Ratio & Proportion — Test Yourself

Is the ratio of 15 kg and 35 kg same as the ratio of 6 years and 14 years?

Answer

Ratio of 15 kg and 35 kg = 15 : 35

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

$\dfrac{15 \div 5}{35 \div 5} = \dfrac{3}{7} = 3 : 7$

Ratio of 6 years and 14 years = 6 : 14

H.C.F. of 6 and 14 is 2.

$\dfrac{6 \div 2}{14 \div 2} = \dfrac{3}{7} = 3 : 7$

Since both ratios are equal to 3 : 7, they are the same.

Hence, yes, the two ratios are the same.

A bus travels 180 km in 3 hours and a train travels 450 km in 5 hours. Find the ratio of the speed of train to the speed of bus.

Answer

Given,

Speed of bus = $\dfrac{180}{3}$ = 60 km/h

Speed of train = $\dfrac{450}{5}$ = 90 km/h

Ratio of speed of train to speed of bus = 90 : 60

H.C.F. of 90 and 60 is 30.

$\dfrac{90 \div 30}{60 \div 30} = \dfrac{3}{2} = 3 : 2$

Hence, the ratio of the speed of the train to the speed of the bus is 3 : 2.

In winters, a school opens at 10 a.m. and closes at 3.30 p.m. If the lunch interval is of 30 minutes, find the ratio of lunch interval to total time of the class periods.

Answer

Given,

School opens at 10 a.m. and closes at 3.30 p.m.

Total time the school is open = 5 hours 30 minutes = 5 × 60 + 30 = 330 minutes

Lunch interval = 30 minutes

Total time of the class periods = 330 − 30 = 300 minutes

Ratio of lunch interval to total time of the class periods = 30 : 300

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

$\dfrac{30 \div 30}{300 \div 30} = \dfrac{1}{10} = 1 : 10$

Hence, the ratio of lunch interval to total time of the class periods is 1 : 10.

Rohit goes to his school by car at 60 km per hour and Manoj goes to the same school by scooty at 40 km per hour. If they both live in the same locality, find the ratio between the time taken by Rohit and Manoj to reach their school.

Answer

Given,

Speed of Rohit = 60 km/h

Speed of Manoj = 40 km/h

Since both cover the same distance, time is inversely proportional to speed.

∴ Ratio of time taken by Rohit to Manoj = $\dfrac{1}{60} : \dfrac{1}{40}$

Multiplying both terms by the L.C.M. of 60 and 40, i.e. 120 :

$= \dfrac{1}{60} \times 120 : \dfrac{1}{40} \times 120 = 2 : 3$

Hence, the ratio between the time taken by Rohit and Manoj is 2 : 3.

An alloy of zinc and copper weighs $12\dfrac{1}{2}$ kg. If, in the alloy, the ratio of zinc and copper is 1 : 4, find the weight of copper in it.

Answer

Given,

Total weight of the alloy = $12\dfrac{1}{2}$ kg = $\dfrac{25}{2}$ kg = 12.5 kg

Ratio (Zinc : Copper) = 1 : 4

Let the weights of zinc and copper be x and 4x respectively.

⇒ x + 4x = 12.5

⇒ 5x = 12.5

⇒ x = $\dfrac{12.5}{5}$

⇒ x = 2.5 kg

Weight of copper = 4x = 4 × 2.5 = 10 kg.

Hence, the weight of copper in the alloy is 10 kg.

Find the fourth term of the proportion whose first, second and third terms are 18, 27 and 32 respectively.

Answer

Let the fourth term be x.

∴ 18, 27, 32 and x are in proportion, i.e. 18 : 27 : : 32 : x

In a proportion, product of extremes = product of means.

⇒ 18 × x = 27 × 32

⇒ 18x = 864

⇒ x = $\dfrac{864}{18}$ = 48

Hence, the fourth term is 48.

The ratio of the length and the width of a school ground is 5 : 2. Find the length, if the width is 40 metres.

Answer

Given,

Ratio (Length : Width) = 5 : 2

Width = 40 m

Since 2 parts represent the width, 2 parts = 40 m.

1 part = $\dfrac{40}{2}$ = 20 m.

Length = 5 parts = 5 × 20 = 100 m.

Hence, the length of the school ground is 100 m.

Two numbers are in the ratio 9 : 2. If the smaller number is 320, find the larger number.

Answer

Given,

Ratio of the two numbers = 9 : 2

The smaller number corresponds to 2 parts.

Since 2 parts represent the smaller number, 2 parts = 320.

1 part = $\dfrac{320}{2}$ = 160.

Larger number = 9 parts = 9 × 160 = 1440.

Hence, the larger number is 1440.

Ratio of the distance of the school from A's house and from B's house is 2 : 1. If both A and B have the same speed, complete the following table :

Answer

Given,

Ratio of the distance of the school from A's house to that from B's house = 2 : 1.

