Ratio and Proportion

18 : 30

We have : 18 = 2 × 3² and 30 = 2 × 3 × 5

∴ H.C.F. of 18 and 30 = 2 × 3 = 6.

So, $18 : 30 = \frac{18}{30} = \frac{18 \div 6}{30 \div 6} = \frac{3}{5}$.

Hence, 18 : 30 in simplest form is 3 : 5.

7.5 : 9

Multiplying both the terms by 10 to remove the decimal, we get :

$7.5 : 9 = \frac{7.5}{9} = \frac{7.5 \times 10}{9 \times 10} = \frac{75}{90}$

We have : 75 = 3 × 5² and 90 = 2 × 3² × 5

∴ H.C.F. of 75 and 90 = 3 × 5 = 15.

So, $\frac{75}{90} = \frac{75 \div 15}{90 \div 15} = \frac{5}{6}$.

Hence, 7.5 : 9 in simplest form is 5 : 6.

$6\frac{2}{3} : 7\frac{1}{2}$

We have :

$6\frac{2}{3} : 7\frac{1}{2} = \frac{20}{3} : \frac{15}{2}$

Calculating L.C.M. of 3 and 2:

$$\begin{array}{l|rr} 2 & 3, & 2 \\ \hline 3 & 3, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 3 and 2 = 2 × 3 = 6.

Multiplying both the terms by 6, we get :

$$\begin{aligned}\Rightarrow \left(\frac{20}{3} \times 6\right) : \left(\frac{15}{2} \times 6\right) \\[1em] \Rightarrow 40 : 45\end{aligned}$$

We have : 40 = 2³ × 5 and 45 = 3² × 5

∴ H.C.F. of 40 and 45 = 5.

$\Rightarrow \frac{40}{45} = \frac{40 \div 5}{45 \div 5} = \frac{8}{9}$

Hence, $\mathbf{6\frac{2}{3} : 7\frac{1}{2}}$ in simplest form is 8 : 9.

$\frac{1}{6} : \frac{1}{9} : \frac{1}{12}$

Calculating L.C.M. of 6, 9 and 12:

$$\begin{array}{l|rrr} 2 & 6, & 9, & 12 \\ \hline 2 & 3, & 9, & 6 \\ \hline 3 & 3, & 9, & 3 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 6, 9 and 12 = 2 × 2 × 3 × 3 = 36.

Multiplying each term by 36, we get :

$\left(\frac{1}{6} \times 36\right) : \left(\frac{1}{9} \times 36\right) : \left(\frac{1}{12} \times 36\right) = 6 : 4 : 3$

Hence, $\mathbf{\frac{1}{6} : \frac{1}{9} : \frac{1}{12}}$ in simplest form is 6 : 4 : 3.

$5 : 7 : \frac{9}{2}$

Multiplying each term by 2, we get :

$(5 \times 2) : (7 \times 2) : \left(\frac{9}{2} \times 2\right) = 10 : 14 : 9$

Hence, $\mathbf{5 : 7 : \frac{9}{2}}$ in simplest form is 10 : 14 : 9.

$3\frac{1}{5} : 5\frac{1}{3} : 6\frac{2}{3}$

We have :

$3\frac{1}{5} : 5\frac{1}{3} : 6\frac{2}{3} = \frac{16}{5} : \frac{16}{3} : \frac{20}{3}$

Calculating L.C.M. of 5, 3 and 3:

$$\begin{array}{l|rrr} 3 & 5, & 3, & 3 \\ \hline 5 & 5, & 1, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 5, 3 and 3 = 3 × 5 = 15.

Multiplying each term by 15, we get :

$\left(\frac{16}{5} \times 15\right) : \left(\frac{16}{3} \times 15\right) : \left(\frac{20}{3} \times 15\right) = 48 : 80 : 100$

We have : 48 = 2⁴ × 3, 80 = 2⁴ × 5 and 100 = 2² × 5²

∴ H.C.F. of 48, 80 and 100 = 2² = 4.

$\Rightarrow 48 : 80 : 100 = \frac{48}{4} : \frac{80}{4} : \frac{100}{4} = 12 : 20 : 25$

Hence, $\mathbf{3\frac{1}{5} : 5\frac{1}{3} : 6\frac{2}{3}}$ in simplest form is 12 : 20 : 25.

We know that,

Ratio can be defined between two quantities of the same kind taken in same units.

75 paise : 4 rupees

₹ 1 = 100 paise.

∴ 4 rupees = (4 × 100) paise = 400 paise.

∴ Ratio between 75 paise and 4 rupees = 75 : 400

We have : 75 = 3 × 5² and 400 = 2⁴ × 5²

∴ H.C.F. of 75 and 400 = 5² = 25.

So, $\frac{75}{400} = \frac{75 \div 25}{400 \div 25} = \frac{3}{16}$.

Hence, 75 paise : 4 rupees in simplest form is 3 : 16.

We know that,

Ratio can be defined between two quantities of the same kind taken in same units.

1 m 8 cm : 72 cm

1 m = 100 cm.

∴ 1 m 8 cm = (100 + 8) cm = 108 cm.

∴ Ratio between 1 m 8 cm and 72 cm = 108 : 72

We have : 108 = 2² × 3³ and 72 = 2³ × 3²

∴ H.C.F. of 108 and 72 = 2² × 3² = 36.

So, $\frac{108}{72} = \frac{108 \div 36}{72 \div 36} = \frac{3}{2}$.

Hence, 1 m 8 cm : 72 cm in simplest form is 3 : 2.

We know that,

Ratio can be defined between two quantities of the same kind taken in same units.

