Ratio and Proportion

By prime factorization,

22 = 2 × 11

66 = 2 × 3 × 11

H.C.F. of 22 and 66 = 2 × 11 = 22.

$$\begin{aligned}\Rightarrow 22 : 66 \\[1em] \Rightarrow \frac{22}{66} \\[1em] \Rightarrow \frac{22 \div 22}{66 \div 22} \\[1em] \Rightarrow \frac{1}{3} \\[1em] \Rightarrow 1 : 3.\end{aligned}$$

Hence, 22 : 66 = 1 : 3.

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.

$$\begin{aligned}\Rightarrow 15 : 25 \\[1em] \Rightarrow \frac{15}{25} \\[1em] \Rightarrow \frac{15 \div 5}{25 \div 5} \\[1em] \Rightarrow \frac{3}{5} \\[1em] \Rightarrow 3 : 5.\end{aligned}$$

Hence, 1.5 : 2.5 = 3 : 5.

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

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

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

By prime factorization,

25 = 5 × 5

50 = 2 × 5 × 5

H.C.F. of 25 and 50 = 5 × 5 = 25.

$$\begin{aligned}\Rightarrow 25 : 50 \\[1em] \Rightarrow \frac{25}{50} \\[1em] \Rightarrow \frac{25 \div 25}{50 \div 25} \\[1em] \Rightarrow \frac{1}{2} \\[1em] \Rightarrow 1 : 2.\end{aligned}$$

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

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.

$$\begin{aligned}\Rightarrow 40 : 100 \\[1em] \Rightarrow \frac{40}{100} \\[1em] \Rightarrow \frac{40 \div 20}{100 \div 20} \\[1em] \Rightarrow \frac{2}{5} \\[1em] \Rightarrow 2 : 5.\end{aligned}$$

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

₹ 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.

$$\begin{aligned}\Rightarrow 10 : 100 \\[1em] \Rightarrow \frac{10}{100} \\[1em] \Rightarrow \frac{10 \div 10}{100 \div 10} \\[1em] \Rightarrow \frac{1}{10} \\[1em] \Rightarrow 1 : 10.\end{aligned}$$

Hence, 10 paise : ₹ 1 = 1 : 10.

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.

$$\begin{aligned}\Rightarrow 200 : 5000 \\[1em] \Rightarrow \frac{200}{5000} \\[1em] \Rightarrow \frac{200 \div 200}{5000 \div 200} \\[1em] \Rightarrow \frac{1}{25} \\[1em] \Rightarrow 1 : 25.\end{aligned}$$

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

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.

$$\begin{aligned}\Rightarrow 3 : 24 \\[1em] \Rightarrow \frac{3}{24} \\[1em] \Rightarrow \frac{3 \div 3}{24 \div 3} \\[1em] \Rightarrow \frac{1}{8} \\[1em] \Rightarrow 1 : 8.\end{aligned}$$

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

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

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

By prime factorization,

6 = 2 × 3

16 = 2 × 2 × 2 × 2

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

$$\begin{aligned}\Rightarrow 6 : 16 \\[1em] \Rightarrow \frac{6}{16} \\[1em] \Rightarrow \frac{6 \div 2}{16 \div 2} \\[1em] \Rightarrow \frac{3}{8} \\[1em] \Rightarrow 3 : 8.\end{aligned}$$

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

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

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

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

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

Given:

Total length = 64 cm

Ratio = 5 : 3

Let the parts be 5x and 3x.

Total = 3x + 5x = 8x

Now, 8x = 64

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

First part = 5x = 5 × 8 = 40 cm

Second part = 3x = 3 × 8 = 24 cm

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

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

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

x's share = 4a = 4 × ₹ 80 = ₹ 320

y's share = 5a = 5 × ₹ 80 = ₹ 400

Hence, x gets ₹ 320 and y gets ₹ 400.

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°.

Given:

Length = 100 m

Breadth = 80 m

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

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

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

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.

