Exercise 4.1

Question: 1

Write down the numerator of each of the following rational numbers:

(i) -7/5 

(ii)  15/(-4) 

(iii)  (- 17)/(- 21) 

(iv) 8/9   

(v)  5

Solution:

Numerators are:

(i) -7

(ii)  15

(iii) -17

(iv) 8

(v) 5

Question: 2

Write down the denominator of each of the following rational numbers:

(i) – 4/5

(ii)  11/(-34)

(iii)  15/82

(iv) 15

(v)  0

Solution:

Denominators are:

(i)  5

(ii) – 34

(iii)  – 82

(iv) 1

(v)  1

Question: 3

Write down the rational number whose numerator is (-3) ×4, and whose denominator is (34 – 23) ×(7 – 4).

Solution:

According to the question:

Numerator = (-3) x 4 = – 12

Denominator = (34 – 23) x (7 – 4) =11 x 3 = 33

Therefore, Rational number = -12/32

Question: 4

Write the following rational numbers as integers:

Solution:

Integers are 7, -12, 34, -73 and 95.

Question: 5

Write the following integers as rational numbers with denominator 1:

-15, 17, 85, -100

Solution:

Rational numbers of given integers with denominator 1 are:

Question: 6

Write down the rational whose numerator is the smallest three digit number and denominator is the largest four digit number.

Solution:

Smallest three digit number = 100

Largest four digit number = 9999

Therefore rational number = 100/9999

Question: 7

Seperate positive and negative rational numbers from the following rational numbers:

Solution:

Given rational numbers can be rewritten as:

Thus, positive rational numbers are:

Negative rational numbers are:

Question: 8

Which of the following rational numbers are positive:

(i)  (- 8)/7

(ii)   9/8

(iii)  19/13

(iv) (- 21)/13

Solution:

The numbers can be rewritten as:

(i)  (- 8)/7

(ii)   9/8

(iii)  19/13

(iv) -21/13

Positive rational numbers are (ii) and (iii), i.e., 9/8 and (-19)/(-13)

Question: 9

Which of the following rational numbers are negative?

(i)  (- 3)/7

(ii) (- 5)/(- 8)

(iii) 9/(- 83) 

(iv) 115/197 

Solution:

The numbers can be rewritten as:

(i)  (- 3)/7

(ii)   (-5)/(- 8)

(iii)  9/(- 83)

(iv) 115/197

Negative rational numbers are (i) and (iii).

Exercise 4.2

Question: 1

Express each of the following as rational number with positive denominator:

(i)  (-15)/(-28)

(ii)  6/(-9)

(iii)   (-28)/(-11)

(iv)   19/(- 7)

Solution:

Rational number with positive denominators:

(i) Multiplying the number by -1, we get: -15 – 28 = -15 x -1 – 28 x – 1 = 1528

(ii) Multiplying the number by -1, we get:  6 – 9 = 6x – 1 – 9x – 1 = – 69

(iii) Multiplying the number by -1, we get: – 28 – 11= – 28 x – 1 – 11 x – 1 = 2811

(iv) Multiplying the number by -1, we get: 19 – 7 = 19 x – 1 – 7 x – 1= -197

Question: 2

Express 3/5 as a rational number with numerator:

(i)  6

(ii) -15

(iii)  21

(iv) – 27

Solution:

Rational number with numerator:

(i) 6 is:

(Multiplying numerator and denominator by 2)

(ii) -15 is:

(Multiplying numerator and denominator by – 5)

(iii) 21 is:

(Multiplying numerator and denominator by 7)

(iv) – 27 is:

(Multiplying numerator and denominator by – 9)

Question: 3

Express 5/7 as a rational number with denominator:

(i) -14

(ii) 70

(iii) -28

(iv) – 84

Solution:

5/7 as a rational number with denominator:

(1) -14 is:

(Multiplying numerator and denominator by -2)

(ii) 70 is:

(Multiplying numerator and denominator by 10)

(iii) – 28 is:

(Multiplying numerator and denominator by – 4)

(iv) – 84 is:

(Multiplying numerator and denominator by -12)

Question: 4

Express 3/4 as a rational number with denominator:

(i) 20

(ii) 36

(iii) 44

(iv) – 80

Solution:

3/4 as rational number with denominator:

(i) 20 is:

(Multiplying numerator and denominator by 5)

(ii) 36 is:

(Multiplying numerator and denominator by 9)

(iii)  44 is:

(Multiplying numerator and denominator by 11)

(iv) – 80 is:

(Multiplying numerator and denominator by – 20)

Question: 5

Express 2/5 as a rational number with numerator:

(i) -56

(ii) 154

(iii) -750

(iv)  – 80

Solution:

2/5 as a rational number with numerator:

(i)  – 56 is:

(Multiplying numerator and denominator by -28)

(ii) 154 is:

(Multiplying numerator and denominator by 77)

(iii) – 750 is:

(Multiplying numerator and denominator by -375)

(iv) 500 is:

(Multiplying numerator and denominator by 250)

Question: 6

Express (-192)/108 as a rational number with numerator:

(i) 64

(ii) -16

(iii) 32

(iv) – 48

Solution:

Rational number with numerator:

(i)  64 as numerator:

(Dividing the numerator and denomintor by -3)

(ii) -16 as numerator:

(Dividing the numerator and denomintor by 12)

(iii) 32 as numerator:

(Dividing the numerator and denomintor by – 6)

(iv) – 48 as numerator:

(Dividing the numerator and denomintor by 4)

Question: 7

Express  168/(-294)  as a rational number with denominator:

(i) 14

(ii) -7

(iii) – 49

(iv) 1470

Solution:

Rational number with denominator:

(i) 14 as denominator:

(Dividing the numerator and denomintor by -21)

(ii) -7 as denominator:

(Dividing the numerator and denomintor by 42)

(iii) – 49 as denominator:

(Dividing the numerator and denomintor by 6)

(iv) 1470 as denominator:

(Multiplying the numerator and denomintor by – 5)

Question: 8

Write (-14)/42 form so that numerator is equal to:

(i) -2

(ii)  7

(iii) 42

(iv) -70

Solution:

Rational number with numerator:

(i) -2 is:

(Dividing numerator and denominator by 7)

(ii) 7 is:

(Dividing numerator and denominator by -2)

(iii) 42 is:

(Multiplying numerator and denominator by -3)

(iv) -70 is:

(Multiplying numerator and denominator by 5)

Question: 9

Select those rational numbers which can be written as a rational number with numerator 6:

Solution:

Given rational numbers that can be written as a rational number with numerator 6 are:

1/22 (On multiplying by 6) = 6/132, 2/3 (On multiplying by 3)

= 6/9, 3/4 (On multiplying by 2)

= 6/8, (-6)/7 (On multiplying by -1)

= 6/(-7)

Question: 10

Select those rational numbers which can be written as a rational number with denominator 4:

Solution:

Given rational numbers that can be written as a rational number with denominator 4 are:

7/8 (On dividing by 2) = 3.5/4,

64/16 (On dividing by 4) = 16/4,

36/-12(On dividing by 3) = 12/-4 = -12/4,

16/17 can’t be expressed with a denominator 4.

5/- 4(On multiplying by -1) = – 5/4

140/28(On dividing by 7) = 20/4

Question: 11

In each of the following, find an equivalent form of the rational number having

common denominator:

(i)  3/4 and 5/12

(ii)   2/3, 7/6 and 11/12

(iii)   5/7, 3/8, 9/14 and 20/21<

Solution:

Equivalent forms of the rational number having common denominator are:

(i) 3/4 = (3 × 3)/(4 × 3) = 9/12 and 512.

(ii) 2/3 = (2 × 4)/(3 × 4) = 8/12 and 7/6 = (7×2)/(6×2) = 14/12 and 11/12

Forms are 8/12, 14/12 and 11/12

(iii) 5/7 = (5 × 24)/(7 × 24) = 120/168, 3/8 = (3 × 21)/(8 × 21) = 63/168, 9/14

= (9 × 12)/(14 × 12)

= 108/168 and 20/21

= (20 × 8)/(21×8)

=160/168

Forms are 120/168, 63/168, 108/168 and 160/168.

Exercise 4.3

Question: 1

Determine whether the following rational numbers are in the lowest form or not:

(i)  65/84

(ii)  (-15)/32

(iii) 24/128

(iv) (-56)/(-32)

Solution:

(i) We observe that 65 and 84 have no common factor their HCF is 1.

Thus, 65/84 is in its lowest form.

(ii) We observe that -15 and 32 have no common factor i.e., their HCF is 1.

Thus, -15/32 is in its lowest form.

(iii) HCF of 24 and 128 is not 1.

Thus, given rational number is not in its simplest form.

(iv) HCF of 56 and 32 is 8.

Thus, given rational number is not in its simplest form.

Question: 2

Express each of the following numbers to the lowest form:

(i)  4/22

(ii)  (- 36)/180

(iii)  132/428

(iv) 32/56

Solution:

Lowest form of:

(i)  4/22 is:

4 = 2 × 2, 22 = 2 × 11

HCF of 4 and 22 is 2.

Dividing the fraction by 2, we get 2/11.

(ii) -36/180 is:

36 = 3 × 3 × 2 × 2, 180 = 5 × 3 × 3 × 2 × 2

HCF of 36 and 180 is 36.

Dividing the fraction by 36, we get -1/5.

(iii) 132/- 428 is:

132 = 2 × 3 × 2 × 11, 428 = 2 × 2 × 107

HCF of 132 and 428 is 4.

Dividing the fraction by 4, we get 33/-107.

(iv) – 32/ – 56 is:

32 =2 × 2 × 2 × 2 × 2, 56 = 2 × 2 × 2 × 7

HCF of 32 and 56 is 8.

Dividing the fraction by 8, we get 4/7.

Question: 3

Fill in the blanks:

Solution:

Exercise 4.4

Question: 1

Write each of the following rational numbers in the standard form:

(i) 2/10

(ii) – 8/36

(iii) 4/-16

(iv) -15/-35

(v) 299/-161

(vi) – 63/- 210

(vii) 68/- 119

(viii) – 195/275

Solution:

(i) The denominator is positive and HCF of 2 and 10 is 2.

Therefore, Dividing the numerator and denominator by 2, we get:

2/10 = 2/2, 10/2 = 1/5

(ii) The denominator is positive and HCF of 8 and 36 is 4.

Therefore, Dividing the numerator and denominator by 4, we get: (- 8)/36

= -8/4, 36/4 = -2/9

(iii) The denominator is negative.

(4x-1)/(-16x-1) = -4/16

HCF of 4 and 16 is 4.

Therefore, Dividing the numerator and denominator by 4, we get: (-4)/4, 16/4

= -1/4

(iv) -15/-35

The H.C.F. of -15 and -35 is 5.

Dividing the Nr and Dr of -15/-35 by 5, we get:

(v) 299/-161

The H.C.F. of 299 and -161 is 23.

Dividing the Nr and Dr of 299/-161 by 23, we get:

(vi) – 63/-210

The H.C.F of 63 and 210 is 21.

Dividing the Nr and Dr of -63/-210 by 21, we get:

(vii) 68/-119

The H.C.F. of 68 and 119 is 17.

Dividing the Nr and Dr of 68/-119 by 17, we get

(viii) -195/275

The H. C. F. of 195 and 275 is 5, we get:

Exercise 4.5

Question: 1

Which of the following numbers are equal?

(i) – 9/12 and 8/-12

(ii)  -16/20 and 20/-25

(iii)  -7/21 and 3/-9

(iv)  – 8/-14 and 13/21

Solution:

(i) The standard form of -9/12 is -9/3, 12/3 = -34

The standard form of 8/-12 is 8/-4, 12/-4 = -2/3

Since, the standard forms of two rational numbers are not same. Hence, they are not equal.

(ii) Since, LCM of 20 and 25 is 100.

Therefore making the denominators equal,

Therefore, -16/20 = 20/-25 .

(iii) Since, LCM of 21 and 9 is 63.

Therefore making the denominators equal,

(iv) Since, LCM of 14 and 21 is 42.

Therefore making the denominators equal,

Question: 2

If each of the following pairs represents a pair of equivalent rational numbers,

find the values of x:

(i)  2/3 and 5/x

(ii) -3/7 and x/4

(iii)  3/5 and x/-25

(iv) 13/6 and – 65/x

Solution:

(i) 2/3 = 5/x, then x = (5 × 3)/2 = 15/2

(ii) -3/7 = x/4, then x = (-3/7) × 4 = -12/7

(iii)  3/5 = x/-25, then x = 3/5 x (-25) = -75/5 = -15

(iv) 13/6 = – 65/x , then x = 6/13 x (- 65) = 6 x (-5) = -30

Question: 3

In each of the following, fill in the blanks so as to make the statement true:

(i)  A number which can be expressed in the form p/q, where p and q are integers and q is not equal to zero , is called a ………..

(ii)  If the integers p and q have no common divisor other than 1 and q is positive, then the rational number p/q is said to be in the ….

(iii) Two rational numbers are said to be equal, if they have the same …. form

(iv) If m is a common divisor of a and b,

(v)  If p and q are positive Integers, then p/q is a …..rational number and  p/(-q)  is a …… rational number .

(vi)  The standard form of -1 is …

(vii)  If p/q is a rational number, then q cannot be ….

(viii)  Two rational numbers with different numerators are equal, if their numerators are in the same …. as their denominators .

Solution:

(i)  rational number

(ii)  standard rational number

(iii)  standard form

(iv)  a/b = (a ÷ m)/(b ÷ m )

(v)  positive rational number, negative rational number

(vi) -1/1

(vii)  Zero

(viii)  ratio

Question: 4

 In each of the following state if the statement is true (T) or false (F):

(i) The quotient of two integers is always an integer.

(ii) Every integer is a rational number.

(iii) Every rational number is an integer.

(iv) Every traction is a rational number.

(v)  Every rational number is a fraction.

(vi) If a/b is a rational number and m any integer,

(vii)  Two rational numbers with different numerators cannot be equal.

(viii)  8 can be written as a rational number with any integer as denominator.

(ix) 8 can be written as a rational number with any integer as numerator.

(x)  2/3 is equal to 4/6.

Solution:

(i)  False; not necessary

(ii)  True; every integer can be expressed in the form of p/q, where q is not zero.

(iii)  False; not necessary

(iv) True; every fraction can be expressed in the form of p/q, where q is not zero.

(v) False; not necessary

(vi) True

(vii) False; they can be equal, when simplified further.

(viii) False

(ix) False

(x)  True; in the standard form, they are equal.

Exercise 4.6

Question: 1

Draw the number line and represent the following rational numbers on it:

(i) 2/3

(ii) 3/4

(iii) 3/8

(iv) -5/8

(v) -3/16

(vi) -7/3

(vii) 22/-7

(viii) -31/3

Solution:

Question: 2

Which of the two rational numbers in each of the following pairs of rational numbers is greater?

(i) -3/8, 0

(ii) 5/2, 0

(iii) – 4/11, 3/11

(iv) – 7/12, 5/- 8

(v)  4/9, – 3/- 7

(vi) – 5/8, 3/- 4

(vii) 5/9, -3/- 8

(viii)  5/- 8, -7/12

Solution:

(i)  We know that every positive rational number is greater than zero and every negative

 rational number is smaller than zero. Thus, – 3/8 > 0

(ii)  5/2 > 0.Because every positive rational number is greater than zero and every negative rational number is smaller than zero.

(iii) – 4/11 < 3/11. Because every positive rational number is greater than zero and every negative rational number is smaller than zero.

Question: 3

 Fill in the blanks by the correct symbol out >, =, or < :

Solution:

Question: 4

Fill in the blanks by the correct symbol out of >, =, or < :

Solution:

(i) Because every positive number is greater than a negative number, (- 6)/7 < 7/13.

(iv) Because every positive number is greater than a negative number, 0 > (-2)/5.

Question: 5

Arrange the following rational numbers in ascending order:

Solution:

(i) Ascending order:

Since, LCM of 5, -30, -15, 10 is 30.

Multiplying the numerators and denominators to get the denominator equal to the LCM 3/5

(ii) Since, LCM of 9, -12, -18, 3 is 36.

Multiplying the numerators and denominators to get the denominator to get the denominator equal to the LCM ,

Question: 6

 Arrange the following rational numbers in descending order:

Solution:

We have to arrange them in descending order.

(i) Since, LCM of 8, 16, -12, -4, 28 is 336.

Multiplying the numerators and denominators, to get the denominator equal to the LCM, 7/8

(ii) Since, LCM of 10, -30, -15, 20 is 60.

Multiplying the numerators and denominators, to get the denominator equal to LCM,

Question: 7

Which of the following statements are true:

(i) The rational number 29/23 lies to the left of zero on the number line.

(ii) The rational number-12/-17 lies to the left of zero on the number line.

(iii) The rational number 3/4 lies to the right of zero on the number line.

(iv) The rational number (-12)/(-5) and (- 7)/(-17) are on the opposite side of zero on the number line.

(v) The rational number -2/15 and 7/(-31)  are on the opposite side of zero on the number line .

(vi) The rational number (- 3)/(-5)  is on the right of (- 4)/7 on the number line.

Solution:

(i) False; it lies to the right of zero because it is a positive number.

(ii) False; it lies to the right of zero because it is a positive number.

(iii) True

(iv) True; they are of opposite signs.

(v) False; they both are of same signs.

(vi) True; they both are of opposite signs and positive number is greater than the negative number.

Thus, it is on the right of the negative number.

Exercise 3.1

Question: 1

Write each of the following as decimals:

(i)  8/100

Solution:

(i)  8/100

Mark the decimal point two places from right to left

Mark the decimal point one place from right to left

9/10 = 0.9

4/100

Mark the decimal point two places from right to left

4/100

= 0.04

Convert 2/10 and 6/1000 into decimals

2/10

Mark the decimal point one place from right to left

2/10

= 0.2

6/1000

Mark the decimal point three places from right to left

6/1000

= 0.006

23, 0.2, 0.006 are unlike decimals. So we convert them into like decimals.

Question: 2

Convert each of the following into fractions in the lowest form:

(i) 0.04

(ii) 2.34

(iii) 0.342

(iv) 17.38

Solution:

(i) 0.04

= 0.04/1

= 4/100

= 1/25

(ii) 2.34

= 2.34/1

= 234/100

= 117/50

(iii) 0.342

= 0.342/1

= 342/1000

= 171/500

(iv) 17.38

=17.38/1

= 1738/100

= 869/50

Question: 3

Express the following fractions as decimals:

(i)  23/10

Solution:

(i)  23/10

= 23/10

= 2.3

Question: 4

Add the following:

(i) 41.8, 39.24, 5.01 and 62.6

 (ii) 18.03, 146.3, 0.829 and 5.324

Solution:

(i) 148.65

(ii) 170.483

Question: 5

Find the value of:

(i) 9.756 – 6.28

(ii) 48.1- 0.37

(iii) 108.032 – 86.8

(iv) 100 – 26.32

Solution:

(i) 3.476

(ii) 47.73

(iii) 21.232

(iv) 73.68

Question: 6

Take out 3.547 from 7.2

Solution:

3.653

Question: 7

What is to be added to 36.85 to get 59.41?

Solution:

x + 36.85 = 59.41

x = 59.41-36.85

x = 22.56

Therefore 22.56 is added to 36.85 to get 59.41

Question: 8

What is to be subtracted from 17.1 to get 2.051?

Solution:

17.1- x = 2.051

17.1 = x + 2.051

x = 17.1 – 2.051

x = 15.049

Therefore 15.049 is subtracted from 17.1 to get 15.049

Question: 9

By how much should 34.79 be increased to get 70.15?

Solution:

34.79 + x = 70.15

x = 70.15 – 34.79

x = 35.36

Therefore 35.36 is increased to 70.15

Question: 10

By how much should 59.71 be decreased to get 34.58?