1 hour 15 minutes : 45 minutes

1 hour = 60 minutes.

∴ 1 hour 15 minutes = (60 + 15) minutes = 75 minutes.

∴ Ratio between 1 hour 15 minutes and 45 minutes = 75 : 45

We have : 75 = 3 × 5² and 45 = 3² × 5

∴ H.C.F. of 75 and 45 = 3 × 5 = 15.

So, $\frac{75}{45} = \frac{75 \div 15}{45 \div 15} = \frac{5}{3}$.

Hence, 1 hour 15 minutes : 45 minutes in simplest form is 5 : 3.

We know that,

Ratio can be defined between two quantities of the same kind taken in same units.

2 kg 750 g : 3 kg

1 kg = 1000 g.

∴ 2 kg 750 g = (2 × 1000 + 750) g = 2750 g and 3 kg = (3 × 1000) g = 3000 g.

∴ Ratio between 2 kg 750 g and 3 kg = 2750 : 3000

We have : 2750 = 2 × 5³ × 11 and 3000 = 2³ × 3 × 5³

∴ H.C.F. of 2750 and 3000 = 2 × 5³ = 250.

So, $\frac{2750}{3000} = \frac{2750 \div 250}{3000 \div 250} = \frac{11}{12}$.

Hence, 2 kg 750 g : 3 kg in simplest form is 11 : 12.

We know that,

Ratio can be defined between two quantities of the same kind taken in same units.

1 year 9 months : 2 years 4 months

1 year = 12 months.

∴ 1 year 9 months = (12 + 9) months = 21 months and 2 years 4 months = (2 × 12 + 4) months = 28 months.

∴ Ratio between 1 year 9 months and 2 years 4 months = 21 : 28

We have : 21 = 3 × 7 and 28 = 2² × 7

∴ H.C.F. of 21 and 28 = 7.

So, $\frac{21}{28} = \frac{21 \div 7}{28 \div 7} = \frac{3}{4}$.

Hence, 1 year 9 months : 2 years 4 months in simplest form is 3 : 4.

(4 : 9) or (3 : 7)

We have : $4 : 9 = \frac{4}{9}$ and $3 : 7 = \frac{3}{7}$.

Calculating L.C.M. of 9 and 7:

$$\begin{array}{l|rr} 3 & 9, & 7 \\ \hline 3 & 3, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 9 and 7 = 3 × 3 × 7 = 63.

$\frac{4}{9} = \frac{4 \times 7}{9 \times 7} = \frac{28}{63} \quad \text{and} \quad \frac{3}{7} = \frac{3 \times 9}{7 \times 9} = \frac{27}{63}$

Clearly, $\frac{28}{63} \gt \frac{27}{63}$.

Hence, (4 : 9) is greater than (3 : 7).

$\left(2\frac{1}{3} : 3\frac{1}{3}\right)$ or (3.6 : 4.8)

We have :

$2\frac{1}{3} : 3\frac{1}{3} = \frac{7}{3} : \frac{10}{3} = \frac{7}{3} \times \frac{3}{10} = \frac{7}{10}$

And, $3.6 : 4.8 = \frac{3.6}{4.8} = \frac{36}{48} = \frac{3}{4}$.

Now, we compare $\frac{7}{10}$ and $\frac{3}{4}$.

Calculating L.C.M. of 10 and 4:

$$\begin{array}{l|rr} 2 & 10, & 4 \\ \hline 2 & 5, & 2 \\ \hline 5 & 5, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 10 and 4 = 2 × 2 × 5 = 20.

$\frac{7}{10} = \frac{7 \times 2}{10 \times 2} = \frac{14}{20} \quad \text{and} \quad \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$

Clearly, $\frac{15}{20} \gt \frac{14}{20}$.

Hence, (3.6 : 4.8) is greater than $\mathbf{\left(2\frac{1}{3} : 3\frac{1}{3}\right)}$.

$\left(\frac{1}{2} : \frac{1}{3}\right)$ or $\left(\frac{1}{6} : \frac{1}{4}\right)$

We have :

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

And,

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

Now, we compare $\frac{3}{2}$ and $\frac{2}{3}$.

Calculating L.C.M. of 2 and 3:

$$\begin{array}{l|rr} 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 2 and 3 = 2 × 3 = 6.

$\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \quad \text{and} \quad \frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

Clearly, $\frac{9}{6} \gt \frac{4}{6}$.

Hence, $\mathbf{\left(\frac{1}{2} : \frac{1}{3}\right)}$ is greater than $\mathbf{\left(\frac{1}{6} : \frac{1}{4}\right)}$.

$\left(3\frac{1}{3} : 4\frac{1}{6}\right)$ or (0.9 : 1)

We have :

$3\frac{1}{3} : 4\frac{1}{6} = \frac{10}{3} : \frac{25}{6} = \frac{10}{3} \times \frac{6}{25} = \frac{60}{75} = \frac{4}{5}$

And, $0.9 : 1 = \frac{0.9}{1} = \frac{9}{10}$.

Now, we compare $\frac{4}{5}$ and $\frac{9}{10}$.

Calculating L.C.M. of 5 and 10:

$$\begin{array}{l|rr} 2 & 5, & 10 \\ \hline 5 & 5, & 5 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 5 and 10 = 2 × 5 = 10.

$\frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10} \quad \text{and} \quad \frac{9}{10} = \frac{9}{10}$

Clearly, $\frac{9}{10} \gt \frac{8}{10}$.

Hence, (0.9 : 1) is greater than $\mathbf{\left(3\frac{1}{3} : 4\frac{1}{6}\right)}$.

