − 15 = 9984

Thus, 9984 is the largest four-digit number that is divisible by 16.

Page No 56:

Question 11:

Divide the largest 5 digit number by 653. Check your answer by the division algorithm.

ANSWER:

Largest five-digit number =99999
  
Dividend = 99999, Divisor = 653, Quotient = 153 and Remainder = 90
Check: Divisor ×Quotient + Remainder
         = 653 × 153 + 90
         = 99909 + 90
         = 99999
         = Dividend

∴ Dividend = Divisor × Quotient + Remainder                  
   Verified.

Page No 56:

Question 12:

Find the least 6-digit number exactly divisible by 83.

ANSWER:

Least six-digit number = 100000
Here, dividend = 100000 and divisor = 83


 
In order to find the least 6-digit number exactly divisible by 83, we have to add 83 − 68 = 15 to the dividend.
I.e., 100000 + 15 = 100015

So, 100015 is the least six-digit number exactly divisible by 83.

Page No 56:

Question 13:

1 dozen bananas cost Rs 29. How many dozens can be purchased for Rs 1392?

ANSWER:

Cost of 1 dozen bananas = Rs 29
Number of dozens purchased for Rs 1392 = 1392 ÷ 29




Hence, 48 dozen of bananas can be purchased with Rs. 1392.

Page No 56:

Question 14:

19625 trees have been equally planted in 157 rows. Find the number of trees in each row.

ANSWER:

Number of trees planted in 157 rows = 19625
Trees planted in 1 row = 19625 ÷ 157



∴ 125 trees are planted in each row.                                  

Page No 56:

Question 15:

The population of a town is 517530. If one out of every 15 is reported to be literate, find how many literate persons are there in the town.

ANSWER:

Population of the town = 517530
(115)115 of the population is reported to be literate, i.e., (115)115 × 517530 = 517530 ÷÷ 15


∴ There are 34502 illiterate persons in the given town.

Page No 56:

Question 16:

The cost price of 23 colour television sets is Rs 570055. Determine the cost price of each TV set if each costs the same.

ANSWER:

Cost price of 23 colour TV sets = Rs 5,70,055
Cost price of 1 TV set  = Rs 570055 ÷ 23

                        
∴ The cost price of one TV set is Rs 24,785.

Page No 56:

Exercise 3F

Question 1:

The smallest whole number is
(a) 1
(b) 0
(c) 2
(d) none of these

ANSWER:

(b) 0

The smallest whole number is 0.

Page No 56:

Question 2:

The least number of 4 digits which is exactly divisible by 9 is
(a) 1018
(b) 1026
(c) 1009
(d) 1008

ANSWER:

(d) 1008


(a)

Hence, 1018 is not exactly divisible by 9.

(b)

Hence, 1026 is exactly divisible by 9.
(c)

Hence, 1009 is not exactly divisible by 9.

(d)

Hence, 1008 is exactly divisible by 9.

(b) and (d) are exactly divisible by 9, but (d) is the least number which is exactly divisible by 9.

Page No 57:

Question 3:

The largest number of 6 digits which is exactly divisible by 16 is
(a) 999980
(b) 999982
(c) 999984
(d) 999964

ANSWER:

(c) 999984

(a)

Hence, 999980 is not exactly divisible by 16.
(b)

Hence, 999982 is not exactly divisible by 16.
(c)

Hence, 999984 is exactly divisible by 16.
(d)

Hence, 999964 is not exactly divisible by 16.

The largest six-digit number which is exactly divisible by 16 is 999984.

          

Page No 57:

Question 4:

What least number should be subtracted from 10004 to get a number exactly divisible by 12?
(a) 4
(b) 6
(c) 8
(d) 20

ANSWER:

(c) 8

Here we have to tell what least number should be subtracted from 10004 to get a number exactly divisible by 12
So, we will first divide 10004 by 12.


Remainder = 8
So, 8 should be subtracted from 10004 to get the number exactly divisible by 12.
i.e., 10004 − 8 = 9996


  Hence, 9996 is exactly divisible by 12.

Page No 57:

Question 5:

What least number should be added to 10056 to get a number exactly divisible by 23?
(a) 5
(b) 18
(c) 13
(d) 10

ANSWER:

(a) 18

Here , we have to tell that what least number must be added to 10056 to get a number exactly divisible by 23
So, first we will divide 10056 by 23


Remainder = 5
Required number = 23 − 5 = 18

So, 18 must be added to 10056 to get a number exactly divisible by 23.
i.e., 10056 + 18 = 10074
  
Hence, 10074 is exactly divisible by 23 .

Page No 57:

Question 6:

What whole number is nearest to 457 which is divisible by 11?
(a) 450
(b) 451
(c) 460
(d) 462

ANSWER:

(d) 462

(a)

Hence, 450 is not divisible by 11.
(b)

Hence, 451 is divisible by 11.
(c)

Hence, 460 is not divisible by 11.
(d)

Hence, 462 is divisible by 11.

Here, the numbers given in options (b) and (d) are divisible by 11. However, we want a whole number nearest to 457 which is divisible by 11.
So, 462 is whole number nearest to 457 and divisible by 11.    

Page No 57:

Question 7:

How many whole numbers are there between 1018 and 1203?
(a) 185
(b) 186
(c) 184
(d) none of these

ANSWER:

(c) 184

Number of whole numbers = (1203 − 1018) − 1
                                             = 185 − 1
                                              =  184

Page No 57:

Question 8:

A number when divided by 46 gives 11 as quotient and 15 as remainder. The number is
(a) 491
(b) 521
(c) 701
(d) 679

ANSWER:

(b) 521

Divisor = 46
Quotient = 11
Remainder = 15
Dividend = divisor × quotient + remainder
             = 46 × 11 + 15
             = 506 + 15
             = 521

Page No 57:

Question 9:

In a division sum, we have dividend = 199, quotient = 16 and remainder = 7. The divisor is
(a) 11
(b) 23
(c) 12
(d) none of these

ANSWER:

(c) 12

Dividend = 199
Quotient = 16
Remainder = 7
According to the division algorithm, we have: 
Dividend = divisor × quotient + remainder
⇒ 199 = divisor × 16 + 7
⇒ 199 − 7 = divisor × 16
⇒ Divisor = 192 ÷ 16
            

Page No 57:

Question 10:

7589 − ? = 3434
(a) 11023
(b) 4245
(c) 4155
(d) none of these

ANSWER:

(a) 11023

7589 − ? = 3434
⇒ 7589 − x = 3434
⇒ x = 7589 + 3434
⇒ x = 11023

Page No 57:

Question 11:

587 × 99 = ?
(a) 57213
(b) 58513
(c) 58113
(d) 56413

ANSWER:

(c) 58113

587 × 99
= 587 × (100 − 1)
= 587 × 100 − 587 × 1         [Using distributive property of multiplication over subtraction]
= 58700 − 587
= 58113

Page No 57:

Question 12:

4 × 538 × 25 = ?
(a) 32280
(b) 26900
(c) 53800
(d) 10760

ANSWER:

(c) 53800

4 × 538 × 25
= (4 × 25) × 538
=  100 × 538  
= 53800

Page No 57:

Question 13:

24679 × 92 + 24679 × 8 = ?
(a) 493580
(b) 1233950
(c) 2467900
(d) none of these

ANSWER:

(c) 2467900

By using the distributive property, we have:
24679 × 92 + 24679 × 8  
= 24679 ×  (92 + 8)
= 24679 × 100
= 2467900

Page No 57:

Question 14:

1625 × 1625 − 1625 × 625 = ?
(a) 1625000
(b) 162500
(c) 325000
(d) 812500

ANSWER:

(a) 1625000

By using the distributive property, we have:

1625 × 1625 − 1625 × 625
= 1625 × (1625 − 625)
=1625 × 1000
= 1625000

Page No 57:

Question 15:

1568 × 185 − 1568 × 85 = ?
(a) 7840
(b) 15680
(c) 156800
(d) none of these

ANSWER:

(c) 156800

By using the distributive property, we have:
1568 × 185 − 1568 × 85
= 1568 × (185 − 85)
= 1568 × 100
= 156800

Page No 57:

Question 16:

(888 + 777 + 555) = (111 × ?)
(a) 120
(b) 280
(c) 20
(d) 140

ANSWER:

(c) 20

(888 + 777 + 555) = (111 × ?)
⇒ (888 + 777 + 555) = 111 × (8 + 7 + 5)          [By taking 111 common]
                                   = ​111 × (20) = 2220

Page No 57:

Question 17:

The sum of two odd numbers is
(a) an odd number
(b) an even number
(c) a prime number
(d) a multiple of 3

ANSWER:

(b) an even number

The sum of two odd numbers is an even number.

Example: 5 + 3 = 8

Page No 57:

Question 18:

The product of two odd numbers is
(a) an odd number
(b) an even number
(c) a prime number
(d) none of these

ANSWER:

(a) an odd number

The product of two odd numbers is an odd number.

Example: 5 × 3 = 15

Page No 57:

Question 19:

If a is a whole number such that a + a = a, then a = ?
(a) 1
(b) 2
(c) 3
(d) none of these

ANSWER:

(d) none of these

Given: a is a whole number such that a + a = a.

If a = 1, then 1+ 1 = 2 ≠ 1 
If a =2, then 2 + 2 = 4 ≠ 2
If a =3, then 3 + 3 = 6 ≠ 3

Page No 57:

Question 20:

The predecessor of 10000 is
(a) 10001
(b) 9999
(c) none of these

ANSWER:

(b) 9999

Predecessor of 10000 = 10000 − 1 = 9999

Page No 57:

Question 21:

The successor of 1001 is
(a) 1000
(b) 1002
(c) none of these

ANSWER:

(b) 1002

Successor of 1001 = 1001 + 1 = 1002

Page No 57:

Question 22:

The smallest even whole number is
(a) 0
(b) 2
(c) none of these

ANSWER:

(b) 2

The smallest even whole number is 2. Zero (0) is neither an even number nor an odd number.

Page No 59:

Exercise 3G

Question 1:

How many whole numbers are there between 1064 and 1201?

ANSWER:

Number of whole numbers between 1201 and 1064 = ( 1201 − 1064 ) − 1
                                                                                    = 137 − 1
                                                                                     = 136

Page No 59:

Question 2:

Fill in the blanks.
     1000000  −****1*7042*      1000000  -****1*7042* 

ANSWER:

   1000000 
−      ****1
                      
      *7042*  

Then, we have:

   1000000
−    29571
                      
     970429

Page No 59:

Question 3:

Use distributive law to find the value of            1063 × 128 − 1063 × 28.

ANSWER:

Using distributive law, we have:
 1063 × 128 − 1063 × 28
= 1063 × (128 − 28)
= 1063 × 100
= 106300

Page No 59:

Question 4:

Find the product of the largest 5-digit number and the largest 3-digit number using distributive law.

ANSWER:

Largest five-digit number = 99999
Largest three-digit number = 999

By using distributive law, we have:

Product = 99999 × 999               
= 99999 × (1000 − 1)                                  [By using distributive law]
= 99999 × 1000 −  99999 × 1
= 99999000 − 99999
= 99899001

OR

999 × 99999
= 999 × ( 100000 − 1)                                   [By using distributive law]
= 999 × 100000 − 999 × 1
= 99900000 − 999
= 99899001

Page No 59:

Question 5:

Divide 53968 by 267 and check the result by the division algorithm.

ANSWER:

 

Dividend = 53968, Divisor = 267, Quotient = 202 and Remainder = 34

Check: Quotient × Divisor + Remainder
         = 267  × 202 + 34
         =  53934 + 34
          = 53968
           = Dividend

∴ Dividend = Quotient × Divisor + Remainder

Verified.

Page No 59:

Question 6:

Find the largest 6-digit number divisible by 16.

ANSWER:

Largest six-digit number = 999999
  
 Remainder = 15

Largest six-digit number divisible by 16 = 999999 − 15 = 999984

∴ 999984 is divisible by 16.

Page No 59:

Question 7:

The cost price of 23 TV sets is Rs 570055. Find the cost of each such set.

ANSWER:

Cost price of 23 TV sets = Rs 5,70,055
Cost price of 1 TV set = 570055 ÷ 23



∴ The cost of one TV set is Rs 24,785.

Page No 59:

Question 8:

What least number must be subtracted from 13801 to get a number exactly divisible by 87?

ANSWER:

We have to find the least number that must be subtracted from 13801 to get a number exactly divisible by 87
So, first we will divide 13801 by 87
   

Remainder = 55
The number 55 should be subtracted from 13801 to get a number divisible by 87.
i.e., 13801 − 55 = 13746

∴ 13746 is divisible by 87.

Page No 59:

Question 9:

The value of (89 × 76 + 89 × 24) is
(a) 890
(b) 8900
(c) 89000
(d) 10420

ANSWER:

(b) 8900

(89 × 76 + 89 × 24)
= 89 × (76 + 24)       [Using distributive property of multiplication over addition]
= 89 × 100
= 8900   

Page No 59:

Question 10:

On dividing a number by 53 we get 8 as quotient and 5 as remainder. The number is
(a) 419
(b) 423
(c) 429
(d) none of these

ANSWER:

(c) 429

Divisor = 53, Quotient = 8, Remainder = 5 and Dividend = ?

Now, Dividend = Quotient × Divisor +Remainder
                         = 8 × 53 + 5 
                          = 429

Page No 59:

Question 11:

The whole number which has no predecessor is
(a) 1
(b) 0
(c) 2
(d) none of these

ANSWER:

(b) 0

The whole number which has no predecessor is 0.

i.e., 0 − 1 = −1, which is not a whole number.

Page No 59:

Question 12:

67 + 33 = 33 + 67 is an example of
(a) closure property
(b) associative property
(c) commutative property
(d) distributive property

ANSWER:

(c) Commutative property

67 + 33 = 33 + 67 is an example of​ commutative property of addition.

