Ratio and Proportion — Exercise 7(A)

Express the given ratio in its simplest form:

22 : 66

Answer

By prime factorization,

22 = 2 × 11

66 = 2 × 3 × 11

H.C.F. of 22 and 66 = 2 × 11 = 22.

$\Rightarrow 22 : 66 \\[1em] \Rightarrow \dfrac{22}{66} \\[1em] \Rightarrow \dfrac{22 \div 22}{66 \div 22} \\[1em] \Rightarrow \dfrac{1}{3} \\[1em] \Rightarrow 1 : 3.$

Hence, 22 : 66 = 1 : 3.

Express the given ratio in its simplest form:

1.5 : 2.5

Answer

Multiplying both the terms by 10 to remove the decimals,

1.5 : 2.5 = 15 : 25

By prime factorization,

15 = 3 × 5

25 = 5 × 5

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

$\Rightarrow 15 : 25 \\[1em] \Rightarrow \dfrac{15}{25} \\[1em] \Rightarrow \dfrac{15 \div 5}{25 \div 5} \\[1em] \Rightarrow \dfrac{3}{5} \\[1em] \Rightarrow 3 : 5.$

Hence, 1.5 : 2.5 = 3 : 5.

Express the given ratio in its simplest form:

$6\dfrac{1}{4} : 12\dfrac{1}{2}$

Answer

$6\dfrac{1}{4} : 12\dfrac{1}{2} = \dfrac{25}{4} : \dfrac{25}{2}$

Multiplying both the terms by 4 (the L.C.M. of the denominators 4 and 2),

$\dfrac{25}{4} \times 4 : \dfrac{25}{2} \times 4 = 25 : 50$

By prime factorization,

25 = 5 × 5

50 = 2 × 5 × 5

H.C.F. of 25 and 50 = 5 × 5 = 25.

$\Rightarrow 25 : 50 \\[1em] \Rightarrow \dfrac{25}{50} \\[1em] \Rightarrow \dfrac{25 \div 25}{50 \div 25} \\[1em] \Rightarrow \dfrac{1}{2} \\[1em] \Rightarrow 1 : 2.$

Hence, $6\dfrac{1}{4} : 12\dfrac{1}{2}$ = 1 : 2.

Express the given ratio in its simplest form:

40 kg : 1 quintal

Answer

1 quintal = 100 kg.

⇒ 40 kg : 1 quintal = 40 : 100

By prime factorization,

40 = 2 × 2 × 2 × 5

100 = 2 × 2 × 5 × 5

H.C.F. of 40 and 100 = 2 × 2 × 5 = 20.

$\Rightarrow 40 : 100 \\[1em] \Rightarrow \dfrac{40}{100} \\[1em] \Rightarrow \dfrac{40 \div 20}{100 \div 20} \\[1em] \Rightarrow \dfrac{2}{5} \\[1em] \Rightarrow 2 : 5.$

Hence, 40 kg : 1 quintal = 2 : 5.

Express the given ratio in its simplest form:

10 paise : ₹ 1

Answer

₹ 1 = 100 paise.

⇒ 10 paise : ₹ 1 = 10 : 100

By prime factorization,

10 = 2 × 5

100 = 2 × 2 × 5 × 5

H.C.F. of 10 and 100 = 2 × 5 = 10.

$\Rightarrow 10 : 100 \\[1em] \Rightarrow \dfrac{10}{100} \\[1em] \Rightarrow \dfrac{10 \div 10}{100 \div 10} \\[1em] \Rightarrow \dfrac{1}{10} \\[1em] \Rightarrow 1 : 10.$

Hence, 10 paise : ₹ 1 = 1 : 10.

Express the given ratio in its simplest form:

200 m : 5 km

Answer

1 km = 1000 m

⇒ 5 km = 5 × 1000 m = 5000 m.

⇒ 200 m : 5 km = 200 : 5000

By prime factorization,

200 = 2 × 2 × 2 × 5 × 5

5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5

H.C.F. of 200 and 5000 = 2 × 2 × 2 × 5 × 5 = 200.