(2 : 3), (5 : 9) and (11 : 15)

Given ratios are $\frac{2}{3}, \frac{5}{9}$ and $\frac{11}{15}$.

Calculating L.C.M. of 3, 9 and 15:

$$\begin{array}{l|rrr} 3 & 3, & 9, & 15 \\ \hline 3 & 1, & 3, & 5 \\ \hline 5 & 1, & 1, & 5 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 3, 9 and 15 = 3 × 3 × 5 = 45.

Now,

$$\begin{aligned}\Rightarrow \frac{2}{3} = \frac{2 \times 15}{3 \times 15} = \frac{30}{45}; \\[1em] \Rightarrow \frac{5}{9} = \frac{5 \times 5}{9 \times 5} = \frac{25}{45}; \\[1em] \Rightarrow \frac{11}{15} = \frac{11 \times 3}{15 \times 3} = \frac{33}{45}\end{aligned}$$

Clearly, $\frac{25}{45} \lt \frac{30}{45} \lt \frac{33}{45}$.

Hence, (5 : 9) < (2 : 3) < (11 : 15).

(5 : 7), (9 : 14), (20 : 21) and (3 : 8)

Given ratios are $\frac{5}{7}, \frac{9}{14}, \frac{20}{21}$ and $\frac{3}{8}$.

Calculating L.C.M. of 7, 14, 21 and 8:

$$\begin{array}{l|rrrr} 2 & 7, & 14, & 21, & 8 \\ \hline 2 & 7, & 7, & 21, & 4 \\ \hline 2 & 7, & 7, & 21, & 2 \\ \hline 3 & 7, & 7, & 21, & 1 \\ \hline 7 & 7, & 7, & 7, & 1 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 7, 14, 21 and 8 = 2 × 2 × 2 × 3 × 7 = 168.

Now,

$$\begin{aligned}\Rightarrow \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168}; \quad \frac{9}{14} = \frac{9 \times 12}{14 \times 12} = \frac{108}{168}; \\[1em] \Rightarrow \frac{20}{21} = \frac{20 \times 8}{21 \times 8} = \frac{160}{168}; \quad \frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168}\end{aligned}$$

Clearly, $\frac{63}{168} \lt \frac{108}{168} \lt \frac{120}{168} \lt \frac{160}{168}$.

Hence, (3 : 8) < (9 : 14) < (5 : 7) < (20 : 21).

Given,

Ratio = $\frac{1}{4} : \frac{1}{5}$.

Calculating L.C.M. of 4 and 5:

$$\begin{array}{l|rr} 2 & 4, & 5 \\ \hline 2 & 2, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 4 and 5 = 2 × 2 × 5 = 20.

Multiplying both terms of the ratio by 20, we get:

$\Rightarrow \frac{1}{4} : \frac{1}{5} = \left(\frac{1}{4} \times 20\right) : \left(\frac{1}{5} \times 20\right) = 5 : 4$

Sum of ratio terms = (5 + 4) = 9.

Gagan's share = ₹ $\left(142.20 \times \frac{5}{9}\right)$ = ₹ 79.

Mukesh's share = ₹ $\left(142.20 \times \frac{4}{9}\right) = ₹\ 63.20.$

Hence, Gagan's share = ₹ 79 and Mukesh's share = ₹ 63.20.

Given ratio = $\frac{1}{3} : \frac{1}{4} : \frac{1}{6}$.

Calculating L.C.M. of 3, 4 and 6:

$$\begin{array}{l|rrr} 2 & 3, & 4, & 6 \\ \hline 2 & 3, & 2, & 3 \\ \hline 3 & 3, & 1, & 3 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 3, 4 and 6 = 2 × 2 × 3 = 12.

Multiplying each term of the ratio by 12, we get:

$\Rightarrow \frac{1}{3} : \frac{1}{4} : \frac{1}{6} = \left(\frac{1}{3} \times 12\right) : \left(\frac{1}{4} \times 12\right) : \left(\frac{1}{6} \times 12\right) = 4 : 3 : 2$

Sum of ratio terms = (4 + 3 + 2) = 9.

A's share = ₹ $\left(3726 \times \frac{4}{9}\right)$ = ₹ 1,656.

B's share = ₹ $\left(3726 \times \frac{3}{9}\right)$ = ₹ 1,242.

C's share = ₹ $\left(3726 \times \frac{2}{9}\right)$ = ₹ 828.

Hence, A's share = ₹ 1,656, B's share = ₹ 1,242 and C's share = ₹ 828.

Given ratio = $\frac{1}{4} : \frac{2}{5} : 1\frac{3}{8} = \frac{1}{4} : \frac{2}{5} : \frac{11}{8}$.

Calculating L.C.M. of 4, 5 and 8:

$$\begin{array}{l|rrr} 2 & 4, & 5, & 8 \\ \hline 2 & 2, & 5, & 4 \\ \hline 2 & 1, & 5, & 2 \\ \hline 5 & 1, & 5, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 4, 5 and 8 = 2 × 2 × 2 × 5 = 40.

Multiplying each term of the ratio by 40, we get:

$\Rightarrow \frac{1}{4} : \frac{2}{5} : \frac{11}{8} = \left(\frac{1}{4} \times 40\right) : \left(\frac{2}{5} \times 40\right) : \left(\frac{11}{8} \times 40\right) = 10 : 16 : 55$

Sum of ratio terms = (10 + 16 + 55) = 81.

A's share = ₹ $\left(810 \times \frac{10}{81}\right)$ = ₹ 100.

B's share = ₹ $\left(810 \times \frac{16}{81}\right)$ = ₹ 160.