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

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

Given:

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

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

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

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

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

Now, 1639x = 4917

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

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.

Given:

Ratio = 3 : 4

Let the quantities be 3x and 4x.

First quantity = ₹ 810

3x = 810

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

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

Hence, the second quantity is ₹ 1080.

Given:

Ratio = 5 : 7

Let the numbers be 5x and 7x.

Difference = 7x − 5x = 2x

Now, 2x = 10

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

First number = 5x = 5 × 5 = 25

Second number = 7x = 7 × 5 = 35

Hence, the numbers are 25 and 35.

Given:

Ratio = 10 : 11

Let the numbers be 10x and 11x.

Sum = 10x + 11x = 21x

Now, 21x = 168

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

First number = 10x = 10 × 8 = 80

Second number = 11x = 11 × 8 = 88

Hence, the numbers are 80 and 88.

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.

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

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.

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 = $\frac{8x}{15x} = \frac{8}{15}$

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

Given:

Total population = 180,000

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

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

= 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.

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

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

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.

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

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.

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

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

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.

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

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

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

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.

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

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

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.

Given:

Total area = 600 sq m

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

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

Let the shares be 3x and 5x.

Total = 3x + 5x = 8x

Now, 8x = 600

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

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.

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.

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

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

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

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

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

First number = 4x = 4 × 6 = 24

Second number = 7x = 7 × 6 = 42

Hence, the numbers are 24 and 42.

Given:

Total money = ₹ 300

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

(i) A : B = $\frac{1}{2}$ : 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 = $\frac{300}{3}$ = 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.

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.

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 = $\frac{1600}{80}$ = 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.

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 = $\frac{45000}{3}$ = 15000

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

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

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 3x × 56 = 168x

Product of means = 7x × 24 = 168x

Since product of extremes = product of means, the quantities are in proportion.

Hence, 3x, 7x, 24 and 56 form a proportion.

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 0.8 × 9 = 7.2

Product of means = 3 × 2.4 = 7.2

Since product of extremes = product of means, the quantities are in proportion.

Hence, 0.8, 3, 2.4 and 9 form a proportion.

Four quantities form a proportion if product of extremes = product of means.

$1\frac{1}{2}, 3\frac{1}{4}, 4\frac{1}{2} \text{ and } 9\frac{3}{4} = \frac{3}{2}, \frac{13}{4}, \frac{9}{2} \text{ and } \frac{39}{4}$

Product of extremes = $\frac{3}{2} \times \frac{39}{4} = \frac{117}{8}$

Product of means = $\frac{13}{4} \times \frac{9}{2} = \frac{117}{8}$

Since product of extremes = product of means, the quantities are in proportion.

Hence, $1\frac{1}{2}, 3\frac{1}{4}, 4\frac{1}{2} \text{ and } 9\frac{3}{4}$ form a proportion.

Four quantities form a proportion if product of extremes = product of means.

Product of extremes = 0.4 × 3.5 = 1.4

Product of means = 0.5 × 2.9 = 1.45

Since product of extremes ≠ product of means, the quantities are not in proportion.

Hence, 0.4, 0.5, 2.9 and 3.5 do not form a proportion.

Four quantities form a proportion if product of extremes = product of means.

$2\frac{1}{2}, 5\frac{1}{2}, 3.0 \text{ and } 6.0 = \frac{5}{2}, \frac{11}{2}, 3 \text{ and } 6$

Product of extremes = $\frac{5}{2} \times 6 = 15$

Product of means = $\frac{11}{2} \times 3 = \frac{33}{2} = 16.5$

Since product of extremes ≠ product of means, the quantities are not in proportion.

Hence, $2\frac{1}{2}, 5\frac{1}{2}, 3.0 \text{ and } 6.0$ do not form a proportion.

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 3, b = 12, c = 4

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x}\\[1em] \Rightarrow x = \frac{b \times c}{a}\\[1em] = \frac{12 \times 4}{3} \\[1em] = \frac{48}{3}\\[1em] = 16\end{aligned}$$

Hence, the fourth proportional is 16.