Page No 59:

Question 13:

Additive inverse of 36 is
(a) 136136
(b) 0
(c) −36
(d) none of these

ANSWER:

(c) -36
The additive inverse of 36 is −36.

i.e., 36 + (−36) = 0

Page No 59:

Question 14:

Which of the following is not zero?
(a) 0 × 0
(b) 0202
(c) (8 − 8)28 – 82
(d) 2 + 0

ANSWER:

(d) 2+0

(a) 0 × 0 = 0
(b) 0/2 = 0 
(c) (8−8)2=028-82=02 =0
(d) 2 + 0 = 2

Page No 59:

Question 15:

The predecessor of the smallest 3-digit number is
(a) 999
(b) 100
(c) 101
(d) 99

ANSWER:

(d) 99

Smallest three-digit number = 100
∴ Predecessor of 100 = 100 − 1 = 99

Page No 59:

Question 16:

The number of whole numbers between the smallest whole number and the greatest 2-digit number is
(a) 88
(b) 98
(c) 99
(d) 101

ANSWER:

(b) 98
Smallest whole number = 0
Greatest two-digit number = 99
Number of whole numbers between 0 and 99 = (99 − 0 ) − 1 = 98

Page No 59:

Question 17:

Fill in the blanks.
(i) The smallest natural number is …… .
(ii) The smallest whole number is …… .
(iii) Division by …… is not defined.
(iv) …… is a whole number which is not a natural number.
(v)…… is a whole number which is not a natural number.

ANSWER:

(i) The smallest natural number is 1.
(ii) The smallest whole number is 0.
(iii) Division by 0 is not defined.
(iv) 0 is a whole number which is not a natural number.
(v) 1 is the multiplicative identity for whole numbers.

Page No 60:

Question 18:

Write ‘T’ for true and ‘F’ for false in each of the following:
(i) 0 is the smallest natural number.
(ii) Every natural number is a whole number.
(iii) Every whole number is a natural number.
(iv) 1 has no predecessor in whole numbers.

ANSWER:

(i)  F (false). 0 is not a natural number.
​(ii) T (true).  
(iii) F (false). 0 is a whole number but not a natural number.
(iv) F (false). 1 − 1 = 0 is a predecessor of 1, which is a whole number.

Page No 60:

Question 19:

Match the following columns on whole numbers:

column A	column B
(a) 137 + 63 = 63 + 137	(i) Associativity of multiplication
(b) (16 × 25) is a number	(ii) Commutativity of multiplication
(c) 365 × 18 = 18 × 365	(iii) Distributive law of multiplication over addition
(d) (86 × 14) × 25 = 86 × (14 × 25)	(iv) Commutativity of addition
(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)	(v) Closure property for multiplication

ANSWER:

Column A	Column B
(a) 137 + 63 = 63 + 137	(iv) Commutativity of addition
(b) (16 × 25) is a number	(v) Closure property for multiplication
(c) 365 × 18 = 18 × 365	(ii) Commutativity of multiplication
(d) (86 × 14) × 25 = 86 × (14 × 25)	(i) Associativity of multiplication
(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)	(iii) Distributive law of multiplication over addition

Exercise 2A

Page No 25:

Question 1:

Define: (i) factor (ii) multiple. Give five examples of each.

ANSWER:

Factor: A factor of a number is an exact divisor of that number.
Multiple: A multiple of a number is a number obtained by multiplying it by a natural number.

Example 1: We know that 15 = 1 × 15 and 15 = 3 × 5

∴ 1, 3, 5 and 15 are the factors of 15
In other words, we can say that 15 is a multiple of 1, 3, 5 and 15.
 
Example 2: We know that 8 = 8 × 1, 8 = 2 × 4 and 8 =  4 × 2

∴ 1, 2, 4 and 8 are the factors of 8.
In other words, we can say that 8 is a multiple of 1, 2, 4 and 8.
 
Example 3: We know that 30 = 30 × 1, 30 = 5 × 6 and 30 = 6 × 5

∴ 1, 5, 6 and 30 are factors of 30.
In other words, we can say that 30 is a multiple of 1, 5, 6 and 30.
 
Example 4: We know that 20 = 20 × 1, 20 = 4 × 5 and 20 = 5 × 4

∴ 1, 4, 5 and 20 are factors of 20.
In other words, we can say that 20 is a multiple of 1, 4, 5 and 20.
 
Example 5: We know that 10 = 10 × 1, 10 = 2 × 5 and 10 = 5 × 2

∴ 1, 2, 5 and 10 are factors of 10.
In other words, we can say that 10 is a multiple of 1, 2, 5 and 10.

Page No 25:

Question 2:

Write down all the factors of
(i) 20
(ii) 36
(iii) 60
(iv) 75

ANSWER:

(i) 20
20 = 1 × 20; 20 = 10 × 2 and 20 = 4 × 5
The factors of 20 are 1, 2, 4, 5, 10 and 20.

(ii) 36
36 = 1 × 36; 36 = 2 × 18; 36 = 3 × 12 and 36 = 4 × 9
The factors of 36 are 1, 2, 3, 4, 6, 9, 12 and 36.

(iii) 60
60 = 1 × 60; 60 = 2 × 30; 60 = 3 × 20; 60 = 4 × 15 and 60 = 5 × 12
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15 and 60.

(iv) 75
75 = 1 × 75; 75 = 3 × 25 and 75 = 5 × 15
The factors of 75 are 1, 3, 5, 15, 25 and 75.

Page No 25:

Question 3:

Write the first five multiples of each of the following numbers:
(i) 17
(ii) 23
(iii) 65
(iv) 70

ANSWER:

(i) 17
17 × 1 = 17; 17 × 2 = 34; 17 × 3 = 51; 17 × 4 = 68 and 17 × 5 = 85
∴ The first five multiples of 17 are 17, 34, 51, 68 and 85.

(ii) 23
23 × 1=23; 23 × 2 = 46; 23 × 3 = 69; 23 × 4 = 92 and 23 × 5 = 115
∴ The first five multiples of 23 are 23, 46, 69, 92 and 115.

(iii) 65
65 × 1 = 65; 65 × 2 = 130; 65 × 3 = 195; 65 × 4 = 260 and 65 × 5 = 325
∴ The first five multiples of  65 are 65, 130, 195, 260 and 325.

(iv) 70
70 × 1=70; 70 × 2 = 140; 70 × 3 = 210; 70 × 4 = 280 and 70 × 5 = 350
∴ The first five multiples of 70 are 70, 140, 210, 280 and 350.

Page No 25:

Question 4:

Which of the following numbers are even and which are odd?
(i) 32
(ii) 37
(iii) 50
(iv) 58
(v) 69
(vi) 144
(vii) 321
(viii) 253

ANSWER:

(i) 32
Since 32 is a multiple of 2, it is an even number.
(ii) 37
Since 37 is not a multiple of 2, it is an odd number.
(iii) 50
Since 50 is a multiple of 2, it is an even number.
(iv) 58
Since 58 is a multiple of 2, it is an even number.
(v) 69
Since 69 is not a multiple of 2, it is an odd number.
(vi) 144
Since 144  is a multiple of 2, it is an even number.
(vii) 321
Since 321  is not a multiple of 2, it is an odd number.
(viii) 253
Since 253 is not a multiple of 2, it is an odd number.

Page No 25:

Question 5:

What are prime numbers? Give ten examples.

ANSWER:

Prime number: A number is called a prime number if it has only two factors, namely 1 and itself .

Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 are prime numbers.

Page No 25:

Question 6:

Write all the prime numbers between
(i) 10 and 40
(ii) 80 and 100
(iii) 40 and 80
(iv) 30 and 40

ANSWER:

(i) All prime numbers between 10 and 40 are 11, 13, 17, 19, 23, 29, 31 and 37.
(ii) All prime numbers between 80 and 100 are 83, 89 and 97.
(iii) All prime numbers between 40 and 80 are 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.
(iv) All prime numbers between 30 and 40 are 31 and 37.

Page No 25:

Question 7:

(i) Write the smallest prime number.
(ii) List all even prime numbers.
(iii) Write the smallest odd prime number.

ANSWER:

(i) The smallest prime number is 2.
(ii) There is only one even prime number, i.e., 2.
(iii) The smallest odd prime number is 3.

Page No 25:

Question 8:

Find which of the following numbers are primes:
(i) 87
(ii) 89
(iii) 63
(iv) 91

ANSWER:

(i) 87
The divisors of 87 are 1, 3, 29 and 87 i.e. 87 has more than 2 factors. Therefore 87 is not a prime number.

(ii) 89
The divisors of 89 are 1 and 89. Therefore 89 is a prime number.

(iii) 63
The divisors of 63 are 1, 3, 7, 9, 21 and 63 i.e. 63 has more than 2 factors. Therefore 63 is not a prime number.

(iv) 91
The divisors of 91 are 1, 7, 13 and 91 i.e. 91 has more than 2 factors. Therefore 91 is not a prime number.

Page No 25:

Question 9:

Make a list of seven consecutive numbers, none of which is prime.

ANSWER:

90, 91, 92, 93, 94, 95 and 96 are seven consecutive numbers and none of them is a prime.

Page No 25:

Question 10:

(i) Is there any counting number having no factor at all?
(ii) Find all the numbers having exactly one factor.
(iii) Find numbers between 1 and 100 having exactly three factors.

ANSWER:

(i) No, there are no counting numbers with no factors at all because every number has at least two factors, i.e., 1 and itself.
(ii) There is only one number that has exactly one factor, i.e, 1.
(iii) The numbers between 1 and 100 that have exactly three factors are 4, 9, 25 and 49.

Page No 25:

Question 11:

What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.

ANSWER:

The numbers that have more than two factors are known as composite numbers.
Yes, a composite number can be odd.
The smallest odd composite number is 9.

Page No 25:

Question 12:

What are twin primes? Write all the pairs of twin primes between 50 and 100.

ANSWER:

Two consecutive odd prime numbers are called twin primes.
The pairs of twin primes between 50 to 100 are (59, 61) and (71, 73).

Page No 25:

Question 13:

What are co-primes? Give examples of five pairs of co-primes. Are co-primes always primes? If no, illustrate your answer by an example.

ANSWER:

 If two numbers do not have a common factor other than 1, they are said to be co-primes.

Five pairs of co primes: (i) 2 and 3 (ii) 3 and 4 (iii) 4 and 5 (iv) 4 and 9 (v) 8 and 15

No, co–primes are not always primes.

For example, 3 and 4 are co-prime numbers, where 3 is a prime number and 4 is not a prime number.

Page No 25:

Question 14:

Express each of the following numbers as the sum of two odd primes:
(i) 36
(ii) 42
(iii) 84
(iv) 98

ANSWER:

(i) 36
36 as the sum of two odd prime numbers is (36 = 31 + 5).
(ii) 42
42 as the sum of two odd prime numbers is (42 = 31 + 11).
(iii) 84
84 as the sum of two odd prime numbers is (84 = 41 + 43).
(iv) 98
98 as the sum of two odd prime numbers is (98 = 31 + 67).

Page No 25:

Question 15:

Express each of the following odd numbers as the sum of three odd prime numbers:
(i) 31
(ii) 35
(iii) 49
(iv) 63

ANSWER:

(i) 31
31 can be expressed as the sum of three odd prime numbers as (31 = 5 + 7 + 19).
(ii) ) 35
35 can be expressed as the sum of three odd prime numbers as (35 = 17 + 13 + 5).
(iii) 49
49 can be expressed as the sum of three odd prime numbers as (49 = 13 + 17 + 19).
(iv)  63
63 can be expressed as the sum of three odd prime numbers as (63 = 29 + 31 + 3).

Page No 25:

Question 16:

Express each of the following numbers as the sum of twin primes:
(i) 36
(ii) 84
(iii) 120
(iv) 144

ANSWER:

(i) 36
36 can be expressed as the sum of twin primes as (36 = 17 + 19).
(ii) 84
84 can be expressed as the sum of twin primes as (84 = 41 + 43).
(iii) 120
120 can be expressed as the sum of twin primes as (120 = 59 + 61).
(iv) 144
144 can be expressed as the sum of twin primes as (144 = 71 + 73).

Page No 26:

Question 17:

Which of the following statements are true?
(i) 1 is the smallest prime number.
(ii) If a number is prime, it must be odd.
(iii) The sum of two prime numbers is always a prime number.
(iv) If two numbers are co-primes, at least one of them must be a prime number.

ANSWER:

(i) False. 2 is the smallest prime number.
(ii) False. 2 is an even prime number.
(iii) False. 3 and 7 are two prime numbers and their sum is 10, which is even.
(iv) False. 4 and 9 are co-primes but neither of them is a prime number.

Page No 29:

Exercise 2B

Question 1:

Test the divisibility of the following numbers by 2:
(i) 2650
(ii) 69435
(iii) 59628
(iv) 789403
(v) 357986
(vi) 367314

ANSWER:

A number is divisible by 2 if its ones digit is 0, 2, 4, 6 or 8.
(i)   Since the digit in the ones place in 26250 is 0, it is divisible by 2
(ii)  Since the digit in the ones place in 69435 is not 0, 2, 4, 6 or 8, it is not divisible by 2.
(iii) Since the digit in the ones place in 59628 is 8, it is divisible by 2.
(iv) Since the digit in the ones place in 789403 is not 0, 2, 4, 6, or 8, it is not divisible by 2.
(v) Since the digit in the ones place in  357986 is 6, it is divisible by 2.
(vi) Since the digit in the ones place in 367314 is 4, it is divisible by 2.