Solution:

59.71 – x = 34.58

59.71 – 34.58 = x

x = 25.13

Therefore 25.13 is decreased to get 34.58

Exercise 3.2

Question: 1

Find the product:

(i) 4.74 x 10

(ii) 0.45 x 10

(iii) 0.0215 x 10

(iv) 0.0054 x 100

Solution:

(i) 4.74 x 10

Shifting the decimal point by one place to the right

4.74 x 10 = 47.4

(ii) 0.45 x 10

Shifting the decimal point by one place to the right

0.45 × 10 = 4.5

(iii) 0.0215 x 10

Shifting the decimal point by one place to the right

0.0215 × 10 = 0.215

(iv) 0.0054 x 100

Shifting the decimal point by two places to the right

0.0054 × 100 = 0.054

Question: 2

Find the product:

(i) 35.853 x 100

(ii) 42.5 x 100

(iii) 12.075 x 100

(iv) 100 x 0.005

Solution:

(i) 35.853 x 100

Shifting the decimal point by two places to the right

35.853 = 3585.3

(ii) 42.5 x 100

Shifting the decimal point by two places to the right

42.5 × 100 = 4250

(iii) 12.075 x 100

Shifting the decimal point by two places to the right

12.075 × 100 = 1207.5

(iv) 100 x 0.005

Shifting the decimal point by two places to the right

0.005 × 100 = 0.5

Question: 3

Find the product:

(i) 2.506 x 1000

(ii) 20.708 x 1000

(iii) 0.0529 x 1000

(iv) 1000 x 0.1

Solution:

(i) 2.506 x 1000

Shifting the decimal point by three places to the right

2.506 × 1000 = 2506

(ii) 20.708 x 1000

Shifting the decimal point by three places to the right

20.708 × 1000 = 20708

(iii) 0.0529 x 1000

Shifting the decimal point by three places to the right

0.0529 × 1000 = 52.9

(iv) 1000 x 0.1

Shifting the decimal point by three places to the right

0.1 × 1000 = 100

Question: 4

Find the product:

(i) 3.4 x 17

(ii) 0.745 x 12

(iii) 28.73 x 47

(iv) 0.0415 x 59

Solution:

(i) 3.4 x 17

Multiply the number without looking into the decimal points

3.4 × 17 = 578

Mark the decimal point in the product to have one place of decimal as there in the given decimal

= 57.8

(ii) 0.745 x 12

Multiply the number without looking into the decimal points

745×12 = 8940

Mark the decimal point in the product to have three places of decimal as there in the given decimal

0.745 x 12= 8.940

(iii) 28.73 x 47

Multiply the number without looking into the decimal points

2873 × 47 = 135031

Mark the decimal point in the product to have two places of decimal as there in the given decimal

28.73 × 47 = 1350.31

(iv) 0.0415 x 59

Multiply the number without looking into the decimal points

415 × 59 = 24485

Mark the decimal point in the product to have two places of decimal as there in the given decimal

0.0415 × 59 = 2.4485

Question: 5

Find:

(i) 1.07 x 0.02

(ii) 211.9 x 1.13

(iii) 10.05 x 1.05

(iv) 13.01 x 5.01

Solution:

(i) 1.07 x 0.02

Multiply the number without looking into the decimal points

107× 2 = 214

Sum of the decimal places in the given decimals is 2 + 2 = 4

Mark the decimal point in the product to have four places of decimals

1.07 × 0.02 = 0.0214

(ii) 211.9 x 1.13

Multiply the number without looking into the decimal points

2119 × 113 = 239447

Sum of the decimal places in the given decimals is 1+ 2 = 3

Mark the decimal point in the product to have three places of decimals

211.9 × 1.13 = 239.447

(iii) 10.05 x 1.05

Multiply the number without looking into the decimal points

1005 × 105 = 105525

Sum of the decimal places in the given decimals is 2 + 2 = 4

Mark the decimal point in the product to have four places of decimals

10.05 × 1.05 = 10.5525

(iv) 13.01 x 5.01

Multiply the number without looking into the decimal points

1301 × 501 = 651801

Sum of the decimal places in the given decimals is 2+ 2 = 4

Mark the decimal point in the product to have four places of decimals

13.01 × 5.01 = 65.1801

Question: 6

Find the area of a rectangle whose length is 5.5 m and breadth is 3.4 m.

Solution:

We have

Length of rectangle = 5.5 m

Breadth of rectangle = 3.4 m

Area of rectangle = Length × Breadth

= 5.5 x 3.4

=18.7 m²

Question: 7

If the cost of a book is Rs 25.75, find the cost of 24 such books.

Solution:

Cost of one book = Rs.25.75

Therefore cost of 24 books = 25.75 x 24

= Rs.618.00

Question: 8

A car covers a distance of 14.75 km in one litre of petrol. How much distance will it cover in 15.5 litres of petrol?

Solution:

We have,

Distance covered in one litre of petrol = 14.75 km

Distance covered in 15.5 litres of petrol = 14.75 x 15.5

= 228.625 km

Question: 9

One kg of rice costs Rs 42.65. What will be the cost of 18.25 kg of rice?

Solution:

Cost of one kg rice = 42.65

Cost of 18.25kg = 42.65 x 18.25

= Rs.778.3625

Question: 10

One metre of cloth costs Rs 152.50. What is the cost of 10.75 metres of cloth?

Solution:

We have,

One metre of cloth cost = Rs.152.50

Cost of 10.75 metres =10.75 x 152.50

= Rs.1639.375

Exercise 3.3

Question: 1

Divide:

(i) 142.45 by 10

(ii) 54.25 by 10

(iii) 3.45 by 10

(iv) 0.57 by 10

(v) 0.043 by 10

(vi) 0.004 by 10

Solution:

(i) 142.45 by 10

Shifting the decimal point by one place to the left

142.45/10

= 14.245

 (ii) 54.25 by 10

Shifting the decimal point by one place to the left

54.25/10

= 5.425

(iii) 3.45 by 10

Shifting the decimal point by one place to the left

3.45/10

0.345

(iv) 0.57 by 10

Shifting the decimal point by one place to the left

0.57/10

= 0.057

 (v) 0.043 by 10

Shifting the decimal point by one place to the left

0.043/10

= 0.0043

(vi) 0.004 by 10

Shifting the decimal point by one place to the left

0.004/10

= 0.0004

Question: 2

Divide:

(i) 459.5 by 100

(ii) 74.3 by 100

(iii) 5.8 by 100

(iv) 0.7 by 100

(v) 0.48 by 100

(vi) 0.03 by 100

Solution:

(i) 459.5 by 100

Shifting the decimal point by two places to the left

459.5/100

= 4.595

(ii) 74.3 by 100

Shifting the decimal point by two places to the left

74.3/100

= 0.743

(iii) 5.8 by 100

Shifting the decimal point by two places to the left

5.8/100

= 0.058

(iv) 0.7 by 100

Shifting the decimal point by two places to the left

0.7/100

= 0.007

 (v) 0.48 by 100

Shifting the decimal point by two places to the left

0.48/100

= 0.0048

(vi) 0.03 by 100

Shifting the decimal point by two places to the left

0.03/100

= 0.0003

Question: 3

Divide:

(i) 235.41 by 1000

(ii) 29.5 by 1000

(iii) 3.8 by 1000

(iv) 0.7 by 1000

Solution:

(i) 235.41 by 1000

Shifting the decimal point by three places to the left

235.41/1000

= 0.23541

(ii) 29.5 by 1000

Shifting the decimal point by three places to the left

29.51/000

= 0.0295

(iii) 3.8 by 1000

Shifting the decimal point by three places to the left

3.8/1000

= 0.0038

(iv) 0.7 by 1000

Shifting the decimal point by three places to the left

0.7/1000

= 0.007

Question: 4

Divide:

(i) 0.45 by 9

(ii) 217.44 by 18

(iii) 319.2 by 2.28

(iv) 40.32 by 9.6

(v) 0.765 by 0.9

(vi) 0.768 by 1.6

Solution:

(i) 0.45 by 9

= 0.45/9

= 0.05

(ii) 217.44 by 18

= 217.44/18

= 12.08

(iii) 319.2 by 2.28

= 319.2/2.28

(iv) 40.32 by 9.6

= 40.32/9.6

= 403.2/96

= 4.2

= 403.2/96 = 4.2

(v) 0.765 by 0.9

= 0.765/0.9

= 7.65/9

= 0.85

= 7.65/0.9 = 0.85

(vi) 0.768 by 1.6

= 0.768/1.6

= 7.68/16

= 0.48

Question: 5

Divide: 
(i) 16.64 by 20
(ii) 0.192 by 12
(iii) 163.44 by 24
(iv) 403.2 by 96
(v) 16.344 by 12
(vi) 31.92 by 228

Solution:

(i) 16.64 by 20

= 16.64/20

= 1.664/2

= 0.832

(ii) 0.192 by 12

= 0.192/12

= 0.016

(iii) 163.44 by 24

= 163.44/24

= 6.81

 (iv) 403.2 by 96

= 403.2/96

= 4.2

 (v) 16.344 by 12

= 16.344/12

= 1.362

(vi) 31.92 by 228

= 31.92/228

= 0.14

Question: 6

Divide:

(i) 15.68 by 20
(ii) 164.6 by 200
(iii) 403.80 by 30

Solution:

(i) 15.68 by 20 

=15.68/20

=1.568/2

= 0.784

(ii) 164.6 by 200

Question: 7

(i) 76 by 0.019

(ii) 88 by 0.08

(iii) 148 by 0.074

(iv) 7 by 0.014

(iii) 403.80 by 30

Solution:

(i) 76 by 0.019

= 76/0.019

= 76000/19

= 4000

(ii) 88 by 0.08

= 88/0.08

= 8800/8

=1100

(iii) 148 by 0.074

= 148/0.074

= 148000/74

= 2000

(iv) 7 by 0.014

= 70.014

= 7000/14

= 0.823

(iii) 403.80 by 30

= 403.80/30

Question: 8

Divide:

(i) 20 by 50

(ii) 8 by 100

(iii) 72 by 576

(iv) 144 by 15

Solution:

(i) 20 by 50 
= 20/50
 = 0.4
 (ii) 8 by 100 
= 8/100
By shifting the decimal point to the left
= 8/100
= 0.08
(iii) 72 by 576 
= 72/576
= 0.125

(iv)144 by 15 
= 144/15

= 9.6

Question: 9

A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it travel in 1 litre of petrol?

Solution:

Distance covered in 2.4 litres of petrol = 43.2 km

Distance covered in 1 litre of petrol = 43.2/2.4

= 18km

The distance travelled in 1 litre of petrol is 18 km

Question: 10

The total weight of some bags of wheat is 1743 kg. If each bag weights 49.8 kg, how many bags are there?

Solution:

Total weight of bags of wheat = 1743 kg

Each bag weight = 49.8 kg

No of bags = 1743/49.8

= 35

Therefore the total numbers of bags are 35

Question: 11

Shikha cuts 50 m of cloth into pieces of 1.25 m each. How many pieces does she get?

Solution:

Total length of cloth = 50 m

Length of each piece of cloth = 1.25 m

Number of pieces = 50/1.25

= 5000/125

= 40 pieces

Therefore Shikha got 40 pieces

Question: 12

Each side of a rectangular polygon is 2.5 cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have?

Solution:

Length of each side of rectangular polygon = 5.2 cm
Perimeter of polygon = 12.5 cm
No of sides polygon has = 12.5 cm
No of sides polygon have = 12.5/2.5

= 5

Therefore the sides of the polygon is 5

Question: 13

The product of two decimals is 42.987. If one of them is 12.46, find the other.

Solution:

We have,

The product of the given decimals = 42.987

 one decimal = 12.46

The other decimal = 42.987/12.46

= 3.45

The number is 3.45

Question: 14

The weight of 34 bags of sugar is 3483.3 kg. If all bags weigh equally, find the weight of each bag.

Solution:

Total weight of sugar = 3483.3kg

No of bags = 34

Weight of each bag = 3483.3/34

= 102.45 kg

Therefore weight of each bag is 102.45 kg

Question: 15

How many buckets of equal capacity can be filled from 586.5 litres of water, if each bucket has capacity of 8.5 litres?

Solution:

Capacity of each bucket = 805 litres

Total water available = 586.5 litres

Number of buckets = 805/586.5

= 69

Total number of buckets is 69

Exercise 2.1

Question: 1

 Compare the following fractions by using the symbol > or < or =;

Solution:

(i) We have,

Taking the LCM of 9 and 13, we get,

9 x 13 = 117

Now, we convert the given fractions to equivalent fractions by making the denominators 117,

As we know, 91 > 72

Hence, 79 > 813

(ii)  We have,

The given fractions are equivalent fractions as the denominators are equal,

And we know that, 11 > 5

Therefore, 119 > 59

(iii)  We have,

Taking the LCM of 41 and 30, we get,

41 x 30 = 1230

Now, we convert the given fractions to equivalent fractions by making the denominators 1230,

Now, we clearly know 1110 > 779

Hence, 3741 > 1930

(iv) We have,

Taking the LCM of 15 and 105, we get,

5 x 3 x 7 = 105

Now, we convert the given fractions to equivalent fractions by making the denominators 105,

Question: 2

Arrange the following fractions in ascending order:

Solution:

(i)  We have,

Taking the LCM of 8, 6, 8, 4 and 3, we get,

2 x 4 x 3 = 24

Now, we convert the given fractions to equivalent fractions by making the denominators 24,

We know that, 8 < 9 < 12 < 18 < 20

Hence, 13 < 38 < 24 < 68 < 56

(ii) We have,

Taking the LCM of 6, 8, 12 and 16, we get,

2 x 2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48,

We know that, 12 < 15 < 18 < 32

Question: 3

Arrange the following fractions in descending order:

Solution:

(i)  We have,

Taking the LCM of 5, 10, 15 and 20, we get,

5 x 2 x 2 x 3 = 60

Now, we convert the given fractions to equivalent fractions by making the denominators 48

As we know 51 > 48 > 44 > 42

(ii) We have,

Taking the LCM of 7, 35, 14 and 28, we get,

7 x 5 x 2 x 2 = 140

Now, we convert the given fractions to equivalent fractions by making the denominators 140

As we know 40 > 44 > 65 > 90

Question: 4

Write the equivalent fractions of 3/5

Solution:

Multiplying or dividing both the numerator and denominator by the same number,

so that the fraction keeps its value.

So the equivalent fractions of 3/5 are

are the five equivalent fractions of 3/5

Question: 5

Find the sum:

Solution:

(i)  We have,

Taking the LCM of 8 and 10, we get,

2 x 4 x 5 = 40

Now, we convert the given fractions to equivalent fractions by making the denominators 40

(ii) We have,

Taking out the LCM of 4 and 5, we get,

4 x 5 = 20

Now, we convert the given fractions to equivalent fractions by making the denominators 20

(iii)  We have,

Taking out the LCM of 6 and 4, we get,

2 x 2 x 3 = 12

Now, we convert the given fractions to equivalent fractions by making the denominators 12

(iv) We have,

Taking out the LCM of 5, 10 and 15, we get,

5 x 2 x 3 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

Question: 6

Find the difference of

Solution:

(i)  We have,

Taking out the LCM of 24 and 16, we get,

2 x 2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48

(ii)  We have, 6 and 23/3

The difference between 6 and 23/3

(iii)  We have,

Taking out the LCM of 25 and 20, we get,

5 x 5 x 4 = 100

Now, we convert the given fractions to equivalent fractions by making the denominators 100

The difference between both the fractions are

(iv) We have,

Taking out the LCM of 10 and 15, we get,

2 x 3 x 5 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

The difference between both the fractions are

Question: 7

Find the difference:

Solution:

(i)  We have,

Taking out the LCM of 7 and 11, we get,

7 x 11 = 77

Now, we convert the given fractions to equivalent fractions by making the denominators 77

The difference between both the fractions are

(ii)  We have,

(iii) We have,

(iv) We have,

Taking out the LCM of 10 and 15, we get,

2 x 3 x 5 = 30

Now, we convert the given fractions to equivalent fractions by making the denominators 30

Question: 8

Simplify:

Solution:

(i) We have,

Taking out the LCM of 3, 6 and 9, we get,

3 x 3 x 2 = 18

Now, we convert the given fractions to equivalent fractions by making the denominators 18, we get,

(ii)  We have,

(iii) We have,

Taking out the LCM of 6, 8 and 12, we get,

2 x 2 x 2 x 3 = 48

Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,

Question: 9

What should be added to to get 12?

Solution:

We have,

Let x be the number added to 38/7 to get 12

Therefore,

x + 387 = 12

Question: 10

 What should be added to?

Solution:

Let x be the number added to 79/15 to get 63/5

Taking out the LCM of 5 and 15, we get,

3 x 5 = 15

Now, we convert the given fractions to equivalent fractions by making the denominators 15, we get,

Question: 11

Suman studies forhours daily. She devoteshours of her time for science and mathematics. How much time does she devote for other subjects?

Solution:

Given,

Suman studies forhours daily

She devoteshours of her time for science and mathematics. Let x be time she devotes for other subjects.

Taking out the LCM of 3 and 5, we get,

3 x 5 = 15

Now, we convert the given fractions to equivalent fractions by making the denominators 48, we get,

Question: 12

A piece of wire of lengthIf it is cut into two pieces in such a way that the length of one piece iswhat is the length of the other piece?

Solution:

A piece of wire of lengthone piece is

Let the length of other piece be x m.

Question: 13

A rectangular piece of paper islong andwide. Find its perimeter?

Solution:

Given,

A rectangular piece of paper islong andwide

Perimeter = 2(length + width)

Question: 14

In a “magic square”, the sum of numbers in each row, in each column and along the diagonal is same. Is this a “magic square”?

4/11	9/11	2/11
3/11	5/11	7/11
8/11	1/11	6/11

Solution:

Given,

4/11	9/11	2/11
3/11	5/11	7/11
8/11	1/11	6/11

Therefore, the sum of numbers in each row, in each column and along the diagonal is same and the sum is 15/11.

Question: 15

The cost of Mathematics book is Rsand that of science book isWhich costs more and by how much?

Solution:

Given,

The cost of Mathematics book is Rsand that of science book is

We need to compare the cost of mathematics and science book,

Taking out the LCM of 4 and 2, we get,

2 x 2 = 4

Now, we convert the given fractions to equivalent fractions by making the denominators 4, we get

= 103/4

As we know, 103 > 82

Hence, the cost of mathematics book is more than that of the cost of the science book.

Question: 16

 Provide the number in the box [] and also give its simplest form in each of the following:

Solution:

Exercise 2.2

Question: 1

Multiply

Solution:

(i)  We have,

(ii)  We have,

(iii)  We have,

(iv) We have,

Question: 2

Find the product:

Solution:

(i)  We have,

(ii) We have,

(iv) We have,

Question: 3

Simplify:

Solution:

(i)  We have,

(ii) We have,

(iii) We have,

Question: 4

Find:

Solution:

(i) We have,

(ii) We have,

(iii) We have,

Question: 5

Which is greater?

Solution:

While comparing two fractions, when the numerators of both the fractions are same, then the denominator having higher value shows the fraction has lower value.

So, 6/14 is greater.

Therefore, 1/2 of 6/7 is greater.

Question: 6

(i)  7/11 of 330

(ii) 5/9 of 108 meters

(iii) 3/7 of 42 litres

(iv) 1/12 of an hour

(v) 5/6 of an year

(vi) 3/20 of a Kg

(vii) 7/20 of a litres

(viii) 5/6 of a day

(ix)  2/7 of a week

Solution:

(i)  We have,

(ii)  We have,

5/9 of 108 meters

= 5/9 × 108 meters

= 5 × 12 meters

60 meters

(iii) We have,

3/7 of 42 litres

= 3/7× 42 litres

= 3 × 6 litres

=18 litres

(iv) We have,

1/12 of an hour

An hour = 60 minutes

Therefore, 1/12 × 60 minutes

= 5 minutes

(v) We have,

5/6 of an year

I Year = 12 months

Therefore,

5/6 × 12 months

= 5 × 2 months

= 10 months

(vi) We have,

3/20 of a kg

1 Kg = 1000 gms

Therefore,

3/20 × 1000 gms

= 3 × 50 gms

= 150 gms

(vii) We have, 

7/20 of a litre

1 litre = 1000 ml

Therefore,

7/20 × 1000 ml

= 7 × 50 ml

= 350 ml

(viii) We have,

5/6 of a day

I day = 24 hours

Therefore,

5/6 × 24 hours

= 5 × 4 hours

= 20 hours

(ix) We have,

2/7 of a week

I week = 7 days

Therefore,

2/7 × 7 days = 2 days

Question: 7

Shikha plans 5 saplings in a row in her garden. The distance between two adjacent saplings is 3/4 m. Find the distance between first and last sapling.

Solution:

There are 4 adjacent spacing for 5 saplings.

Given, the distance between two adjacent saplings is 3/4 m.

4 adjacent spacing for 5 saplings = 3/4 × 4 = 3 m

Therefore, the distance between first and last sapling is 3 m.

Question: 8

Ravish reads 1/3 part of a book in one hour. How much part of the book will be read in 2(1/5) hours?