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 5, b = 9, c = 45

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x}\\[1em] \Rightarrow x = \frac{b \times c}{a}\\[1em] = \frac{9 \times 45}{5} \\[1em] = \frac{405}{5} \\[1em] = 81\end{aligned}$$

Hence, the fourth proportional is 81.

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 2.1, b = 1.5, c = 8.4

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x}\\[1em] \Rightarrow x = \frac{b \times c}{a}\\[1em] = \frac{1.5 \times 8.4}{2.1} \\[1em] = \frac{12.6}{2.1} \\[1em] = 6.0\end{aligned}$$

Hence, the fourth proportional is 6.0

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where $a = \frac{1}{3}, b = \frac{2}{5}, c = 8.4$

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x}\\[1em] \Rightarrow x = \frac{b \times c}{a}\\[1em] = \frac{\frac{2}{5} \times 8.4}{\frac{1}{3}} \\[1em] = \frac{2}{5} \times 8.4 \times 3 \\[1em] = \frac{2 \times 8.4 \times 3}{5} \\[1em] = \frac{50.4}{5} \\[1em] = 10.08\end{aligned}$$

Hence, the fourth proportional is 10.08.

Convert into minutes:

4 hours 40 minutes = (4 × 60) + 40 = 280 minutes

1 hour 10 minutes = (1 × 60) + 10 = 70 minutes

16 hours = 16 × 60 = 960 minutes

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 280, b = 70, c = 960

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x}\\[1em] \Rightarrow x = \frac{b \times c}{a}\\[1em] = \frac{70 \times 960}{280} \\[1em] = \frac{67200}{280} \\[1em] = 240 \text{ minutes} = 4 \text{ hours}\end{aligned}$$

Hence, the fourth proportional is 4 hours.

Let the third proportional to a and b be x, then a : b :: b : x, where a = 27, b = 9

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x}\\[1em] \Rightarrow x = \frac{b \times b}{a} = \frac{b^2}{a}\\[1em] = \frac{9^2}{27} \\[1em] = \frac{81}{27} \\[1em] = 3\end{aligned}$$

Hence, the third proportional is 3.

2 m 40 cm = 240 cm.

Let the third proportional to a and b be x, then a : b :: b : x, where a = 240, b = 40

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x}\\[1em] \Rightarrow x = \frac{b \times b}{a} = \frac{b^2}{a}\\[1em] = \frac{40^2}{240} \\[1em] = \frac{1600}{240} \\[1em] = \frac{20}{3} = 6\frac{2}{3} \text{ cm}\end{aligned}$$

Hence, the third proportional is $6\frac{2}{3}$ cm.

Let the third proportional to a and b be x, then a : b :: b : x, where a = 1.8, b = 0.6

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x}\\[1em] \Rightarrow x = \frac{b \times b}{a} = \frac{b^2}{a}\\[1em] = \frac{0.6^2}{1.8} \\[1em] = \frac{0.36}{1.8} \\[1em] = 0.2\end{aligned}$$

Hence, the third proportional is 0.2.

Let the third proportional to a and b be x, then a : b :: b : x, where $a = \frac{1}{7}, b = \frac{3}{14}$

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x}\\[1em] \Rightarrow x = \frac{b \times b}{a} = \frac{b^2}{a}\\[1em] = \frac{\left(\frac{3}{14}\right)^2}{\frac{1}{7}} \\[1em] = \frac{9}{196} \times 7 \\[1em] = \frac{63}{196} \\[1em] = \frac{9}{28}\end{aligned}$$

Hence, the third proportional is $\frac{9}{28}$.

Let the third proportional to a and b be x, then a : b :: b : x, where a = 1.6, b = 0.8

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x}\\[1em] \Rightarrow x = \frac{b \times b}{a} = \frac{b^2}{a}\\[1em] = \frac{0.8^2}{1.6} \\[1em] = \frac{0.64}{1.6} \\[1em] = 0.4\end{aligned}$$

Hence, the third proportional is 0.4.