Page No 29:

Question 2:

Test the divisibility of the following numbers by 3:
(i) 733
(ii) 10038
(iii) 20701
(iv) 524781
(v) 79124
(vi) 872645

ANSWER:

A number is divisible by 3 if the sum of its digits is divisible by 3.
(i)  733 is not divisible by 3 because the sum of its digits, 7 + 3 + 3, is 13, which is not divisible by 3.
(ii) 10038 is divisible by 3 because the sum of its digits, 1 + 0 + 0 + 3 + 8, is 12, which is divisible by 3.
 (iii) 20701 is not divisible by 3 because the sum of its digits, 2 + 0 + 7 + 0 + 1, is 10, which is not divisible by 3.
(iv) 524781 is divisible by 3 because the sum of its digits, 5 + 2 + 4 + 7 + 8 + 1, is 27, which is divisible by 3.
(v) 79124 is not divisible by 3 because the sum of its digits, 7 + 9 + 1 + 2 + 4, is 23, which is not divisible by 3.
(vi) 872645 is not divisible by 3 because the sum of its digits, 8 + 7 + 2 + 6 + 4 + 5, is 32, which is not divisible by 3.

Page No 29:

Question 3:

Test the divisibility of the following numbers by 4:
(i) 618
(ii) 2314
(iii) 63712
(iv) 35056
(v) 946126
(vi) 810524

ANSWER:

A number is divisible by 4 if the number formed by the digits in its tens and units place is divisible by 4.

(i)  618 is not divisible by 4 because the number formed by its tens and ones digits is 18, which is not divisible by 4.
(ii) 2314 is not divisible by 4 because the number formed by its tens and ones digits is 14, which is not divisible by 4.
(iii) 63712 is divisible by 4 because the number formed by its tens and ones digits is 12, which is divisible by 4.
(iv) 35056 is divisible by 4 because the number formed by its tens and ones digits is 56, which is divisible by 4.
(v)  946126 is not divisible by 4 because the number formed by its tens and ones digits is 26, which is not divisible by 4.
(vi) 810524 is divisible by 4 because the number formed by its tens and ones digits is 24, which is divisible by 4.

Page No 29:

Question 4:

Test the divisibility of the following numbers by 5:
(i) 4965
(ii) 23590
(iii) 35208
(iv) 723405
(v) 124684
(vi) 438750

ANSWER:

A number is divisible by 5 if its ones digit is either 0 or 5.

(i) 4965 is divisible by 5, because the digit at its ones place is 5.
  
(ii) 23590 is divisible by 5, because the digit at its ones place is 0.
    
(iii) 35208 is not divisible by 5, because the digit at its ones place is 8.
     
 (iv) 723405 is divisible by 5, because the digit at its ones place is 5.
     
(v) 124684 is not divisible by 5, because the digit at its ones place is 4.
     
(vi) 438750 is divisible by 5, because the digit at its ones place is 0.
    

Page No 30:

Question 5:

Test the divisibility of the following numbers by 6:
(i) 2070
(ii) 46523
(iii) 71232
(iv) 934706
(v) 251780
(vi) 872536

ANSWER:

A number is divisible by 6 if it is divisible by both 2 and 3.

i)  Since 2070 is divisible by 2 and 3, it is divisible by 6.
     Checking the divisibility by 2: Since the number 2070 has 0 in its units place, it is divisible by 2.
     Checking the divisibility by 3: The sum of the digits of 2070, 2 + 0 + 7 + 0, is 9, which is divisible by 3. So, it is divisible by 3.

   
(ii) Since 46523 is not divisible by 2, it is not divisible by 6.

Checking the divisibility by 2: Since the number 46523 has 3 in its units place, it is not divisible by 2.
  
(iii) Since 71232 is divisible by both 2 and 3, it is divisible by 6.
     
Checking the divisibility by 2: Since the number has 2 in its units place, it is divisible by 2.
Checking the divisibility by 3: The sum of the digits of the number, 7 + 1 + 2 + 3 + 2, is 15, which  is divisible by 3. So, the number is divisible by 3. 
   
(iv) Since 934706 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: Since the sum of the digits of the number, 9 + 3 + 4 + 7 + 0 + 6, is 29, which is not divisible by 3. So, the number is not divisible by 3.

(v) Since 251780 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 2 + 5 + 1 + 7 + 8 + 0, is 23, which is not divisible by 3. So, the number is not divisible by 3.

(vi)  Since 872536 is not divisible by 3, it is not divisible by 6.
Checking the divisibility by 3: The sum of the digits of the number, 8 + 7 + 2 + 5 + 3 + 6, is 31, which is not divisible by 3. So, the number is not divisible by 3.

Page No 30:

Question 6:

Test the divisibility of the following numbers by 7:
(i) 826
(ii) 117
(iii) 2345
(iv) 6021
(v) 14126
(vi) 25368

ANSWER:

To determine if a number is divisible by 7, double the last digit of the number and subtract it from the number formed by the remaining digits. If their difference is a multiple of 7, the number is divisible by 7.

(i) 826 is divisible by 7.
 We have 82 − 2 × 6 = 70, which is a multiple of 7.

(ii) 117 is not divisible by 7.
      We have 11 − 2 × 7 = −3, which is not a multiple of 7.

(iii) 2345 is divisible by 7.
 We have  234 − 2 × 5 = 224, which is a multiple of 7.

(iv) 6021 is divisible by 7. 
 We have 602 − 2 × 1 = 600, which is not a multiple of  7.

(v)  14126 is divisible by 7.
    We have 1412 − 2 × 6 = 1400, which is a multiple of 7.

(vi)  25368 is divisible by 7.
      We have 2536 − 2 × 8 = 2520, which is a multiple of 7.

Page No 30:

Question 7:

Test the divisibility of the following numbers by 8:
(i) 9364
(ii) 2138
(iii) 36792
(iv) 901674
(v) 136976
(vi) 1790184

ANSWER:

A number is divisible by 8 if the number formed by the last three digits (digits in the hundreds, tens and units places) is divisible by 8.

(i) 9364 is not divisible by 8.
  It is because the number formed by its hundreds, tens and ones digits, i.e., 364, is not divisible by 8.

(ii) 2138 is not divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 138, is not divisible by 8.

(iii) 36792 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 792, is divisible by 8.

(iv) 901674 is not divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 674, is not divisible by 8.

(v) 136976 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 976, is divisible by 8.

(vi) 1790184 is divisible by 8.
 It is because the number formed by its hundreds, tens and ones digits, i.e., 184, is divisible by 8.

Page No 30:

Question 8:

Test the divisibility of the following numbers by 9:
(i) 2358
(ii) 3333
(iii) 98712
(iv) 257106
(v) 647514
(vi) 326999

ANSWER:

A number is divisible by 9 if the sum of its digits is divisible by 9.

(i)  2358 is divisible by 9, because the sum of its digits, 2 + 3 + 5 + 8, is 18, which is divisible by 9.

(ii)  3333 is not divisible by 9, because the sum of its digits, 3 + 3 + 3 + 3, is 12, which is not divisible by 9.

(iii)  98712 is divisible by 9, because the sum of its digits, 9 + 8 + 7 + 1 + 2, is 27, which is divisible by 9.

(iv)  257106 is not divisible by 9, because the sum of its digits, 2 + 5 + 1 0 + 6, is 21, which is not divisible by 9.

(v)  647514 is divisible by 9, because the sum of its digits, 6 + 4 + 7 + 5 + 1 + 4, is 27, which is divisible by 9.

(vi)  326999 is not divisible by 9, because the sum of its digits, 3 + 2 + 6 + 9 + 9 + 9, is 38, which is not divisible by 9. 

Page No 30:

Question 9:

Test the divisibility of the following numbers by 10:
(i) 5790
(ii) 63215
(iii) 55555

ANSWER:

A number is divisible by 10 if its ones digit is 0.

(i) 5790 is divisible by 10, because its ones digit is 0.
(ii) 63215 is not divisible by 10, because its ones digit is 5, not 0.
(iii) 55555 is not divisible by 10, because its ones digit is 5, not 0.

Page No 30:

Question 10:

Test the divisibility of the following numbers by 11:
(i) 4334
(ii) 83721
(iii) 66311
(iv) 137269
(v) 901351
(vi) 8790322

ANSWER:

A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.

(i)  4334 is divisible by 11.
           
Sum of the digits at odd places = (4 + 3) = 7
Sum of the digits at even places = (3 + 4) = 7
    Difference of the two sums = (7 − 7) = 0, which is divisible by 11.

(ii)  83721 is divisible by 11.
Sum of the digits at odd places = (1 + 7 + 8) = 16
Sum of the digits at even places = (2 + 3) = 5
     Difference of the two sums = (16 − 5) = 11, which is divisible by 11.

(iii) 66311 is not divisible by 11.
          
   Sum of the digits at odd places = (1 + 3 + 6) = 10
   Sum of the digits at even places = (1 + 6) = 7
   Difference of the two sums = (10 − 7) = 3, which is not divisible by 11.
         
(iv) 137269 is divisible by 11.
           
   Sum of the digits at odd places = (9 + 2 + 3) = 14
   Sum of the digits at even places = (6 + 7 + 1) = 14
     Difference of the two sums = (14 − 14) = 0, which is a divisible by 11.

(v)  901351 is divisible by 11.
           
  Sum of the digits at odd places = (0 + 3 + 1) = 4
 Sum of the digits at even places = (9 + 1 + 5) = 15
 Difference of the two sums = (4 − 15) = −11, which is divisible by 11.
          
(vi) 8790322 is not divisible by 11.
           
    Sum of the digits at odd places = (2 + 3 + 9 + 8) = 22
    Sum of the digits at even places = (2 + 0 + 7) = 9
    Difference of the two sums = (22 − 9) = 13, which is not divisible by 11.

Page No 30:

Question 11:

In each of the following numbers, replace * by the smallest number to make it divisible by 3:
(i) 27*4
(ii) 53*46
(iii) 8*711
(iv) 62*35
(v) 234*17
(vi) 6*1054

ANSWER:

(i) 2724
 Here, 2 + 7 + * + 4 = 13 + * should be a multiple of 3.
 To be divisible by 3, the least value of * should be 2, i.e., 13 + 2 = 15, which is a multiple of 3.
∴ * = 2

(ii) 53046
Here, 5 + 3 + * + 4 + 6 = 18 + * should be a multiple of 3.
As 18 is divisible by 3, the least value of  * should be 0, i.e., 18 + 0 = 18.
 ∴ * = 0

 (iii) 81711
Here, 8+ * + 7 + 1 + 1 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18 , which is a multiple of 3.
∴ * = 1

 (iv) 62235
Here, 6 + 2 + * + 3 + 5 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
∴ * = 2
 
(v) 234117
Here, 2+ 3 +4 + * + 1 + 7 = 17 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 1, i.e., 17 + 1 = 18, which is a multiple of 3.
∴ * =1

(vi) 621054
Here, 6 + * +1 + 0 + 5 + 4 = 16 + * should be a multiple of 3.
To be divisible by 3, the least value of  * should be 2, i.e., 16 + 2 = 18, which is a multiple of 3.
∴ * =2

Page No 30:

Question 12:

In each of the following numbers, replace * by the smallest number to make it divisible by 9:
(i) 65*5
(ii) 2*135
(iii) 6702*
(iv) 91*67
(v) 6678*1
(vi) 835*86

ANSWER:

(i) 6525
Here, 6 + 5+ * + 5 = 16 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 2, i.e., 16 + 2 = 18, which is a multiple of 9.
∴ * =2

(ii) 27135
Here, 2 + * + 1 + 3 + 5 = 11 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 7, i.e., 11 + 7 = 18, which is a multiple of 9.
∴ * = 7

(iii) 67023
Here, 6 + * + 7 + 0 + 2 = 15 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 3, i.e., 15 + 3 = 18, which  is a multiple of 9.
∴ * = 3

(iv) 91467
Here, 9 + 1 * + 6 + 7 = 23 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 4, i.e., 23 + 4 = 27, which is a multiple of 9.
∴ * = 4

(v) 667881
Here,  6 + 6 + 7 + 8 + * + 1 = 28 + * should be a multiple of 9.
To be divisible by 9, the least value of * should be 8, i.e., 28 + 8 = 36, which is a multiple of 9.
∴ * = 8
 
(vi) 835686
Here, 8 + 3 + 5 + * + 8 + 6 = 30 + * should be a multiple of 9.
To be divisible of 9, the least value of * should be 6, i.e.,  30 + 6 = 36, which is a multiple of 9.
∴ * = 6

Page No 30:

Question 13:

In each of the following numbers, replace * by the smallest number to make it divisible by 11:
(i) 26*5
(ii) 39*43
(iii) 86*72
(iv) 467*91
(v) 1723*4
(vi) 9*8071

ANSWER:

(i) 26*5
Sum of the digits at odd places = 5 + 6 = 11
Sum of the digits at even places = * + 2
Difference = sum of odd terms – sum of even terms
= 11 – (* + 2)  
= 11 – * – 2
 = 9 – *
Now, (9 – *) will be divisible by 11 if * = 9.
i.e., 9 – 9 = 0
0 is divisible by 11.
∴ * = 9
Hence, the number is 2695.

(ii) 39*43
Sum of the digits at odd places = 3 + * + 3 = 6 + *
Sum of the digits at even places = 4 + 9 = 13
Difference = sum of odd terms – sum of even terms
= 6 + * – 13  
= * – 7
Now, (* – 7) will be divisible by 11 if * = 7.
i.e., 7 – 7 = 0
0 is divisible by 11.
∴ * = 7
Hence, the number is 39743.
 
(iii) 86*72
Sum of the digits at odd places  2 + * + 8 = 10 + *
Sum of the digits at even places  6 + 7 = 13
Difference = sum of odd terms – sum of even terms
= 10 + * – 13  
= * – 3
  Now, (* – 3) will be divisible by 11 if * = 3.
i.e., 3 – 3 = 0
0 is divisible by 11.
∴ * = 3
 Hence, the number is 86372.