Solution:

Let x be the full part of book.

Given, Ravish reads 1/3 part of a book in one hour

1 hour = (1/3) x

Part of the book will he read in 2(1/5) hours

Question: 9

Lipika reads a book for 1(3/4) hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Solution:

Given,

Time taken by Lipika to read a book per day = 1(3/4) = 7/4 hours

Time taken by Lipika to read a book for 6 days

Question: 10

Find the area of a rectangular park which is 41(2/3)m long and 18(3/5)m broad.

Solution:

Area of a rectangular park = (length x breadth)

Question: 11

If milk is available at Rs 17(3/4) per litre, find the cost of 7(2/5) litres of milk.

Solution:

Given,

Question: 12

Sharda can walk 8(1/3) km in one hour. How much distance will she cover in 2(2/5) hours.

Solution:

Given,

Distance covered by Sharda in one hour = 25/3 km

Question: 12

A sugar bag contains 30 kg of sugar. After consuming 2/3 of it, how much sugar is left in the bag

Solution:

Given, A sugar bag contains 30 kg of sugar

After consuming 23 of it, the amount of sugar left in the bag

Question: 14

Each side of a square is 6 (2/3)m long. Find its area.

Solution:

Given,

Question: 15

There are 45 students in a class and 3/5 of them are boys. How many girls are. there in the class?

Solution:

Given,

There are 45 students in a class,

And 3/5 of them are boys.

Therefore, no of girls in the class = 45 –3/5 × 45

= 45 – 27

= 18

Exercise 2.3

Question: 1

Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers

(i) 3/7

(ii) 5/8

(iii) 9/7

(iv) 6/5

(v) 12/7

(vi) 1/8

Solution:

(i)  3/7

7/3 = improper number

(ii)  5/8

8/5 = improper number

(iii)  9/7 

7/9 = proper number

(iv) 6/5

5/6 = proper number

(v) 12/7

7/12 = proper number

(vi) 1/8

8 = whole number

Question: 2

Divide:

Solution:

Question: 3

Divide:

Solution:

Question: 4

Simplify:

Solution:

Question: 5

A wire of length 12(1/2)m is cut into 10 pieces of equal length. Find the length of each piece.

Solution:

Question: 6

The length of a rectangular plot of area 65(1/3) m² is 12(1/4) m. What is the width of the plot?

Solution:

Given,

The length of a rectangular plot of area 65(1/3) m² is 12(1/4) m. 

Question: 7

By what number 6(2/9) be multiplied to get 4(4/9)?

Solution:

Given,

Let x be the number which needs to be multiplied by 56/9,

Now,

Question: 8

The product of two numbers is 25(5/6). If one of the numbers is 6(2/3), find the other?

Solution:

Given,

The product of two numbers is 25(5/6). If one of the numbers is 6(2/3),

Let the other number be x.

Question: 9

The cost of 6(1/4) kg of apples is Rs 400. At what rate per kg are the apples being sold?

Solution:

Given,

Question: 10

By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630. How many dozens of oranges does he sell?

Solution:

Given,

By selling oranges at the rate of Rs 5(1/4) per orange, a fruit seller get Rs 630.

5(1/4) = 2(1/4)

12 apples =1 dozen

Therefore, 120 apples = 10 dozen

Question: 11

In mid-day meal scheme 3/10 litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every-day in the school, how many students are there in the school?

Solution:

Given,

3/10  litre of milk is given to each student of a primary school.

30 litres of milk is distributed everyday in the school

Number of students given 3/10 litres of milk = 1

Number of students given 1 litre of milk = 10/3

Number of students given 30 litres of milk = 10/3 × 30 = 100 Students

Question: 12

In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50(3/4), how many tickets were sold?

Solution:

Given,

Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50(3/4),

Exercise 1.1

Question: 1

Determine each of the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (- 25) × (- 9)

(iv) (125) × (- 8)

Solution:

(i) We have,

12 × 7 = 84  [The product of two integers of like signs is equal to the product of their absolute value]

(ii) We have,

 (- 15) × 8 [The product of two integers of opposite

= (- 15 × 8) signs is equal to the additive inverse of the

= –120 [product of their absolute values]

(iii) We have,

(-25) × (-9)

= + (25 × 9)

= 225

(iv) We have,

(125) × (- 8)

= – (125 × 8)

= –1000

Question: 2

Find each of the following products:

(i) 3 × (- 8) × 5

(ii) 9 × (- 3) × (- 6)

(iii) (- 2) × 36 × (- 5)

(iv) (- 2) × (- 4) × (- 6) × (- 8)

Solution:

(i) We have,

3 × (- 8) × 5

= – (3 × 8) × 5

= (- 24) × 5

= – (24 × 5)

= – 120

(ii) We have,

9 × (-3) × (- 6)

= – (9 × 3) × (- 6)

= (- 27) × (- 6)

= + (27 × 6)

= 162

(iii) We have,

(-2) × 36 × (- 5)

= – (2 × 36) × (- 5)

= (- 72) × (- 5)

= (72 × 5)

= 360

(iv) We have,

(- 2) × (- 4) × (- 6) × (- 8)

= (2 × 4) × (6 × 8)

= (8 × 48)

= 384

Question: 3

Find the value of:

(i) 1487 × 327 + (- 487) × 327

(ii) 28945 × 99 – (- 28945)

Solution:

(i) We have,

1487 × 327 + (- 487) × 327

= 486249 – 159249

= 327000

(ii) We have,

28945 × 99 – (- 28945)

= 2865555 – 28945

= 2894500

Question: 4

Complete the following multiplication table:

Second Number

X	– 4	-3	– 2	-1	0	1	2	3	4
– 4
First Number
– 2
-1
0
1
2
3
4

Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?

Solution:

Second number

X	– 4	– 3	– 2	– 1	0	1	2	3	4
– 4	16	12	8	4	0	-4	-8	-12	-16
First number	12	9	6	3	0	-3	-6	-9	-12
-3
-2	8	6	4	2	0	-2	-4	-6	-8
-1	4	3	2	1	0	-1	-2	-3	-4
0	0	0	0	0	0	0	0	0	0
1	-4	-3	-2	-1	0	1	2	3	4
2	-8	-6	-4	-2	0	2	4	6	8
3	-12	-9	-6	-3	0	3	6	9	12
4	-16	-12	-8	-4	0	4	8	12	16

Question: 5

Determine the integer whose product with ‘-1’ is

(i) 58

(ii) 0

(iii) – 225

Solution:

(i) 58 x (–1) = – (58 x 1)

= – 58

(ii) 0 x (–1) = 0

(iii) (–225) x (–1) = + (225 x 1)

= 225

Question: 6

What will be the sign of the product if we multiply together

(i) 8 negative integers and 1 positive integer?

(ii) 21 negative integers and 3 positive integers?

(iii) 199 negative integers and 10 positive integers? 

Solution:

(i) Positive ∵ [- ve × – ve = + ve]

(ii) Negative ∵ [- ve × + ve = – ve]

(iii) Negative

Question: 7

State which is greater: 

(i) (8 + 9) × 10 and 8 + 9 × 10

(ii) (8 – 9) × 10 and 8 – 9 × 10 

(iii) ((-2) – 5) × – 6 and (-2) – 5 × (- 6)

Solution:

(i) (8 + 9) × 10 = 17 × 10

= 170

8 + 9 × 10 = 8 + 90 = 98

(8 + 9) × 10 > 8 + 9 × 10

(ii) (8 – 9) × 10 = – 1 × 10

= – 10

8 – 9 × 10 = 8 – 90 = – 82

(8 – 9) × 10 > 8 – 9 × 10

(iii)  ((-2) – 5) × – 6 = (- 7) × (- 6)

= (7 x 6)

= 42

(– 2) – 5 x (– 6) = – 2 + (5 x 6)

= 30 – 2

= 28

Therefore, ((-2) – 5×(- 6)) > (- 2) – 5 × (- 6)

Question: 8

(i) If a× (-1) = – 30, is the integer a positive or negative?

(ii) If a × (-1) = 30, is the integer a positive or negative? 

Solution:

(i) When multiplied by ‘a’ negative integer, a gives a negative integer. Hence, ‘a’ should be

a positive integer.

a = 30

(ii) When multiplied by ‘a’ negative integer, a gives a positive integer. Hence, ‘a’ should be

a negative integer.

a = – 30

Question: 9

 Verify the following:

(i) 19 × (7 + (-3)) = 19 × 7 + 19 × (-3)

(ii) (-23)[(-5)+ (+19)] = (-23) × (- 5) + (- 23) × (+19)

Solution:

(i) L.H.S = 19 × (7+ (-3))

= 19 × (7-3)

= 19 × 4

= 76

R.H.S = 19 × 7 + 19 × (-3)

= 133 – 57

= 76

Therefore, L.H.S = R.H.S

(ii)  L.H.S = (-23)[(-5) + (+19)]

= (-23)[-5 + 19]

= (-23)[14]

= – 322

R.H.S = (-23) × (-5) + (-23) × (+19)

= 115 – 437

= –322

Therefore, L.H.S = R.H.S

Question: 10

Which of the following statements are true?

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

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

(iii) Of the two integers, if one is negative, then their product must be positive.

(iv) For all non-zero integers a and b, a × b is always greater than

either a or b.

(v) The product of a negative and a positive integer may be zero.

(vi) There does not exist an integer b such that for a >1, a × b = b × a = b. <

Solution:

(i) True

(ii) True

(iii) False

(iv) False

(v) False

(vi) True

Exercise 1.2

Question: 1

Divide:

(i) 102 by 17

(ii) –85 by 5

(iii) –161 by –23

(iv) 76 by –19

(v) 17654 by –17654

(vi) (–729) by (–27)

(vii) 21590 by – 10

(viii) 0 by –135

Solution:

(i) We have,

(ii) We have,

(iii) We have,

(iv) We have,

(v) We have,

(vi) We have,

(vii) We have,

(viii) We have,

Question: 2

 Fill in the blanks:

 (i) 296 ÷ …. = – 148

(ii) – 88 ÷ …. = 11

(iii) 84 ÷ …. = 12

(v) …. ÷ 156 = – 2

(vi)  …. ÷ 567 = – 1

Solution:

(i)  We have,

(ii) We have,

(iii) 84/12 = 7

(iv)  …. ÷ -5 = 25

25 × (- 5) = – 125

(v) …. ÷ 156 = – 2

– (156 × 2)

= – 312

(vi) x/567 = – 1

⇒ x = – (567 × 1)

= – 567

∴ – (567/567) = -1

Question: 3

Which of the following statements are true?

(i) 0 ÷ 4 = 0

(ii) 0 ÷ (-7) = 0

(iii) -15 ÷ 0 = 0

(iv) 0 ÷ 0 = 0

(v) (- 8) ÷ (- 1) = – 8

(vi) – 8 ÷ (- 2) = 4<

Solution:

(i) True

(ii) True

(iii) False

(iv) False

(v) False

(vi) True

Exercise 1.3

Question: 1

Find the value of 36 ÷ 6 + 3.

Solution:

36 ÷ 6 + 3 = 6 + 3

= 9

Question: 2

Find the value of 24 + 15 ÷ 3.

Solution:

24 + 15 ÷ 3 = 24 + 5

= 29

Question: 3

Find the value of 120 – 20 ÷ 4.

Solution:

120 – 20 ÷ 4 = 120 – 5

= 115

Question: 4

Find the value of 32 – (3 × 5) + 4.

Solution:

32 – (3 x 5) + 4 = 32 – 15 + 4

= 17 + 4

= 21

Question: 5

Find the value of  3 – (5 – 6 ÷ 3).

Solution:

3 – (5 – 6 ÷ 3) = 3 – (5 – 2)

= 3 – 3

= 0

Question: 6

Find the value of 21 – 12 ÷ 3 × 2.

Solution:

21 – 12 ÷ 3 × 2 = 21 – 123 × 2

= 21 – 4 × 2

= 21 – 8

= 13

Question: 7

Find the value of 16 + 8 ÷ 4 – 2 × 3.

Solution:

16 + 8 ÷ 4 – 2 × 3

= 16 + 2 – 6

= 18 – 6

= 12

∴ 16 + 8 ÷ 4 – 2 × 3 = 12

Question: 8

Find the value of 28 – 5 × 6 + 2.

Solution:

28 – 5 × 6 + 2 = 28 – (5 × 6) + 2

= 28 – 30 + 2

= 30 – 30

= 0

Question: 9

Find the value of  (-20) × (-1) + (-28) ÷ 7.

Solution:

(-20) × (-1) + (-28) ÷ 7 = 20 + ∣-28∣ ∣7∣

= 20 – 287

= 20 – 4

= 16

Question: 10

Find the value of  (-2) + (-8) ÷ (- 4).

Solution:

(-2) + (-8) ÷ (- 4) = – 2 + ∣- 8∣ ∣- 4∣

= – 2 + 2

= 0

Question: 11

Find the value of (- 15) + 4 ÷ (5 – 3).

Solution:

-15 + 4 ÷ (5 – 3) = – 15 + 4 ÷ 2

= – 15 + 2

= – 13

-15 + 4 ÷ (5 – 3) = – 13

Question: 12

Find the value of (- 40) × (-1) + (-28) ÷ 7.

Solution:

(- 40) × (-1) + (-28) ÷ 7 = 40 + (- 4)

= 40 – 4

= 36

Question: 13

Find the value of (-3) + (-8) ÷ (-4) – 2× (-2).

Solution:

(-3) + (-8) ÷ (-4) – 2 × (-2) = (-3) + (-8)(-4) – 2×(-2)

= – 3 + 2 + 4

= 6 – 3

= 3

Question: 14

Find the value of (-3) × (-4) ÷ (-2) + (-1).

Solution:

(-3) × (-4) ÷ (-2) + (-1) = 12 ÷ (-2) + (-1)

= – 6 – 1

= – 7

∴ (-3) × (-4) ÷ (-2) + (-1) = – 7

Exercise 1.4

Question: 1

Simplify

3 – (5 – 6 ÷ 3)

Solution:

3 – (5 – 6 ÷ 3)

= 3 – (5 – 2)

= 3 – 3

= 0

∴ 3 – (5 – 6 ÷ 3) = 0

Question: 2

Simplify

– 25 + 14 ÷ (5 – 3)

Solution:

-25 + 14 ÷ (5 – 3) = – 25 + 14 ÷ (2)

= – 25 + 14/2

= –25 + 7

= –18

∴ – 25 + 14 ÷ (5 – 3) = – 18

Question: 3

Simplify

Solution:

Question: 4

Simplify

Solution:

Question: 5

Simplify

36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]

Solution:

36 – [18 – (14 – (15 – 4 ÷ 2 × 2))]

= 36 – [18 – (14 – (11 ÷ 2 × 2))]

= 36 – [18 – (14 – 11/2 × 2))]

= 36 – [18 – (14 – 11)]

= 36 – [18 – 3]

= 36 – 15

= 21

∴ 36 – [18 – (14 – (15 – 4 ÷ 2 × 2))] = 21

Question: 6

Simplify

45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]

Solution:

45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)]

= 45 – [38 – (20 – (6 – 3) ÷ 3)]

= 45 – [38 – (20 – 3 ÷ 3)]

= 45 – [38 – (20 – 1)]

= 45 – [38 – 19]

= 45 – [19]

= 26

∴ 45 – [38 – (60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3)] = 26

Question: 7

Simplify

Solution:

= 23 – [23 – (23 – (23 – 0))]

= 23 – [23 – (23 – 23)]

= 23 – [23 – 0]

= 23 – 23

= 0

Question: 8

Simplify

Solution:

= 2550 – [510 – (270 – (90 – 150))]

= 2550 – [510 – (270 – (–60))]

= 2550 – [510 – 330]

= 2550 – [180]

= 2550 – 180

= 2370

Question: 9

Simplify

Solution:

Question: 10

Simplify

Solution:

Question: 11

Simplify

Solution:

Question: 12

Simplify

[29 – ( – 2)(6 – (7 – 3))] ÷ [3 × (5 + ( – 3) × (-2))]

Solution:

[29 – (-2)(6 – (7 – 3))] ÷ [3 × (5 + (- 3) × (- 2))]

= [29 – (- 2)(6 – 4)] ÷ [3 × (5 + (3 × 2))]

= [29 – (-2)(2)] ÷ [3 × (5 + 6)]

= [29 + 4] ÷ [3 × 11]

= [33] ÷ [33]

= 1

∴ [29 – (-2) (6 -(7- 3))] ÷ [3 × (5 + (- 3) × (-2))] = 1

Question: 13

Using brackets, write a mathematical expression for each of the following:

(i) Nine multiplied by the sum of two and five.

(ii) Twelve divided by the sum of one and three.

(iii) Twenty divided by the difference of seven and two.

(iv) Eight subtracted from the product of two and three.

(v) Forty divided by one more than the sum of nine and ten.

(vi) Two multiplied by one less than the difference of nineteen and six.

Solution:

(i) 9 (2 + 5)

(ii) 12 ÷ (1 + 3)

(iii) 20 ÷ (7 – 2)

(iv) 2 × 3 – 8

(v) 40 ÷ [1 + (9 + 10)]

(vi) 2 × [(19 – 6) – 1]

Exercise 22.1

Question 1.
Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.
Solution:
Diameter of the base of cylinder = 7 cm
∴ Radius (r) = 72 cm
Height (h) = 60m
∴ carved surface area = 2πh
= 2 x 227 x 72 x 60cm² = 1320cm²

Question 2.
The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm.
Solution:
Curved surface area =132 cm²
Radius (r) = 0.35 cm
Let h be the length of the rod Then 2πrh = 132

Question 3.
The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder.
Solution:
Let r be the radius of the base of the cylinder, then
Area of the base = πr²

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
Solution:
Height of the cylinder (h) = 15 cm
Circumference of the base = 88 cm
Let r be the radius of the base, the circumference = 2πr
∴ 2πr = 88 cm …(i)

Question 5.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.
Solution:
Dimensions of rectangular strip = 25 cm x 7 cm
By rotating the strip along longer side, a solid is formed whose radius = 7 cm
and height = 25 cm

Question 6.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the the total surface area of the cylinder thus generated.
Solution:
By rolling along length wire, we get a cylinder whose circumference of its base = 20 cm

Question 7.
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3

Question 8.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. Prove that its height and radius are equal.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr² = 2πr (h + r)
But their ratio is 1 : 2

Hence their radius and height are equal.