C's share = ₹ $\left(810 \times \frac{55}{81}\right)$ = ₹ 550.

Hence, A's share = ₹ 100, B's share = ₹ 160 and C's share = ₹ 550.

Given,

Ratio = $2\frac{2}{3} : 6\frac{2}{3} = \frac{8}{3} : \frac{20}{3}$.

Multiplying both the terms by 3, we get :

$\frac{8}{3} : \frac{20}{3} = 8 : 20$

We have : 8 = 2³ and 20 = 2² × 5

∴ H.C.F. of 8 and 20 = 2² = 4.

So, $8 : 20 = \frac{8}{20} = \frac{8 \div 4}{20 \div 4} = \frac{2}{5} = 2 : 5$.

Sum of ratio terms = (2 + 5) = 7.

Geeta's share = ₹ $\left(1050 \times \frac{2}{7}\right)$ = ₹ 300.

Renu's share = ₹ $\left(1050 \times \frac{5}{7}\right)$ = ₹ 750.

Hence, Geeta's share = ₹ 300 and Renu's share = ₹ 750.

Let 4A = 5B = 7C = $K$. Then,

$\text{A} = \frac{K}{4}, \quad \text{B} = \frac{K}{5} \quad \text{and} \quad \text{C} = \frac{K}{7}$.

Calculating L.C.M. of 4, 5 and 7:

$$\begin{array}{l|rrr} 2 & 4, & 5, & 7 \\ \hline 2 & 2, & 5, & 7 \\ \hline 5 & 1, & 5, & 7 \\ \hline 7 & 1, & 1, & 7 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 4, 5 and 7 = 2 × 2 × 5 × 7 = 140.

$$\begin{aligned}\Rightarrow \text{A : B : C} = \frac{K}{4} : \frac{K}{5} : \frac{K}{7} = \left(\frac{K}{4} \times \frac{140}{K}\right) : \left(\frac{K}{5} \times \frac{140}{K}\right) : \left(\frac{K}{7} \times \frac{140}{K}\right) \\[1em] \Rightarrow \text{A : B : C} = 35 : 28 : 20\end{aligned}$$

Sum of ratio terms = (35 + 28 + 20) = 83.

A's share = ₹ $\left(747 \times \frac{35}{83}\right)$ = ₹ 315.

B's share = ₹ $\left(747 \times \frac{28}{83}\right)$ = ₹ 252.

C's share = ₹ $\left(747 \times \frac{20}{83}\right)$ = ₹ 180.

Hence, A's share = ₹ 315, B's share = ₹ 252 and C's share = ₹ 180.

Let the number of one rupee, 50 p and 25 p coins be $5x, 6x$ and $8x$ respectively. Then,

Value of one rupee coins = ₹ $(5x \times 1) = ₹\ 5x$.

Value of 50 p coins = ₹ $\left(6x \times \frac{50}{100}\right) = ₹\ \frac{6x}{2}$.

Value of 25 p coins = ₹ $\left(8x \times \frac{25}{100}\right) = ₹\ \frac{8x}{4}$.

$$\begin{aligned}\Rightarrow 5x + \frac{6x}{2} + \frac{8x}{4} = 210 \\[1em] \Rightarrow 5x + 3x + 2x = 210 \\[1em] \Rightarrow 10x = 210 \\[1em] \Rightarrow x = \frac{210}{10} = 21\end{aligned}$$

Number of one rupee coins = (5 × 21) = 105,

Number of 50 p coins = (6 × 21) = 126,

Number of 25 p coins = (8 × 21) = 168.

Hence, the bag contains 105 one rupee coins, 126 fifty paise coins and 168 twenty five paise coins.

(i) A : B = 2 : 5 and B : C = 7 : 9

Calculating LCM of two values of B, namely 5 and 7:

$$\begin{array}{l|rr} 5 & 5, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 5 and 7 = 5 × 7 = 35.

A : B = 2 : 5 = (2 × 7) : (5 × 7) = 14 : 35

B : C = 7 : 9 = (7 × 5) : (9 × 5) = 35 : 45

Hence, A : B : C = 14 : 35 : 45.

(ii) A : B = 3 : 4 and B : C = 6 : 11

Calculating L.C.M. of two values of B, namely 4 and 6:

$$\begin{array}{l|rr} 2 & 4, & 6 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 4 and 6 = 2 × 2 × 3 = 12.

A : B = 3 : 4 = (3 × 3) : (4 × 3) = 9 : 12

B : C = 6 : 11 = (6 × 2) : (11 × 2) = 12 : 22

Hence, A : B : C = 9 : 12 : 22.

(iii) A : B = $\frac{1}{2} : \frac{1}{3}$ and B : C = $\frac{1}{4} : \frac{1}{5}$

Calculating L.C.M. of 2 and 3:

$$\begin{array}{l|rr} 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 2 and 3 = 2 × 3 = 6.

Calculating L.C.M. of 4 and 5:

$$\begin{array}{l|rr} 2 & 4, & 5 \\ \hline 2 & 2, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 4 and 5 = 2 × 2 × 5 = 20.

We have :

A : B = $\frac{1}{2} : \frac{1}{3} = \left(\frac{1}{2} \times 6\right) : \left(\frac{1}{3} \times 6\right) = 3 : 2$

B : C = $\frac{1}{4} : \frac{1}{5} = \left(\frac{1}{4} \times 20\right) : \left(\frac{1}{5} \times 20\right) = 5 : 4$

Calculating L.C.M. of two values of B, namely 2 and 5:

$$\begin{array}{l|rr} 2 & 2, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 2 and 5 = 2 × 5 = 10.