(iv) 467*91
Sum of the digits at odd places 1  + * + 6 = 7 + *
Sum of the digits at even places  9 + 7 + 4 = 20
Difference = sum of odd terms – sum of even terms
= (7 + *) − 20  
   = * − 13
  Now, (* −13) will be divisible by 11 if * = 2.
i.e., 2− 13 = −11
−11 is divisible by 11.
∴ * = 2
 Hence, the number is 467291.

(v) 1723*4
Sum of the digits at odd places 4+ 3+ 7= 14
Sum of the digits at even places  *+2+1 = 3 + *
Difference = sum of odd terms – sum of even terms
= 14 – (3 + *)  
   = 11 − *
  Now, (11 − *) will be divisible by 11 if * = 0.
i.e., 11 − 0 = 11
11 is divisible by 11.
∴ * = 0
 Hence, the number is 172304.

(vi) 9*8071
Sum of the digits at odd places 1+0+*  = 1 + *
Sum of the digits at even places  7 + 8 + 9 = 24
Difference = sum of odd terms – sum of even terms
=1 + *  – 24  
   =   * − 23
  Now, (* − 23) will be divisible by 11 if * = 1.
i.e., 1 − 23 = −22
−22 is divisible by 11.
∴  * = 1
Hence, the number is 918071.

Page No 30:

Question 14:

Test the divisibility of:
(i) 1000001 by 11
(ii) 19083625 by 11
(iii) 2134563 by 9
(iv) 10001001 by 3
(v) 10203574 by 4
(vi) 12030624 by 8

ANSWER:

(i) 10000001 by 11
10000001 is divisible by 11.
Sum of digits at odd places = (1 + 0 + 0 + 0) = 1
Sum of digits at even places = (0 + 0 + 0 + 1) = 1
Difference of the two sums = (1 − 1) = 0, which is divisible by 11.
 
 
(ii) 19083625 by 11
19083625 is divisible by 11.
Sum of digits at odd places = (5 + 6 + 8 + 9) = 28
Sum of digits at even places = (2 + 3 + 0 + 1) = 6
 Difference of the two sums = (28 − 6) = 22, which is divisible by 11.

(iii) 2134563 by 9
    2134563 is not divisible by 9.
It is because the sum of its digits, 2 + 1 + 3 + 4 + 5 + 6 + 3, is 24, which is not divisible by 9.
          
(iv) 10001001 by 3
10001001 is divisible by 3. 
It is because the sum of its digits, 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1, is 3, which is divisible by 3.

(v) 10203574 by 4                  
  10203574 is not divisible by 4.
It is because the number formed by its tens and the ones digits is 74, which is not divisible by 4.
(vi) 12030624 by 8
     12030624 is divisible by 8.
It is because the number formed by its hundreds, tens and ones digits is 624, which is divisible by 8.

Page No 30:

Question 15:

Which of the following are prime numbers?
(i) 103
(ii) 137
(iii) 161
(iv) 179
(v) 217
(vi) 277
(vii) 331
(viii) 397

ANSWER:

A number between 100 and 200 is a prime number if it is not divisible by any prime number less than 15.
Similarly, a number between 200 and 300 is a prime number if it is not divisible by any prime number less than 20.

(i) 103 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(ii) 137 is a prime number, because it is not divisible by 2, 3, 5, 7 and 11.
(iii) 161 is a not prime number, because it is divisible by 7.
(iv) 179 is a prime number, because it is not divisible by 2, 3, 5, 7, 11 and 13.
(v) 217 is a not prime number, because it is divisible by 7.
(vi) 277 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(vii) 331 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.
(viii) 397 is a prime number, because it is not divisible by 2, 3, 5, 7, 11, 13, 17 and 19.

Page No 30:

Question 16:

Give an example of a number
(i) which is divisible by 2 but not by 4.
(ii) which is divisible by 4 but not by 8.
(iii) which is divisible by both 2 and 8 but not by 16.
(iii) which is divisible by both 3 and 6 but not by 18.

ANSWER:

(i) 14 is divisible by 2, but not by 4.
(ii) 12 is divisible by 4, but not by 8.
(iii) 24 is divisible by both 2 and 8, but not by 16.
(iv) 30 is divisible by both 3 and 6, but not by 18.

Page No 30:

Question 17:

Write (T) for true and (F) for false against each of the following statements:
(i) If a number is divisible by 4, it must be divisible by 8.
(ii) If a number is divisible by 8, it must be divisible by 4.
(iii) If a number divides the sum of two numbers exactly, it must exactly divide the numbers separately.
(iv) If a number is divisible by both 9 and 10, it must be divisible by 90.
(v) A number is divisible by 18 if it is divisible by both 3 and 6.
(vi) If a number is divisible by 3 and 7, it must be divisible by 21.
(vii) The sum of two consecutive odd numbers is always divisible by 4.
(viii) If a number divides two numbers exactly, it must divide their sum exactly.

ANSWER:

(i) If a number is divisible by 4, it must be divisible by 8. False
Example: 28 is divisible by 4 but not divisible by 8.

(ii) If a number is divisible by 8, it must be divisible by 4. True
Example: 32 is divisible by both 8 and 4.

(iii) If a number divides the sum of two numbers exactly, it must exactly divide the numbers separately. False
Example: 91 (51 + 40) is exactly divisible by 13. However, 13 does not exactly divide 51 and 40.

(iv) If a number is divisible by both 9 and 10, it must be divisible by 90. True
Example: 900 is both divisible by 9 and 10. It is also divisible by 90.

(v)  A number is divisible by 18 if it is divisible by both 3 and 6. False
A number has to be divisible by 9 and 2 to be divisible by 18.
Example: 48 is divisible by 3 and 6, but not by 18.

 (vi) If a number is divisible by 3 and 7, it must be divisible by 21. True
Example: 42 is divisible by both 3 and 7. It is also divisible by 21.

(vii) The sum of consecutive odd numbers is always divisible by 4. True
 Example: 11 and 13 are consecutive odd numbers.
11 + 13 = 24, which is divisible by 4.  

(viii) If a number divides two numbers exactly, it must divide their sum exactly.  True
Example: 42 and 56 are exactly divisible by 7.
42+56 = 98, which is exactly divisible by 7. 

Page No 32:

Exercise 2C

Question 1:

Give the prime factorization of each of the following numbers:
12

ANSWER:

We use the division method as shown below:

212      26        33           1212      26        33           1
∴ 12 = 2 × 2 × 3
            = 2² × 3

Page No 32:

Question 2:

Give the prime factorization of each of the following numbers:
18

ANSWER:

We will use the division method as shown below:
218      39        33           1218      39        33           1

∴ 18 = 2 × 3 × 3
         = 2 × 3²

Page No 32:

Question 3:

Give the prime factorization of each of the following numbers:
48

ANSWER:

We will use the division method as shown below:

248        224        212        26          33                 1  248        224        212        26          33                 1  

∴ 48 = 2 × 2 × 2 × 2 × 3
         =2⁴×3

Page No 32:

Question 4:

Give the prime factorization of each of the following numbers:
56

ANSWER:

We will use the division method as shown below:

  256      228       214           7256      228       214           7
∴ 56 = 2 × 2 × 2 × 7
           = 2³×7

Page No 32:

Question 5:

Give the prime factorization of each of the following numbers:
90

ANSWER:

We will use the division method as shown below:

  290        345         315         55         1         290        345         315         55         1       

∴ 90 = 2 × 3 × 3 × 5
         = 2 × 3² × 5

Page No 32:

Question 6:

Give the prime factorization of each of the following numbers:
136

ANSWER:

 We will use the division method as shown below:
  
      2136         268        234     1717            1  2136         268        234     1717            1
∴ 136 = 2 × 2 × 2 × 17
              = 2³ × 17

Page No 32:

Question 7:

Give the prime factorization of each of the following numbers:
252

ANSWER:

We will use the division method as shown below:

2252       2126        363        321       77             12252       2126        363        321       77             1

∴ 252 = 2 × 2 × 3 × 3 × 7 × 1
           = 2² × 3² × 7 × 1                         

Page No 32:

Question 8:

Give the prime factorization of each of the following numbers:
420

ANSWER:

We will use the division method as shown below:

2420       2210        3105       735       55            12420       2210        3105       735       55            1
∴ 420 = 2 × 2 × 3 × 7 ×5 × 1
            =2²×3×5×7

Page No 32:

Question 9:

Give the prime factorization of each of the following numbers:
637

ANSWER:

We will use the division method as shown below:

7637       791        1313              17637       791        1313              1

∴ 637 = 7 ×7 × 13
              = 7² × 13

Page No 32:

Question 10:

Give the prime factorization of each of the following numbers:
945

ANSWER:

We will use the division method as shown below:

3945       3315        3105       535       77            1       3945       3315        3105       535       77            1       

∴ 945 = 3 × 3 × 3 × 5 × 7 × 1
            = 3³ × 5 × 7

Page No 32:

Question 11:

Give the prime factorization of each of the following numbers:
1224

ANSWER:

We will use the division method as shown below:

     21224           2612             2306            3153            351         17 17                 1     21224           2612             2306            3153            351         17 17                 1
∴ 1224 = 2 × 2 × 2 × 3 ×3 × 17
                 = 2³ × 3² × 17

Page No 32:

Question 12:

Give the prime factorization of each of the following numbers:
1323

ANSWER:

We will use the division method as shown below:

31323      3441       3147       749       77             131323      3441       3147       749       77             1
∴ 1323 = 3 × 3 × 3 ×7 × 7 × 1
                  =3³ × 7²

Page No 32:

Question 13:

Give the prime factorization of each of the following numbers:
8712

ANSWER:

We will use the division method as shown below:

 28712       24356       22178       31089         3363        11121       1111                1 28712       24356       22178       31089         3363        11121       1111                1
∴ 8712 = 2 × 2 × 2 × 3 × 3 × 11 × 11
                 = 2³ × 3² × 11²

Page No 32:

Question 14:

Give the prime factorization of each of the following numbers:
9317

ANSWER:

 We will use the division method as shown below:

  79317      111331      11121      1111              1  79317      111331      11121      1111              1
∴ 9317 = 7 × 11 × 11 × 11
             = 7 × 11³

Page No 32:

Question 15:

Give the prime factorization of each of the following numbers:
1035

ANSWER:

  We will use the division method as shown below:

   31035        3345        5115      2323              1   31035        3345        5115      2323              1
∴ 1035 =3 × 3 × 5 × 23
                  = 3² × 5 × 23

Page No 32:

Question 16:

Give the prime factorization of each of the following numbers:
1197

ANSWER:

We will use the division method as shown below:

  31197       3399         7133      1919            1  31197       3399         7133      1919            1
∴ 1197 = 3 × 3 × 7 × 19
                 = 3² × 7 × 19

Page No 32:

Question 17:

Give the prime factorization of each of the following numbers:
4641

ANSWER:

 We will use the division method as shown below:

  34641        71547      13221     1717             1  34641        71547      13221     1717             1
∴ 4614 = 3 × 7 × 13 × 17

Page No 32:

Question 18:

Give the prime factorization of each of the following numbers:
4335

ANSWER:

 We will use the division method as shown below:

  34335        51445    17289     1717             1  34335        51445    17289     1717             1
∴ 4335 = 3 × 5 × 17 × 17
                  = 3 × 5 × 17²

Page No 32:

Question 19:

Give the prime factorization of each of the following numbers:
2907

ANSWER:

We will use the division method as shown below:

  32907       3969       17323     1919             1  32907       3969       17323     1919             1
∴ 2907 = 3 × 3 × 17 × 19
              = 3² × 17 × 19

Page No 32:

Question 20:

Give the prime factorization of each of the following numbers:
13915

ANSWER:

  We will use the division method as shown below:

    513915       112783        11253     23 23        1          513915       112783        11253     23 23        1      
∴ 13915 = 5 × 11 × 11 × 23
                    = 3 × 11² × 23

Page No 36:

Exercise 2D

Question 1:

Find the HCF of the numbers in each of the following, using the prime factorization method:
84, 98

ANSWER:

The given numbers are 84 and 98.

We have:

284        242     321      77           1284        242     321      77           1298      749      77          1298      749      77          1
 
84 = 2 × 2 × 3 × 7 = 2² × 3 × 7
98 = 2 × 7 × 7 = 2 × 7²
  
∴ HCF of the given numbers = 2 × 7 = 14

Page No 36:

Question 2:

Find the HCF of the numbers in each of the following, using the prime factorization method:
170, 238

ANSWER:

The given numbers are 170 and 238.

We have:

  2170          585     1717           1  2170          585     1717           1  2238          7119     1717              1         2238          7119     1717              1       

170 = 2 × 5 × 17
238 = 2 × 7 × 17

∴ H.C.F. of the given numbers = 2 × 17 = 34

Page No 36:

Question 3:

Find the HCF of the numbers in each of the following, using the prime factorization method:
504, 980

ANSWER:

The given numbers are 504 and 980.

We have:

2504        2252     2126      363        321        77               1      2504        2252     2126      363        321        77               1             2980        2490     5245      749        77               1      2980        2490     5245      749        77               1      

    504 = 2 × 2 ×2 × 3 × 3 × 7 = 2³ × 3² × 7
    980 = 2 × 2 × 5× 7 × 7 = 2² × 5 × 7²
∴ HCF of the given numbers = 2² × 7 = 28

Page No 36:

Question 4:

Find the HCF of the numbers in each of the following, using the prime factorization method:
72, 108, 180

ANSWER:

The given numbers are 72, 108 and 180

We have:

272        236     218      39        33             1      272        236     218      39        33             1         2108        254     327      39        33          1      2108        254     327      39        33          1          2180        2 90        345          315          5 5               12180        2 90        345          315          5 5               1
Now, 72=2 × 2 × 2 × 3 × 3 = 2³ × 3²
108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³
180 = 2 × 2 × 3 ×3 × 5 = 2² × 3² × 5
∴ HCF =2²×3² =36

Page No 36:

Question 5:

Find the HCF of the numbers in each of the following, using the prime factorization method:
84, 120, 138

ANSWER:

The given numbers are 84, 120 and 138.