Question 9.
The curved surface area of a cylinder is 1320 cm² and its base has diameter 21 cm. Find the height of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm²
Diameter of its base (d) = 21 cm 21
Radius (r) = 212 cm
Let h be the height of the cylinder

Question 10.
The height of a right circular, cylinder is 10.5 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.
Solution:
Height of cylinder = 10.5 m
Let r be the radius and h be the height of a right circular cylinder, then Area of its two circular faces = 2π²
and area of curved surface = 2πrh
Now, according to the condition:

Question 11.
Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.
Solution:
Diameter of the top of a cylindrical well = 6m

∴ Radius (r) = 62 = 3 m
and depth (h) = 21 m
∴ Curved surface area = 2πrh = 2 x 227 x 3 x 21 m² = 396 m²
Rate of plastering = Rs 9.50 per m²
∴ Total cost of plastering = Rs 9.50 x 396 = Rs 3,762

Question 12.
A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of the tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.
Solution:
Diameter of the vessel = 20 cm

Question 13.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.
Solution:
Diameter of the well = 3.5 m

Question 14.
The diameter of a roller is 84 cm and its length is T20 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground ?
Solution:
Diameter of the roller = 84 cm

Question 15.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre ?
Solution:
Diameter of each pillar = 0.50 m

Question 16.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of the base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder opened from both sides = 4620 cm²
Area of base ring = 115.5 cm²
Height of cylinder (h) = 7 cm
Let R be the outer radius and r be the inner radius

Question 17.
The sum of the radius of the base and height of a solid cylinder is 37 m, if the total surface area of the solid cylinder is 1628 m², find the circumference of its base.
Solution:
Let r be the radius and h be the height of the solid cylinder, then r + h = 37 m …(i)
Total surface area = 1628
⇒ 2πr (r + h) = 1628

Question 18.
Find the ratio between the total surface area of a cylinder to is curved surface area,given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius (r) of cylinder = 3.5 cm
and height (h) = 7.5 cm
∴ Curved surface area = 2πrh
and total surface area = 2πr (h + r)
∴ Ratio = 2πr (h + r)- 2πrh = (h + r): h = 7.5 + 3.5 : 7.5
⇒ 11 : 7.5

Question 19.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm².
Solution:
Radius of the vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
∴ Area of inner and outer curved surfaces and bases = 2 x 2πrh + 2πr²

Exercise 22.2

Question 1.
Find the volume of a cuboid whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m
Solution:
(i) Radius (r) = 3.5 cm
Height (h) = 40 cm
Volume of cylinder = πr²h

Question 2.
Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are:
(i) d= 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m.
Solution:
(i) Diameter (d) = 21 cm

Question 3.
The area of the base of a right circular cylinder is 616 cm3 and its height is 25 cm. Find the volume of the cylinder.
Solution:
Base area of cylinder = 616 cm²
Height (h) = 25 cm.
∴ Volume = Area of base x height
= 616 x 25 cm³ = 15400 cm³

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder.
Solution:
Circumference of the base of cylinder = 88 cm
Let r be the radius

Question 5.
A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.
Solution:
Length (Height) of hollow cylindrical pipe = 21 dm = 210 cm
Inner diameter = 6 cm
Outer diameter = 10 cm

Question 6.
Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm
Solution:
Radius of the cylider (r) = 7 cm
and height (h) = 15 cm
(i) Curved surface area = 2πrh

Question 7.
The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder.
Solution:
Diameter of the base of cylinder = 42 cm

Question 8.
Find the volume of a cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres.
Solution:
Diameter of cylinder = 7 cm

Question 9.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the volume of the solid, thus generated.
Solution:
Size of rectangular strip = 25 cm x 7 cm
By rotating, about the longer side, we find a right circular cylinder,
then the radius of cylinder (r) = 7 cm

Question 10.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.
Solution:
Size of sheet of paper = 44 cm x 20 cm
By rolling along length, a cylinder is formed in which circumference of its base = 20 cm
and height (h) = 44 cm

Question 11.
The volume and the curved surface area of a cylinder are 1650 cm3 and 660 cm2 respectively. Find the radius and height of the cylinder.
Solution:
Volume of cylinder = 1650 cm³
and curved surface area = 660 cm²
Let r be the radius and h be the height of the cylinder, then

Question 12.
The radii of two cylinders are in the ratio 2 :3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes.
Solution:
Ratio in radii = 2:3
and in heights = 5:3
Let r₁,h₁ and r₂, h₂ are the radii and heights of two cylinders respectively.

Question 13.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm².
Solution:
Ratio in curved surface are and total surface area =1 : 2
Let r be the radius and h be the height.
Then 2πrh : 2πr (h + r)= 1 : 2

Question 14.
The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the volume of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm²
Diameter of base = 21 cm

Question 15.
The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm³.
Solution:
Ratio between radius and height of a cylinder = 2:3
Volume = 1617 cm³
Let r be the radius and A be the height of the cylinder, then

Question 16.
The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the diameter and the height of the pillar.
Solution:
Let r be the radius and A be the height of the cylinder, then
2πrh = 264 …(i)
and πr2h= 924 …(ii)
Dividing (ii) by (i),

Question 17.
Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.
Solution:
Volumes of two cylinders are equal.
Let r1, r2 are the radii and h₁, h₂ are their heights, then

Now, volume of first cylinder = πr₁²h₁
and Volume of second cylinder = πr₂²h₂
Their volumes are equal

Question 18.
The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.
Solution:
Height of cylinder (A) = 10.5 m.
Let r be the radius, then
Sum of areas of two circular faces = 2π²
and curved surface area = 2πrh
According to the condition,

Question 19.
How many cubic metres of earth must be dug-out to sink a well 21 m deep and 6 m diameter ?
Solution:
Diameter of well = 6 m

Question 20.
The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk.
Solution:
Circumference of cylindrical trunk = 176 cm

Question 21.
A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad, what is the height of the platform so formed ?
Solution:
Diameter of the well = 7 m
∴ Radius (r) = 72 m
Depth (h) = 20 m
= Volume of earth dug out = πr²h

Question 22.
A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 14 m

Question 23.
A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.
Solution:
Diameter of the base of a cylindrical container = 56 cm

Dimensions of the rectangular solid = 32 cm x 22 cm x 14 cm
∴ Volume of solid = 32 x 22 x 14 cm³ = 9856 cm³
∴Volume of water rose up = 9856 cm³
Let h be the height of water, then πr²h = 9856

Question 24.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.
Solution:
Size of paper = 30 cm and 18 cm.

(i) By rolling length wise,
The circumference of base = 30 cm

Question 25.
The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what is the rise of water level in the tank due to it.
Solution:
Length of roof (l) = 18 m
Breadth (b) = 16.5 m
Height of water on the roof = 10 cm
∴ Volume of water collected

Question 26.
A Piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawnout into a wire of diameter 1 mm. What will be the length of the wire so formed ?
Solution:
Diameter of ductile metal = 1 cm
and length (h) = 5 cm
and radius (r) = 12 cm

Question 27.
Find the length of 13.2 kg of copper wire of diameter 4 mm when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Weight of wire = 13.2 kg
Diameter of wire = 4 mm

Question 28.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Volume of brass = 2.2 cu.dm
= 2.2 x 10 x 10 x 10 = 2200 cm²
Diameter of wire = 0.25 cm

Question 29.
The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.
Solution:
Let r and R be the radii of inner and outer surfaces of a cylindrical tube,

Question 30.
Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes.
Solution:
Diameter of pipe = 2 cm
∴ Radius (r) = 22 = 1 cm = 0.01 m
Length of flow of water in 1 second = 6 m Length of flow in 30 minutes = 6 x 30 x 60 m = 10800 m
∴ Volume of water = πr²h

Question 31.
A cylindrical tube, open at both ends, is to made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal tube is 8 mm everywhere. Calculate the volume of the metal.
Solution:
Length of metal tube (h) = 25 cm
Internal diameter = 10.4 cm
∴ Internal radius (r) = 10.42 = 5.2 cm
Thickness of metal = 8 mm = 0.8 cm
∴ Outer radius = 5.2 + 0.8 = 6 cm
Now volume of the metal

Question 32.
From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.
Solution:
Inner radius of pipe (r) = 0.75 cm
Rate of water flow = 7 m per second
∴ Length of water flow in 1 hour (h)
= 7 x 3600 m = 25200 m
∴ Volume of water in 1 hour

Question 33.
A cylindrical water tank of diameter 1. 4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled ?
Solution:
Diameter of cylindrical tank = 1.4 m
∴ Radius (r) = 1.42 = 0.7 m
Height (h) = 2.1m.

Question 34.
A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways Le. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinder thus formed.
Solution:
In first case,
By rolling the paper along its length,
the circumference of the base = 30 cm
and height (h) = 18 cm

Note: See Q.No. 24 this exercise

Question 35.
How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec ?
Solution:
Speed of water = 30 cm/sec
∴ Water flow in 1 minute = 30 cm x 60 = 1800 cm
Area of cross-section = 5 cm²
∴ Volume of water = 1800 x 5 = 9000 cm³
Capacity of water in litres = 9000 x l ml

Question 36.
A solid cylinder has a total surface area of 231 cm². Its curved surface area is 23 of the total surface area. Find the volume of the cylinder.
Solution:
Total surface area of a solid cylinder = 231 cm²
Curved surface area = 23 of 231 cm² = 2 x 77 = 154 cm²
and area of two circular faces = 231-154 = 77 cm²
Let r be the radius, then

Question 37.
Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre.
Solution:
Diameter of well = 3 m
∴ Radius (r) = 32 m
and depth (h) = 280 m
(i) Volume of earth dug out = πr²h

Rate of sinking the well = Rs 3.60 per m³
∴ Total cost of sinking = Rs 1980 x 3.60 = Rs 7,128
(ii) Inner curved surface area = 2πrh

Rate of cementing = Rs 2.50 per m²
∴ Total cost of cementing = Rs 2.50 x 2640
= Rs 6,600

Question 38.
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.
Solution:
Note : See Q.No. 27 of this exercise

Question 39.
2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.
Solution:
Note: See Q.No. 28 of this exercise.

Question 40.
A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 10 m

Question 41.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Width of roller (h) = 63 cm.

Outer circumference of roller = 440 cm

Question 42.
What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm ?
Solution:
Length of hollow cylinder (h) = 16 cm
External diameter = 20 cm
∴ External radius (R) = 202 = 10 cm
Thickness of iron = 2.5 mm
∴ Internal radius (r) = 10 – 0.25 = 9.75 cm
∴ Volume of iron = πh (R² – r²)

Question 43.
In the middle of rectangular field measuring 30 m x 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Diameter of well = 7 m
∴ Radius (r) = 72 m
Depth (h) = 10 m.
∴ Volume of earth dug out = πr²h

Exercise 21.1

Question 1.
Find the volume of a cuboid whose
(i) length = 12 cm, breadth = 8 cm, height = 6 cm
(ii) length = 1.2 m, breadth = 30 cm, height = 15 cm
(iii) length = 15 cm, breadth = 2.5 dm, height = 8 cm

Solution:
In a cuboid,
(i) Length (l) = 12 cm
Breadth (b) = 8 cm
Height (h) = 6 cm
∴ Volume = Ibh = 12 x 8 x 6 cm³ = 576 cm³
(ii) Length (l) = 1.2 m = 120 cm
breadth (6) = 30 cm
Height (h) = 15 cm
∴ Volume = Ibh = 120 x 30 x 15 cm³ = 54000 cm³
(iii) Length (l) = 15 cm
Breadth (b) = 2.5 dm = 25 cm
Height (h) = 8 cm
∴ Volume = Ibh
= 15 x 25 x 8 cm³ = 3000 cm²

Question 2.
Find the volume of the cube whose side is
(i) 4 cm
(ii) 8 cm
(iii) 1.5 dm
(iv) 1.2 m
(v) 25 mm.
Solution:
(i) Side of a cube (a) = 4 cm
∴ Volume = a3 = (4)³ cm³ = 4 x 4 x 4 = 64 cm³

(ii) Side of cube (a) = 8 cm
∴ Volume = a3 = (8)³ 4 cm
= 8 x 8 x 8 cm³ = 512 cm³
(iii) Side of cube (a) = 1.5 dm = 15 cm
∴ Volume = a³ = (1.5)3 dm² = (15)³ cm³
= 15 x 15 x 15 = 3375 cm³
(iv) Side of cube (a) = 1.2 m = 120 cm
∴ Volume = a³ = (120)³ cm³
= 120 x 120 x 120 = 1728000 cm³
(v) Side of cube (a) = 25 mm = 2.5 cm.
∴ Volume = a³ = (2.5)³ cm³
= 2.5 x 2.5 x 2.5 cm³ = 15.625 cm³

Question 3.
Find the height of a cuboid of volume 100 cm3 whose length and breadth are 5 cm and 4 cm respectively.
Solution:
Volume of a cuboid =100 cm³
Length (1) = 5 cm
and breadth (b) = 4 cm

Question 4.
A cuboidal vessel is 10 cm long and 5 cm wide, how high it must be made to hold 300 cm3 of a liquid ?
Solution:
Volume of the liquid in the vessel = 300 cm³
Length (l)= 10 cm
Breadth (b) = 5 cm

Question 5.
A milk container is 8 cm long and 50 cm wide. What should be its height so that it can hold 4 litres of milk ?
Solution:
Capacity of milk = 4 litres
∴ Volume of the container = 4 x 1000 cm³ = 4000 cm³
Length (l) = 8 cm
Width (b) = 50 cm

Question 6.
A cuboidal wooden block contains 36 cm³ wood. If it be 4 cm long and 3 cm wide, find its height.
Solution:
Volume of wooden cuboid block = 36 cm³
Length (l) = 4 cm
Breadth (b) = 3 cm

Question 7.
What will happen to the volume of a cube, if its edge is (i) halved (ii) trebled ?
Solution:
Let side of original cube = a cm
∴ Volume = a³ cm³
(i) In first case,

(ii) In second case, when side (edge) is trebled, then side = 3a
∴ Volume = (3a)³ = 27a³
∴ It will be 27 times

Question 8.
What will happen to the volume of a cuboid if its (i) Length is doubled, height is same and breadth is halved ? (ii) Length is doubled, height is doubled and breadth is same ?
Solution:
Let l, b and h be the length, breadth and height of the given cuboid respectively.
∴ Volume = lbh.
(i) Length is doubled = 21

∴ The volume will be the same.
(ii) Length is doubled = 21
breadth is same = b height is doubled = 2h
∴ Volume = 2l x b x 2h = 4 lbh
∴ Volume will be 4 times

Question 9.
Three cuboids of dimensions 5 cm x 6 cm x 7 cm, 4 cm x 7 cm * 8 cm and 2 cm x 3 cm x 13 cm are melted and a cube is made. Find the side of cube.
Solution:
Dimensions of first cuboid = 5 cm x 6 cm x 7 cm
∴ Volume = 5 x 6 x 7 = 210 cm³
Dimensions of second cuboid = 4 cm x 7 cm x 8 cm
∴ Volume = 4x 7 x 8 = 224 cm³
Dimensions of third cuboid = 2 cm x 3 cm x 13 cm
∴ Volume = 2 x 3 x 13 = 78 cm³
Total volume of three cubes = 210 + 224 + 78 cm3 = 512 cm³
∴ Volume of cube = 512 cm³

Question 10.
Find the weight of solid rectangular iron piece of size 50 cm x 40 cm x 10 cm, if 1 cm3 of iron weighs 8 gm.
Solution:
Dimension of cuboidal iron piece = 50 cm x 40 cm x 10 cm
∴ Volume = 50 x 40 x 10 = 20000 cm³
Weight of 1 cm³ = 8 gm
∴ Total weight of piece = 20000 x 8 gm

Question 11.
How many wooden cubical blocks of side 25 cm can be cut from a log of wood of size 3 m by 75 cm by 50 cm, assuming that there is no wastage ?
Solution:
Length of log (l) = 3 m = 300 cm.
Breadth (b) = 75 cm
and height (h) = 50 cm
∴ Volume of log = lbh = 300 x 75 x 50 cm3 = 1125000 cm³
Side of cubical block = 25 cm
∴ Volume of one block = a² = 25 x 25 x 25 cm3 = 15625 cm³
∴ Number of blocks to be cut out

Question 12.
A cuboidal block of silver is 9 cm long, 4 cm broad and 3.5 cm in height. From it, beads of volume 1.5 cm² each are to be made. Find the number of beads that can be made from the block ?
Solution:
Length of block (l) = 9 cm
Breadth (b) = 4 cm
and height (h) = 3.5 cm
∴ Volume = l x b x h = 9 x 4 x 3.5 cm³ = 126 cm³
Volume of one bead = 1.5 cm³
∴ Number of beads = 126105 = 84

Question 13.
Find the number of cuboidal boxes measuring 2 cm by 3 cm by 10 cm which can be stored in a carton whose dimensions are 40 cm, 36 cm and 24 cm.
Solution:
Length of cuboidal box (l) = 2 cm
breadth (b) = 3 cm
and height (h) = 10 cm
∴ Volume = lx b x h = 2 x 3 x 10 = 60 cm³
Volume of carton = 40 x 36 x 24 cm³
= 34560 cm³
∴ Number of boxes to be height in the carton

Question 14.
A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from the block ? Assume cutting causes no wastage.
Solution:
Dimensions of block = 50 cm, 45 cm, 34 cm
∴ Volume = 50 x 45 x 34 = 76500 cm³
Size of cuboid = 5 cm x 3 cm x 2 cm
∴ Volume of cuboid =  5 x 3 x 2 = 30 cm³

Question 15.
A cube A has side thrice as long as that of cube B ? What is the ratio of the volume of cube A to that of cube B ?
Solution:
Let side of cube B = a
Then Volume = a³
and side of cube A = 3a
Volume = (3a)³ = 3a x 3a x 2a = 27a³
∴ Ratio of volume’s A and B = 27a³ : a³
= 27 : 1

Question 16.
An ice-cream brick measures 20 cm by 10 cm by 7 cm. How many such bricks can be stored in a deep fridge whose inner dimensions are 100 cm by 50 cm by 42 cm ?
Solution:
Dimensions of ice cream brick = 20 cm x 10 cm x 7 cm
∴ Volume = 20 x 10 x 7 cm³ = 1400 cm³
Dimensions of inner of fridge = 100 cm x 50 cm x 42 cm = 210000 cm³
∴ Number of bricks to be kept in the fridge

Question 17.
Suppose that there are two cubes, having edges 2 cm and 4 cm, respectively. Find the volume V₁ and V₂ of the cubes and compare them.
Solution:
Side of first cube (a) = 2 cm
∴ Volume (V₁) = a³ = (2) = 8 cm³
Similarly side of second cube = 4 cm
and volume (V₂) = (4)³ = 64 cm³
Now V₂ = 64 cm³ = 8 x 8 cm³
= 8 x V₁
⇒ V₂ = 8V₁

Question 18.
A tea-packet measures 10 cm x 6 cm x 4 cm.How many such tea-packets can be placed in a cardboard box of dimensions 50 cm x 30 cm x 0.2 m ?
Solution:
Dimensions of tea-packet = 10cm x 6cm x 4 cm
∴ Volume =10 x 6 x 4 = 240 cm³
Dimensions of box = 50 cm x 30 cm x 0.2 m
= 50 cm x30 cm x20 cm
∴ Volume = 50 x 30 x 20 = 30000 cm³
∴ Number of tea-packets to be kept = 30000240

Question 19.
The weight of a metal block of size 5 cm by 4 cm by 3 cm is 1 kg. Find the weight of a block of the same metal of size 15 cm by 8 cm by 3 cm.
Solution:
Dimensions of a metal block = 5 cm x 4 cm x 3 cm = 5 x 4 x 3 = 60 cm³
Dimensions of a second block = 15 cm x 8 cm x 3 cm = 15 x 8 x 3 = 360 cm³
But weight of first block = 1 kg
∴ Weight of second block
= 116 x 360 = 6 kg

Question 20.
How many soap cakes can be placed in a box of size 56 cm x 0.4 m x 0.25 m, it the size of soap cake is 7 cm x 5 cm x 2.5 cm ?
Solution:
Size of box = 56 cm x 0.4 m x 0.25 m = 56 cm x 40 cm x 25 cm
∴ Volume = 56 x 40 x 25 cm³ = 56000 cm³
Size of a soap cake = 7 cm x 5 cm x 2.5 cm
∴ Volume = 7 x 5 x 2.5 cm³ = 87.5 cm³
∴ Number of cakes to be kept in the box
= 5600087.5 = 640

Question 21.
The volume of a cuboid box is 48 cm3. If its height and length are 3 cm and 4 cm respectively, find its breadth.
Solution:
Volume of cuboid box = 48 cm³
Length (l) = 4 cm
Height = (h) = 3 cm

Exercise 21.2

Question 1.
Find the volume in cubic metres (cu.m) of each of the cuboids whose dimensions are :
(i) length = 12 cm, breadth = 10 m, height = 4.5 m
(ii) length = 4 m, breadth = 2.5 m, height = 50 cm
(iii) length = 10 m, breadth = 25 dm, height = 25 cm.
Solution:
(i) Length of cuboid (l) = 12 m
Breadth (b) = 10m
and height (h) = 4.5 m
∴Volume = l x b x h = 12 x 10 x 4.5 m³
= 540 m³
(ii) Length of cuboid (l) = 4 m
Breadth (b) = 2.5m
Height (h) = 50 cm = 0.5 m
∴ Volume = l x b x h = 4 x 2.5 x 0.5 = 5 m³
(iii) Length of cuboid (l) = 10 m
Breadth (b) = 25 dm = 2.5 m
Height (h) = 25 cm 0.25 m
∴ Volume = l x b x h = 10 x 2.5 x 0.25 m³ = 6.25 m³

Question 2.
Find the volume in cubic decimetre of each of the cubes whose side is
(i) 1.5 m
(it) 75 cm
(iii) 2 dm 5 cm
Solution:
(i) Side of cube (a) = 1.5 m
∴ Volume = a³ = (1.5)³ m³
= 1.5 x 1.5 x 1.5 m³ = 3.375 m³
= 3.375 x 1000 = 3375 dm³
(ii) Side of cube (a) = 75 cm = 7.5 dm
∴ Volume = a³ = (7.5)3 dm³
= 421.875 dm³
(iii) Side of cube (a) = 2 dm 5 cm = 2.5 dm
∴ Volume = (a)³ = (2.5)³ dm³
= 15.625 dm³