A : B = 3 : 2 = (3 × 5) : (2 × 5) = 15 : 10

B : C = 5 : 4 = (5 × 2) : (4 × 2) = 10 : 8

Hence, A : B : C = 15 : 10 : 8.

We have : A : B = 4 : 9 ⇒ B : A = 9 : 4 and A : C = 2 : 3.

$\Rightarrow \frac{\text{B}}{\text{C}} = \left(\frac{\text{B}}{\text{A}} \times \frac{\text{A}}{\text{C}}\right) = \left(\frac{9}{4} \times \frac{2}{3}\right) = \frac{18}{12} = \frac{3}{2}$

∴ B : C = 3 : 2.

Now, A : B = 4 : 9 and A : C = 2 : 3 = (2 × 2) : (3 × 2) = 4 : 6.

Since the value of A is the same (i.e., 4) in both the ratios,

Hence, B : C = 3 : 2 and A : B : C = 4 : 9 : 6.

We have : A : C = 5 : 8 and B : C = 5 : 6 ⇒ C : B = 6 : 5.

$\Rightarrow \frac{\text{A}}{\text{B}} = \left(\frac{\text{A}}{\text{C}} \times \frac{\text{C}}{\text{B}}\right) = \left(\frac{5}{8} \times \frac{6}{5}\right) = \frac{6}{8} = \frac{3}{4}$

∴ A : B = 3 : 4.

Now, calculating L.C.M. of two values of C, namely 8 and 6:

$$\begin{array}{l|rr} 2 & 8, & 6 \\ \hline 2 & 4, & 3 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 8 and 6 = 2 × 2 × 2 × 3 = 24.

A : C = 5 : 8 = (5 × 3) : (8 × 3) = 15 : 24

B : C = 5 : 6 = (5 × 4) : (6 × 4) = 20 : 24

Hence, A : B = 3 : 4 and A : B : C = 15 : 20 : 24.

Let the required numbers be $6x$ and $11x$.

Given,

Adding 2 to the first and 7 to the second, their ratio becomes 8 : 15.

$$\begin{aligned}\therefore \frac{6x + 2}{11x + 7} = \frac{8}{15} \\[1em] \Rightarrow 15(6x + 2) = 8(11x + 7) \\[1em] \Rightarrow 90x + 30 = 88x + 56 \\[1em] \Rightarrow 90x - 88x = 56 - 30 \\[1em] \Rightarrow 2x = 26 \\[1em] \Rightarrow x = 13\end{aligned}$$

So, the required numbers are (6 × 13) and (11 × 13), i.e., 78 and 143.

Hence, the required numbers are 78 and 143.

Let the total amount be ₹ $x$.

Sum of ratio terms = (5 + 7) = 12.

∴ Sohan's share = ₹ $\left(x \times \frac{7}{12}\right) = ₹\ \frac{7x}{12}$

But, Sohan's share = ₹ 665.

$$\begin{aligned}\Rightarrow \frac{7x}{12} = 665 \\[1em] \Rightarrow x = \left(665 \times \frac{12}{7}\right) = 1140\end{aligned}$$

Hence, the total amount is ₹ 1,140.

Let the required numbers be $7x$ and $4x$. Then,

$$\begin{aligned}\Rightarrow 7x - 4x = 72 \\[1em] \Rightarrow 3x = 72 \\[1em] \Rightarrow x = \frac{72}{3} = 24\end{aligned}$$

So, the required numbers are (7 × 24) and (4 × 24), i.e., 168 and 96.

Hence, the required numbers are 168 and 96.

Let the total amount be ₹ $x$.

Sum of ratio terms = (5 + 6 + 7) = 18.

∴ A's share = ₹ $\left(x \times \frac{5}{18}\right) = ₹\ \frac{5x}{18}$

But, A's share = ₹ 175.

$$\begin{aligned}\Rightarrow \frac{5x}{18} = 175 \\[1em] \Rightarrow x = \left(175 \times \frac{18}{5}\right) = 630\end{aligned}$$

∴ Total amount = ₹ 630.

B's share = ₹ $\left(630 \times \frac{6}{18}\right)$ = ₹ 210.

C's share = ₹ $\left(630 \times \frac{7}{18}\right)$ = ₹ 245.

Hence, the total amount is ₹ 630, B's share = ₹ 210 and C's share = ₹ 245.

Total number of students = 1440 and ratio of boys to girls = 7 : 5.

Sum of ratio terms = (7 + 5) = 12.

∴ No. of boys = $\left(1440 \times \frac{7}{12}\right) = 840$

No. of girls = $\left(1440 \times \frac{5}{12}\right) = 600$

Let $x$ new girls be admitted.

Then, number of boys = (840 + 40) = 880 and number of girls = (600 + $x$).

$$\begin{aligned}\Rightarrow \frac{840 + 40}{600 + x} = \frac{4}{3} \\[1em] \Rightarrow \frac{880}{600 + x} = \frac{4}{3} \\[1em] \Rightarrow 4(600 + x) = 880 \times 3 \\[1em] \Rightarrow 2400 + 4x = 2640 \\[1em] \Rightarrow 4x = 240 \\[1em] \Rightarrow x = 60\end{aligned}$$

Hence, 60 new girls may be admitted.

Let the numbers be $a, b$ and $c$. Then,

$\frac{a}{b} = \frac{13}{16} \quad \text{and} \quad \frac{b}{c} = \frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$

∴ $a : b : c$ = 13 : 16 : 24.