We have:

284        242     321      77           1      284        242     321      77           1             2120        260     230      315        55             1      2120        260     230      315        55             1                2138          369     2323           1  2138          369     2323           1

Now, 84 = 2 × 2 ×3 × 7
120 = 2 × 2× 2 ×3 × 5
138 = 2 × 3 × 23
∴ HCF = 2 × 3 = 6

Page No 36:

Question 6:

Find the HCF of the numbers in each of the following, using the prime factorization method:
106, 159, 371

ANSWER:

The given numbers are 106, 159 and 371.
We have:

  2106        5353          1        2106        5353          1        3159        5353          1        3159        5353          1        7371        5353          1        7371        5353          1      
Now, 106 = 2 × 53
159 = 3 × 53
371 = 7 × 53
∴ HCF = 53

Page No 36:

Question 7:

Find the HCF of the numbers in each of the following, using the prime factorization method:
272, 425

ANSWER:

Given numbers are 272 and 425.

We have:

  2272          2136       268        234        1717             1        2272          2136       268        234        1717             1                5425          585     1717           1        5425          585     1717           1      
Now, 272 = 2 × 2× 2 × 2 × 17
425 = 5 × 5 × 17
∴ The required HCF is 17.

Page No 36:

Question 8:

Find the HCF of the numbers in each of the following, using the prime factorization method:
144, 252, 630

ANSWER:

The given numbers are 144, 252 and 630.
We have:

2144        272      236      218        39         33             1      2144        272      236      218        39         33             1                2252        2126        363        321        77              1      2252        2126        363        321        77              1              2630       3315      3105      535        77             1      2630       3315      3105      535        77             1         

Now, 144 = 2 × 2× 2 × 2× 3 × 3
252 = 2 ×2 × 3× 3 × 7
630 = 2 ×3 × 3× 5 × 7
∴ HCF = 2 × 3 × 3 =18

Page No 36:

Question 9:

Find the HCF of the numbers in each of the following, using the prime factorization method:
1197, 5320, 4389

ANSWER:

The given numbers are 1197, 5320 and 4389.

We have:

   31197           3399           7133         1919               1         31197           3399           7133         1919               1              25320          22660           21330           5665             7133          1919                1        25320          22660           21330           5665             7133          1919                1                34389         71463         19209         1111                 1   34389         71463         19209         1111                 1 
Now, 1197 = 3 × 3× 7× 19 = 3² × 7 × 19
5320 = 2 × 2 × 2× 5 × 7 × 19 = 2³ × 5 × 7 × 19
4389 = 3 ×7 × 19 × 11
∴ Required HCF = 19 × 7 = 133

Page No 36:

Question 10:

Find the HCF of the numbers in each of the following, using the division method:
58, 70

ANSWER:

We have:


∴ The HCF of 58 and 70 is 2.

Page No 36:

Question 11:

Find the HCF of the numbers in each of the following, using the division method:
399, 437

ANSWER:

The given numbers are 399 and 437.

We have:

∴ The HCF is 19.

Page No 36:

Question 12:

Find the HCF of the numbers in each of the following, using the division method:
1045, 1520

ANSWER:

The given numbers are 1045 and 1520.
We have:


∴ The HCF of 1045 and 1520 is 95.

Page No 36:

Question 13:

Find the HCF of the numbers in each of the following, using the division method:
1965, 2096

ANSWER:

The given numbers are 1965 and 2096.

We have:
 

∴ The HCF is 131.

Page No 36:

Question 14:

Find the HCF of the numbers in each of the following, using the division method:
2241, 2324

ANSWER:

The given numbers are 2241and 2341.
We have:


∴ HCF = 83

Page No 36:

Question 15:

Find the HCF of the numbers in each of the following, using the division method:
658, 940, 1128

ANSWER:

The given numbers are 658, 940 and 1128.

First we will find the HCF of 658 and 940.


Thus, the HCF of 658 and 940 is 94.

Now, we will find the HCF of 94 and 1128.


Thus, the HCF of 94 and 1128 is 94.

∴ The HCF of 658, 940 and 1128 is 94.

​

Page No 36:

Question 16:

Find the HCF of the numbers in each of the following, using the division method:
754, 1508, 1972

ANSWER:

The given numbers are 754, 1508 and 1972.
 
First, we will find the HCF of 754 and 1508.

So, the HCF of 754 and 1508 is 754.

Now, we will find the HCF of 754 and 1972.


So, the HCF of 754 and 1972 is 58.

∴ The HCF of 754, 1058 and 1972 is 58.

Page No 36:

Question 17:

Find the HCF of the numbers in each of the following, using the division method:
391, 425, 527

ANSWER:

The given numbers are 391, 425 and 527.
First, we will find the HCF of 391 and 425.

So, the HCF of 391 and 425 is 17.
Now, we will find the HCF of 17 and 527.


So, the HCF of 17 and 527 is 17.
∴ The HCF of 391, 425 and 527 is 17.

Page No 36:

Question 18:

Find the HCF of the numbers in each of the following, using the division method:
1794, 2346, 4761

ANSWER:

The given numbers are 1794, 2346 and 4761.
First, we will find the HCF of 1794 and 2346.

So, the HCF of 1794 and 2346 is 138.
Now, we will find the HCF of 138 and 4761.

So, the HCF of 138 and 4761 is 69.

∴ The HCF of 1794, 2346 and 4761 is 69.

Page No 36:

Question 19:

Show that the following pairs are co-primes:
59, 97

ANSWER:

The given numbers are 59 and 97.

59=59×1
97=97×1

∴ HCF = 1

Since 59 and 97 does not have any common factor other than 1, the two numbers are co-primes.

Page No 36:

Question 20:

Show that the following pairs are co-primes:
161, 192

ANSWER:

The given numbers are 161 and 192.
We have:
  7161  2323        1    7161  2323        1   2192      296        248        224        212        26               32192      296        248        224        212        26               3

Now, 161 = 7 × 23 × 1
192 = 2 × 2× 2 ×2 × 2×  2 × 3 = 2⁶  × 3 × 1
∴ HCF = 1
Hence, 161 and 192 are co-primes.

Page No 36:

Question 21:

Show that the following pairs are co-primes:
343, 432

ANSWER:

The given numbers are 343 and 432.
We have:
7343      749        77              1 7343      749        77              1 2432      2216        2108        254        327        39           33             1           2432      2216        2108        254        327        39           33             1           

Now, 343 = 7 × 7× 7 × 1 = 7³ × 1
432 = 2 × 2× 2 ×2 × 3× 3 ×3 = 2⁴ × 3³ × 1
∴ HCF =1
Hence, 343 and 432 are co-primes.

Page No 36:

Question 22:

Show that the following pairs are co-primes:
512, 945

ANSWER:

Given numbers are 512 and 945.
We have:
2512      2256        2128        264        232        216           28           24           22                1    2512      2256        2128        264        232        216           28           24           22                1    
3945        3315      3105     535         Unknown node type: brUnknown node type: br3945        3315      3105     535         
77       77       
   1         1         

512 = 2 × 2 ×2 × 2 × 2× 2 × 2× 2 × 2 = 2⁹
 945 = 3 × 3 × 3 × 5 × 7 = 3³ × 5 × 7
Thus, the HCF of 512 and 945 is 1.
∴ 512 and 945 are co-primes.

Page No 36:

Question 23:

Show that the following pairs are co-primes:
385, 621

ANSWER:

The given numbers are 385 and 621.
  5385          777        1111           1          5385          777        1111           1           3621          3207          369        2323                1           3621          3207          369        2323                1        

385 = 5 × 7 × 11 × 1
621 = 3 × 3 × 3 × 23 = 3³ × 23 × 1
∴ HCF = 1

Hence, they are co-primes.

Page No 36:

Question 24:

Show that the following pairs are co-primes:
847, 1014

ANSWER:

The given numbers are 847 and 1014.

  7847        11121        1111            1          7847        11121        1111            1          21014          3507        13169        1313           1          21014          3507        13169        1313           1        

847 = 7 × 11 × 11 × 1 = 7 × 11² × 1
1014 = 2 × 3 × 13 × 13 × 1
∴ HCF = 1
Hence, 847 and 1014 are co-primes.

Page No 36:

Question 25:

Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case.

ANSWER:

Because the remainder is 6, we have to find the number that exactly divides (615 – 6) and (963 – 6).

Required number = HCF of 609 and 957
6091  957−609        348609(1           −348                       261348(1             −261––––           87261(3              −261––––                    06091  957−609        348609(1           −348                       261348(1             −261_           87261(3              −261_                    0

Therefore, the required number is 87. 

Page No 36:

Question 26:

Find the greatest number which divides 2011 and 2623, leaving remainders 9 and 5 respectively.

ANSWER:

Clearly, we have to find the number which exactly divides (2011 − 9) and (2623 − 5).
So, the required number is the HCF of 2002 and 2618.
20021    2618−2002                   6162002(3                     −1848                                      154616(4                              −616                                   020021    2618−2002                   6162002(3                     −1848                                      154616(4                              −616                                   0

∴ The required number is 154.

Page No 36:

Question 27:

Find the greatest number that will divide 445, 572 and 699, leaving remainders 4, 5, 6 respectively.

ANSWER:

Since the respective remainders of 445, 572 and 699 are 4, 5 and 6, we have to find the number which exactly divides (445-4), (572-5) and (696-6).

So, the required number is the HCF of 441, 567 and 693.
Firstly, we will find the HCF of 441 and 567.

∴ HCF = 63

Now, we will find the HCF of 63 and 693.

∴ HCF = 63

Hence, the required number is 63.

Page No 36:

Question 28:

Reduce each of the following fractions to the lowest terms:
(i) 161207161207
(ii) 517799517799
(iii) 296481296481

ANSWER:

(i) 161207161207
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
 
Now, we will find the HCF of 161 and 207.

∴ HCF = 23

Dividing the numerator and the denominator by the HCF, we get:

161÷23207÷23=79161÷23207÷23=79


(ii) 517799517799
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 517 and 799.

∴ HCF = 47

Dividing the numerator and the denominator by the HCF, we get:

517÷47799÷47=1117517÷47799÷47=1117

(iii) 296481296481
To reduce the given fraction to its lowest term, we will divide the numerator and the denominator by their HCF.
Now, we will find the HCF of 296 and 481.



∴ HCF = 37

Dividing the numerator and the denominator by the HCF, we get:

296÷37481÷37=813296÷37481÷37=813

Page No 36:

Question 29:

Three pieces of timber, 42-m, 49-m and 63-m long, have to be divided into planks of the same length. What is the greatest possible length of each plank?

ANSWER:

The lengths of the three pieces of timber are 42 m, 49 m and 63 m.
The greatest possible length of each plank will be given by the HCF of 42, 49 and 63.

Firstly, we will  find the HCF of 42 and 49 by division method.

∴ The HCF of 42 and 49 is 7.
Now, we will find the HCF of 7 and 63.
​​

∴ The HCF of 7 and 63 is 7.
Therefore, HCF of all three numbers is 7
Hence, the greatest possible length of each plank is 7 m.

Page No 36:

Question 30:

Three different containers contain 403 L, 434 L and 465 L of milk respectively. Find the capacity of a container which can measure the milk of all the containers in an exact number of times.

ANSWER:

Three different containers contain 403 L, 434 L and 465 L of milk.

The capacity of the container that can measure the milk in an exact number of times will be given by the HCF of 403, 434 and 465.


∴ HCF = 31

Now, we will find the HCF of 31 and 465.


∴ HCF = 31

Hence, the capacity of the required container is 31 L.

Page No 36:

Question 31:

There are 527 apples, 646 pears and 748 oranges. These are to be arranged in heaps containing the same number of fruits. Find the greatest number of fruits possible in each heap. How many heaps are formed?

ANSWER:

Number of apples = 527
Number of pears = 646
Number of oranges = 748
The fruits are to be arranged in heaps containing the same number of fruits.
The greatest number of fruits possible in each heap will be given by the HCF of 527, 646 and 748.

Firstly, we will find the HCF of 527 and 646.


∴ HCF of 527, 646 and 748 = 17

So, the greatest number of fruits in each heap will be 17.

Page No 36:

Question 32:

Determine the longest tape which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.

ANSWER:

7 m = 700 cm
3 m 85 cm = 385 cm
12 m 95 cm = 1295 cm

The required length of the tape that can measure the lengths 700 cm, 385 cm and 1295 cm will be given bu the HCF of 700 cm, 385 cm and 1295 cm.

Evaluating the HCF of 700, 385 and 1295 using prime factorisation method, we have:
 
700 = 2 × 2 × 5 × 5 × 7 = 2² × 5² × 7

385 = 5 × 11 × 7

1295 = 5 × 7 × 37

∴ HCF = 5 ×7 = 35

Hence, the longest tape which can measure the lengths 7 m, 3 m 85 cm and 12 m 95 cm exactly is of 35 cm.

Page No 36:

Question 33:

A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

ANSWER:

Length of the courtyard = 18 m 72 cm = 1872 cm
Breadth of the courtyard = 13 m 20 cm = 1320 cm

Now, maximum edge of the square tile is given by the HCF of 1872 cm and 1320 cm.