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 16, b = 4

$$\begin{aligned}\Rightarrow \frac{a}{x} = \frac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{16 \times 4} \\[1em] = \sqrt{64} \\[1em] = 8\end{aligned}$$

Hence, the mean proportional is 8.

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 3, b = 27

$$\begin{aligned}\Rightarrow \frac{a}{x} = \frac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{3 \times 27} \\[1em] = \sqrt{81} \\[1em] = 9\end{aligned}$$

Hence, the mean proportional is 9.

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 0.9, b = 2.5

$$\begin{aligned}\Rightarrow \frac{a}{x} = \frac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{0.9 \times 2.5} \\[1em] = \sqrt{2.25} \\[1em] = 1.5\end{aligned}$$

Hence, the mean proportional is 1.5.

Let the mean proportional between a and b be x, then a : x :: x : b, where a = 0.6, b = 9.6

$$\begin{aligned}\Rightarrow \frac{a}{x} = \frac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{0.6 \times 9.6} \\[1em] = \sqrt{5.76} \\[1em] = 2.4\end{aligned}$$

Hence, the mean proportional is 2.4.

Let the mean proportional between a and b be x, then a : x :: x : b, where $a = \frac{1}{4}, b = \frac{1}{16}$

$$\begin{aligned}\Rightarrow \frac{a}{x} = \frac{x}{b}\\[1em] \Rightarrow x^2 = a \times b \\[1em] \Rightarrow x = \sqrt{a \times b}\\[1em] = \sqrt{\frac{1}{4} \times \frac{1}{16}} \\[1em] = \sqrt{\frac{1}{64}} \\[1em] = \frac{1}{8}\end{aligned}$$

Hence, the mean proportional is $\frac{1}{8}$.

A : B = 3 : 5 and B : C = 4 : 7

To combine, make the value of B the same. B appears as 5 and 4; their L.C.M. is 20.

A : B = 3 : 5 = (3 × 4) : (5 × 4) = 12 : 20

B : C = 4 : 7 = (4 × 5) : (7 × 5) = 20 : 35

∴ A : B : C = 12 : 20 : 35

Hence, A : B : C = 12 : 20 : 35.

x : y = 2 : 3 and y : z = 5 : 7

Make the value of y the same. y appears as 3 and 5; their L.C.M. is 15.

x : y = 2 : 3 = (2 × 5) : (3 × 5) = 10 : 15

y : z = 5 : 7 = (5 × 3) : (7 × 3) = 15 : 21

∴ x : y : z = 10 : 15 : 21

Hence, x : y : z = 10 : 15 : 21.

$$\begin{aligned}\text{Since, } m : n = 4 : 9 \Rightarrow \frac{m}{n} = \frac{4}{9} \text{ and } n : s = 3 : 7 \Rightarrow \frac{n}{s} = \frac{3}{7} \\[1em] \Rightarrow \frac{m}{s} = \frac{m}{n} \times \frac{n}{s} \\[1em] = \frac{4}{9} \times \frac{3}{7} \\[1em] = \frac{12}{63} \\[1em] = \frac{4}{21}\end{aligned}$$

i.e., m : s = 4 : 21

Hence, m : s = 4 : 21.

