Ratio & Proportion — Exercise 9(A)

Express the ratio in its simplest form :

48 : 54

Answer

The given ratio is 48 : 54.

H.C.F. of 48 and 54 is 6.

∴ 48 : 54 = $\dfrac{48 \div 6}{54 \div 6} = \dfrac{8}{9}$ = 8 : 9

Hence, 48 : 54 = 8 : 9.

Express the ratio in its simplest form :

5 kg : 800 g

Answer

Both quantities must be in the same unit.

5 kg = 5 × 1000 g = 5000 g

∴ 5 kg : 800 g = 5000 g : 800 g = 5000 : 800

H.C.F. of 5000 and 800 is 200.

$\dfrac{5000 \div 200}{800 \div 200} = \dfrac{25}{4}$ = 25 : 4

Hence, 5 kg : 800 g = 25 : 4.

Express the ratio in its simplest form :

3 m : 90 cm

Answer

Both quantities must be in the same unit.

3 m = 3 × 100 cm = 300 cm

∴ 3 m : 90 cm = 300 cm : 90 cm = 300 : 90

H.C.F. of 300 and 90 is 30.

$\dfrac{300 \div 30}{90 \div 30} = \dfrac{10}{3}$ = 10 : 3

Hence, 3 m : 90 cm = 10 : 3.

Express the ratio in its simplest form :

2 years : 9 months

Answer

Both quantities must be in the same unit.

2 years = 2 × 12 months = 24 months

∴ 2 years : 9 months = 24 months : 9 months = 24 : 9

H.C.F. of 24 and 9 is 3.

$\dfrac{24 \div 3}{9 \div 3} = \dfrac{8}{3}$ = 8 : 3

Hence, 2 years : 9 months = 8 : 3.

Express the ratio in its simplest form :

$1\dfrac{1}{2} : 2\dfrac{1}{2}$

Answer

Converting the mixed fractions into improper fractions :

$1\dfrac{1}{2} : 2\dfrac{1}{2} = \dfrac{3}{2} : \dfrac{5}{2}$

Dividing the first term by the second term :

$= \dfrac{3}{2} \times \dfrac{2}{5} = \dfrac{3}{5}$ = 3 : 5

Hence, $1\dfrac{1}{2} : 2\dfrac{1}{2}$ = 3 : 5.

Express the ratio in its simplest form :

$3\dfrac{1}{2} : 7$

Answer

Converting the mixed fraction into an improper fraction :

$3\dfrac{1}{2} : 7 = \dfrac{7}{2} : 7$

Dividing the first term by the second term :

$= \dfrac{7}{2} \times \dfrac{1}{7} = \dfrac{1}{2}$ = 1 : 2

Hence, $3\dfrac{1}{2} : 7$ = 1 : 2.

Express the ratio in its simplest form :

2.5 : 1.5

Answer

Multiplying both terms by 10 to remove the decimals :

2.5 : 1.5 = 25 : 15

H.C.F. of 25 and 15 is 5.

$\dfrac{25 \div 5}{15 \div 5} = \dfrac{5}{3}$ = 5 : 3

Hence, 2.5 : 1.5 = 5 : 3.

If the ratio between $x$ and $3x$ − 4 is 2 : 5, find the value of $x$.

Answer

Given,

$x : (3x − 4)$ = 2 : 5

$\Rightarrow \dfrac{x}{3x − 4} = \dfrac{2}{5}$

By cross multiplication :

$\Rightarrow 5x = 2(3x - 4) \\[1em] \Rightarrow 5x = 6x - 8 \\[1em] \Rightarrow 8 = 6x - 5x \\[1em] \Rightarrow x = 8$

Hence, the value of $\bm{x = 8}$.

If the ratio between $2x − 5$ and $x + 5$ is 7 : 5, find :

(i) $x$

(ii) $2x − 5$

(iii) $x + 5$

Answer

Given,

$(2x − 5) : (x + 5)$ = 7 : 5

$\Rightarrow \dfrac{2x − 5}{x + 5} = \dfrac{7}{5}$

By cross multiplication :

$\Rightarrow 5(2x − 5) = 7(x + 5) \\[1em] \Rightarrow 10x − 25 = 7x + 35 \\[1em] \Rightarrow 10x − 7x = 35 + 25 \\[1em] \Rightarrow 3x = 60 \\[1em] \Rightarrow x = \dfrac{60}{3} = 20$

Substituting value of $x$, we get :

$2x - 5 = 2 \times 20$ - 5 = 40 - 5 = 35.

$x + 5$ = 20 + 5 = 25.

(i) Hence, $\bm{x = 20}$.

(ii) Hence, $\bm{2x - 5 = 35}$.