13201   1872−1320                    5521320(2                      −1104               _____                                      216552(2                              −432____                                 120216(1                                    −120_____                                         96120(1                                            −96____                                                 2496 (4                                                   −96___                                                        013201   1872−1320                    5521320(2                      −1104               _____                                      216552(2                              −432____                                 120216(1                                    −120_____                                         96120(1                                            −96____                                                 2496 (4                                                   −96___                                                        0

HCF of 1872 and 1320 = 24
∴ maximum edge of the square tile = 24 cm

Required number of tiles = area of courtyardarea of each square tile=1872×132024×24=4290Required number of tiles = area of courtyardarea of each square tile=1872×132024×24=4290

Page No 36:

Question 34:

Find the HCF of
(i) two prime numbers
(ii) two consecutive numbers
(iii) two co-primes
(iv) 2 and an even number

ANSWER:

(i) 2 and 3 are two prime numbers.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
∴ HCF = 1

(ii) 4 and 5 are two consecutive numbers.
Now, HCF of 4 and 5 is as follows:
4 = 2 × 2 × 1 = 2² × 1
5 = 5 × 1
∴ HCF = 1

(iii) 2 and 3 are two co-primes.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
∴ HCF = 1

(iv) 2 and 4 are two even numbers.
Now, HCF of 2 and 4 is as follows:​
  2 = 2 × 1
  4 = 2 × 2 × 1 
∴ HCF = 2 × 1 = 2

Page No 40:

Exercise 2E

Question 1:

Find the LCM of the numbers given below:
42, 63

ANSWER:

The given numbers are 42 and 63.

We have:

742,6336,9    32,3     22,1         1,1742,6336,9    32,3     22,1         1,1

∴ LCM =7 × 3 × 3 × 2 × 1
            =126

Page No 40:

Question 2:

Find the LCM of the numbers given below:
60, 75

ANSWER:

The given numbers are 60 and 75.

We have:

360,75520,2554,5    24,1    22,1        1,1 360,75520,2554,5    24,1    22,1        1,1 

∴  LCM = 3 × 5× 5 × 2 × 2
              = 300

Page No 40:

Question 3:

Find the LCM of the numbers given below:
12, 18, 20

ANSWER:

The given numbers are 12, 18 and 20.
We have:

212,18,202 6,9,10   33,9,5      31,3,5       51,1,5              1,1,1 212,18,202 6,9,10   33,9,5      31,3,5       51,1,5              1,1,1 
∴  LCM = 2 × 2 × 3× 3 × 5
              = 180

Page No 40:

Question 4:

Find the LCM of the numbers given below:
36, 60, 72

ANSWER:

The given numbers are 36, 60 and 72.

We have:

236,60,72218,30,36 39,15,18   33,5,6       51,5,2        21,1,2             1,1,1236,60,72218,30,36 39,15,18   33,5,6       51,5,2        21,1,2             1,1,1
∴  LCM = 2 × 2 × 2 × 3 × 3 × 5
             = 360

Page No 40:

Question 5:

Find the LCM of the numbers given below:
36, 40, 126

ANSWER:

The given numbers are 36, 40 and 126.

We have:

236,40,126318,20,63  36,20,21  22,20,7       21,10,7       51,5,7        71,1,7             1,1,1236,40,126318,20,63  36,20,21  22,20,7       21,10,7       51,5,7        71,1,7             1,1,1
∴  LCM = 2 × 3 × 3 ×2 × 2 × 5 × 7
                   = 2520

Page No 40:

Question 6:

Find the LCM of the numbers given below:
16, 28, 40, 77

ANSWER:

The given numbers are 16, 28, 40 and 77.
We have:

 216,28,40,77 78,14,20,77    28,2,20,11      24,1,10,11       22,1,5,11        51,1,5,11      111,1,1,11          1,1,1,1       216,28,40,77 78,14,20,77    28,2,20,11      24,1,10,11       22,1,5,11        51,1,5,11      111,1,1,11          1,1,1,1      

∴  LCM = 2 × 7 × 2 × 2 × 2 × 5 × 11
             = 6160

Page No 40:

Question 7:

Find the LCM of the numbers given below:
28, 36, 45, 60

ANSWER:

The given numbers are 28, 36, 45 and 60.

We have:

228,36,45,602 14,18,45,3037,9,45,15     37,3,15,5       57,1,5,5          77,1,1,1               1,1,1,1    228,36,45,602 14,18,45,3037,9,45,15     37,3,15,5       57,1,5,5          77,1,1,1               1,1,1,1    

∴  LCM = 2 × 2 × 3 × 3 × 5 × 7
            = 1260

Page No 40:

Question 8:

Find the LCM of the numbers given below:
144, 180, 384

ANSWER:

The given numbers are 144, 180 and 384.
We have:
2144,180,3842 72,90,192   236,45,96      218,45,48       39,45,24         33,15,8            21,5,8              21,5,4              21,5,2              51,5,1                  1,1,1                 2144,180,3842 72,90,192   236,45,96      218,45,48       39,45,24         33,15,8            21,5,8              21,5,4              21,5,2              51,5,1                  1,1,1                 

∴  LCM = 2⁷ × 3² × 5
              = 5760

Page No 40:

Question 9:

Find the LCM of the numbers given below:
48, 64, 72, 96, 108

ANSWER:

The given numbers are 48, 64, 72, 96 and 108.
We have:

248,64,72,96,108224,32,36,48,54 212,16,18,24,27  26,8,9,12,27      33,4,9,6,27       21,4,3,2,9       21,2,3,1,9       31,1,3,1,9      31,1,1,1,3           1,1,1,1,1248,64,72,96,108224,32,36,48,54 212,16,18,24,27  26,8,9,12,27      33,4,9,6,27       21,4,3,2,9       21,2,3,1,9       31,1,3,1,9      31,1,1,1,3           1,1,1,1,1

∴  LCM = 2⁶ × 3³
              = 1728

Page No 40:

Question 10:

Find the HCF and LCM of
117, 221

ANSWER:

The given numbers are 117 and 221.

We have:

  3117  339 1313       1  3117  339 1313       1    132211717     1132211717     1

Now,
117 = 3 × 3 × 13
221 = 13 × 17

∴  HCF = 13 × 1

Now, LCM = 13 × 17 × 3 × 3
                  = 1989

Page No 40:

Question 11:

Find the HCF and LCM of
234, 572

ANSWER:

The given numbers are 234 and 572.

We have:

   2234 3117 339 1313      1   2234 3117 339 1313      1      2572  228613143 1111      1  2572  228613143 1111      1

Now, we have:

234 = 2 × 3 × 3 × 13
572 = 2 × 2 × 1 3 × 11

∴  LCM = 13 × 2  × 2 × 11 × 9
              = 5148
Also, HCF = 13 × 2 = 26

Page No 40:

Question 12:

Find the HCF and LCM of
693, 1078

ANSWER:

The given numbers are 693 and 1078.

We have:

 3693 3231 777 1111      1 3693 3231 777 1111      1  21078  7539  777 1111     1  21078  7539  777 1111     1

Now, we have:

693 = 3 × 3 ×7 × 11
1078 = 2 × 7× 7 × 11

∴  HCF = 7 × 11= 77
Also, LCM = 2 × 3 × 3 ×7 × 7 × 11 = 9702

Page No 40:

Question 13:

Find the HCF and LCM of
145, 232

ANSWER:

The given numbers are 145 and 232.
We have:
  51452929     1  51452929     1             2232  2116  258 2929       1     2232  2116  258 2929       1  

Now, we have:

145 = 5 × 29
232 = 2 ×2 × 2 × 29

∴  HCF = 29
Also, LCM = 29 × 2 × 2 × 2 × 5 = 1160

Page No 40:

Question 14:

Find the HCF and LCM of
861, 1353

ANSWER:

The given numbers are 861 and 1353.

We have:

  3861  72874141      1  3861  72874141      1  31353114514141      1  31353114514141      1

Now, we have:

861 = 3 × 41 × 7
1353 = 41 × 11 × 3

∴ HCF = 41 × 3 = 123
Also, LCM = 41 × 3 × 11 × 7 = 9471

Page No 40:

Question 15:

Find the HCF and LCM of
2923, 3239

ANSWER:

HCF of 2923 and 3239:

∴ HCF = 79

We know that product of two numbers = HCF × LCM

⇒ LCM = Product of two numbersHCF ⇒ LCM = 2923×323979∴ LCM = 119843⇒ LCM = Product of two numbersHCF ⇒ LCM = 2923×323979∴ LCM = 119843

Page No 40:

Question 16:

For each pair of numbers, verify that their product = (HCF × LCM).
(i) 87, 145
(ii) 186, 403
(iii) 490, 1155

ANSWER:

(i) 87 and 145


We have:
87 = 3 × 29
145 = 5 × 29

HCF = 29
LCM = 29 × 15 × 1 = 435

Now, HCF × LCM = 29 × 435 = 12615
Product of the two numbers = 87 × 145 = 12615

∴ HCF × LCM = Product of the two numbers
Verified.

(ii)186 and 403

186 = 2 × 3 × 31
403 = 31 × 13

HCF = 31
LCM = 31 × 13 × 6 = 2418

Now, HCF × LCM = 31 × 2418 = 74958
Product of the two numbers = 186 × 403 = 74958

∴ HCF × LCM = Product of the two numbers
 Verified.

(iii) 490 and 1155

490 = 7 × 7 × 2 × 5
1155 = 5 × 7 ×3 × 11

HCF = 7 × 5 = 35
LCM = 7 × 5 ×7 × 2 × 3 × 11 = 16170

Now, HCF × LCM = 35 × 16170 = 565950
Product of the two numbers = 490 × 1155 = 565950

∴ HCF × LCM = Product of the two numbers
Verified.

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Question 17:

The product of two numbers is 2160 and their HCF is 12. Find their LCM.

ANSWER:

Product of the two numbers = 2160
HCF = 12

We know that LCM × HCF = Product of the two numbers
       
∴ LCM = 216012216012 = 180

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Question 18:

The product of two numbers is 2160 and their LCM is 320. Find their HCF.

ANSWER:

Product of the two numbers = 2560
LCM = 320

We know that

LCM × HCF = Product of the two numbers

∴ HCF = 25603202560320 = 8         

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Question 19:

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other.

ANSWER:

HCF = 145
LCM = 2175
One of the number = 725

We know that
HCF × LCM = Product of two numbers
∴ Other number = 145×2175725145×2175725 = 435

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Question 20:

The HCF and LCM of two numbers are 131 and 8253 respectively. If one of the numbers is 917, find the other.

ANSWER:

HCF = 131
LCM = 8253
One of the number = 917

We know that
LCM × HCF = Product of two numbers
Other number = 8253×1319178253×131917

∴ The other number is 1179.

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Question 21:

Find the least number divisible by 15, 20, 24, 32 and 36.

ANSWER:

The given numbers are 15, 20, 24, 32 and 36.

The smallest number divisible by the numbers given above will be their LCM.

215,20,24,32,36315,10,12,16,18 55,10,4,16,6  21,2,4,16,6   21,1,2,8,3       21,1,1,4,3      21,1,1,2,3       31,1,1,1,3           1,1,1,1,1  215,20,24,32,36315,10,12,16,18 55,10,4,16,6  21,2,4,16,6   21,1,2,8,3       21,1,1,4,3      21,1,1,2,3       31,1,1,1,3           1,1,1,1,1  

LCM = 2⁵ × 3² × 5 
            =  1440
∴ The least number divisible by 15, 20, 24, 32 and 36 is 1440.
 

​

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Question 22:

Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case.

ANSWER:

25, 40 and 60 exactly divides the least number that is equal to their LCM.
So, the required number that leaves 9 as a remainder will be LCM + 9.

Finding the LCM:

225,40,60225,20,30 225,10,15  325,5,15   525,5,5       55, 1, 1        1, 1, 1    225,40,60225,20,30 225,10,15  325,5,15   525,5,5       55, 1, 1        1, 1, 1    
LCM = 2³ × 3 × 5² = 600
∴ Required number = 600 + 9 = 609 

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Question 23:

Find the least number of five digits that is exactly divisible by 16, 18, 24 and 30.

ANSWER:

LCM of 16, 18, 24 and 30:

216,18,24,3028,9,12,15    24,9,6,15    22,9,3,15      31,9,3,15      31,3,1,5        51,1,1,5             1,1,1,1216,18,24,3028,9,12,15    24,9,6,15    22,9,3,15      31,9,3,15      31,3,1,5        51,1,1,5             1,1,1,1

LCM = 2⁴ × 3² × 5 = 720

We have to find the least five-digit number that is exactly divisible by 16, 18, 24 and 30.
But LCM=720 is a three digit number.

The least five digit number = 10000
Dividing 10000 by 720, we get:
72013  10000       -720       2800          − 2160    640    The greatest four-digit number exactly divisible by 720 = 10000-640                             =9360    So, the least five-digit number exactly divisible by 720 = 9360 + 720                                                                                                            = 10080   72013  10000       -720       2800          – 2160    640    The greatest four-digit number exactly divisible by 720 = 10000-640                             =9360    So, the least five-digit number exactly divisible by 720 = 9360 + 720                                                                                                            = 10080   

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Question 24:

Find the greatest number of five digits exactly divisible by 9, 12, 15, 18 and 24.

ANSWER:

First, we will find the LCM of 9, 12, 15, 18 and 24.
29,12,15,18,2429,6,15,9,1229,3,15,9,639,3,15,9,333,1,5,3,151,1,5,1,1  1,1,1,1,1 ∴ LCM of the numbers = 23 × 32 × 5                                         = 360The least six-digit number = 100000The greatest five-digit number divisible by 360 will be the quotient of 100000360 mutiplied by 360.          360277100000         72028002520280        So, the greatest five-digit number exactly divisible               by the given numbers will be360×277 = 9972029,12,15,18,2429,6,15,9,1229,3,15,9,639,3,15,9,333,1,5,3,151,1,5,1,1  1,1,1,1,1 ∴ LCM of the numbers = 23 × 32 × 5                                         = 360The least six-digit number = 100000The greatest five-digit number divisible by 360 will be the quotient of 100000360 mutiplied by 360.          360277100000         72028002520280        So, the greatest five-digit number exactly divisible               by the given numbers will be360×277 = 99720

Page No 40:

Question 25:

Three bells toll at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?

ANSWER:

Three bells toll at intervals of 9, 12 and, 15 minutes.
The time when they will toll together again is given by the LCM of 9, 12 and 15.