Sum of ratio terms = (13 + 16 + 24) = 53.

$$\begin{aligned}\Rightarrow a = \left(212 \times \frac{13}{53}\right) = 52 \\[1em] \Rightarrow b = \left(212 \times \frac{16}{53}\right) = 64 \\[1em] \Rightarrow c = \left(212 \times \frac{24}{53}\right) = 96\end{aligned}$$

Hence, the required numbers are 52, 64 and 96.

Let the required number be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{27 + x}{35 + x} = \frac{4}{5} \\[1em] \Rightarrow 5(27 + x) = 4(35 + x) \\[1em] \Rightarrow 135 + 5x = 140 + 4x \\[1em] \Rightarrow 5x - 4x = 140 - 135 \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, the required number is 5.

Let the present ages of Mr. Sen and his son be $17x$ years and $9x$ years respectively.

Then, 9 years ago their ages were ($17x - 9$) years and ($9x - 9$) years respectively.

$$\begin{aligned}\Rightarrow \frac{17x - 9}{9x - 9} = \frac{7}{3} \\[1em] \Rightarrow 3(17x - 9) = 7(9x - 9) \\[1em] \Rightarrow 51x - 27 = 63x - 63 \\[1em] \Rightarrow 63 - 27 = 63x - 51x \\[1em] \Rightarrow 12x = 36 \\[1em] \Rightarrow x = 3\end{aligned}$$

∴ Present age of Mr. Sen = (17 × 3) years = 51 years

and present age of his son = (9 × 3) years = 27 years.

Hence, Mr. Sen's present age is 51 years and his son's present age is 27 years.

Volume of acetic acid = 705 ml.

Volume of water = 4.7 litres = (4.7 × 1000) ml = 4700 ml.

∴ Ratio by volume of acetic acid to water = 705 : 4700

We have : 705 = 3 × 5 × 47 and 4700 = 2² × 5² × 47

∴ H.C.F. of 705 and 4700 = 5 × 47 = 235.

$\Rightarrow 705 : 4700 = \frac{705}{4700} = \frac{705 \div 235}{4700 \div 235} = \frac{3}{20}$

Hence, the ratio by volume of acetic acid to water is 3 : 20.

Let the salaries of A, B and C (initially) be ₹ $2x, ₹ 3x$ and ₹ $5x$ respectively.

A's new salary = 115% of $2x = \left(\frac{115}{100} \times 2x\right) = \frac{23x}{10}$

B's new salary = 110% of $3x = \left(\frac{110}{100} \times 3x\right) = \frac{33x}{10}$

C's new salary = 120% of $5x = \left(\frac{120}{100} \times 5x\right) = \frac{60x}{10}$

∴ New ratio of their salaries

$= \frac{23x}{10} : \frac{33x}{10} : \frac{60x}{10} = 23 : 33 : 60$

Hence, the new ratio of their salaries is 23 : 33 : 60.

Ratio of marks obtained to maximum marks in Physics = $58 : 75 = \frac{58}{75}$.

Ratio of marks obtained to maximum marks in Maths = $97 : 120 = \frac{97}{120}$.

Calculating L.C.M. of 75 and 120:

$$\begin{array}{l|rr} 2 & 75, & 120 \\ \hline 2 & 75, & 60 \\ \hline 2 & 75, & 30 \\ \hline 3 & 75, & 15 \\ \hline 5 & 25, & 5 \\ \hline 5 & 5, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 75 and 120 = 2 × 2 × 2 × 3 × 5 × 5 = 600.

$\frac{58}{75} = \frac{58 \times 8}{75 \times 8} = \frac{464}{600} \quad \text{and} \quad \frac{97}{120} = \frac{97 \times 5}{120 \times 5} = \frac{485}{600}$

Clearly, $\frac{485}{600} \gt \frac{464}{600}$, i.e., (97 : 120) > (58 : 75).

Hence, Anuj performed better in Maths.

Ratio of A : B = 2 : 1.

Sum of ratio terms = (2 + 1) = 3.

Quantity of component A in the mixture = $\left(60 \times \frac{2}{3}\right) \text{ litres} = 40 \text{ litres}$

∴ Quantity of component B in the mixture = (60 − 40) litres = 20 litres.

Let $x$ litres of component B be added to the mixture.

$$\begin{aligned}\Rightarrow \frac{40}{20 + x} = \frac{1}{2} \\[1em] \Rightarrow 20 + x = 40 \times 2 \\[1em] \Rightarrow 20 + x = 80 \\[1em] \Rightarrow x = 60\end{aligned}$$

Hence, 60 litres of component B has to be added to the mixture.

27 : 36 = 4.5 : 6

We have :

27 = 3³ and 36 = 2² × 3²

∴ H.C.F. of 27 and 36 = 3² = 9.

So, $27 : 36 = \frac{27}{36} = \frac{27 \div 9}{36 \div 9} = \frac{3}{4} = 3 : 4$.

And, $4.5 : 6 = \frac{4.5}{6} = \frac{45}{60}$.

45 = 3² × 5 and 60 = 2² × 3 × 5

∴ H.C.F. of 45 and 60 = 3 × 5 = 15.

So, $\frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4} = 3 : 4$.

Since both the ratios are equal,

Hence, the given statement is true.

$\frac{3}{4} : \frac{15}{16} = \frac{2}{3} : \frac{5}{6}$

We have :

$\frac{3}{4} : \frac{15}{16} = \frac{3}{4} \times \frac{16}{15} = \frac{48}{60}$

48 = 2⁴ × 3 and 60 = 2² × 3 × 5

∴ H.C.F. of 48 and 60 = 2² × 3 = 12.

So, $\frac{48}{60} = \frac{48 \div 12}{60 \div 12} = \frac{4}{5} = 4 : 5$.