Question 3.
How much clay is dug out in digging a well measuring 3m by 2m by 5m?
Solution:
Length of well (l) = 3m
breadth (b) = 2 m
and height (depth) (h) = 5 m
Volume of earth dug out = l x b x h = 3 x 2 x 5 = 30m³

Question 4.
What will be the height of a cuboid of volume 168 m³, if the area of its base is 28 m² ?
Solution:
Volume of a cuboid = 168 m³
Area of its base l.e., l x b = 28 m³

Question 5.
A tank is 8 m long, 6 m broad and 2 m high. How much water can it contain ?
Solution:
Length of tank (l) = 8 m
Breadth (b) = 6 m
Height (h) = 2 m
∴ Volume of water in the tank = l x b x h = 8 x 6 x 2 = 96 m³
= 96 x 1000 = 96000litres (∵1m³ = 1000litre)

Question 6.
The capacity of a certain cuboidal tank is 50000 litres of water. Find the breadth of the tank if its height and length are 10 m and 2.5 m respectively.
Solution:
Capacity of water in the tank = 50000 litres
∴ Volume of water = 50000 x 11000 = 50 m³ (1000 l = 1 m³)
Height of tank (h)= 10 m
and length (l) = 2.5 m
Volume 50

Question 7.
A rectangular diesel tanker is 2 m long, 2 m wide and 40 cm deep. How many litres of diesel can it hold ?
Solution:
Length of tanker (l) = 2 m
Breadth (b) = 2m
Depth (h) = 40 cm = 0.4 m
∴ Volume = l x bx h = 2 x 2 x 0.4=1.6m³
Quantity of diesel = 1.6 x 1000 litres (1 m³= 1000 l)
= 1600 litres

Question 8.
The length, breadth and height of a room are 5 m, 4.5 m and 3 m, respectively. Find the volume of the air it contains.
Solution:
Length of room (l) = 5 m
Breadth (6) = 4.5 m
and height (h) = 3 m
∴ Volume of air it contains
= l x b x h = 5 x 4.5 x 3 m³
= 67.5 m³

Question 9.
A water tank is 3 m long, 2 m broad and 1 m deep. How many litres of water can it hold ?
Solution:
Length of tank (l) = 3 m
Breadth (b) = 2 m
and depth (h) = 1 m
∴ Volume of tank = l x b x h
= 3 x 2 x 1 = 6 m³
∴ Quantity of water it can contains
= 6 x 1000 litres = 6000 litres (1 m³= 1000 litres)

Question 10.
How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick ?
Solution:
Length of wooden block (l) = 6 m
Width (b) = 75 cm = 0.75 m
Thickness (h) = 45 cm = 0.45 m
∴ Volume = l x b x h = 6 x 0.75 x 0.45 m³
Length of plank (l) = 3 m
Breadth (b) = 15 cm = 0.15 m
Thickness (h) = 5 cm = 0.05 m
∴ Volume = 3 x 0.15 x 0.05 m³
Number of planks

Question 11.
How many bricks each of size 25 cm x 10 cm x 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick assuming that the volume of sand and cement used in the construction is negligible ?
Solution:
Size of one brick = 25 cm x 10 cm x 8 cm
∴ Volume of one brick = 25 x 10 x 8 cm³

Length of wall (l) = 5 m
Width (b) = 0.16 m
Height (h) = 3 m
∴ Volume of wall = l x b x h
= 5 x 0.16 x 3 m³ = 2.4 m³
∴ Number of bricks required

Question 12.
A village, having a population of 4000 requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days the water of this tank will last ?
Solution:
Total population of a village = 4000
Water required for each person for one day = 150 litres
∴ Water required for 4000 persons for one day = 150 x 4000 = 600000 litres
Length of tank (l) = 20 m
Breadth (b) = 15 m
Height (h) = 6 m
∴ Volume of tank = l x b x h = 20 x 15 x 6 m³ = 1800 m³
Capacity of water in the tank = 1800 x 1000 l= 1800000l (1 m³ = 1000 l)
∴ Number of days, the water will last

Question 13.
A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m x 8 m x 6 m is dug outside the field and the earth dugout from this well is spread evenly on the field. How much will the earth level rise ?
Solution:
Length of well (l) = 14 m
Breadth (A) = 8m
Depth (A) = 6m
∴ Volume of earth dugout = l x bx h
= 14 x 8 x 6 = 672 m³ Length of field = 70 m
and breadth = 60 m
Let h be the height of earth spread over
Then 70 x 60 x h = 672
⇒ h = 67270×60 = 0.16m
∴ Height of earth = 0.16 m = 16 cm

Question 14.
A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water.
Solution:
Volume of water = 3250 m³
Length of pool (l) = 250 m
Breadth (b)= 130 m
∴ Height of water level

Question 15.
A beam 5 m long and 40 cm wide contains 0.6 cubic metres of wood. How thick is the beam?
Solution:
Volume of wood of the beam = 0.6 m³ = 600000
Length of beam (l) = 5 m = 500 cm
Breadth (b) = 40 cm

Question 16.
The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day ?
Solution:
Area of the field = 3 hectares
= 3 x 10000 square metres
= 30000 square metres
Height of rainfall = 6 cm = m³

Question 17.
An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge.
Solution:
Length of cuboidal beam (l) = 8 m = 800 cm
Number of cubical sliced = 4000
Edge of each cube = 1 cm
Volume of beam = 4000 (1)³ cm³ = 4000 cm³
One edge of the beam = 0.5 m = 50 cm.

Question 18.
The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed ?
Solution:
Dimensions of metal block = 2.25 m x 1.5 m x 27 cm
∴ Volume = 2.25 x 1.5 x 0.27 m³
= 225 x 150 x 27 cm³ = 911250 cm³
Side of each cube (a) = 45 cm
∴ Volume of one cube = a³ = (45)³ cm³ = 91125 cm³
∴ Number of cubes = 91125091125 = 10

Question 19.
A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece if 1 cm3 of iron weighs 8 gm.
Solution:
Dimensions of a piece of rectangular iron = 6m x 6cm x 2cm
∴ Volume = 600 x 6 x 2 cm³ = 7200 cm³
Weight of 1 cm³ = 8 gm
∴ Total weight of the piece = 7200 x 8 gm
= 57600 gm = 576001000 kg = 57.6 kg

Question 20.
Fill in the blanks in each of the following so as to make the statement true :
(i) 1 m³ = ……… cm³
(ii) 1 litre = …….. cubic decimetre
(iii) 1 kl = …… m³
(iv) The volume of a cube of side 8 cm is …….. .
(v) The volume of wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is …….. cm
(vi) 1 cu.dm = …….. cu.mm
(vii) 1 cu.km = ……cu.m
(viii) 1 litre =…….. cu.cm
(ix) 1 ml = ……… cu.cm
(x) 1 kl = ……… cu.dm = ……. cu.cm.
Solution:
(i) 1 m³ = 1000000 or 10⁶ cm³
(ii) 1 litre = 1 cubic decimetre
(iii) 1 kl = 1 m³
(iv) The volume of a cube of side 8 cm is 512 cm³ (V = a³ = 8 x 8 x 8 = 512 cm³)
(v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm³. The height of the cuboid is 50 cm

(vi) 1 cu.dm = 1000000 cu mm = 10⁶ cu.mm
(vii) 1 cu.km = 1000 x 1000 x 1000 cu.m = 109 cu.m
(viii) 1 litre = 1000 cu.cm = 10³ cu.cm
(ix) 1 ml = 1 cu.cm
(x) 1 kl = 1000 cu.dm = 100 x 100 x 100 cu.cm = 10⁶ cu.cm

Exercise 21.3

Question 1.
Find the surface area of a cuboid whose :
(i) length = 10 cm, breadth = 12 cm and height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2 m, breadth = 4 m and height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.
Solution:
(i) Length of cuboid (l) = 10 cm
Breadth (b) = 12 cm
Height (h) = 14 cm
∴ Surface area = 2(1 × b + b × h + h × l)
= 2(10 x 12 + 12 x 14 + 14 x 10) cm²
= 2(120+ 168 + 140) cm²
= 2 x 428 = 856 cm²
(ii) Length of cuboid (l) = 6 dm
Breadth (b) = 8 dm
Height (h) = 10 dm
∴ Surface area = 2 ( l × b + b x h + h× l)
= 2(6 x 8 + 8 x 10 + 10 x 6) dm²
= 2(48 + 80 + 60) dm² = 2 x 188 = 376 dm²
(iii) Length of cuboid (l) = 2 m
Breadth (b) = 4 m
Height (h) = 5 m
∴ Surface area = 2(l × b + b × h + h × l)
= 2(2 x 4 + 4 x 5 + 5 x 2) m2
= 2(8 + 20 + 10) m2 = 76 m2
(iv) Length of cuboid (l) = 3.2 m = 32 dm
Breadth (b) = 30 dm
Height (h) = 250 cm = 25 dm
∴ Surface area = 2(1 x b + b x h + h x l)
= 2(32 x 30 + 30 x 25 + 25 x 32) dm²
= 2(960 + 750 + 800) dm²
= 2 x 2510 = 5020 dm²

Question 2.
Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1m
Solution:
(i) Edge of the cube (a) = 1.2 m
∴ Surface area = 6a²= 6 x (1,2)² m²
= 6 x 1.44 = 8.64 m²
(ii) Edge of cube (a) = 27 cm
∴ Surface area = 6a² = 6 x (27)² m²
= 6 x 729 = 4374 m²
(iii) Edge of cube (a) = 3 cm
Surface area = 6a² = 6 x (3)² m²
= 6×9 cm² = 54 cm²
(iv) Edge of cube (a) = 6 m
∴ Surface area = 6a² = 6 x (6)2 m²
= 6 x 6 x 6 = 216 m²
(v) Edge of the cube (a) = 2.1 m
∴ Surface area = 6a2 = 6 x (2.1)2 m²
= 6 x 4.41 = 26.46 m²

Question 3.
A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.
Solution:
Length of cuboid box (l) = 5 cm
Breadth (b) = 5 cm
and height (h) = 4 cm
∴ Surface area = 2 (l x b + b x h + h x l)
= 2 (5 x 5 + 5 x 4 + 4 x 5) cm²
= 2 (25 + 20 + 20)
= 2 x 65 cm²
= 130 cm²

Question 4.
Find the surface area of a cube whose volume is :
(i) 343 m³
(ii) 216 dm³.
Solution:
(i) Volume of a cube = 343 m³

Question 5.
Find the volume of a cube whose surface area is
(i) 96 cm²
(ii) 150 m².
Solution:
(i) Surface area of a cube = 96 cm²

Question 6.
The dimensions of a cuboid are in the ratio 5:3:1 and its total surface area is 414 m². Find the dimensions.
Solution:
Ratio in .dimensions = 5 : 3 : 1
Let length (l) = 5x
breadth (b) = 3x
and height (h) = x
∴ Surface area = 2(1 x b + b x h + h x l)
= 2(5x x 3x + 3x x x + x x 5x)
= 2(15×2 + 3×2 + 5×2) = 2 x 23×2 = 46×2

Question 7.
Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.
Solution:
Length of cardboard (l) = 25 cm
Breadth (b) = 0.5 m = 50 cm
Height (h)= 15 cm.
∴ Surface area of cardboard = 2 (l x b + b x h + h x l)
= 2(25 x 50 + 50 x 15 + 15 x 25) cm²
= 2(1250+ 750+ 375) cm²
= 2(2375)
= 4750 cm²

Question 8.
Find the surface area of a wooden box whose shape is of a cube and if the edge of the box is 12 cm.
Solution:
Edge of cubic wooden box = 12 cm
∴ Surface area = 6a² = 6(12)² cm²
= 6 x 144 = 864 cm²

Question 9.
The dimensions of an oil tin are 26 cm x 26 cm x 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs 10, find the cost of the tin sheet used for these 20 tins.
Solution:
Length of tin (l) = 26 cm = 0.26 m
Breadth (b) = 26 cm = 0.26 m
Height (h) = 45 cm = 0.45 m
∴ Surface area = 2(l x b + b x h +h xl)
= 2(0.26 x 0.26 + 0.26 x 0.45 + 0.45 x 0.26) m²
= 2(0.0676 + 0.117 + 0.117) m²
= 2(0.3016) = 0.6032 m²
Sheet required for such 20 tins
= 0.6032 x 20= 12.064 m²
Cost of 1 m² tin sheet = 10 m
∴ Total cost = Rs 12.064 x 10 = Rs 120.64
and area of sheet = 12.064 m² = 120640 cm²

Question 10.
A classroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows etc.)
Solution:
Length of room (l) = 11 m
Width (b) = 8 m
and height (h) = 5 m
Area of floor = l x b = 11 x8 = 88m2
Area of four walls = 2 (l + b) x h
= 2(11 + 8) x 5 m² = 2 x 19×5 = 190 m²
∴ Total area = 88 m² + 190 m² = 278 m²

Question 11.
A swimming pool is 20 m long, 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs 25 per square metre.
Solution:
Length of pool (l) = 20 m
Breadth (b) = 15 m
and Depth (h) = 3 m.
Area of floor = l x b = 20 x 15 = 300 m²
and area of its walls = 2(l + b) x h
= 2(20 + 15) x 3 = 2 x 35 x 3 m² = 210 m²
∴ Total area = 300 + 210 = 510 m²
Rate of repairing it = Rs 25 per sq. me

Question 12.
The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room.
Solution:
Perimeter of floor = 30 m
i.e. 2(1 + b) = 30 m
Height = 3 m
∴ Area of four walls = Perimeter x height = 30 x 3 = 90 m²

Question 13.
Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.
Solution:
Let length of the room = l
and breadth = b
and height = h
Volume = l x b x h
Area of floor = l x b = lb.
Area of two adjacent walls = hl x bh.
∴ Product of areas of floor and two adjacent walls of the room = lb (hi x bh)
= l²b²h² = (l.b.h)² = (Volume)²
Hence proved

Question 14.
The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5, 3m and 350 cm, respectively. Find the cost of plastering at the rate of Rs 8 per square metre.
Solution:
Length of room (l) = 4.5 m
Width (b) = 3 m
and height (h) = 350 cm = 3.5 m
∴ Area of walls = 2(l + b) x h
= 2(4.5 + 3) x 3.5 m² = 2 x 7.5 x 3.5 m² = 52.5 m²
Area of ceiling = l x b = 4.5 x 3 = 13.5 m²
∴ Total area = 52.5 + 13.5 m² = 66 m²
Rate of plastering = Rs 8 per sq. m
∴ Total cost = Rs 8 x 66 = Rs 528

Question 15.
A cuboid has total surface area of 50 m² and lateral surface area its 30 m². Find the area of its base.
Solution:
Total surface area of cuboid = 50 m²
Lateral surface area = 30 m²
∴ Area of floor and ceiling = 50 – 30 = 20 m²
But area of floor = area of ceiling
∴ Area of base (floor) = 202 = 10 m²

Question 16.
A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m². What is the cost of white-washing the walls at the rate of Rs 1.50 per m².
Solution:
Length of room (l) = 7 m
Breadth (b) = 6 m
and height (h) = 3.5 m
∴ Area of four walls = 2(1 + b) x h
= 2(7 + 6) x 3.5 m² = 2 x 13 x 3.5 = 91 m²
Area of doors and windows = 17 m2
∴ Remaining area of walls = 91 – 17 = 74 m²
Rate of whitewashing = Rs 1.50 per m²
∴ Total cost = 74 x Rs 1.50 = Rs 111

Question 17.
The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m x 1.5 m and 10 windows each of size 1.5 m x l m. If the cost of the white-washing the walls of the hall at the rate of Rs 1.20 per m² is Rs 2385.60, find the breadth of the hall.
Solution:
Length of hall (l) = 80 m
Height (h) = 8 m
Size of each door = 3 m x 1.5 m
∴ Area of 10 doors = 3 x 1,5 x 10 m²
= 45 m²
A size of each windows = 1.5 m x 1 m
∴ Area of 10 windows = 1.5 m x 1 x 10= 15 m²
Total cost of whitewashing the walls = Rs 2385.60
Rate of whitewashing = Rs 1.20 per m²
∴ Area of walls which are whitewashed

Exercise 21.4

Question 1.
Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.
Solution:
Length of room (l) = 12m
Breadth (b) = 9 m
Height (h) = 8 m

Longest rod to be kept in the room

Question 2.
If V is the volume of the cuboid of dimensions a, b, c and S its the surface area then prove that

Solution:
∵ a, b, c are the dimensions of a cuboid
∴ Volume (V) = abc
Surface area (S) = 2(ab + bc + ca)
Now

Question 3.
The areas of three adjacent faces of a cuboid are .v, y and z. If the volume is V¹ prove that V² = xyz.
Solution:
Let length of cuboid = l
Breadth = b
and height = h
Volume = Ibh
∴ x = lb,y = bh and z = hl
Now x.y.z = lb.bh.hl
= l² b² h² = (Ibh)² = V²
∴ V² = xyz Hence proved

Question 4.
A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.
Solution:
Volume of the water in reservoir = 105 m²
Length (l)= 12 m
and breadth (b) = 3.5 m

Question 5.
Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.
Solution:
Edge of cube A = 18 cm
∴ Volume = a² = (18)3 cm³ = 5832 cm³
Edge of cube B = 24 cm
∴ Volume = (24)³ = 13824 cm³
Edge of cube C = 30 cm
∴Volume = (30)³ = 27000 cm³
Volume of A, B, C cubes
= 5832+ 138-24+ 27000 = 46656 cm³
Volume of cube D = 46656 cm³

Question 6.
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu.dm. Find its dimensions.
Solution:
Volume of room = 512 cu.dm
Let height of the room (h) = x
Then breadth (b) = 2x
and length (l) = 2x x 2 = 4x.
∴ Volume = l x b x h = 4x x 2x x x = 8×3

Question 7.
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.
Solution:
Length of iron tank (l) = 12 m
Breadth (b) = 9 m
Depth (h) = 4 m
∴ Surface area of the tank = 2(l x b + b x h + h x l)
= 2(12 x 9 + 9 x 4 + 4 x 12) m²
= 2(108 + 36 + 48) = 2 x 192 m²
= 384 m²
Width of sheet used = 2 m

Question 8.
A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m x 8 m x 6 m, find the cost of iron sheet at Rs 17.50 per metre.
Solution:
Dimensions of the open iron tank = 12mx 8m.x 6m
∴ Surface area (without top)
= 2(1 x b) x h + lb
= 2(12 + 8) x 6+12 x 8m²
= 2 x 20 x 6 + 96 = 240 + 96 m² = 336 m²
Width of sheet used = 4 m
∴ Length of sheet = Areab = 3364 m = 84 m b 4
Rate of sheet = Rs 17.50 per m.
∴ Total cost = Rs 17.50 x 84 = Rs 1470

Question 9.
Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let edge of each equal cubes = x
Then, surface area of one cube = 6x²

and surface area of three cubes = 3 x 6x² = 18x²
By placing the cubes in a row,
The length of newly formed cuboid (l) = 3x
Breadth (b) = x
and height (h) = x
∴ Surface area of the cuboid so formed

Question 10.
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre.
Solution:
Dimensions of a room = 12.5 m x 9 m x 7 m
∴ Total surface area of the walls = 2(1 + b) x h = 2(12.5 + 9) x 7 m²
= 2 x 21.5 x 7 = 301 0 m²
Area of 2 doors = 2 x (2.5 x 1.2) m² = 2 x 3.00 = 6 m²
Area of 4 windows = 4 x (1.5 x 1) m²
4 x 1.5 = 6 m²
∴ Remaining area of the walls = 301 -(6 + 6) m²
= 301 – 12 = 289 m²
∴ Rate of painting the walls = Rs 3.50 per m²
∴ Total cost = Rs 3.50 x 289 = Rs 1011.50

Question 11.
A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much the level of field is raised ?
Solution:
Length of the plot (l) = 50 m
Width (b) = 30 m
and depth (h) = 8 m
∴ Volume of the earth dug out = l x b x h = 50 x 30 x 8 = 12000 m³
Length of the field = 150 m
and breadth = 100 m
∴ Height of the earth spread out on the field

Question 12.
Two cubes, each of volume 512 cm³ are joined end to end, find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 512 cm³

Now by joining the two equal cubes of side 8 cm, the length of so formed cuboid (l)
= 2 x 8 = 16 cm
Breadth (b) = 8 cm
and height (h) = 8 cm
∴ Surface area = 2( l x b + b x h + h x l)
= 2(16 X 8 + 8 X 8 + 8X16) cm²
= 2(128 + 64 + 128) cm²
= 2 x 320 = 640 cm²