$\Rightarrow 200 : 5000 \\[1em] \Rightarrow \dfrac{200}{5000} \\[1em] \Rightarrow \dfrac{200 \div 200}{5000 \div 200} \\[1em] \Rightarrow \dfrac{1}{25} \\[1em] \Rightarrow 1 : 25.$

Hence, 200 m : 5 km = 1 : 25.

Express the given ratio in its simplest form:

3 hours : 1 day

Answer

1 day = 24 hours.

⇒ 3 hours : 1 day = 3 : 24

By prime factorization,

24 = 2 × 2 × 2 × 3

H.C.F. of 3 and 24 = 3.

$\Rightarrow 3 : 24 \\[1em] \Rightarrow \dfrac{3}{24} \\[1em] \Rightarrow \dfrac{3 \div 3}{24 \div 3} \\[1em] \Rightarrow \dfrac{1}{8} \\[1em] \Rightarrow 1 : 8.$

Hence, 3 hours : 1 day = 1 : 8.

Express the given ratio in its simplest form:

6 months : $1\dfrac{1}{3}$ years

Answer

$1\dfrac{1}{3} \text{ years} = \dfrac{4}{3} \text{ years} = \dfrac{4}{3} \times 12 \text{ months} = 16 \text{ months}$

⇒ 6 months : $1\dfrac{1}{3}$ years = 6 : 16

By prime factorization,

6 = 2 × 3

16 = 2 × 2 × 2 × 2

H.C.F. of 6 and 16 = 2.

$\Rightarrow 6 : 16 \\[1em] \Rightarrow \dfrac{6}{16} \\[1em] \Rightarrow \dfrac{6 \div 2}{16 \div 2} \\[1em] \Rightarrow \dfrac{3}{8} \\[1em] \Rightarrow 3 : 8.$

Hence, 6 months : $1\dfrac{1}{3}$ years = 3 : 8.

Express the given ratio in its simplest form:

$1\dfrac{1}{3} : 2\dfrac{1}{4} : 2\dfrac{1}{2}$

Answer

$1\dfrac{1}{3} : 2\dfrac{1}{4} : 2\dfrac{1}{2} = \dfrac{4}{3} : \dfrac{9}{4} : \dfrac{5}{2}$

Multiplying each term by 12 (the L.C.M. of the denominators 3, 4 and 2),

$\dfrac{4}{3} \times 12 : \dfrac{9}{4} \times 12 : \dfrac{5}{2} \times 12 = 16 : 27 : 30$

Hence, $1\dfrac{1}{3} : 2\dfrac{1}{4} : 2\dfrac{1}{2}$ = 16 : 27 : 30.

Divide 64 cm long string into two parts in the ratio 5 : 3.

Answer

Given:

Total length = 64 cm

Ratio = 5 : 3

Let the parts be 5x and 3x.

Total = 3x + 5x = 8x

Now, 8x = 64

$x = \dfrac{64}{8} = 8$

First part = 5x = 5 × 8 = 40 cm

Second part = 3x = 3 × 8 = 24 cm

Hence, the two parts are 40 cm and 24 cm.

₹ 720 is divided between x and y in the ratio 4 : 5. How many rupees will each get?

Answer

Given:

Total money = ₹ 720

Ratio = 4 : 5

Let x's share be ₹ 4a and y's share be ₹ 5a.

Total = 4a + 5a = 9a

Now, 9a = 720

$a = \dfrac{720}{9} = 80$

x's share = 4a = 4 × ₹ 80 = ₹ 320

y's share = 5a = 5 × ₹ 80 = ₹ 400

Hence, x gets ₹ 320 and y gets ₹ 400.

The angles of a triangle are in the ratio 3 : 2 : 7. Find each angle.

Answer

Given:

Ratio of angles = 3 : 2 : 7

Let the angles be 3x, 2x and 7x.

We know that the sum of the angles of a triangle is 180°.

⇒ 3x + 2x + 7x = 180°

⇒ 12x = 180°

⇒ x = 15°

First angle = 3x = 3 × 15° = 45°

Second angle = 2x = 2 × 15° = 30°

Third angle = 7x = 7 × 15° = 105°

Hence, the angles of the triangle are 45°, 30° and 105°.