P : Q = $\frac{1}{2} : \frac{1}{3}$ = (multiplying both terms by 6) = 3 : 2

Q : R = $1\frac{1}{2} : 1\frac{1}{3} = \frac{3}{2} : \frac{4}{3}$ = (multiplying both terms by 6) = 9 : 8

$$\begin{aligned}\text{Since, } \text{ P : Q }= 3 : 2 \Rightarrow \frac{\text{ P }}{\text{ Q }} = \frac{3}{2} \text{ and } \text{ Q : R }= 9 : 8 \Rightarrow \frac{\text{ Q }}{\text{ R }} = \frac{9}{8} \\[1em] \Rightarrow \frac{\text{ P }}{\text{ R }} = \frac{\text{ P }}{\text{ Q }} \times \frac{\text{ Q }}{\text{ R }} \\[1em] = \frac{3}{2} \times \frac{9}{8} \\[1em] = \frac{27}{16}\end{aligned}$$

i.e., P : R = 27 : 16

Hence, P : R = 27 : 16.

a : b = 1.5 : 3.5 = 15 : 35 = 3 : 7

$$\begin{aligned}\text{Since, } a : b = 3 : 7 \Rightarrow \frac{a}{b} = \frac{3}{7} \text{ and } b : c = 5 : 6 \Rightarrow \frac{b}{c} = \frac{5}{6} \\[1em] \Rightarrow \frac{a}{c} = \frac{a}{b} \times \frac{b}{c} \\[1em] = \frac{3}{7} \times \frac{5}{6} \\[1em] = \frac{15}{42} \\[1em] = \frac{5}{14}\end{aligned}$$

i.e., a : c = 5 : 14

Hence, a : c = 5 : 14.

p : q = $1\frac{1}{4} : 2\frac{1}{3} = \frac{5}{4} : \frac{7}{3}$ = (multiplying both terms by 12) = 15 : 28

q : r = $4\frac{1}{2} : 5\frac{1}{4} = \frac{9}{2} : \frac{21}{4}$ = (multiplying both terms by 4) = 18 : 21 = 6 : 7

$$\begin{aligned}\text{Since, } p : q = 15 : 28 \Rightarrow \frac{p}{q} = \frac{15}{28} \text{ and } q : r = 6 : 7 \Rightarrow \frac{q}{r} = \frac{6}{7} \\[1em] \Rightarrow \frac{p}{r} = \frac{p}{q} \times \frac{q}{r} \\[1em] = \frac{15}{28} \times \frac{6}{7} \\[1em] = \frac{90}{196} \\[1em] = \frac{45}{98}\end{aligned}$$

i.e., p : r = 45 : 98

Hence, p : r = 45 : 98.

Given:

2 : x = 3 : 8

Product of extremes = product of means:

2 × 8 = 3 × x

x = $\frac{16}{3}$

Now, x : y = 5 : 4

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{5}{4} \\[1em] \Rightarrow y = \frac{4x}{5}\\[1em] \Rightarrow y = \frac{4}{5} \times \frac{16}{3} \\[1em] \Rightarrow y = \frac{64}{15} = 4\frac{4}{15}\end{aligned}$$

Hence, the value of y is $\frac{64}{15}, \text { i.e. } 4\frac{4}{15}$.

Given:

2.5 : 4 = x : 7.5

$\Rightarrow \frac{2.5}{4} = \frac{x}{7.5}$

Product of extremes = Product of means

⇒ 2.5 × 7.5 = 4 × x

$$\begin{aligned}\Rightarrow 4x = 18.75 \\[1em] \Rightarrow x = \frac{18.75}{4} \\[1em] = \frac{1875}{400} \\[1em] = \frac{75}{16} \\[1em] = 4\frac{11}{16}\end{aligned}$$

Hence, the value of x is $4\frac{11}{16}$.

Three quantities a, b and c are in continued proportion if a : b :: b : c, i.e. if b² = a × c.

Here, a = 2, b = 12 and c = 72.

b² = 12² = 144

a × c = 2 × 72 = 144

Since b² = a × c (144 = 144), the quantities are in continued proportion.

Hence, 2, 12 and 72 are in continued proportion.

As 1 m = 100 cm

2 m = 2 × 100 cm = 200 cm.

Ratio of 2 m to 400 cm = 200 : 400 = 1 : 2.

Hence, option 1 is the correct option.

$\frac{1}{2} : \frac{1}{3}$

Multiply both terms by 6

$= \frac{1}{2} \times 6 : \frac{1}{3} \times 6$

= 3 : 2.