 39,12,1533,4,5    51,4,5    21,4,1    21,2,1       1,1,139,12,1533,4,5    51,4,5    21,4,1    21,2,1       1,1,1

Required time = 2² × 3² × 5
                          = 180 minutes
                          =3 h
If they start tolling together, they will toll together again after 3 h.

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Question 26:

Three boys step off together from the same place. If their steps measure 36 cm, 48 cm and 54 cm, at what distance from the starting point will they again step together?

ANSWER:

From the starting point, they will step together again when they travel a distance that is exactly divisible by the lengths of their steps. 
The least distance from the starting point where they will step together will be given by the LCM of 36, 48 and 54.

236,48,54218,24,2739,12,2733,4,931,4,321,4,1 21,2,1     1,1,1236,48,54218,24,2739,12,2733,4,931,4,321,4,1 21,2,1     1,1,1

The required distance = 2 × 2 ×3 × 3 × 3 × 2 × 2
                                          = 16 × 27
                                          = 432 cm
∴ They will step together again at a distance of 432 cm from the starting point.

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Question 27:

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously?

ANSWER:

The time when the lights will change simultaneously again will be quantity which is exactly divisible by 48, 72 and 108. The least time when they change simultaneously will be given by their LCM. 

248,72,108224,36,54212,18,2726,9,2733,9,2731,3,931,1,3  1,1,1248,72,108224,36,54212,18,2726,9,2733,9,2731,3,931,1,3  1,1,1
Required time = 2⁴ × 3³
                      =  432 seconds
                      = 7 min 12 seconds
So, the lights will change simultaneously at 8:07:12 a.m.

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Question 28:

Three measuring rods are 45 cm, 50 cm and 75 cm in length. What is the least length (in metres) of a rope that can be measured by the full length of each of these three rods?

ANSWER:

The length of the required rope must be such that it is exactly divisible by 45, 50 and 75. The least length will be given by the LCM of 45, 50 and 75. 

245,50,75345,25,75315,25,2555,25,2551,5,5   1,1,1245,50,75345,25,75315,25,2555,25,2551,5,5   1,1,1

Required length = 3 × 3 ×5 × 5 × 2
                              = 450 cm
So, the minimum length of the rope that can be measured by the full length of each of the three rods is 450 cm.

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Question 29:

An electronic device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together?

ANSWER:

The LCM of the time intervals of the beeps will give the time when the electronic devices will beep together.

LCM of 15 and 20:

515,2033,421,421,2   1,1515,2033,421,421,2   1,1

Required time = 5 × 3 × 2 × 2
                            = 60 min
So, they will beep simultaneously after 60 min or 1 h. 

∴ They will beep together again at 7:00 a.m.

Page No 41:

Question 30:

The circumferences of four wheels are 50 cm, 60 cm, 75 cm and 100 cm. They start moving simultaneously. What least distance should they cover so that each wheel makes a complete number of revolutions?

ANSWER:

Distance covered by a wheel for one complete revolution = circumference of the wheel 

All the wheels will make complete numbers of revolutions when the distances covered by them is equal to their LCM. 
550,60,75,100510,12,15,2022,12,3,421,6,3,231,3,3,1   1,1,1,1 550,60,75,100510,12,15,2022,12,3,421,6,3,231,3,3,1   1,1,1,1 
Required least distance = 5 × 5 ×2 × 2 × 3
                                             = 25 × 4 × 3
                                             = 300 cm = 3 m
So, each wheel will make a complete number of revolutions after travelling 3 m.

Page No 41:

Exercise 2F

Question 1:

Which of the following numbers is divisible by 3?
(a) 24357806
(b) 35769812
(c) 83479560
(d) 3336433

ANSWER:

(c) 83479560

A number is divisible by 3 if the sum of its digits is divisible by 3.

a) Consider the number 24357806.
Sum of its digits = 2 + 4 + 3 + 5+ 7 + 8 + 0 + 6 = 35, which is not divisible by 3.
So, 2357806 is not divisible by 3.

b) Consider the number 35769812.
Sum of its digits = 3 + 5 + 7 + 6 +9 + 8 + 1 + 2 = 41, which is not divisible by 3.
So, 35769812 is not divisible by 3.

c) Consider the number 83479560.
Sum of its digits = 8 + 3 + 4+ 7 + 9 + 5 + 6 + 0 = 42, which is divisible by 3.
So, 2357806 is divisible by 3.

d) Consider the number 3336433.
Sum of its digits = 3 + 3 +3 + 6 +4 + 3 +3 = 25, which is not divisible by 3.
So, 3336433  is not divisible by 3.

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Question 2:

Which of the following numbers is divisible by 9?
(a) 8576901
(b) 96345210
(c) 67594310
(d) none of these

ANSWER:

 (a) 8576901

A number is divisible by 9 if the sum of its digits is divisible by 9.

a) Consider the number 8576901.
Sum of its digits = 8 + 5 +7 + 6 + 9+ 0 + 1 = 36, which is divisible by 9.
So, 8576901 is divisible by 9.

b) Consider the number 96345210.
Sum of its digits = 9 + 6 + 3+ 4 + 5+ 2 + 1 + 0 = 30, which is not divisible by 9.
So, 96345210 is not divisible by 9.
 
c) Consider the number 67594310.
Sum of its digits = 6 + 7 + 5 + 9 + 4 + 3 + 1 + 0 = 35, which is not divisible by 9.
So, 67594310 is not divisible by 9.

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Question 3:

Which of the following numbers is divisible by 4?
(a) 78653234
(b) 98765042
(c) 24689602
(d) 87941032

ANSWER:

(d)87941032

A number is divisible by 4 if the number formed by its digits in the tens and ones places is divisible by 4.

(a) 78653234
Consider the number 78653234.
Here, the number formed by the tens and the ones digit is 34, which is not divisible by 4.
Therefore, 78653234 is not divisible by 4.
 
(b) 98765042
Consider the number 98765042.
Here, the number formed by the tens and the ones digit is 42, which is not divisible by 4.
Therefore, 98765042 is not divisible by 4.
 
(c) 24689602
Consider the number 24689602.
Here, the number formed by the tens and the ones digit is 02, which is not divisible by 4.
Therefore, 24689602 is not divisible by 4
 
 
(d) 87941032 
Consider the number 87941032.
Here, the number formed by the tens and ones digit is 32, which is divisible by 4.
Therefore, 87941032 is divisible by 4.

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Question 4:

Which of the following numbers is divisible by 8?
(a) 96354142
(b) 37450176
(c) 57064214
(d) none of these

ANSWER:

 (b) 37450176

A number is divisible by 8 if the number formed by its digits in hundreds, tens and ones places is divisible by 8.

(a) 96354142
Consider the number 96354142.
Here, the number formed by the digits in hundreds, tens and ones places is 142, which is clearly not divisible by 8.
Therefore, 96354142 is not divisible by 8.
 
(b) 37450176
Consider the number 37450176.
The number formed by the digits in hundreds, tens and ones places is 176, which is clearly divisible by 8.
Therefore, 37450176 is divisible by 8.
 
(c) 57064214
Consider the number 57064214.
Here, the number formed by the digits in hundreds, tens and ones places is 214, which is clearly not divisible by 8.
Therefore, 57064214 is not divisible by 8.

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Question 5:

Which of the following numbers is divisible by 6?
(a) 8790432
(b) 98671402
(c) 85492014
(d) none of these

ANSWER:

(a) 8790432 and (c) 85492014

A number is divisible by 6, if it is divisible by both 2 and 3.

(a) 8790432
Consider the number 8790432.
The number in the ones digit is 2.
Therefore, 8790432 is divisible by 2.
Now, the sum of its digits (8+7+9+0+2+3+2) is 33. Since 33 is divisible by 3, we can say that 8790432 is also divisible by 3.
Since 8790432 is divisible by both 2 and 3, it is also divisible by 6.

(b) 98671402
Consider the number 98671402.
The number in the ones digit is 2.
Therefore, 98671402 is divisible by 2.
Now, the sum of its digits (9+8+6+7+1+4+0+2) is 37. Since 37 is not divisible by 3, we can say that 98671402 is also not divisible by 3.
Since 98671402 is not divisible by both 2 and 3, it is not divisible by 6.

(c) 85492014
Consider the number 85492014.
The number in the ones digit is 4.
Therefore, 85492014 is divisible by 2.
Now, the sum of its digits (8+5+4+9+2+0+1+4) is 33. Since 33 is divisible by 3, we can say that 85492014 is also divisible by 3.
Since 85492014 is divisible by both 2 and 3, it is also divisible by 6.

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Question 6:

Which of the following numbers is divisible by 11?
(a) 3333333
(b) 1111111
(c) 22222222
(d) none of these

ANSWER:

(c) 22222222
A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of the digits in even places (starting from ones place) is either 0 or a multiple of 11.

(a) 3333333
Consider the number 3333333.
Sum of its digits in odd places (3 + 3 + 3 + 3) = 12
Sum of its digits in even places (3 + 3 + 3) = 9
Difference of the two sums = 12 − 9 = 3
Since this number (3) is not divisible by 11, 3333333 is not divisible by 11.

(b) 1111111
Consider the number 1111111.
Sum of its digits in odd places (1 + 1 + 1 + 1) = 4
Sum of its digits in even places (1 + 1 + 1) = 3
Difference of the two sums = 4 − 3 = 1
Since this number (1) is not divisible by 11, 1111111 is also not divisible by 11.

(c) 22222222
Consider the number 22222222.
Sum of its digits in odd places (2 + 2 + 2 + 2)= 8
Sum of its digits in even places (2 + 2 + 2 + 2) = 8
Difference of the two sums = 8 − 8 = 0
Since this number (0) is divisible by 11, 22222222 is also divisible by 11.

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Question 7:

Which of the following is a prime number?
(a) 81
(b) 87
(c) 91
(d) 97

ANSWER:

 (d) 97

(a) 81 is not a prime number because 81 can be written as 9×9.
(b) 87 is not a prime number because 87 can be written as 29×3.
(c) 91 is not a prime number because 91 can be written as 13×7.
(d) 97 is a prime number.

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Question 8:

Which of the following is a prime number?
(a) 117
(b) 171
(c) 179
(d) none of these

ANSWER:

 (c) 179

(a) 117 is not a prime number because 117 can be written as 3 × 39.
(b) 171 is not a prime number because 171 can be written as 19×9.
(c) 179 is prime number.

Page No 41:

Question 9:

Which of the following is a prime number?
(a) 323
(b) 361
(c) 263
(d) none of these

ANSWER:

(c)263

(a) 323 is not a prime number because 323 can be written as 17 × 19.
(b) 361 is not a prime number because 361 can be written as 19 × 19.
(c) 263 is a prime number.

Page No 41:

Question 10:

Which of the following are co-primes?
(a) 8, 12
(b) 9, 10
(c) 6, 8
(d) 15, 18

ANSWER:

 (b) 9, 10

(a) 8, 12 are not co-primes as they have a common factor 4.
(b) 9, 10 are co-primes as they do not have a common factor.
(c) 6, 8 are not co-primes as they have a common factor 2.
(d)15,18 are not co-primes as they have a common factor 3.

Page No 41:

Question 11:

Which of the following is a composite number?
(a) 23
(b) 29
(c) 32
(d) none of these

ANSWER:

 (c) 32

(a) 23 is not a composite number as it cannot be broken into factors.
(b) 29 is not a composite number as it cannot be broken into factors.
(c) 32 is a composite number as it can be broken into factors, which are 2 × 2 × 2 × 2 × 2.

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Question 12:

The HCF of 144 and 198 is
(a) 9
(b) 12
(c) 6
(d) 18

ANSWER:

(d) 2 × 3² = 18
We first factorise the two numbers:

   144 = 2 × 2 × 2 × 2 × 3 × 3 =  2⁴ × 3²
   198 = 2 × 3 × 3 × 11  = 2 × 3² × 11
Here, 18 (2 × 3² = 18) is the highest common factor of the two numbers.

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Question 13:

The HCF of 144, 180 and 192 is
(a) 12
(b) 16
(c) 18
(d) 8

ANSWER:

(a) 2²×3= 12
We will first factorise the two numbers:


144 = 2 × 2 × 2 × 2 × 2 × 3 × 3 =  2⁴ × 3²
180 = 2 × 2 × 3 × 3 × 5  = 2² × 3² × 5
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁶ × 3

Here, 12 (i.e. 2²× 3 = 12) is the highest common factor of the three numbers.

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Question 14:

Which of the following are co-primes?
(a) 39, 91
(b) 161, 192
(c) 385, 462
(d) none of these

ANSWER:

 (b) 161 and 192

(a) 39 and 91 are not co-primes as 39 and 91 have a common factor, i.e. 13.
(b) 161 and 192 are co-primes as 161 and 192 have no common factor other than 1.
 (c) 385 and 462 are not co-primes as 385 and 462 have common factors 7 and 11.

Page No 41:

Question 15:

289391289391 when reduced to the lowest terms is
(a) 11231123
(b) 13311331
(c) 17311731
(d) 17231723

ANSWER:

(d) 17231723

289391289391

H.C.F.=17

Dividing both the numerator and the denominator by the H.C.F. of 289 & 391:

289÷17391÷17=1723289÷17391÷17=1723

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Question 16:

The greatest number which divides 134 and 167 leaving 2 as remainder in each case is
(a) 14
(b) 17
(c) 19
(d) 11

ANSWER:

(d) 11
Since we need 2 as the remainder, we will subtract 2 from each of the numbers.
167 − 2 = 165
134 − 2 = 132
Now, any of the common factors of 165 and 132 will be the required divisor.
On factorising:
165 =  3 × 5 × 11
132 = 2 × 2 × 3 × 11
Their common factors are 11 and 3.
So, 11 is the required divisor.