Question 13.
Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.
Solution:
Edge of first cube = 3 cm
∴ Volume = a³ = (3)³ 27 cm³
Edge of second cube = 4 cm
∴Volume = a³ = (4)³ = 64 cm³
Edge of third cube = 5 cm
∴ Volume = a³ = (5)³ = 125 cm³
Volume of three cubes together = 27 + 64+ 125 = 216 cm³
∴ Volume of the new cube = 216 cm³

Question 14.
The cost of preparing the’walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.
Solution:
Length of the room (l) = 12 m
Rate of matting the floor = 85 paise per m²
Total cost of matting = Rs 91.80

Question 15.
The length of a hall is 18 m and width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.
Solution:
Length of hall (l) = 18 m
and breadth (b) = 12 m
∴ Area of floor = l x b = 18 x 12 = 216 m²
and area of roof = 216 m²
Total area of floor and roof
= (216 + 216) m² = 432 m²
∴ Area of four walls = 432 m²
But area of 4 walls = 2(l + b) x h
∴ 2h (l + b) = 432
⇒ 2h (18 + 12) = 432
⇒ 2h x 30 = 432 432
⇒ h = 43260 = 7.2m
∴ Height of the wall = 7.2 m

Question 16.
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
Edge of metal bigger cube = 12 cm
∴ Volume = (12)³ = 1728 cm³
∴ Sum of volumes of 3 smaller cubes = 1728 cm³
Edge of first smaller cube = 6 cm
∴ Volume = (6)³ = 216 cm³
Edge of second smaller cube = 8 cm
∴ Volume = (8)³ = 512 cm³
Sum of volumes of two smaller cubes = 216+ 512 = 728 cm³
∴ Volume of third smaller cube = 1728-728 cm³ = 1000 cm³

Question 17.
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall if each person requires 150 m3 of air ?
Solution:
Length of cinema hall (l) = 100 m
Breadth (b) = 50 m
and height (h) = 18 m
∴ Volume of air of the hall = l x b x h
= 100 x 50 x 18 m³
= 90000 m³
Each person requires air = 150 m³
∴ Number of persons = 90000150= 600

Question 18.
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box ?
Solution:
Outer dimensions of a closed wooden box = 48 cm x 36 cm x 30 cm
Thickness of wood = 1.5 cm.
∴ Inner length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm
Breadth(b) = 36-2 x 1.5 = 36-3 = 33 cm
Height (h) = 30 – 2 x 1.5 = 30 – 3 = 27 cm
∴ Volume of inner box = l x b x h = 45 x 33 x 27 cm³ = 40095 cm³
Volume of one brick of size 6 cm x 3 cm x 0.75 cm
= 6 x 3 x 0.75 = 6 x 3 x 34 cm³ = 272 cm³
∴ Number of bricks = 40095×227
= 1485 x 2 = 2970 bricks

Question 19.
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs 1,248. Find the dimensions of the box.
Solution:
Ratio in the dimensions of a box =2:3:4
Difference in total cost = Rs 1,248
Difference in rates = Rs 9.50 – Rs 8 = Rs 1.50
Let length (l) = 2x
Then breadth (b) = 3x
and height (h) = 4x
∴ Surface area = 2 (l x b + b x h + h x l)
= 2(2x 3x  + 3x x 4x + 4x x 2x)
= 2(6x² + 12x² + 8 x²) = 2 x 26x² = 52x²
First rate of paper = Rs 9.50 per m²
and second rate = 8.00 per m²
∴ First cost = Rs 52x² x 9.50
and second cost = Rs 52x² x 8
∴ 52x² x 9.50 – 52x² x 8= 1248
⇒ 52x² (9.50 – 8) = 1248
⇒ 52x²(1.50) = 1248

Exercise 20.1

Question 1.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m² ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²

Question 2.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot.

Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = b2 = 282 = 14 m
Area = 12 πr²
= 12 x 227 x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²

Question 3.
A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 227).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m

= 227 x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
breadth (b) = 15 m

Area of rectangular piece = l x b = 20 x 15 = 300 m²
Radius of each quadrant (r) = 3.5 m
Total area of 4 quadrants

Area of the remaining portion = 300 – 38.5 m² = 261.5 m²

Question 5.
The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

Solution:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m

Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = 227)

Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm

Question 7.
The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = 227)
Solution:
Diameter of the wheel (d) = 90 cm.

Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d₁) = 16 cm
Second diagonal (d₂)

Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d₁) = 7.5 cm
and diagonal (d₂) = 12 cm

Question 10.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
In quadrilateral shaped field ABCD,
diagonal AC = 24 m

and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = 12 x AC x (BL + DM)
= 12 x 24 x (13 + 8) m²
= 12 x 21 = 252 m²

Question 11.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm

Question 12.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm

Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810

Question 13.
A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m

Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50

Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d₁) = 24 cm

Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = 12 AC = 12 x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD

Question 15.
The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²

Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)

Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²

Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.

Question 18.
In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.
Solution:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m

Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m

Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960

Question 21.
A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ?
Solution:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m

Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = 12 x base x height
= 12 x bh= 12 x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d₁ = 22 cm
Second diagonal (d₂)

Exercise 20.2

Question 1.
Find the area, in square metres, of the trapezium whose bases and altitude are as under:
(i) bases = 12 dm and 20 dm, altitude =10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:

Question 2.
Find the area of trapezium with base 15 cm and height 8 cm. If the side parallel to the given base is 9 cm long.
Solution:
In the trapezium ABCD,

Question 3.
Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.
Solution:
Length of parallel sides of a trapezium are 16 dm and 22 dm i.e.
b₁ = 16 dm, b₂ = 22 dm
and height (h) = 12 dm

Question 4.
Find the height of a trapezium, the sum of lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm²
Solution:
Sum of parallel sides (b₁ + b₂) = 60 cm
Area of trapezium = 600 cm²

Question 5.
Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm.
Solution:
Area of a trapezium = 65 cm²
Bases are 13 cm and 26 cm
i.e. b₁ = 13 cm, b₂ = 26 cm.

Question 6.
Find the sum of the lengths of the bases of trapezium whose area is 4.2 m² and whose height is 280 cm.
Solution:
Area of trapezium = 4.2 m²
Height (h) = 280 cm = 2.8 m.

Question 7.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate the area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle id the sum of the areas of two triangles.
Solution:
In trapezium ABCD, parallel sides or bases are 10 cm and 15 cm and height = 6 cm
Area of trapezium

Area of trapezium = 90 – 15 = 75 cm²
= area of rectangle – areas of two triangles.

Question 8.
The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them:
Solution:
Area of trapezium = 960 cm²
Parallel sides are 34 cm and 46 cm
b₁ + b₂ = 34 + 46 = 80 cm

Distance between parallel sides = 24 cm

Question 9.
Find the area of the figure as the sum of the areas of two trapezium and a rectangle.

Solution:
In the figure,
One rectangle is ABCD whose sides are 50 cm and 10 cm.
Two trapezium of equal size in which parallel sides are 30 cm and 10 cm and height

Question 10.
Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:
Top of a table is of trapezium in shape whose parallel sides are 1 m and 1.2 m and distance between them (h) = 0.8 m
Area of trapezium = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x height

Question 11.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom, and the area of the cross-section is 72 m², determine its depth.
Solution:
Area of cross-section = 72 m²
Parallel sides of the trapezium = 10 m and 6 m

Question 12.
The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Solution:
Area of trapezium = 91 cm²
Height (h) = 7 cm.
Sum of parallel sides

One parallel side = 9 cm
and second side = 9 + 8 = 17 cm
Hence parallel sides are 17 cm, 9 cm

Question 13.
The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.
Solution:
Area of trapezium = 384 cm²
Perpendicular distance (h) = 12 cm
Sum of parallel sides

First parallel side = 8 x 3 = 24 cm
Second side = 8 x 5 = 40 cm

Question 14.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:
Area of the trapezium shaped field = 10500 m²
and perpendicular distance between them (h) = 100 m.

Question 15.
The area of trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.
Solution:
Area of a trapezium = 1586 cm²
and distance between the parallel sides (h) = 26

Question 16.
The parallel sides of a trapezium are 25 cm and 13 cm ; Its nonparallel sides are equal each being 10 cm, find the area of the trapezium.
Solution:
Parallel sides of a trapezium ABCD are 25 cm and 13 cm
i.e. AB = 25 cm, CD = 13 cm
and each non-parallel side = 10 cm
i.e., AD = BC = 10 cm
From C, draw CE || DA and draw CL ⊥ AB
CE = DA = CB = 10 cm
and EB = AB – AE = AB – DC = 25 – 13 = 12 cm
Perpendicular CL bisects base EB of an isosceles ∆CED

Question 17.
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm each.
Solution:
In trapezium ABCD, parallel sides are AB and DC.

AB = 25 cm, CD = 13 cm
and other sides are 15 cm each i.e. AD = CB = 15 cm
From C, draw CE || DA and CL ⊥ AB
AE = DC = 13 cm
and EB = AB – AE = 25 – 13 = 12 cm
Perpendicular CL bisects the base EB of the isosceles triangle CEB

Question 18.
If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.
Solution:
Area of trapezium = 28 cm²
Altitude (h) = 4 cm.

One of the parallel side = 6 cm
Second parallel side = 14 – 6 = 8 cm

Question 19.
In the figure, a parallelogram is drawn in a trapezium the area of the parallelogram is 80 cm², find the area of the trapezium.

Solution:
Area of parallelogram (AECD) = 80 cm²
Side AE (b) = 10 cm

Question 20.
Find the area of the field shown in the figure by dividing it into a square rectangle and a trapezium.

Solution:
Produce EF to H to meet AB at H and draw DK || EH
HF = 4 cm, KD = HE = 4 + 4 = 8 cm
HK = ED = 4 cm,
KB = 12 – (8) = 4 cm
Now, area of square AGFH = 4 x 4 = 16 cm²
area of rectangle KDEH = l x b = 8 x 4 = 32 cm²
and area of trapezium BCDK.

Total area of the figure = 16 + 32 + 22 = 70 cm²

Exercise 20.3

Question 1.
Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3

Solution:
In the figure, here are three triangles and one trapezium.

Question 2.
Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium:

Solution:
(i) In the figure ABCDEF,
Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm

Area of the square = 18 x 18 cm² = 324 cm²
Area of trapezium FCDE = 12 (CF + ED) x 8 cm²
= 12 (18 + 7) x 8
= 12 x 25 x 8 cm²
= 100 cm²
Total area of fig. ABCDEF = 324 + 100 = 424 cm²
(ii) In the figure ABCDEF,
Join BE.
The figure consists of one rectangle BCDE and one trapezium ABEF

Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm²
Area of trapezium ABEF,
= 12 (BE + AF) x height
= 12 (15 + 6) x 8 cm²
= 12 x 21 x 8 cm²
= 84 cm²
Area of the figure ABCDEF = 300 + 84 = 384 cm²
(iii) In the figure ABCDEFGH,

HC= AB = EF = 6 cm
AH = BC = 4 cm
DE = GF = 5 cm
Join HC.
In right ∆CDE,
ED² = CD² + CE²
⇒ (5)² = (4)² + (CE)²
⇒ 25 = 16 + (CE)²
⇒ (CE)² = 25 – 16 = 9 = (3)²
CE = 3 cm
The figure consist a rectangle and a trapezium
Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm²
Area of trapezium GDEF,
= 12 (GD + EF) x CE 1
= 12 (GH + HC + CD + EF) x CE
= 12 (4 + 6 + 4 + 6) x 3 cm²
= 12 x 20 x 3 cm²
= 30 cm²
Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²

Question 3.
There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.


Find the area of this park using both ways. Can you suggest some another way of finding its area ?
Solution:
In first case, the figure ABCDE is divided into 2 trapezium of equal area.
Now area of trapezium DFBC

Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m²
In second case, the figure ABCDE is divided into two parts, namely one square and other triangle.

Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²

Question 4.
Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.

Solution:
In the figure ABCDEF,
AD = 90 cm
BL = 30 cm
AO = 60 cm
CN = 40 cm
AN = 50 cm
EO = 60 cm
AM = 20 cm
FM = 20 cm
AL = 10 cm



Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²

Question 5.
Find the area of the following regular hexagon:

Solution:
In the regular hexagon MNOPQR There are two triangles and one rectangle.
Join MQ, MO and RP
NQ = 23 cm,
NA = BQ = 102 = 5 cm
MR = OP = 13 cm

In right ∆BDQ,
PQ² = BQ² + BP²
⇒ (13)² = (5)² + BP²
⇒ 169 = 25 + BP²
⇒ BP² = 169 – 25 = 144 = (12)²
BP = 12 cm
PR = MO = 2 x 12 = 24 cm
Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm²
Area of ∆PRQ = 12 x PR x BQ
= 12 x 24 x 5 = 60 cm²
Similarly area ∆MON = 60 cm²
Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²

Exercise 18.1

Question 1.
Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 3.8 cm and BD = 6.6 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.4 cm.

(ii) With centre A and radius 3.8 cm and with centre B and radius 6.6 cm, draw arcs intersecting each other at D.
(iii) With centre B and radius 4 cm, and with centre D and radius 6.4 cm, draw arcs intersecting each other at C on the other side of BD.
(iv) Join AD, BD, BC and DC.
The ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD such that AB = BC = 5.5 cm, CD = 4 cm, DA = 6.3 cm and AC = 9.4 cm. Measure BD.
Solution:
(i) Draw a line segment AC = 9.4 cm.
(ii) With centre A and C and radius 5.5 cm, draw arcs intersecting each other at B.
(iii) Join AB and CB.
(iv) Again with centre A and radius 6.3 cm, and with centre C and radius 4 cm, draw arcs intersecting each other at D below the line segment AC.

(v) Join AD and CD.
Then ABCD is the required quadrilateral. On measuring BD, it is 5 cm.

Question 3.
Construct a quadrilateral XYZW in which XY = 5 cm, YZ = 6 cm, ZW = 7 cm, WX = 3 cm and XZ = 9 cm.
Solution:
Steps of construction :
(i) Draw a line segment XZ = 9 cm.
(ii) With centre X and radius 3 cm and with centre Z and radius 7 cm, draw arcs intersecting each other at W.

(iii) Join XW and ZW.
(iv) Again with centre X and radius 5 cm and with centre Z and radius 6 cm, draw arcs, intersecting each other at Y below the line segment XZ.
(v) Join XY and ZY.
Then XYZW is the required quadrilateral.

Question 4.
Construct a parallelogram PQRS such that PQ = 5.2 cm, PR = 6.8 cm and QS = 8.2 cm.
Solution:
Steps of construction:
In a parallelogram, diagonals bisect each other. Now
(i) Draw a line segment PQ = 5.2 cm.
(ii) With centre P and radius 3.4 cm (12 of PR) and with centre Q and radius 4.1 cm (12 of QS) draw arcs intersecting each other at O.

(iii) Join PQ and QO and produced them to R and S respectively such that PO = OR and QO = OS.
(iv) Join PS, SR and RQ.
Then PQRS is the required parallelogram.

Question 5.
Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal.
Solution:
Steps of construction :
Sides of a rhombus are equal.
(i) Draw a line segment AC = 8 cm.

(ii) With centres A and C and radius 6 cm, draw two arcs above the line segment AC and two below the line segment AC, intersecting each other at D and B respectively.
(iii) Join AB, AD, BC and CD.
Then ABCD is the required rhombus.
JoinBD.
On measuring BD, it is approximately 9 cm

Question 6.
Construct a kite ABCD in which AB = 4 cm, BC = 4.9 cm and AC = 7.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment AC = 7.2 cm.

(ii) With centre A and radius 4 cm draw an arc.
(iii) With centre C and radius 4.9 cm, draw another arc which intersects the first arc at B and D.
(iv) Join AB, BC, CD and DA.
Then ABCD is the required kite.

Question 7.
Construct, if possible, a quadrilateral ABCD given, AB = 6 cm BC = 3.7 cm, CD = 5.7 cm, AD = 5.5 cm and BD = 6.1 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction :
(i) Draw a line segment BD = 6.1 cm.
(ii) With centre B and radius 6 cm and with centre D and radius 5.5 cm, draw arcs intersecting at A.
(iii) Join AB and AD.
(iv) Again with centre B and radius 3.7 cm and with centre D and radius 5.7 cm, draw two arcs intersecting each other at C below the BD.

(v) Join BC and DC.
Then ABCD is the required quadrilateral.

Question 8.
Construct, if possible a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5. cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction:
It is not possible to construct this quadrilateral ABCD because
AD + DC = 5.5 cm + 3 cm = 8.5 cm
and AC = 11 cm
AD + DC < AC.
But we know that in a triangle,
Sum of two sides is always greater than its third side.

Exercise 18.2

Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.0 cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) With centre A and radius 2.3 cm and with centre B and radius 3.8 cm draw arcs intersecting each other at D.

(iii) Join AD and BD.
(iv) Again with centre A and radius 4.5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at C.
(v) Join AC and BC and also CD.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 5 cm.

(ii) With centre C and D and radius 6 cm, draw line segments intersecting each other at A.
(iii) Join AC and AD.
(iv) Again with centre C and radius 7.5 cm and with centre D and radius 10 cm, draw arcs intersecting each other at B.
(v) Join CB, CA, DA, DB and AB.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral ABCD, when AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm.
Solution:
Steps of construction :
This quadrilateral is not possible as
BD = 4 cm, AB = 3 cm and AD = 7.5 cm
The sum of any two sides of a triangle is greater than the third side.
But BD + AD = 4 + 3 = 7 cm
BD + AD < AD

Question 4.
Construct a quadrilateral ABCD given AD = 3.5 cm, BC = 2.5 cm, CD = 4.1 cm, AC = 7.3 cm and BD = 3.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.1 cm.
(ii) With centre C and radius 7.3 cm and with centre D and radius 3.5 cm, draw arcs intersecting each other at A.

(iii) Join AC and AD.
(iv) Again with centre C and radius 2.5 cm and with centre D and radius 3.2 cm, draw arcs intersecting each other at B.
(v) Join CB’, and DB’ and join AB’.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD given AD = 5 cm, AB = 5.5 cm, BC = 2.5 cm, AC = 7.1 cm and BD = 8 cm.
Solution:
Steps of construction:
(i) Draw a line segment AB = 5 cm.

(ii) With centre A and radius 7.1 cm and with centre B and radius 2.5 cm, draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm and with centre B and radius 8 cm, draw arcs which intersect each other at D.
(v) Join AD and BD and CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm and BD = 6.5 cm.
Solution:
Steps of construction:
(i) Draw a line segment CD = 5 cm.

(ii) With centre C and radius 5.6 cm and with centre D and radius 4.5 cm, draw arcs which intersect each other at A.
(iii) Join AC and AD.
(iv) Again with centre C and radius 4 cm and with centre D and radius 6.5 cm, draw arcs which intersect each other at B.
(v) Join BC and BD and AB.
Then ABCD is the required quadrilateral.

Exercise 18.3

Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD = 5 cm and ∠B = 80°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) At B, draw a ray BX making an angle of 80° and cut off BC = 3.4 cm.
(iii) With centre A and radius 5 cm and with centre C and radius 4.5 cm, draw arcs which intersect each other at D.
(iv) Join CD and AD.

ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD given that AB = 8 cm, BC = 8 cm, CD = 10 cm, AD = 10 cm and ∠A = 45°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.

(ii) At A, draw a ray AX making an angle of 45° and cut off BC = 8 cm.
(iii) With centre A and C and radius 10 cm, draw arcs intersecting each other at D.
(iv) Join AD, CD.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.8 cm.

(ii) At C, draw a ray CX making an angle of 120° and cut off CD = 5.1 cm.
(iii) With centre B and radius 7.7 cm and with centre D and radius 3.6 cm draw arcs which intersect each other at A.
(iv) Join AD and AB.
Then ABCD is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and ∠B = 120°
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.

(ii) At B, draw a ray BX making an angle of 120° and cut off BC = 3 cm.
(iii) With centres A and C, and radius 5 cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA = 3.3 cm and ∠A = 60°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 2.8 cm.

(ii) At A draw a ray AX making an angle of 60° and cut off AD = 3.3 cm.
(iii) With centre B and radius 3.1 cm and with centre D and radius 2.6 cm, draw arc which intersect each other at C.
(iv) Join CB and CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and ∠B = 120°.
Solution:
Steps of construction:
The construction is not possible to draw as arcs of radius 4.5 cm from A and C, do not intersect at any point.

Exercise 18.4

Question 1.
Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, ZB = 95° and ∠C = 90°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4 cm.