Hence, option 2 is the correct option.

A : B = 3 : 5 and B : C = 15 : 23

$\frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{3}{5} \times \frac{15}{23} = \frac{45}{115} = \frac{9}{23}$

So, A : C = 9 : 23

Hence, option 3 is the correct option.

Given, A : B : C = 8 : 6 : 3

Let A = 8x, B = 6x and C = 3x.

Given, C = 24

⇒ 3x = 24

⇒ x = $\frac{24}{3}$ = 8

∴ B = 6x = 6 × 8 = 48

Hence, option 1 is the correct option.

Let the original salary be ₹ 4x and the new salary be ₹ 7x.

Original salary = ₹ 2,000

4x = 2000

x = $\frac{2000}{4}$ = 500

New salary = 7x = 7 × ₹ 500 = ₹ 3,500

Hence, option 4 is the correct option.

Given,

8 : x = $\frac{1}{3} : \frac{1}{2}$

Product of extremes = Product of means

$8 \times \frac{1}{2} = x \times \frac{1}{3}$

4 = $\frac{x}{3}$

x = 12.

Hence, option 3 is the correct option.

Let the fourth proportional to a, b and c be x, then a : b :: c : x, where a = 6, b = 8, c = 10.

⇒ 6 : 8 :: 10 : x

Here, the extremes are 6 and x, and the means are 8 and 10.

Product of extremes = Product of means

⇒ 6 × x = 8 × 10

$$\begin{aligned}\Rightarrow x = \frac{8 \times 10}{6} \\[1em] = \frac{80}{6} \\[1em] = \frac{40}{3}\end{aligned}$$

Hence, option 2 is the correct option.

Let the third proportional to a and b be x, then a, b and x are in continued proportion, i.e. a : b :: b : x, where a = 2, b = 1.

⇒ 2 : 1 :: 1 : x

Here, the extremes are 2 and x, and the means are 1 and 1.

Product of extremes = Product of means

⇒ 2 × x = 1 × 1

$$\begin{aligned}\Rightarrow x = \frac{1 \times 1}{2} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, option 3 is the correct option.

Let the length on the map representing 132 m be x cm.

The actual lengths and the lengths on the map are in proportion.

⇒ 22 : 3 :: 132 : x

Product of extremes = Product of means

⇒ 22 × x = 3 × 132

$$\begin{aligned}\Rightarrow x = \frac{3 \times 132}{22} \\[1em] = \frac{396}{22} \\[1em] = 18 \text{ cm}\end{aligned}$$

Since 18 cm is not among options 1, 2 and 3, the correct choice is "none of these".

Hence, option 4 is the correct option.

L.C.M. of the denominators 3, 6 and 8 is 24.

Multiply each term by 24:

$\frac{1}{3} \times 24 : \frac{1}{6} \times 24 : \frac{1}{8} \times 24$

= 8 : 4 : 3

Hence, option 4 is the correct option.

Let A = 2B = 3C = k.

Then A = k, B = $\frac{k}{2} \text{ and } C = \frac{k}{3}$.

A : B : C = $k : \frac{k}{2} : \frac{k}{3} = 1 : \frac{1}{2} : \frac{1}{3}$

Multiply each term by 6:

$1 \times 6 : \frac{1}{2} \times 6 : \frac{1}{3} \times 6$

= 6 : 3 : 2.

Hence, option 3 is the correct option.

Let A = 3x and B = 4x.

A + B = 35

3x + 4x = 35

7x = 35

x = $\frac{35}{7}$ = 5

B = 4x = 4 × 5 = 20

Hence, option 4 is the correct option.

$$\begin{aligned}2 : 3 = \frac{2}{3} \text{ and } 3 : 4 = \frac{3}{4}.\\[1em] \text { L.C.M. of 3 and 4 is 12.}\\[1em] \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}\\[1em] \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}\\[1em]\end{aligned}$$

Clearly, $\frac{9}{12} \gt \frac{8}{12}$, so (3 : 4) > (2 : 3).