And, $\frac{2}{3} : \frac{5}{6} = \frac{2}{3} \times \frac{6}{5} = \frac{12}{15}$.

12 = 2² × 3 and 15 = 3 × 5

∴ H.C.F. of 12 and 15 = 3.

So, $\frac{12}{15} = \frac{12 \div 3}{15 \div 3} = \frac{4}{5} = 4 : 5$.

Since both the ratios are equal,

Hence, the given statement is true.

₹ 14 : ₹ 21 = 2 pens : 3 pens

We have :

14 = 2 × 7 and 21 = 3 × 7

∴ H.C.F. of 14 and 21 = 7.

So, $₹\ 14 : ₹\ 21 = \frac{14}{21} = \frac{14 \div 7}{21 \div 7} = \frac{2}{3} = 2 : 3$.

And, $2 \text{ pens} : 3 \text{ pens} = \frac{2}{3} = 2 : 3$.

Since both the ratios are equal,

Hence, the given statement is true.

6.5 km : 2.6 km = ₹ 60 : ₹ 24

We have :

$6.5 \text{ km} : 2.6 \text{ km} = \frac{6.5}{2.6} = \frac{65}{26}$

65 = 5 × 13 and 26 = 2 × 13

∴ H.C.F. of 65 and 26 = 13.

So, $\frac{65}{26} = \frac{65 \div 13}{26 \div 13} = \frac{5}{2} = 5 : 2$.

60 = 2² × 3 × 5 and 24 = 2³ × 3

∴ H.C.F. of 60 and 24 = 2² × 3 = 12.

So, $₹\ 60 : ₹\ 24 = \frac{60}{24} = \frac{60 \div 12}{24 \div 12} = \frac{5}{2} = 5 : 2$.

Since both the ratios are equal,

Hence, the given statement is true.

We know that four numbers are in proportion if product of extremes = product of means.

8, 12, 18, 24

Product of extremes = 8 × 24 = 192.

Product of means = 12 × 18 = 216.

Since 192 ≠ 216, product of extremes ≠ product of means.

Hence, 8, 12, 18, 24 are not in proportion.

We know that four numbers are in proportion if product of extremes = product of means.

6.4, 3.6, 4.8, 2.7

Product of extremes = 6.4 × 2.7 = 17.28.

Product of means = 3.6 × 4.8 = 17.28.

Since product of extremes = product of means,

Hence, 6.4, 3.6, 4.8, 2.7 are in proportion.

We know that four numbers are in proportion if product of extremes = product of means.

$11\frac{1}{3}, 9\frac{1}{3}, 8\frac{1}{2}$, 7

We have : $11\frac{1}{3} = \frac{34}{3}, 9\frac{1}{3} = \frac{28}{3}$ and $8\frac{1}{2} = \frac{17}{2}$.

Product of extremes = $\frac{34}{3} \times 7 = \frac{238}{3}$

Product of means = $\frac{28}{3} \times \frac{17}{2} = \frac{476}{6} = \frac{238}{3}$

Since product of extremes = product of means,

Hence, $\mathbf{11\frac{1}{3}, 9\frac{1}{3}, 8\frac{1}{2}}$, 7 are in proportion.

We know that four numbers are in proportion if product of extremes = product of means.

0.36, 1.8, 6.4, 32

Product of extremes = 0.36 × 32 = 11.52.

Product of means = 1.8 × 6.4 = 11.52.

Since 11.52 = 11.52, product of extremes = product of means.

Hence, 0.36, 1.8, 6.4, 32 are in proportion.

We know that four numbers are in proportion if product of extremes = product of means.

$\frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{9}{10}$

Product of extremes = $\frac{3}{4} \times \frac{9}{10} = \frac{27}{40}$

Product of means = $\frac{5}{6} \times \frac{7}{8} = \frac{35}{48}$

Since $\frac{27}{40} \ne \frac{35}{48}$, product of extremes ≠ product of means.

Hence, $\mathbf{\frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{9}{10}}$ are not in proportion.

8 : $x$ :: 6 : 27

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow 8 \times 27 = x \times 6 \\[1em] \Rightarrow 6x = 216 \\[1em] \Rightarrow x = \frac{216}{6} = 36\end{aligned}$$

Hence, $\bm{x = 36}$.

5.6 : 3.5 :: $x$ : 1.25

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow 5.6 \times 1.25 = 3.5 \times x \\[1em] \Rightarrow 3.5x = 7 \\[1em] \Rightarrow x = \frac{7}{3.5} = 2\end{aligned}$$

Hence, $\bm{x = 2}$.

$1\frac{4}{5} : 2\frac{4}{5} :: x : 3\frac{1}{2}$

We have : $1\frac{4}{5} = \frac{9}{5}, 2\frac{4}{5} = \frac{14}{5}$ and $3\frac{1}{2} = \frac{7}{2}$.

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow \frac{9}{5} \times \frac{7}{2} = \frac{14}{5} \times x \\[1em] \Rightarrow \frac{63}{10} = \frac{14x}{5} \\[1em] \Rightarrow x = \left(\frac{63}{10} \times \frac{5}{14}\right) \\[1em] \Rightarrow x = \frac{9}{4} \\[1em] \Rightarrow x = 2\frac{1}{4}\end{aligned}$$

Hence, $\bm{x = 2\frac{1}{4}}$.

$\frac{2}{3} : \frac{4}{7} :: 1\frac{5}{6} : x$

We have : $1\frac{5}{6} = \frac{11}{6}$.