A rectangular field is 100 m by 80 m. Find the ratio of:

(i) length to its breadth

(ii) breadth to its perimeter.

Answer

Given:

Length = 100 m

Breadth = 80 m

(i) Ratio of length to breadth = 100 : 80

H.C.F. of 100 and 80 is 20.

$\dfrac{100 \div 20}{80 \div 20} = \dfrac{5}{4} = 5 : 4$

Hence, the ratio of length to breadth is 5 : 4.

(ii) Perimeter = 2 × (length + breadth) = 2 × (100 + 80) = 2 × 180 = 360 m

Ratio of breadth to perimeter = 80 : 360

H.C.F. of 80 and 360 is 40.

$\dfrac{80 \div 40}{360 \div 40} = \dfrac{2}{9} = 2 : 9$

Hence, the ratio of breadth to perimeter is 2 : 9.

The sum of three numbers, whose ratios are $3\dfrac{1}{3} : 4\dfrac{1}{5} : 6\dfrac{1}{8}$ is 4917. Find the numbers.

Answer

Given:

$\text{Ratio} = 3\dfrac{1}{3} : 4\dfrac{1}{5} : 6\dfrac{1}{8} = \dfrac{10}{3} : \dfrac{21}{5} : \dfrac{49}{8}$

L.C.M. of the denominators 3, 5 and 8 is 120. Multiply each term by 120:

$\dfrac{10}{3} \times 120 : \dfrac{21}{5} \times 120 : \dfrac{49}{8} \times 120 = 400 : 504 : 735$

Let the numbers be 400x, 504x and 735x.

Total = 400x + 504x + 735x = 1639x

Now, 1639x = 4917

$x = \dfrac{4917}{1639} = 3$

First number = 400x = 400 × 3 = 1200

Second number = 504x = 504 × 3 = 1512

Third number = 735x = 735 × 3 = 2205

Hence, the numbers are 1200, 1512 and 2205.

The ratio between two quantities is 3 : 4. If the first is ₹ 810, find the second.

Answer

Given:

Ratio = 3 : 4

Let the quantities be 3x and 4x.

First quantity = ₹ 810

3x = 810

$x = \dfrac{810}{3} = 270$

Second quantity = 4x = 4 × ₹ 270 = ₹ 1080

Hence, the second quantity is ₹ 1080.

Two numbers are in the ratio 5 : 7. Their difference is 10. Find the numbers.

Answer

Given:

Ratio = 5 : 7

Let the numbers be 5x and 7x.

Difference = 7x − 5x = 2x

Now, 2x = 10

$x = \dfrac{10}{2} = 5$

First number = 5x = 5 × 5 = 25

Second number = 7x = 7 × 5 = 35

Hence, the numbers are 25 and 35.

Two numbers are in the ratio 10 : 11. Their sum is 168. Find the numbers.

Answer

Given:

Ratio = 10 : 11

Let the numbers be 10x and 11x.

Sum = 10x + 11x = 21x

Now, 21x = 168

$x = \dfrac{168}{21} = 8$

First number = 10x = 10 × 8 = 80

Second number = 11x = 11 × 8 = 88

Hence, the numbers are 80 and 88.

A line is divided into two parts in the ratio 2.5 : 1.3. If the smaller one is 35.1 cm, find the length of the line.

Answer

Given:

Ratio = 2.5 : 1.3

Let the two parts be 2.5x and 1.3x.

The smaller part is 1.3x, so 1.3x = 35.1 cm.

$x = \dfrac{35.1}{1.3} = \dfrac{351}{13} = 27$

Length of the line = 2.5x + 1.3x = 3.8x = 3.8 × 27 = 102.6 cm

Hence, the length of the line is 102.6 cm.

In a class, the ratio of boys to the girls is 7 : 8. What part of the whole class are girls?

Answer

Given:

Ratio of boys to girls = 7 : 8

Let the number of boys be 7x and the number of girls be 8x.