Hence, option 2 is the correct option.

Let the length be 8x and the breadth be 5x.

Breadth = 12.5 cm

5x = 12.5

x = $\frac{12.5}{5}$ = 2.5

Length = 8x = 8 × 2.5 = 20 cm

Hence, option 1 is the correct option.

Number of faces of a cube = 6

Number of edges of a cube = 12

Ratio = 6 : 12 = 1 : 2

Hence, option 1 is the correct option.

Let the third proportional to a and b be x, then a, b and x are in continued proportion, i.e. a : b :: b : x,

Let $a = \frac{3}{4} \text{ and }b = 1\frac{1}{8} = \frac{9}{8}$.

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{x} \\[1em] \Rightarrow x = \frac{b \times b}{a} \\[1em] \Rightarrow x = \frac{b^2}{a} \\[1em] = \frac{\left(\frac{9}{8}\right)^2}{\frac{3}{4}} \\[1em] = \frac{81}{64} \div \frac{3}{4} \\[1em] = \frac{81}{64} \times \frac{4}{3} \\[1em] = \frac{27}{16}\end{aligned}$$

So, Statement 1 is true.

We know that,

If a, c and b are in continued proportion, then a : c :: c : b. Here c is the mean proportional between a and b, while the third proportional to a and c is b. So, Statement 2 is false.

Hence, option 3 is the correct option.

In a proportion, the two terms of each ratio must be of the same kind and in the same unit (though the two ratios may be of different kinds). So, Statement 1 is true.

Solving,

1 kg : 250 g = 1000 g : 250 g = 4 : 1,

140 cm : 35 cm = 4 : 1.

Since both ratios are equal, they form a proportion. So, Statement 2 is true.

Hence, option 1 is the correct option.

Solving,

$1.7 : 0.34 = \frac{1.7}{0.34} = \frac{170}{34} = 5$ and,

$3.4 : 0.68 = \frac{3.4}{0.68} = 5$.

Both ratios equal 5 : 1, so they are equivalent. Assertion (A) is true.

We know that,

Multiplying or dividing both terms of a ratio by the same non-zero number gives an equivalent ratio. Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

Let $a = \frac{1}{3}$, $b = \frac{1}{4}$ and $c = \frac{1}{5}$.

Let the fourth proportional to a, b and c be x, then a : b :: c : x,

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{c}{x} \\[1em] \Rightarrow x = \frac{b \times c}{a} \\[1em] \Rightarrow x = \frac{\frac{1}{4} \times \frac{1}{5}}{\frac{1}{3}} \\[1em] \Rightarrow x = \frac{1}{20} \times 3 \\[1em] \Rightarrow x = \frac{3}{20}\end{aligned}$$

So, $\frac{3}{20}$ is the fourth proportional to $\frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$. Assertion (A) is true.

We know that,

Four quantities a, b, c and d are said to be in proportion if the ratio of the first two is equal to the ratio of the last two, i.e. if

$$\begin{aligned}a : b :: c : d \\[1em] \Rightarrow \frac{a}{b} = \frac{c}{d}\end{aligned}$$

Here a and d are the extremes and b and c are the means, so a × d = b × c.

The fourth term d of such a proportion is called the fourth proportional to a, b and c. Reason (R) is true.

Hence, option 3 is the correct option.

Solving,

a : b = 4 : 5 and b : c = 6 : 7

Making the value of b the same in both the ratios (L.C.M. of 5 and 6 is 30),

a : b = 4 : 5 = (4 × 6) : (5 × 6) = 24 : 30

b : c = 6 : 7 = (6 × 5) : (7 × 5) = 30 : 35

∴ a : b : c = 24 : 30 : 35. Assertion (A) is true.

We know that,

The compounded ratio of a : b and c : d is (a × c) : (b × d) = ac : bd. Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.