(iii) Hence, $\bm{x + 5 = 25}$.

If $m : n$ = 4 : 3, find $3m : 4n$.

Answer

Solving,

$\Rightarrow m : n = 4 : 3 \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{3m}{4n} = \dfrac{3}{4} \times \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{3m}{4n} = \dfrac{1}{1}$

Hence, $\bm{3m : 4n = 1 : 1}$.

If $p : q$ = 5 : 4, find $6p : 5q$.

Answer

Solving,

$\Rightarrow p : q = 5 : 4 \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{5}{4} \\[1em] \Rightarrow \dfrac{6p}{5q} = \dfrac{6}{5} \times \dfrac{p}{q} \\[1em] \Rightarrow \dfrac{6p}{5q} = \dfrac{6}{5} \times \dfrac{5}{4} \\[1em] \Rightarrow \dfrac{6p}{5q} = \dfrac{6}{4} = \dfrac{3}{2}$

Hence, $\bm{6p : 5q = 3 : 2}$.

A field is 80 m long and 60 m wide. Find the ratio of its width to its length.

Answer

Given,

Length of the field = 80 m

Width of the field = 60 m

Ratio of width to length = 60 : 80

H.C.F. of 60 and 80 is 20.

$\dfrac{60 \div 20}{80 \div 20} = \dfrac{3}{4} = 3 : 4$

Hence, the ratio of width to length is 3 : 4.

Find the ratio between 5 dozens and 2 scores. [1 score = 20]

Answer

Given,

5 dozens = 5 × 12 = 60

2 scores = 2 × 20 = 40

Ratio = 60 : 40

H.C.F. of 60 and 40 is 20.

$\dfrac{60 \div 20}{40 \div 20} = \dfrac{3}{2} = 3 : 2$

Hence, the required ratio is 3 : 2.

The monthly income of a person is ₹ 12,000 and his monthly expenditure is ₹ 8,500. Find the ratio of his :

(i) income to expenditure

(ii) expenditure to savings

(iii) savings to income

Answer

Given,

Monthly income = ₹ 12,000

Monthly expenditure = ₹ 8,500

Monthly savings = Income − Expenditure = ₹ 12,000 − ₹ 8,500 = ₹ 3,500

(i) Ratio of income to expenditure = 12000 : 8500

H.C.F. of 12000 and 8500 is 500.

$\dfrac{12000 \div 500}{8500 \div 500} = \dfrac{24}{17} = 24 : 17$

Hence, the ratio of income to expenditure is 24 : 17.

(ii) Ratio of expenditure to savings = 8500 : 3500

H.C.F. of 8500 and 3500 is 500.

$\dfrac{8500 \div 500}{3500 \div 500} = \dfrac{17}{7} = 17 : 7$

Hence, the ratio of expenditure to savings is 17 : 7.

(iii) Ratio of savings to income = 3500 : 12000

H.C.F. of 3500 and 12000 is 500.

$\dfrac{3500 \div 500}{12000 \div 500} = \dfrac{7}{24} = 7 : 24$

Hence, the ratio of savings to income is 7 : 24.

The weekly expenses of a boy have increased from ₹ 1,500 to ₹ 2,250. Find the ratio of :

(i) increase in expenses to original expenses.

(ii) original expenses to increased expenses.

(iii) increased expenses to increase in expenses.

Answer

Given,

Original expenses = ₹ 1,500

Increased (new) expenses = ₹ 2,250

Increase in expenses = ₹ 2,250 − ₹ 1,500 = ₹ 750

(i) Ratio of increase in expenses to original expenses = 750 : 1500

H.C.F. of 750 and 1500 is 750.

$\dfrac{750 \div 750}{1500 \div 750} = \dfrac{1}{2} = 1 : 2$

Hence, the required ratio is 1 : 2.

(ii) Ratio of original expenses to increased expenses = 1500 : 2250

H.C.F. of 1500 and 2250 is 750.

$\dfrac{1500 \div 750}{2250 \div 750} = \dfrac{2}{3} = 2 : 3$

Hence, the required ratio is 2 : 3.

(iii) Ratio of increased expenses to increase in expenses = 2250 : 750

H.C.F. of 2250 and 750 is 750.

$\dfrac{2250 \div 750}{750 \div 750} = \dfrac{3}{1} = 3 : 1$

Hence, the required ratio is 3 : 1.

In a club having 360 members, 40 play carrom, 96 play table tennis, 144 play badminton and the remaining members play volley ball. If no member plays two or more games, find the ratio of members who play :

(i) carrom to the number of those who play badminton.

(ii) badminton to the number of those who play table-tennis.

(iii) table tennis to the number of those who play volley-ball.