Total students = 7x + 8x = 15x

Part of the class that are girls = $\dfrac{8x}{15x} = \dfrac{8}{15}$

Hence, girls are $\dfrac{8}{15}$ of the whole class.

The population of a town is 180,000, out of which males are $\dfrac{1}{3}$ of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.

Answer

Given:

Total population = 180,000

Number of males = $\dfrac{1}{3} \text{ of } 180000$

= $\dfrac{1}{3} \times 180000$

= 60,000

Number of females = Total population − Number of males

Number of females = 180,000 − 60,000 = 120,000

Ratio of females to whole population = 120,000 : 180,000

H.C.F. of 120,000 and 180,000 is 60,000.

$\dfrac{120000 \div 60000}{180000 \div 60000} = \dfrac{2}{3} = 2 : 3$

Hence, the number of females is 120,000 and the ratio of females to the whole population is 2 : 3.

Ten grams of an alloy of metals A and B contains 7.5 g of metal A and the rest is metal B. Find the ratio between:

(i) the weights of metals A and B in the alloy.

(ii) the weight of metal B and the weight of the alloy.

Answer

Given:

Total weight of alloy = 10 g

Weight of metal A = 7.5 g

Weight of metal B = 10 − 7.5 = 2.5 g

(i) Ratio of weights of A and B = 7.5 : 2.5

Multiply both terms by 10,

⇒ A : B = 75 : 25

H.C.F. of 75 and 25 is 25.

$\dfrac{75 \div 25}{25 \div 25} = \dfrac{3}{1} = 3 : 1$

Hence, the ratio of the weights of A and B is 3 : 1.

(ii) Ratio of weight of B to weight of alloy = 2.5 : 10

Multiply both terms by 10: 25 : 100

H.C.F. of 25 and 100 is 25.

$\dfrac{25 \div 25}{100 \div 25} = \dfrac{1}{4} = 1 : 4$

Hence, the ratio of the weight of B to the weight of the alloy is 1 : 4.

The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide ₹ 3,150 in the ratio of their ages.

Answer

Given:

Age of A = 6 years 8 months = (6 × 12) + 8 = 80 months

Age of B = 7 years 4 months = (7 × 12) + 4 = 88 months

Ratio of their ages = 80 : 88

H.C.F. of 80 and 88 is 8.

$\dfrac{80 \div 8}{88 \div 8} = \dfrac{10}{11} = 10 : 11$

Now divide ₹ 3,150 in the ratio 10 : 11.

Let A's share be ₹ 10x and B's share be ₹ 11x.

Total = 10x + 11x = 21x

Now, 21x = 3150

$x = \dfrac{3150}{21} = 150$

A's share = 10x = 10 × ₹ 150 = ₹ 1,500

B's share = 11x = 11 × ₹ 150 = ₹ 1,650

Hence, A gets ₹ 1,500 and B gets ₹ 1,650.

Three persons start a business and spend ₹ 25,000, ₹ 15,000 and ₹ 40,000 respectively. Find the share of each out of a profit of ₹ 14,400 in a year.

Answer

Given:

Investments = ₹ 25,000, ₹ 15,000 and ₹ 40,000

Ratio of investments = 25000 : 15000 : 40000 = 25 : 15 : 40 = 5 : 3 : 8

Total profit = ₹ 14,400

Let the shares be 5x, 3x and 8x.

Total = 5x + 3x + 8x = 16x

Now, 16x = 14400

$x = \dfrac{14400}{16} = 900$

First person's share = 5x = 5 × ₹ 900 = ₹ 4,500

Second person's share = 3x = 3 × ₹ 900 = ₹ 2,700

Third person's share = 8x = 8 × ₹ 900 = ₹ 7,200

Hence, the shares are ₹ 4,500, ₹ 2,700 and ₹ 7,200.

A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifths of what the second gets. Find the share of each.

Answer

Given:

Total area = 600 sq m

The first person gets $\dfrac{3}{5}$ of what the second gets.

So, first person's share : second person's share = $\dfrac{3}{5} : 1 = 3 : 5$

Let the shares be 3x and 5x.