∴ Distance from A's house = 2 × (Distance from B's house), and Distance from B's house = $\dfrac{1}{2}$ × (Distance from A's house).

Completing each column using this relation :

When distance from A's house = 4 km, distance from B's house = $\dfrac{1}{2} \times 4 = 2$ km.

When distance from B's house = 9 km, distance from A's house = 2 × 9 = 18 km.

When distance from A's house = 8 km, distance from B's house = $\dfrac{1}{2} \times 8 = 4$ km.

When distance from B's house = 8 km, distance from A's house = 2 × 8 = 16 km.

When distance from A's house = 6 km, distance from B's house = $\dfrac{1}{2} \times 6 = 3$ km.

The completed table is :

Mr. Gupta divides ₹ 81,000 among his three children, Ashok, Mohit and Geeta, in such a way that Ashok gets four times what Mohit gets and Mohit gets 2.5 times what Geeta gets. Find the share of each of them.

Answer

Given,

Total money = ₹ 81,000

Let Geeta's share = 1 part.

Mohit gets 2.5 times what Geeta gets, so Mohit's share = 2.5 parts.

Ashok gets four times what Mohit gets, so Ashok's share = 4 × 2.5 = 10 parts.

∴ Ratio (Ashok : Mohit : Geeta) = 10 : 2.5 : 1

Multiplying each term by 2 to remove the decimal :

= 20 : 5 : 2

Let the shares of Ashok, Mohit and Geeta be 20x, 5x and 2x respectively.

⇒ 20x + 5x + 2x = 81000

⇒ 27x = 81000

⇒ x = $\dfrac{81000}{27}$

⇒ x = ₹ 3,000

Shares are :

Ashok's share = 20x = 20 × ₹ 3,000 = ₹ 60,000,

Mohit's share = 5x = 5 × ₹ 3,000 = ₹ 15,000,

Geeta's share = 2x = 2 × ₹ 3,000 = ₹ 6,000.

Hence, Ashok gets ₹ 60,000, Mohit gets ₹ 15,000 and Geeta gets ₹ 6,000.

If $(4x + 3y) : (3x + 5y)$ = 6 : 7, find :

(i) $x : y$

(ii) $x$, if $y$ = 10

(iii) $y$, if $x$ = 27

Answer

(i) Given,

$(4x + 3y) : (3x + 5y)$ = 6 : 7

⇒ $\dfrac{4x + 3y}{3x + 5y} = \dfrac{6}{7}$

By cross multiplication :

$\Rightarrow 7(4x + 3y) = 6(3x + 5y) \\[1em] \Rightarrow 28x + 21y = 18x + 30y \\[1em] \Rightarrow 28x − 18x = 30y − 21y \\[1em] \Rightarrow 10x = 9y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{9}{10}$

Hence, $\bm{x : y = 9 : 10}$.

(ii) Since, $\dfrac{x}{y} = \dfrac{9}{10}$, we have $x = \dfrac{9}{10} y$.

$x = \dfrac{9}{10} \times 10$ = 9

Hence, $\bm{x = 9}$ when $\bm{y = 10}$.

(iii) Since, $\dfrac{x}{y} = \dfrac{9}{10}$, we have $y = \dfrac{10}{9} x$.

$y = \dfrac{10}{9} \times 27$ = 10 × 3 = 30

Hence, $\bm{y = 30}$ when $\bm{x = 27}$.

If $\dfrac{2y + 5x}{3y − 5x} = 2\dfrac{1}{2}$, find :

(i) $x : y$

(ii) $x$, if $y$ = 70

(iii) $y$, if $x$ = 33

Answer

(i) Given,

$\dfrac{2y + 5x}{3y − 5x} = 2\dfrac{1}{2} = \dfrac{5}{2}$

By cross multiplication :

$\Rightarrow 2(2y + 5x) = 5(3y − 5x) \\[1em] \Rightarrow 4y + 10x = 15y − 25x \\[1em] \Rightarrow 10x + 25x = 15y − 4y \\[1em] \Rightarrow 35x = 11y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{11}{35}$

Hence, $\bm{x : y = 11 : 35}$.

(ii) Since, $\dfrac{x}{y} = \dfrac{11}{35}$, we have $x = \dfrac{11}{35} y$.

$x = \dfrac{11}{35} \times 70$ = 11 × 2 = 22

Hence, $\bm{x = 22}$ when $\bm{y = 70}$.

(iii) Since, $\dfrac{x}{y} = \dfrac{11}{35}$, we have $y = \dfrac{35}{11} x$.

$y = \dfrac{35}{11} \times 33$ = 35 × 3 = 105

Hence, $\bm{y = 105}$ when $\bm{x = 33}$.