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Question 17:

The LCM of 24, 36, 40 is
(a) 4
(b) 90
(c) 360
(d) 720

ANSWER:

(c)  360

224,36,40212,18,20  26,   9, 10   33,  9,   5     31,  3,   5     51,  1,   5        1,  1,   1    224,36,40212,18,20  26,   9, 10   33,  9,   5     31,  3,   5     51,  1,   5        1,  1,   1    

L.C.M. = 2³ × 3² × 5
            = 360           

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Question 18:

The LCM of 12, 15, 20, 27 is
(a) 270
(b) 360
(c) 480
(d) 540

ANSWER:

 (d) 540

212,15,20,2726,15,10,27  33,15, 5, 27   31, 5, 5,  9   31, 5, 5,  3   51, 5, 5,  1        1, 1, 1, 1    212,15,20,2726,15,10,27  33,15, 5, 27   31, 5, 5,  9   31, 5, 5,  3   51, 5, 5,  1        1, 1, 1, 1    
L.C.M. = 2² × 3³ × 5 = 540 

Page No 42:

Question 19:

The smallest number which when diminished by 3 is divisible by 11, 28, 36 and 45, is
(a) 1257
(b) 1260
(c) 1263
(d) none of these

ANSWER:

 (d) none of these

The smallest number that is exactly divisible by 11, 28, 36 and 45 will be their L.C.M.
So, the required number will be the L.C.M. plus 3.

211,28,36,45211,14,18,45  311, 7,  9 , 45   311, 7, 3,   15   511, 7, 1,  5  711, 7, 1,  1   1111, 1, 1,  1         1, 1, 1, 1    211,28,36,45211,14,18,45  311, 7,  9 , 45   311, 7, 3,   15   511, 7, 1,  5  711, 7, 1,  1   1111, 1, 1,  1         1, 1, 1, 1    

L.C.M. of the three numbers = 2² × 3² × 5 × 7× 11
                                              = 13860
∴ Required number = 13860 + 3 = 13863

Page No 42:

Question 20:

The HCF of two co-primes is
(a) the smaller number
(b) the larger number
(c) 1
(d) none of these

ANSWER:

 (c) 1
H.C.F. of two co-primes is 1.
This is because two co-prime numbers do not have any common factor.
For example, 15 and 16 are co-primes.
Their H.C.F. is 1.

Page No 42:

Question 21:

If a and b are co-primes, then their LCM is
(a) 1
(b) abab
(c) ab
(d) none of these

ANSWER:

 (c) ab
 
If a and b are co-primes then their LCM will be ab.
For example, 4 and 9 are co-primes.
L.C.M. of 4 and 9 is 4×9. 

Page No 42:

Question 22:

The product of two numbers is 2160 and their HCF is 12. The LCM of these numbers is
(a) 12
(b) 25920
(c) 180
(d) none of these

ANSWER:

(c) 180
Here, H.C.F. = 12
Product of two numbers = 2160
 
We know:
L.C.M. × H.C.F. = Product of the two numbers
 
L.C.M. = 2160H.C.F.2160H.C.F.

          =216012216012
          = 180
L.C.M. = 180

Page No 42:

Question 23:

The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, the other number is
(a) 290
(b) 435
(c) 5
(d) none of these

ANSWER:

 (b) 435
 
One of the numbers is 725.
H.C.F. = 145
L.C.M. = 2175
We know:
L.C.M. × H.C.F. = Product of the two numbers
∴ Product of the two numbers = 145 × 2175
                                               = 315375
∴ Other number =315375725315375725
                         = 435

Page No 42:

Question 24:

The least number divisible by each of the numbers 15, 20, 24, 32 and 36 is
(a) 1660
(b) 2880
(c) 1440
(d) none of these

ANSWER:

 (c) 1440
The least number divisible by each of the numbers 15, 20, 24, 32 and 36 is their L.C.M. 
215,20,24,32,36215,10,12,16,18  215, 5,  6 , 8, 9   215, 5,  3,  4, 9   215, 5,  3,  2, 9  315, 5,  3,  1, 9 3 5,  5, 1,  1, 3   5 5,  5, 1,  1, 1        1,  1,  1,  1,  1 215,20,24,32,36215,10,12,16,18  215, 5,  6 , 8, 9   215, 5,  3,  4, 9   215, 5,  3,  2, 9  315, 5,  3,  1, 9 3 5,  5, 1,  1, 3   5 5,  5, 1,  1, 1        1,  1,  1,  1,  1 
L.C.M. = 2⁵ × 3² × 5
            = 1440 

Page No 42:

Question 25:

Three bells toll together at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?
(a) 1 hour
(b) 112 hours112 hours
(c) 212 hours212 hours
(d) 3 hours

ANSWER:

 (d) 3 hours

The L.C.M. of 9, 12 and 15 will give us the minutes after which the bells will next toll together.

29,12,1529,6,15  39, 3, 15   33, 1,  5     51,  1,   5        1,  1,   1    29,12,1529,6,15  39, 3, 15   33, 1,  5     51,  1,   5        1,  1,   1    
L.C.M. = 2² × 3² × 5
            = 180
So,the bells will toll together after 180 min.
On converting into hours:
180/60 = 3 hours

Page No 43:

Exercise 2G

Question 1:

Test the divisibility of 5869473 by 11.

ANSWER:

5869473
A number is divisible by 11 if the the difference of the sums of the digits at the odd places and that at the even places (starting from ones place) is either 0 or a multiple of 11.

Sum of the digits at even places = 7 + 9 + 8
                                                   = 24
Sum of the digits in odd places = 3 + 4 + 6 + 5
                                                  = 18
Difference = 24−18
                 = 6
Since 6 is not divisible by 11, 5869473 is not divisible by 11.

Page No 43:

Question 2:

Test the divisibility of 67529124 by 8.

ANSWER:

67529124
A number is divisible by 8 if the number formed by the hundreds, tens and ones digits is divisible by 8.
Since the digits at the hundred’s, ten’s and unit places are 124, which is not divisible by 8, 67529124 is not divisible by 8. 

Page No 43:

Question 3:

On dividing 5035 by 31, the remainder is 13. Find the quotient.

ANSWER:

Remainder is 13
∴ Number exactly divisible by 31 =  5035 − 13
                                            = 5022
 
So, the required quotient is 162.

Page No 43:

Question 4:

The HCF of two number is 15 and their product is 1650. Find their LCM.

ANSWER:

H.C.F. × L.C.M. = Products of the two numbers
Product of the two numbers = 1650
H.C.F. = 15
Required L.C.M. =165015165015
                          =110                  

Page No 43:

Question 5:

Find the least 5-digit number which is exactly divisible by 20, 25, 30.

ANSWER:

Least five digit number = 10000

L.C.M. of 20,25,30 is 300.
But we want the least five digit number which is divisible by 20, 25, 30.
So, we will multiply the L.C.M. by a number that makes it the least five digit number divisible by 20, 25, 30.
300××31 = 9300
300××32 = 9600
300××33 = 9900
300××34 = 10200
 
So, the least five digit number divisible by 20, 25, 30 is 10200.

Page No 43:

Question 6:

Find the largest number which divides 630 and 940 leaving remainders 6 and 4 respectively.

ANSWER:

Since 6 and 4 are the remainders, the number must exactly divide the following:
630 − 6 = 624
and 
940 − 4 = 936



624 = 2 × 2 × 2 × 2 × 3 × 13
936 = 2 × 2 × 2 × 3 × 3 × 13
H.C.F. of 624 and 936 = 8 × 3 × 13
                                     = 312
So, 312 is the greatest number that divides 630 and 940, leaving 6 and 4 as the respective remainders.

Page No 43:

Question 7:

Find the least number which when divided by 16, 36 and 40 leaves 5 as remainder in each case.

ANSWER:

On subtracting 5 from each number:
16 − 5 = 11
36 − 5 = 31
40 − 5 = 35
The required number will be the least common multiple of 11, 31 and 35.
L.C.M. of 11, 31 and 35 = 11×31×35
                                       = 11935
This is because they do not have any factor in common.
So, 11935 is the required number.

Page No 43:

Question 8:

Write all prime numbers between 50 and 100.

ANSWER:

53, 59, 61, 67, 71, 73, 79, 83, 89, 97 are the prime numbers between 50 and 100.

Page No 43:

Question 9:

Write seven consecutive composite numbers less than 100 having no prime number between them.

ANSWER:

Seven consecutive composite numbers less than 100 having no prime number between them are 90, 91, 92, 93, 94, 95 and 96.

Page No 43:

Question 10:

Can two numbers have 12 as their HCF and 512 as their LCM? Justify your answer.

ANSWER:

No, they cannot have 512 as their L.C.M.

We know that the H.C.F. is one of the factors of the L.C.M. Here, 3, which is a factor of 12, is not a factor of 512. 

Page No 43:

Question 11:

Which of the following are co-primes?
(a) 91 and 72
(b) 34 and 51
(c) 21 and 36
(d) 15 and 20

ANSWER:

The correct option is (a).
The H.C.F. of 72 and 91 is 1.
So, they are co-primes.

Option (b) is not correct because 34 and 51 have 17 as their H.C.F.
Option (c) is not correct because 21 and 56 have 3 as their H.C.F.
Option (d) is not correct because 15 and 20 have 5 as their H.C.F.

Page No 43:

Question 12:

The LCM of two co-prime numbers is their
(a) sum
(b) difference
(c) product
(d) quotient

ANSWER:

The correct option is (c).
The L.C.M of two co-prime numbers is their product.

Page No 43:

Question 13:

The number which is neither prime nor composite is
(a) 0
(b) 1
(c) 2
(d) 3

ANSWER:

The correct option is (b).
1 is neither prime nor composite.

Option (a) is not correct because composite numbers are defined for positive numbers, but 0 is neither a positive number nor a negative number.
Option (c) ​is not correct because 2 is a prime number.
Option (d) is not correct because 3 is a prime number.​

Page No 43:

Question 14:

What least number should be replaced for * so that the number 67301*2 is exactly divisible by 9?
(a) 5
(b) 6
(c) 7
(d) 8

ANSWER:

The correct option is (d).

6 + 7 + 3 + 0 + 1 + * + 2 = 19 + *
8 is the least number that should be added to 19 such that number will be divisible by 9.
Sum of the digits:
6 + 7 + 3 + 0 + 1 + 8 + 2 = 27
27 is divisible by 9.

Page No 43:

Question 15:

Which of the following numbers is divisible by 6?
(a) 67821
(b) 78134
(c) 87432
(d) none of these

ANSWER:

The correct option is (c).
A number is divisible by 6 if it is divisible by both 2 and 3.

Since the ones digit of 87432 is 2, it is divisible by 2.
Now, 8 + 7 + 4 + 3 + 2 = 24
24 is divisible by 3.
Hence, 87432 is divisible by 6 because it is divisible by both 2 and 3.

Option (a) is not correct because 67821 is not divisible by 2.
Option (b)  is not correct because 78134 is not divisible by 3.
7 + 8 + 1 + 3 + 4 = 23
23 is not divisible by 3.

Page No 43:

Question 16:

Which of the following is a prime number?
(a) 143
(b) 131
(c) 147
(d) 161

ANSWER:

The correct option is (b).
To find a prime number between 100 and 200, we have to check whether the given number is divisible by any prime number less than 15. If yes, it is not prime, otherwise it is.
By examining, we find that 131 is a prime number.

Page No 43:

Question 17:

289391289391 when reduced to lowest term is
(a) 13171317
(b) 17191719
(c) 17231723
(d) 17211721

ANSWER:

(c) 17231723

289 = 17 ×× 17
391 = 17 ×× 23
The H.C.F. of 289 and 391 is 17.

Dividing both the numerator and the denominator by 17:289 ÷ 17391 ÷ 17= 1723Dividing both the numerator and the denominator by 17:289 ÷ 17391 ÷ 17= 1723

Page No 43:

Question 18:

Every counting number has an infinite number of
(a) factors
(b) multiples
(c) prime factors
(d) none of these

ANSWER:

The correct option is (b).

Every counting number has an infinite number of multiples. 
If p is a counting number, its multiples are 1p, 2p, 3p…. 

Page No 43:

Question 19:

Fill in the blanks.
(i) 1 is neither …… nor ……. .
(ii) The smallest prime number is …… .
(iii) The smallest composite number is …… .
(iv) The HCF of two consecutive odd numbers is ……. .
(v) Two perfect numbers are …… and …… .

ANSWER:

(i) prime, composite 
(ii) 2
(iii) 4
(iv) 1
(v) 6, 28 

Page No 44:

Question 20:

Write ‘T’ for true and ‘F’ for false statement.
(i) Every prime numbner is odd.
(ii) Every even number is composite.
(iii) The sum of two odd numbers is always odd.
(iv) The sum of two even numbers is always even.
(v) The HCF of two given numbers is always a factor is their LCM.

ANSWER:

(i) F
2 is an even prime number.
(ii) F
2 is an even number, but it is not composite.
(iii) F
The sum of two odd numbers is always even. For example, 9 and 11 are odd numbers, but their sum, i.e. 20, is an even number.
(iv) T
The sum of two even numbers is always even. For example, 4 and 10 are even numbers, and their sum, i.e. 14, is an even number.
(v) T
For example, 4 and 6 are two numbers whose H.C.F is 2 and L.C.M. is 12, but 2 is a factor of 12.   

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) 5^oC 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) -7^oC
(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	−1	0	1	2	3
−3
−2
−1
0
1
2
3

ANSWER:

×	–3	–2	–1	0	1	2	3
–3	9	6	3	0	–3	–6	–9
–2	6	4	2	0	–2	–4	–6
–1	3	2	1	0	–1	–2	–3
0	0	0	0	0	0	0	0
1	–3	–2	–1	0	1	2	3
2	–6	–4	–2	0	2	4	6
3	–9	–6	–3	0	3	6	9

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