(iv) volley ball to the number of those who play other games.

Answer

Given,

Total members = 360

Carrom = 40, Table tennis = 96, Badminton = 144

Volley ball = 360 − (40 + 96 + 144) = 360 − 280 = 80

(i) Ratio of carrom to badminton = 40 : 144

H.C.F. of 40 and 144 is 8.

$\dfrac{40 \div 8}{144 \div 8} = \dfrac{5}{18} = 5 : 18$

Hence, the required ratio is 5 : 18.

(ii) Ratio of badminton to table tennis = 144 : 96

H.C.F. of 144 and 96 is 48.

$\dfrac{144 \div 48}{96 \div 48} = \dfrac{3}{2} = 3 : 2$

Hence, the required ratio is 3 : 2.

(iii) Ratio of table tennis to volley ball = 96 : 80

H.C.F. of 96 and 80 is 16.

$\dfrac{96 \div 16}{80 \div 16} = \dfrac{6}{5} = 6 : 5$

Hence, the required ratio is 6 : 5.

(iv) Members who play other games (carrom, table tennis and badminton) = 40 + 96 + 144 = 280

Ratio of volley ball to other games = 80 : 280

H.C.F. of 80 and 280 is 40.

$\dfrac{80 \div 40}{280 \div 40} = \dfrac{2}{7} = 2 : 7$

Hence, the required ratio is 2 : 7.

₹ 120 is to be divided between Hari and Gopi in the ratio 5 : 3. How much does each get?

Answer

Given,

Total money = ₹ 120

Ratio = 5 : 3

Total parts : 5 + 3 = 8 parts.

The value of 1 part : 120 ÷ 8 = ₹ 15.

Hari's share : 5 × ₹ 15 = ₹ 75

Gopi's share : 3 × ₹ 15 = ₹ 45

Hence, Hari gets ₹ 75 and Gopi gets ₹ 45.

Divide 72 in the ratio $2\dfrac{1}{2} : 1\dfrac{1}{2}$.

Answer

Given,

Ratio = $2\dfrac{1}{2} : 1\dfrac{1}{2} = \dfrac{5}{2} : \dfrac{3}{2}$

Multiplying both terms by 2 :

$= \dfrac{5}{2} \times 2 : \dfrac{3}{2} \times 2 = 5 : 3$

Total parts : 5 + 3 = 8 parts.

The value of 1 part : 72 ÷ 8 = 9.

First part : 5 × 9 = 45

Second part : 3 × 9 = 27

Hence, 72 is divided into 45 and 27.

Divide ₹ 10,400 among A, B and C in the ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$.

Answer

Given ratio = $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$

L.C.M. of the denominators 2, 3 and 4 is 12.

Multiplying each term by 12 :

$\dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 = 6 : 4 : 3$

Let the shares of A, B and C be 6x, 4x and 3x respectively.

⇒ 6x + 4x + 3x = 10400

⇒ 13x = 10400

⇒ x = $\dfrac{10400}{13}$

⇒ x = ₹ 800

Shares are :

A's share = 6x = 6 × ₹ 800 = ₹ 4,800,

B's share = 4x = 4 × ₹ 800 = ₹ 3,200,

C's share = 3x = 3 × ₹ 800 = ₹ 2,400.

Hence, A gets ₹ 4,800, B gets ₹ 3,200 and C gets ₹ 2,400.

The angles of a triangle are in the ratio 3 : 7 : 8. Find the greatest and the smallest angles.

Answer

Given,

Ratio of the angles = 3 : 7 : 8

Let the angles be 3x, 7x and 8x.

The sum of the angles of a triangle = 180°

⇒ 3x + 7x + 8x = 180°

⇒ 18x = 180°

⇒ x = $\dfrac{180\degree}{18}$

⇒ x = 10°

Angles are :

3x = 3 × 10° = 30°,

7x = 7 × 10° = 70°,

8x = 8 × 10° = 80°.

Hence, the greatest angle is 80° and the smallest angle is 30°.

The sides of a triangle are in the ratio 3 : 2 : 4. If the perimeter of the triangle is 27 cm, find the length of each side.

Answer

Given,

Ratio of the sides = 3 : 2 : 4

Perimeter of the triangle = 27 cm

Let the sides be 3x, 2x and 4x.

⇒ 3x + 2x + 4x = 27

⇒ 9x = 27

⇒ x = $\dfrac{27}{9}$

⇒ x = 3 cm

Sides are :

3x = 3 × 3 cm = 9 cm,

2x = 2 × 3 cm = 6 cm,

4x = 4 × 3 cm = 12 cm.

Hence, the lengths of the sides are 9 cm, 6 cm and 12 cm.