Total = 3x + 5x = 8x

Now, 8x = 600

$x = \dfrac{600}{8} = 75$

First person's share = 3x = 3 × 75 = 225 sq m

Second person's share = 5x = 5 × 75 = 375 sq m

Hence, the first person gets 225 sq m and the second person gets 375 sq m.

Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2 : 3. If the height of the smaller pole is 7.5 m, find the height of the other pole.

Answer

Given:

Ratio between the lengths of the shadows = 2 : 3

At the same time, the heights of the poles are in the same ratio as the lengths of their shadows.

So, ratio of heights of the poles = 2 : 3

Let the heights of the poles be 2x and 3x.

The smaller pole is 2x, so 2x = 7.5 m.

$x = \dfrac{7.5}{2} = 3.75$

Height of the other pole = 3x = 3 × 3.75 = 11.25 m

Hence, the height of the other pole is 11.25 m.

Two numbers are in the ratio 4 : 7. If their L.C.M. is 168, find the numbers.

Answer

Given:

Ratio = 4 : 7

Let the numbers be 4x and 7x, where x is their H.C.F.

Since 4 and 7 are co-prime, the L.C.M. of 4x and 7x = 4 × 7 × x = 28x.

Given, 28x = 168

$x = \dfrac{168}{28} = 6$

First number = 4x = 4 × 6 = 24

Second number = 7x = 7 × 6 = 42

Hence, the numbers are 24 and 42.

₹ 300 is divided between A and B in such a way that A gets half of B. Find:

(i) the ratio between the shares of A and B.

(ii) the share of A and the share of B.

Answer

Given:

Total money = ₹ 300

A gets half of B, i.e. A = $\dfrac{1}{2}$ B.

(i) A : B = $\dfrac{1}{2}$ : 1 = 1 : 2

Hence, the ratio between the shares of A and B is 1 : 2.

(ii) Let A's share be ₹ 1x and B's share be ₹ 2x.

Total = 1x + 2x = 3x

Now, 3x = 300

x = $\dfrac{300}{3}$ = 100

A's share = 1x = ₹ 100

B's share = 2x = 2 × ₹ 100 = ₹ 200

Hence, A's share is ₹ 100 and B's share is ₹ 200.

The ratio between two numbers is 5 : 9. Find the numbers, if their H.C.F. is 16.

Answer

Given:

Ratio = 5 : 9

H.C.F. = 16

Let the numbers be 5x and 9x.

Since 5 and 9 are co-prime, the H.C.F. of 5x and 9x is x.

So, x = 16.

First number = 5x = 5 × 16 = 80

Second number = 9x = 9 × 16 = 144

Hence, the numbers are 80 and 144.

A bag contains ₹ 1,600 in the form of ₹ 10 and ₹ 20 notes. If the ratio between the numbers of ₹ 10 and ₹ 20 notes is 2 : 3; find the total number of notes in all.

Answer

Given:

Total money = ₹ 1,600

Ratio of numbers of ₹ 10 notes to ₹ 20 notes = 2 : 3

Let the number of ₹ 10 notes be 2x and the number of ₹ 20 notes be 3x.

Total value = (10 × 2x) + (20 × 3x) = 20x + 60x = 80x

Now, 80x = 1600

x = $\dfrac{1600}{80}$ = 20

Number of ₹ 10 notes = 2x = 2 × 20 = 40

Number of ₹ 20 notes = 3x = 3 × 20 = 60

Total number of notes = 40 + 60 = 100

Hence, the total number of notes is 100.

The ratio between the prices of a scooter and a refrigerator is 4 : 1. If the scooter costs ₹ 45,000 more than the refrigerator, find the price of the refrigerator.

Answer

Given:

Ratio of prices (scooter : refrigerator) = 4 : 1

Let the price of the scooter be ₹ 4x and the price of the refrigerator be ₹ 1x.

The scooter costs ₹ 45,000 more than the refrigerator.

⇒ 4x − 1x = 45000

⇒ 3x = 45000

x = $\dfrac{45000}{3}$ = 15000

Price of the refrigerator = 1x = ₹ 15,000

Hence, the price of the refrigerator is ₹ 15,000.