(ii) At B, draw a ray BX making an angle of 95° and cut off BA = 6 cm.
(iii) At C, draw a ray CY making an angle of 90° and cut off CD = 4 cm.
(iv) Join AD.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, ∠B = 30° and ∠C = 150°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 30° and cut of BA = 4.2 cm.
(iii) At C, draw another ray CY making an angle of 150° and cut off CD = 4.8 cm.

(iv) Join AD.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°.
Solution:
Steps of construction :
(i) Draw a line segment QR = 2.5 cm.

(ii) At Q, draw a ray QX making an angle of 75° and cut off QP = 3.5 cm.
(iii) At R, draw another ray RY making an angle of 120° and cut off RS = 4.1 cm.
(iv) Join PS.
Then PQRS is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and ∠D = 100° and ∠C = 95°.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.4 cm.

(ii) At C, draw a ray CX making an angle of 95° and cut off CB = 6.6 cm
(iii) At D, draw another ray DY making an angle of 100° and cut off DA = 5.6 cm.
(iv) Join AB.
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, ∠A = 125° and ∠B = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB 4.4 cm.

(ii) At A, draw a ray AX making an angle of 125° and cut off AD = 3.5 cm.
(iii) At B, draw another ray BY making an angle of 120° and cut off BC = 4.7 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral PQRS in which ∠Q = 45°, ∠R = 90°, QR = 5 cm, PQ = 9 cm and RS = 7 cm.
Solution:
Steps of construction :
This quadrilateral is not possible to construct as shown in the figure.

Question 7.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, ∠A = 90° and ∠B = 105°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.

(ii) At A, draw a ray AX making an angle of 90° and cut off AD = 5 cm.
(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.

Question 8.
Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm and ∠F = 50° and ∠E = 100°
Solution:
Steps of construction :
(i) Draw a line segment EF = 3.5 cm.
(ii) At E, draw a ray EX making an angle of 100° and cut off ED = 4.5 cm.
(iii) At F, draw another ray FY making an angle of 45° and cut off FB = 6.5 cm.
(iv) Join DB.

Then BDEF is the required quadrilateral

Exercise 18.5

Question 1.
Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4 cm.

(ii) At A draw a ray AX making an angle of 75°.
(iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm.
(iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°.
Solution:
∠A = 60°, ∠B = 105° and ∠D = 90°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 105° + ∠C + 90° = 360°
⇒ 255° + ∠C = 360°
⇒ ∠C = 360° – 255° = 105°
Steps of construction :
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of

(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm.
(iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°.
Solution:
∠P = 105°, ∠R = 105° and ∠S = 75°
But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral)
⇒ 105° + ∠Q + 105° + 75° = 360°
⇒ 285° + ∠Q = 360°
⇒ ∠Q = 360° – 285° = 75°
Steps of construction :
(i) Draw a line segment PQ = 3.5 cm.

(ii) At P, draw a ray PX making an angle of 105°.
(iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm.
(iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S.
Then PQRS is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°.
Solution:
∠A = 70°, ∠B = 110°, ∠D = 85°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 70° + 110° + ∠C + 85° = 360°
⇒ 265° + ∠C = 360°
⇒ ∠C = 360° – 265° = 95°
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) At B, draw a ray BX making an angle of 110°.
(iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm.
(iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A.

Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm.
Solution:
∠A = 65°, ∠B = 105°, ∠C = 75°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 65° + 105° + 75° + ∠D = 360°
⇒ 245° + ∠D = 360°
⇒ ∠D = 360° – 245° = 115°
Steps of construction:
(i) Draw a line segment BC = 5.7 cm.
(ii) At B, draw a ray BX making an angle of

(iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm.
(iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At P, draw a ray PX making an angle of 50°.
(iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm.
(iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S.

Then PQRS is the required quadrilateral.

Exercise 17.1

Question 1.
Given below is a parallelogram ABCD. Complete each statement along with the definition or property used:
(i) AD = ………
(ii) ∠DCB = ……….
(iii) OC’ = …….
(iv) ∠DAB + ∠CDA = …….

Solution:
(i) AD = BC
(ii) ∠DCB = ∠ADC

(iii) OC = OA
(iv) ∠DAB + ∠CDA = 180°

Question 2.
The following figures are parallelograms. Find the degree values of the unknowns x,y, z.


Solution:
In a parallelogram, opposite angles are equal and sum of adjacent angle is 180°.
(i) In parallelogram ABCD,
∠B = 100°
∠A = ∠C = 180° (Co-interior angles)

⇒ x + 100° = 180°
⇒ x = 180° – 100°
⇒ x = 80°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
A = x
⇒ z = x
⇒ z = 80°
and ∠D = ∠B
⇒ y = 100°
x = 80°, y = 100° and z = 80°
(ii) In parallelogram PQRS, side PQ is produced to T.

∠S = 50°
∠PQR = ∠S (Opposite angles)
w = 50°
But ∠P + ∠PQR = 180° (Sum of adjacent angles)
⇒ x + 50° = 180°
⇒ x = 180° – 50° = 130°
⇒ But ∠P = ∠R (Opposite angles)
x = y
⇒ y = 130°
But w + z = 180° (A linear pair)
⇒ 50° + z = 180°
⇒ z = 180° – 50° = 130°
⇒ x = 130°, y = 130°, ∠ = 130
(iii) In parallelogram LMNP, PM is its diagonal ∠NPM = 30°, ∠PMN = 90°

PN || LM (Opposite sides of a parallelogram) and PM is its transversal
∠NPM = ∠PML
⇒ 30° = x
x = 30°
In ∆PMN,
∠P = 30°, ∠M = 90°
But ∠P + ∠M + ∠N = 180° (Sum of angles of a triangle)
⇒ 30° + 90° + z = 180°
⇒ 120° + z = 180°
⇒ z = 180° – 120° = 60°
But ∠L = ∠N (Opposite angles of a parallelogram)
y = z
⇒ y = 60°
Hence x = 30°, y = 60° and ∠ = 60°
(iv) In rhombus ABCD, diagonals AC and BD bisect each other at right angles.

x = 90°
In ∆OCD,
∠O + ∠C + ∠D = 180°
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60°
y = 60°
CD || AB, BD is the transversal
y = z (Alternate angles)
z = 60°
Hence x = 90°, y = 60° and z = 60°
(v) In parallelogram PQRS, side QR is produced to T

∠Q = 80°
∠P + ∠Q = 180° (Sum of adjacent angles)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
∠Q = ∠S (Opposite angles of a parallelogram)
⇒ 80° = y
⇒ y = 80°
PQ || SR and QRT is transversal
∠TRS = ∠RQP (Corresponding angles)
⇒ ∠ = 80°
Hence x = 100°, y = 80° and z = 80°
(vi) In parallelogram TUVW, UW is its diagonal

∠TUW = 40° and ∠V = 112°
∠T = ∠V (Opposite angles)
y = 112°
In ∆TUW,
∠T + ∠V + ∠W = 180° (Sum of angles of a parallelogram)
⇒ y + 40° + x = 180°
⇒ 112° + 40° + x = 180°
⇒ 152° + x = 180°
⇒ x = 180° – 152° = 28°
UV || TW and UW is its transversal
∠WUV = ∠TWU (Alternate angles)
⇒ z = x
⇒ z = 28°
Hence x = 28°, y = 112°, z = 28°

Question 3.
Can the following figures be parallelograms. Justify your answer.

Solution:
(i) In quadrilateral PLEH
∠H = 100°, ∠L = 80°
But there are opposite angles
∠H ≠ ∠L
PLEH is not a parallelogram.
(ii) In quadrilateral GNIR,
RI = 8 cm, GN = 8 cm, RG = 5 cm and IN = 5 cm
PI = GH and RG = IN
But there are opposite sides of the quadrilateral.
GNIR is a parallelogram.
(iii) In quadrilateral BEST,
BS and ET are its diagonals
But these diagonal do not bisect each other.
BEST is not a parallelogram

Question 4.
In the adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the geometrical truths you use to find them.

Solution:
In the figure, HOPE is a parallelogram in which HG is produced and HP is the diagonal

∠EHP = 40° and ∠POQ = 70°
But ∠POQ + ∠POH = 180° (Linear pair)
⇒ 70° + w = 180°
⇒ w = 180° – 70° = 110°
But ∠E = ∠POE (Opposite angles of a parallelogram)
x = 110°
HE || OP and HP is its transversal.
∠EHP = ∠HPO (Alternate angles)
⇒40° = y
⇒ y = 40°
In ∆PHO,
Ext. ∠POQ = ∠PHO + ∠HPO
⇒ 70° = z + y
⇒ 70° = z + 40°
⇒ z = 70° – 40° = 30°
Hence x = 110°, y = 40°, z = 30°

Question 5.
In the following figures GUNS and RUNS are parallelograms. Find x and y.

Solution:
(i) In parallelogram GUNS,
Opposite sides are parallel and equal
3x = 18
⇒ x = 6
and 3y – 1 = 26
⇒ 3y = 26 + 1 = 27
⇒ y = 9
x = 6, y = 9
(ii) Diagonals of a parallelogram bisect each other.
y – 7 = 20
⇒ y = 20 + 7 = 27
and x – 27 = 16
⇒ x – 27 = 16
⇒ x = 16 + 27 = 43
x = 43, y = 27

Question 6.
In the following figure RISK and CLUE are parallelograms. Find the measure of x.

Solution:
In the figure, RISK and CLUE are parallelograms
∠K = 120° and ∠L = 70°
In parallelogram RISK

RK || IS
∠RKS = ∠ISU (Corresponding angles)
⇒ ∠ISU = 120°
In parallelogram CLUE,
∠E = ∠L (Opposite angles of a parallelogram)
∠E = 70° (∠L = 70°)
Now in ∆EOS,
Ext. ∠ISU = x + ∠E
⇒ 120° = x + 70°
⇒ x = 120° – 70° = 50°
x = 50°

Question 7.
Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each angle of the parallelogram.
Solution:
In parallelogram ABCD,

∠A = (3x – 2)° and ∠C = (50 – x)°
∠A = ∠C (Opposite angles of a parallelogram)
⇒ 3x – 2° = 50° – x
⇒ 3x + x = 50° + 2°
⇒ 4x = 52°
x = 13°

Question 8.
If an angle of a parallelogram is two- third of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the parallelogram is ABCD and ∠A of a parallelogram ABCD be x

Hence other angles will be
∠C = ∠A = 108° (Opposite angles)
and ∠D = ∠B = 72° (Opposite angles)
Hence ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 9.
The measure of one angle of a parallelogram is 70°. What are the measures of the remaining angles ?
Solution:
Let in parallelogram ABCD,

∠A = 70°
But ∠A + ∠B= 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70°
⇒ ∠B = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 10.
Two adjacent angles of a parallelogram are as 1 : 2. Find the measures of all the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,

∠A : ∠B = 1 : 2
Let ∠A = x, then ∠B = 2x
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ x + 2x = 180°
⇒ 3x = 180°
∠A = x = 60°
and ∠B = 2x = 2 x 60°= 120°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 60° and ∠D = 120°
Hence ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 11.
In a parallelogram ABCD, ∠D = 135°, determine the measure of ∠A and ∠B.
Solution:
In parallelogram ABCD,

∠D = 135°
But ∠A + ∠D= 180° (Sum of adjacent angles)
∠A + 135° = 180°
∠A = 180° – 135° = 45°
But ∠B = ∠D (Opposite angles)
∠B = 135°
Hence ∠A = 45° and ∠B = 135°

Question 12.
ABCD is a parallelogram in which ∠A = 70°, compute ∠B, ∠C and ∠D.
Solution:
In parallelogram ABCD,

∠A = 70°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 70° + ∠B = 180°
⇒ ∠B = 180° – 70° = 110°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 70° and ∠D = 110°
Hence ∠B = 110°, ∠C = 70° and ∠D = 110°

Question 13.
The sum of two opposite angles of a parallelogram is 130°. Find all the angles of the parallelogram.
Solution:
Let in parallelogram

∠A + ∠C = 130°
But ∠A = ∠C (Opposite angles)
⇒ ∠C = 1302 = 65°
∠B + ∠D = 180° (Sum of adjacent angles)
⇒ 65° + ∠B = 180°
⇒ ∠B = 180° – 65° = 115°
⇒ ∠B = 115°
But ∠D = ∠B (Opposite angles)
∠D = 115°
Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠D = 115°

Question 14.
All the angles of a quadrilateral are equal to each other. Find the measure of each. Is the quadrilateral a parallelogram ? What special type of parallelogram is it ?
Solution:
All the angles of a quadrilateral are equal and sum of the four angles = 360°
Each angle will be = 3604 = 90°
Let in quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
It is a parallelogram as opposite angles are equal
i.e., ∠A = ∠C and ∠B = ∠D.
Each angle is of 90°
This parallelogram is a rectangle.

Question 15.
Two adjacent sides of a parallelogram are 4 cm and 3 cm respectively. Find its perimeter.
Solution:
Length of two adjacent sides = 4 cm and 3 cm
i.e., l = 4 cm and b = 3 cm
Perimeter = 2 (l + b) = 2 (4 + 3) cm = 2 x 7 = 14 cm

Question 16.
The perimeter of a parallelogram is 150 cm. One of its sides is greater ii.au the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:
Perimeter of a parallelogram = 150 cm
Let l he the longer side and b be the shorter side
l = b + 25 cm.
⇒ 2 (l + b) = 150
⇒ l + b = 75
⇒ b + 25 + b = 75
⇒ 2b = 75 – 25 = 50
⇒ b = 25
l = b + 25 = 25 + 25 = 50
Sides are 50 cm, 25 cm

Question 17.
The shorter side of a parallelogram is 4.8 cm and the longer side is half as much again as the shorter side. Find the perimeter of the parallelogram.
Solution:
In a parallelogram shorter side (b) = 4.8 cm.
longer side (l) = 4.8 + 12 x 4.8
= 4.8 + 2.4 = 7.2 cm
Perimeter = 2 (l + b) = 2 (7.2 + 4.8) cm = 2 x 12.0 = 24 cm

Question 18.
Two adjacent angles of a parallelogram are (3x – 4)° and (3x + 10)°. Find the angles of the parallelogram.
Solution:
Let in parallelogram ABCD,

∠A = (3x + 4)° and ∠B = (3x + 10)°
But ∠A + ∠B = 180° (Sum of adj adjacent angles)
⇒ 3x – 4 + 3x + 10 = 180°
⇒ 6x + 6° = 180°
⇒ 6x = 180° – 6° = 174°
⇒ x = 29°
∠A = 3x – 4 = 3 x 29 – 4 = 87° – 4° = 83°
∠B = 3x + 10 = 3 x 29° + 10° = 87° + 10° = 97°
But ∠C = ∠A and ∠D = ∠B (Opposite angles)
∠C = 83° and ∠D = 97°
Hence ∠A = 83°, ∠B = 97°, ∠C = 83° and ∠D = 97°

Question 19.
In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°.
Find : ∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.
Solution:
In parallelogram ABCD, diagonal AC and ED bisect each other at O.

∠ABC = 30°, ∠CAB = 70° and ∠BDC = 10°
∠ADC = ∠ABC = 30° (Opposite angles)
and ∠ADB = ∠ADC – ∠BDC = 30° – 10° = 20°
AB || DC and AC is the transversal
∠ACD = ∠CAB = 70°(Altemate angles)
AB || DC and BD is transversal
∠CDB = ∠ABD = 10°(Altemate angles)
In ∆ABC
∠CAB + ∠ABC + ∠BCA = 180° (Sum of anlges of a triangle)
⇒ 70° + 30° + ∠BCA = 180°
⇒ 100° + ∠BCA = 180°
∠BCA = 180° – 100° = 80°
∠BCD = ∠BCA + ∠ACD = 80° + 70° = 150°
∠BCD = ∠DAB (Opposite angles)
⇒ ∠DAB = 150° and ∠CAD = 150° – 70° = 80°
In ∆OCD,
∠ODC + ∠OCD + ∠COD = 180° (Angles of a triangle)
⇒ ∠ACD + ∠ACD + ∠COD = 180°
⇒ 70° + 10° + ∠COD = 180°
⇒ 80° + ∠COD = 180°
⇒ ∠COD = 180° – 80° = 100°
∠COD = 100°
But ∠AOD + ∠COD = 180° (Linear pair)
⇒ ∠AOD + 100° = 180°
⇒ ∠AOD = 180° – 100° = 80°
⇒ ∠AOD = 80°
But ∠AOB = ∠COD
and ∠BOC = ∠AOD (Vertically opposite angles)
∠AOB = 100° and ∠BOC = 80°
Hence ∠DAB = 150°, ∠ADC = 30°, ∠BCD = 150°
∠AOD = 80° ∠DOC = 100°
∠BOC = 80° ∠AOB = 100°
∠ACD = 70°, ∠CAB = 70°, ∠AD = 20°, ∠ACB = 80°, ∠DBC = 20°, ∠DBA = 10

Question 20.
Find the angles marked with a question mark shown in the figure.

Solution:
In parallelogram ABCD,

CE ⊥ AB and CF ⊥ AD
∠BCE = 40°
In ∆BCE,
∠BCE + ∠CEB + ∠EBC = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠EBC = 180°
⇒ 130° + ∠EBC = 180°
⇒ ∠EBC = 180° – 130° = 50°
or ∠B = 50°
But ∠D = ∠B (Opposite angles)
∠D = 50° or ∠ADC = 50°
Similarly in ∆DCF,
∠DCF + ∠CFD + ∠FDC = 180°
⇒ ∠DCF + 90° + 50° = 180°
⇒ ∠DCF + 140° = 180°
⇒ ∠DCF = 180° – 140° = 40°
But ∠C + ∠B= 180° (Sum of adjacent angles)
⇒ ∠BCE + ∠ECF + ∠DCF + ∠B = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ 40° + ∠ECF + 40° + 50° = 180°
⇒ ∠ECF + 130° = 180°
⇒ ∠ECF = 180° – 130° = 50°
∠ECF = 50°

Question 21.
The angle between the altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.
Solution:
In parallelogram ABCD,

∠A is an obtused angle
AE ⊥ BC and AF ⊥ DC
∠EAF = 60°
In quadrilateral AECF,
∠EAF + ∠F + ∠C + ∠E = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 90° + ∠C + 90° = 360°
⇒ 240° + ∠C = 360°
⇒ ∠C = 360° – 240° = 120°
∠C = 120°
But ∠A = ∠C (Opposite angles)
∠A = 120°
But ∠A + ∠B = 180° (Sum of adjacent angles)
⇒ 120° + ∠B = 180°
⇒ ∠B = 180° – 120° = 60°
∠B = 60°
But ∠D = ∠B (Opposite angles)
∠D = 60°
Hence ∠A = 120°, ∠B = 60°, ∠C = 120° and ∠D = 60°

Question 22.
In the figure, ABCD and AEFG are parallelograms. If ∠C = 55°, what is the measure of ∠F ?

Solution:
In the parallelogram ABCD,
∠A = ∠C …..(i) (Opposite angles of a parallelogram)
Similarly in parallelogram AEFG,
∠A = ∠F …(ii)
From (i) and (ii),
∠C = ∠F = 55° (∠C = 55°)
Hence ∠F = 55°

Question 23.
In the figure, BDEF and DCEF are each a parallelogram. It is true that BD = DC ? Why or why not ?

Solution:
In parallelogram BDEF,
BD = EF ……(i) (Opposite sides of a parallelogram)
Similarly, in parallelogram DCEF
DC = EF
From (i) and (ii),
BD = DC
Hence it is true that BD = DC.

Question 24.
In the figure, suppose it is known that DE = DF. Then is ∆ABC isosceles ? Why or why not ?
Solution:
In parallelogram BDEF,

∠B = ∠E ……(i) (Opposite angles of a parallelogram)
Similarly, in parallelogram DCEF,
∠C = ∠F ……(ii)
But DE = DF (Given)
In ∆DEF
∠E = ∠F
From (i) and (ii),
∠B = ∠C
AC = AB (Sides opposite to equal angles)
∆ABC is an isosceles triangle.

Question 25.
Diagonals of parallelogram ABCD intersect at O as shown in the figure. XY contains O, and X, Y are points on opposite sides of the parallelogram. Give reasons for each of the following:
(i) OB = OD
(ii) ∠OBY = ∠ODX
(iii) ∠BOY = ∠DOX
(iv) ∆BOY = ∆DOX
Now, state if XY is bisected at O.
Solution:
In parallelogram ABCD,

diagonals AC and BP intersect each other at O
O is the mid-point of AC and BD.
Through O, XY is draw such that X lies on AD and Y, on BC.
(i) OB = OD (O is mid-point of BD)
(ii) AD || BC and BD is transversal
∠OBY = ∠ODX (Alternate angles)
(iii) ∠BOY = ∠DOX (Vertically opposite angles)
(iv) Now in ∆BOY and ∆DOX,
OB = OD
∠OBY = ∠ODX
∠BOY = ∠DOX
∆BOY = ∆DOX (ASA axiom)
OY = OX (c.p.c.t.)
Hence XY is bisected at O.

Question 26.
In the figure ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same.
(i) ∠A = ∠C
(ii) ∠FAB = 12 ∠A
(iii) ∠DCE = 12 ∠C
(iv) ∠CEB = ∠FAB
(v) CE || AF.
Solution:
In parallelogram ABCD,
CE is the bisector of ∠C and and AF is the bisector of ∠A.

(i) ∠A = ∠C (Opposite angles of a parallelogram)
(ii) AF is the bisector of ∠A
∠FAB = 12 ∠A
(iii) CE is the bisector of ∠C
∠DCE = 12 ∠C
(iv) From (i), (ii) and (iii)
∠FAB = ∠DCE
(v) ∠FAB = ∠DCE
But these are opposite angles of quadrilateral AECF
AB or AE || DC or FC
AECF is a parallelogram
CE || AF
Hence proved

Question 27.
Diagonals of a parallelogram ABCD intersect at O. AL and CM are drawn perpendiculars to BD such that L and M lie on BD. Is AL = CM ? Why or why not ?
Solution:
In parallelogram ABCD, diagonals AC and BD intersect each other at O.

O is the mid-point of AC and BD.
AL ⊥ BD and CM ⊥ BD.
In ∆ALO and ∆CMO
∠L = ∠M (Each 90°)
∠AOL = ∠COM (Vertically opposite angles)
AO = CO (O is mid-point of AC)
∆ALO = ∆CMO (AAS axiom)
AL = CM (c.p.c.t.)
Hence proved

Question 28.
Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE ?
Solution:
In parallelogram ABCD,

AC is its diagonal. E and F are points on AC such that AE = CF
Join EB, BF, FD and DE .
Join also diagonal BD which intersects AC at O
O is the mid-point of AC and BD
AO = OC
But AE = CF
⇒ AO – AE = CO – CF
⇒ EO = OF
But BO = OD (O is mid-point of BD)
Diagonals EF and BD of quadrilateral bisect each other at O.
BFDE is a parallelogram.

Question 29.
In a parallelogram ABCD, AB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
Solution:
AB || DC

∠DEA = ∠EAB (Alternate angles)
= ∠DAE (EA is bisector of ∠A)
In ∆DAE,
∠DEA = ∠DAE
AD = DE = 6 cm
But DE = AB = 10 cm.
EC = DC – DE = 10 – 6 = 4 cm
AD || BC or BF and AF is transversal
∠DAE = ∠EFC (Alternate angle)
But ∠DAE = ∠DEA Prove
= ∠FEC (DEA = FEC vertically opposite angles)
In ∆ECF,
CE = CF = 4 cm (CE = 4 cm)

Exercise 17.2

Question 1.
Which of the following statements are true for a rhombus ?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal sides.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are at right angles.
(v) Its diagonals bisect each other at right angles.
(vi) Its diagonals are equal and perpendicular.
(vii) it has all its sides of equal lengths.
(viii) It is a parallelogram.
(ix) It is a quadrilateral.
(x) It can be a square.
(xi) It is a square.
Solution:
(i) True
(ii) True
(iii) False (Its all sides are equal)
(iv) False (Opposite angles are equal)
(v) True
(vi) False (Diagonals are not equal)
(vii) True
(viii) True
(ix) True
(x) True (It is a rhombus)
(xi) False

Question 2.
Fill in the blanks, in each of the following, so as to make the statement true:
(i) A rhombus is a parallelogram in which ………..
(ii) A square is a rhombus in which ………..
(iii) A rhombus has all its sides of …….. length.
(iv) The diagonals of a rhombus each ………. other at ………. angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a ………..
Solution:
(i) A rhombus is a parallelogram in which adjacent sides are equal.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A rhombus has all its sides of equal length.
(iv) The diagonals of a rhombus bisect each other at right angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

Question 3.
The diagonals of a parallelogram are not perpendicular. Is it a rhombus ? Why or why not ?
Solution:
By definition of a rhombus, its diagonals bisect each other at right angle.
So, the given parallelogram is not a rhombus.

Question 4.
The diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus. If your answer is ‘No’, draw a figure to justify your answer.
Solution:
The diagonals of a quadrilateral are perpendicular to each.

It is not always possible to be a rhombus. It can be of the diagonals bisect each other at right angles and if not, then it is not rhombus as shown in the figure given above:

Question 5.
ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.
Solution:
In rhombus ABCD, diagonals AC and BD intersect each other at O.

∠ACB = 40°, we have to find ∠ADB.
BD || AD and AC.is its transversal..
∠ACB = ∠CAD (Alternate angles)
Now in ∆AOD
∠OAD + ∠AOD + ∠ADO = 180° (Sum of angles of a triangle)
⇒ 40° + 90° + ∠ADO = 180°
⇒ 130° + ∠ADO = 180°
⇒ ∠ADO = 180° – 130° = 50°
∠ADB = 50°

Question 6.
If the diagonals of a rhombus are 12 cm and 16 cm, find the length of each side.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles.

AC = 16 cm, BD = 12 cm
AO = OC = 162 = 8 cm
BO = OD = 122 = 6 cm.
Now, in right angled ∆AOB,
AB² = AO² + BO² = (8)² + (6)² = 64 + 36 = 100 = (10)²
AB = 10 cm
Each side of rhombus = 10 cm

Question 7.
Construct a rhombus whose diagonals are of length 10 cm and 6 cm.
Solution:
(i) Draw a line segment AC =10 cm.

(ii) Draw its perpendicular bisector and cut off OB = OD = 3 cm (12 of 6 cm).
(iii) Join AB, BC, CD and DA.
Then ABCD is the required rhombus.

Question 8.
Draw a rhombus, having each side of length 3.5 cm and one of the angles as 40°.
Solution:
Steps of construction
(i) Draw a line segment AB = 3.5 cm.

(ii) Draw a ray BX making an angle of 40° at B and cut off BC = 3.5 cm.
(iii) With centres C and A, and radius 3.5 cm. Draw arcs intersecting each other at D.
(iv) Join AD and CD.
Then ABCD is a required rhombus.

Question 9.
One side of a rhombus is of length 4 cm and the length of an altitude is 3.2 cm. Draw the rhombus.
Solution:
(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a perpendicular BX and cut off BL = 3.2 cm.

(iii) At L, draw a line LY parallel to AB.
(iv) With centres B and radius 4 cm, draw an arc intersecting the line LY at C.
(v) With centre C cut off CD = 4 cm.
(vi) Join BC and AD.
Then ABCD is the required rhombus.

Question 10.
Draw a rhombus ABCD if AB = 6 cm and AC = 5 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) With centre A and radius 5 cm and with centre B and radius 6 cm, draw arcs intersecting each other at C.

(iii) Join AC and BC.
(iv) Again with centres C and A and radius 6 cm, draw arcs intersecting each other at D.
(v) Join AD and CD.
Then ABCD is the required rhombus.

Question 11.
ABCD is a rhombus and its diagonals intersect at O.
(i) Is ∆BOC = ∆DOC ? State the congruence condition used ?
(ii) Also state, if ∠BCO = ∠DCO.
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at O.

(i) Now in ∆BOC and ∆DOC
OC = OC (Common)
BC = CD (Sides of rhombus)
OB = OC (Diagonals bisect each other at O)
∆BOC = ∆DOC (SSS. condition)
(ii) ∠BCO = ∠DCO

Question 12.
Show that each diagonal of a rhombus bisects the angle through which it passes:
Solution:
In rhombus ABCD, AC is its diagonal we have to prove that AC bisects ∠A and ∠C.

Now, in ∆ABC and ∆ADC
AC = AC (Common)
AB = CD (Sides of a rhombus)
BC = AD
∆ABC = ∆ADC (SSS condition)
∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)
Hence AC bisects the angle A.
Similarly, by joining BD, we can prove that BD bisects ∠B and ∠D.
Hence each diagonal of a rhombus bisects the angle through which it passes.

Question 13.
ABCD is a rhombus whose diagonals intersect at O. If AB = 10 cm, diagonal BD = 16 cm, find the length of diagonal AC.
Solution:
In rhombus ABCD,

AB = 10 cm, diagonal BD = 16 cm.
Draw diagonal AC which bisects BD at O at right angle.
BO = OD = 8 cm and AO = OC.
Now in ∆AOB,
AB² = AO² + BO²
⇒ (10)² = AO² + (8)²
⇒ 100 = AO² – 64
⇒ AO² = 100 – 64 = 36 = (6)²
AO = 6.
AC = 2AO = 2 x 6 = 12 cm

Question 14.
The diagonals of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each other at right angles, what is the length of each side of the quadrilateral ?
Solution:
In a quadrilateral ABCD, diagonals AC and BD bisect each other at right angles.
AC = 8 cm and BD = 6 cm.

AO = OC = 4 cm and BO = OD = 3 cm
Now, in right ∆AOB,
AB² = AO² + BO² = (4)² + (3)² = 16 + 9 = 25 = (5)²
AB = 5 cm
Hence each side of quadrilateral will be 5 cm.

Exercise 17.3

Question 1.
Which of the following statements are true for a rectangle ?
(i) It has two pairs of equal sides.
(ii) It has all its sides of equal length.
(iii) Its diagonals are equal.
(iv) Its diagonals bisect each other.
(v) Its diagonals are perpendicular.
(vi) Its diagonals are perpendicular and bisect each other.
(vii) Its diagonals are equal and bisect each other.
(viii) Its diagonals are equal and perpendicular, and bisect each other.
(ix) All rectangles are squares.
(x) All rhombuses are parallelograms.
(xi) All squares are rhombuses and also rectangles.
(xii) All squares are not parallelograms.
Solution:
(i) True.
(ii) False. (Only pair of opposite sides is equal)
(iii) True
(iv) True
(v) False (Diagonals are not perpendicular)
(vi) False (Diagonals are not perpendicular to each other)
(vii) True
(viii) False (Diagonals are equal but not perpendicular)
(ix) False (All rectangles are not square but a special type can be a square)
(x) True
(xi) True
(xii) False (All squares are parallelograms because their opposite sides are parallel and equal)

Question 2.
Which of the following statements are true for a square ?
(i) It is a rectangle.
(ii) It has all its sides of equal length.
(iii) Its diagonals bisect each other at right angle.
(iv) Its diagonals are equal to its sides.
Solution:
(i) True
(ii) True
(iii) True
(iv) False (Each diagonal of a square is greater than its side)

Question 3.
Fill in the blanks in each of the following so as to make the statement true :
(i) A rectangle is a parallelogram in which ……..
(ii) A square is a rhombus in which ……….
(iii) A square is a rectangle in which ………
Solution:
(i) A rectangle is a parallelogram in which one angle is right angle.
(ii) A square is a rhombus in which one angle is right angle.
(iii) A square is a rectangle in which adjacent sides are equal.

Question 4.
A window frame has one diagonal longer than the other. Is the window frame a rectangle ? Why or why not ?
Solution:
No, it is not a rectangle as rectangle has diagonals of equal length.

Question 5.
In a rectangle ABCD, prove that ∆ACB = ∆CAD.
Solution:
In rectangle ABCD, AC is its diagonal.

Now in ∆ACB and ∆CAD
AB = CD (Opposite sides of a rectangle)
BC = AD
AC = AC (Common)
∆ACB = ∆CAD (SSS condition)

Question 6.
The sides of a rectangle are in the ratio 2 : 3 and its perimeter is 20 cm. Draw the rectangle.
Solution:
Perimeter of a rectangle = 20 cm
Ratio in the sides = 2 : 3

Let breadth (l) = 2x
Then length (b) = 3x
Perimeter = 2 (l + b)
⇒ 20 = 2 (2x + 3x)
⇒ 4x + 6x = 20
⇒ 10x = 20
⇒ x = 2010 = 2
Length = 3x = 3 x 2 = 6
and breadth = 2x = 2 x 2 = 4 cm
Steps of construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A and B draw perpendicular AX and BY.
(iii) Cut off from AX and BY,
AD = BC = 4 cm.
(iv) Join CD.
Then ABCD is the required rectangle.

Question 7.
The sides of a rectangle are the ratio 4 : 5. Find its sides if the perimeter is 90 cm.
Solution:
Perimeter of a rectangle = 90 cm.

Ratio in sides = 4 : 5
Let first side = 4x
Then second side = 5x
Perimeter = 2 (l + b)
⇒ 2 (4x + 5x) = 90
⇒ 2 x 9x = 90
⇒ 18x = 90
⇒ x = 5
First side = 4x = 4 x 5 = 20 cm
and second side = 5x = 5 x 5 = 25 cm

Question 8.
Find the length of the diagonal of a rectangle whose sides are 12 cm and 5 cm.
Solution:
In rectangle ABCD, AB = 12 cm and AD = 5 cm

BD is its diagonal.
Now, in right angled ∆ABD,
BD² = AB² + AD² (Pythagoras theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
BD = 13 cm
Length of diagonal = 13 cm

Question 9.
Draw a rectangle whose one side measures 8 cm and the length of each of whose diagonals is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm

(ii) At B, draw a perpendicular BX
(iii) With centre A and radius 10 cm, draw an arc which intersects BX at C.
(iv) With centre C and radius equal to AB and with centre A and radius equal to BC, draw arcs which intersect at D.
(v) Join AD, AC, CD and BD.
Then ABCD is the required rectangle.

Question 10.
Draw a square whose each side measures 4.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.8 cm.

(ii) At A and B, draw perpendiculars AX and BY.
(iii) Cut off AD = BC = 4.8 cm
(iv) Join CD.
Then ABCD is the required square.

Question 11.
Identify all the quadrilaterals that have:
(i) Four sides of equal length.
(ii) Four right angles.
Solution:
(i) A quadrilateral whose four sides are equal can be a square or a rhombus.
(ii) A quadrilateral whose four angle are right angle each can be a square or a rectangle.

Question 12.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle ?
Solution:
(i) A square is a quadrilateral as it has four sides and four angles.
(ii) A square is a parallelogram, because its opposite sides are parallel and equal.
(iii) A square is a rhombus because it has all sides equal and opposite sides are parallel.
(iv) A square is a rectangle as its opposite sides are equal and each angle is of 90°.

Question 13.
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisector of each other
(iii) are equal.
Solution:
(i) A quadrilateral whose diagonals bisect each other can be a square, rectangle, rhombus or a parallelogram.
(ii) A quadrilateral whose diagonals are perpendicular bisector of each other can be a square or a rhombus.
(iii) A quadrilateral whose diagonals are equal can be a square or a rectangle.

Question 14.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C.
Solution:
In ∆ABC, ∠B = 90°.
O is the mid-point of AC i.e. OA = OC.

BO is joined.
Now, we have to prove that OA = OB = OC
Produce BO to D such that OD = OB.
Join DC and DA.
In ∆AOB and ∆COD
OA = OC (O is the mid point of AC)
OB = OD (Construction)
∠AOB = ∠COD (Vertically opposite angles)
∆AOB = ∆COD (SAS condition)
AB = CD (c.p.c.t.) …..(i)
Similarly, we can prove that
∆BOC = ∆AOD
BC = AD …….(ii)
From (i) and (ii)
ABCD is a rectangle.
But diagonals of a rectangle bisect each other and are equal in length.
AC and BD bisect each other at O.
OA = OC = OB.
O is equidistant from A, B and C.

Question 15.
A mason has made a concrete slap. He needs it to be rectangular. In what different ways can he make sure that it is a rectangular ?
Solution:
By definition, a rectangle has each angle of 90° and their diagonals are equal.
The mason will check the slab whether it is a rectangular in shape by measuring that
(i) its each angle is 90°
(ii) its both diagonals are equal.

Exercise 14.1

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years

Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year

Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years

Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year

Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years

Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year

Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year

Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 112 years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 112 years = 3 half-years

Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year

Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year

Compound interest for the third half-years or 112 years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters

Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter

Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter

Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter

Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 1212 % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000

Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year

Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250

Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 112 years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 112 years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 112 years or 3 half-years

Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year

Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year

Total compound interest for 3 half-years
or 112 years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years

In second case
Interest for the first year

Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year

Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly

Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter

Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter

Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.

Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.

Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100

Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or 52 % quarterly
Period (T) = 1 year = 4 quarters

Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter

Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240

= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter

Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters

Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter

Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter

Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 512 % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 512 % = 112 % p.a.
Period (T) = 3 years.

Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year

Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year

Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years

Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year

Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year

Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years

Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year

Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Exercise 14.2

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 712 %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years

and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years

and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.

C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years

C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800

C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years

C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.

C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 1212 % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.

Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 1712 % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000


C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 1212 % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096

Question 8.
Find the amount and the compound interest on Rs 8,000 for 112 years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly

and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 1212 % per annum to build a house. Find the amount that she pays LIC after 112 years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000

Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years


C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760


C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 1212 % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters

Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters

C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 214 years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years

= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years

C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,

Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum

Question 21.
Simple interest on a sum of money for 2 years at 612 % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200


Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years

In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years

= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

Exercise 14.3

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 1212 % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years

n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 112 years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then

Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years

Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 634 % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690

Exercise 14.4

Question 1.
The present population of a town is 28,000. If it increases at the rate of 5% per annum, what will be its population after 2 years ?
Solution:
Present population = 28000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
Population after 2 years

Question 2.
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
Solution:
Present population = 125000
Rate of birth = 5.5%
and rate of death = 3.5%
Increase = 5.5 – 3.5 = 2% p.a.
Period = 3 years.
Population after 3 years

Question 3.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Present population = 25000
Increase in first year = 4%
in second year= 5% and
in third year = 8%
Population after 3 years =

Question 4.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Solution:
Three years ago,
Population of a town = 50000
Annual increase in population in first year = 4%
in second year = 5%
and in third year = 3%
Present population

Question 5.
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago ?
Solution:
Let 3 years ago, population = P
Present population = 9261
Rate of increase (R) = 5% p.a.
Period (n) = 3 years.

Question 6.
In a factory, the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
Solution:
Production of scooters 3 years ago (P) = 40000
Present production (A) = 46305
Period (n) = 3 years

Question 7.
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago ?
Solution:
Let 3 years ago, the population of a city = P
Rate of growth (R) = 8% p.a.
Present population = 196830
Period (n) = 3 years

Question 8.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Solution:
Population after 2 years = 22050
Rate of increase = 50 per thousand
Period (n) = 2 years
Let present population = P, then

Question 9.
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?
Solution:
Present count of bacteria = 13125000
In first hour increase = 10%
decrease in second hour = 8%
increase in third hour = 12%
Count of bacteria after 3 hours

Question 10.
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000 ?
Solution:
On the last day of 1998,
Population of a town = 72000
In the first year, increase = 7%
In the second year, decrease = 10%
Population in the last day of 2000

Question 11.
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year ?
Solution:
Number of workers at the beginning = 6400
Period = 4 years.
At the end of 1st year, workers retrenched = 25%
At the end of second year, workers retrenched = 25%
At the end of third year, workers increased = 25%
Total number of workers during the 4 years

Question 12.
Aman started a factory with an initial investment of Rs 1,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
Initial investment = Rs 100000
Loss in the first year = 5%
Profit in the second year = 10%
Profit in the third year = 12%
Investment at the end of 3 years

Question 13.
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago ?
Solution:
Present population (A) = 175760
Increase rate = 40 per 1000
Period = 3 years
Let 3 years ago,
population was = P

Question 14.
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years ?
Solution:
Production of Mixi in 1996 = 8000
Increase in next 2 years = 15%
Decrease in the third year = 5%
Production after 3 years

Question 15.
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Solution:
Population of a city in 1999 = 6760000
Increase = 4%
(i) Population in 2001 is after 2 years

Question 16.
Jitendra set up a factory by investing Rs 25,00,000. During the first two successive years his profits were 5% and f 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.
Solution:
Investment in the beginning = Rs 25,00,000
Profit during the first 2 years = 5% and 10% respectively
Investment after 2 years will be

= Rs 28,87,500
Amount of profit = Rs 28,87,500 – Rs 25,00,000 = Rs 3,87,500

Exercise 14.5

Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years

Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years

Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years

Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years

Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740

Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680

Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001

Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years