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow \frac{2}{3} \times x = \frac{4}{7} \times \frac{11}{6} \\[1em] \Rightarrow \frac{2x}{3} = \frac{44}{42} = \frac{22}{21} \\[1em] \Rightarrow x = \left(\frac{22}{21} \times \frac{3}{2}\right) \\[1em] \Rightarrow x = \frac{11}{7} \\[1em] \Rightarrow x = 1\frac{4}{7}\end{aligned}$$

Hence, $\bm{x = 1\frac{4}{7}}$.

2.8, 14 and 3.5

Let the fourth proportional to 2.8, 14 and 3.5 be $x$. Then,

$$\begin{aligned}\Rightarrow 2.8 : 14 :: 3.5 : x \\[1em] \Rightarrow 2.8 \times x = 14 \times 3.5 \\[1em] \Rightarrow 2.8x = 49 \\[1em] \Rightarrow x = \frac{49}{2.8} = 17.5\end{aligned}$$

Hence, the required fourth proportional is 17.5.

$3\frac{1}{3}, 1\frac{2}{3}$ and $2\frac{1}{2}$

We have : $3\frac{1}{3} = \frac{10}{3}, 1\frac{2}{3} = \frac{5}{3}$ and $2\frac{1}{2} = \frac{5}{2}$.

Let the fourth proportional to $\frac{10}{3}, \frac{5}{3}$ and $\frac{5}{2}$ be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{10}{3} : \frac{5}{3} :: \frac{5}{2} : x \\[1em] \Rightarrow \frac{10}{3} \times x = \frac{5}{3} \times \frac{5}{2} \\[1em] \Rightarrow \frac{10x}{3} = \frac{25}{6} \\[1em] \Rightarrow x = \left(\frac{25}{6} \times \frac{3}{10}\right) \\[1em] \Rightarrow x = \frac{5}{4} \\[1em] \Rightarrow x = 1\frac{1}{4}\end{aligned}$$

Hence, the required fourth proportional is $\mathbf{1\frac{1}{4}}$.

$1\frac{5}{7}, 2\frac{3}{14}$ and $3\frac{3}{5}$

We have : $1\frac{5}{7} = \frac{12}{7}, 2\frac{3}{14} = \frac{31}{14}$ and $3\frac{3}{5} = \frac{18}{5}$.

Let the fourth proportional to $\frac{12}{7}, \frac{31}{14}$ and $\frac{18}{5}$ be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{12}{7} : \frac{31}{14} :: \frac{18}{5} : x \\[1em] \Rightarrow \frac{12}{7} \times x = \frac{31}{14} \times \frac{18}{5} \\[1em] \Rightarrow \frac{12x}{7} = \frac{558}{70} \\[1em] \Rightarrow x = \left(\frac{558}{70} \times \frac{7}{12}\right) \\[1em] \Rightarrow x = \frac{93}{20} \\[1em] \Rightarrow x = 4\frac{13}{20}\end{aligned}$$

Hence, the required fourth proportional is $\mathbf{4\frac{13}{20}}$.

$1\frac{1}{5}, 1\frac{3}{5}$ and 2.1

We have : $1\frac{1}{5} = \frac{6}{5}, 1\frac{3}{5} = \frac{8}{5}$ and $2.1 = \frac{21}{10}$.

Let the fourth proportional to $\frac{6}{5}, \frac{8}{5}$ and $\frac{21}{10}$ be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{6}{5} : \frac{8}{5} :: \frac{21}{10} : x \\[1em] \Rightarrow \frac{6}{5} \times x = \frac{8}{5} \times \frac{21}{10} \\[1em] \Rightarrow \frac{6x}{5} = \frac{168}{50} \\[1em] \Rightarrow x = \left(\frac{168}{50} \times \frac{5}{6}\right) \\[1em] \Rightarrow x = \frac{28}{10} \\[1em] \Rightarrow x = 2.8\end{aligned}$$

Hence, the required fourth proportional is 2.8.

12, 16

Let the third proportional to 12 and 16 be $x$. Then,

$$\begin{aligned}\Rightarrow 12 : 16 :: 16 : x \\[1em] \Rightarrow 12 \times x = 16 \times 16 \\[1em] \Rightarrow 12x = 256 \\[1em] \Rightarrow x = \frac{256}{12} \\[1em] \Rightarrow x = \frac{64}{3} \\[1em] \Rightarrow x = 21\frac{1}{3}\end{aligned}$$

Hence, the required third proportional is $\mathbf{21\frac{1}{3}}$.

4.5, 6

Let the third proportional to 4.5 and 6 be $x$. Then,

$$\begin{aligned}\Rightarrow 4.5 : 6 :: 6 : x \\[1em] \Rightarrow 4.5 \times x = 6 \times 6 \\[1em] \Rightarrow 4.5x = 36 \\[1em] \Rightarrow x = \frac{36}{4.5} = 8\end{aligned}$$

Hence, the required third proportional is 8.

$5\frac{1}{2}, 16\frac{1}{2}$

We have : $5\frac{1}{2} = \frac{11}{2}$ and $16\frac{1}{2} = \frac{33}{2}$.

Let the third proportional to $\frac{11}{2}$ and $\frac{33}{2}$ be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{11}{2} : \frac{33}{2} :: \frac{33}{2} : x \\[1em] \Rightarrow \frac{11}{2} \times x = \frac{33}{2} \times \frac{33}{2} \\[1em] \Rightarrow \frac{11x}{2} = \frac{1089}{4} \\[1em] \Rightarrow x = \left(\frac{1089}{4} \times \frac{2}{11}\right) \\[1em] \Rightarrow x = \frac{99}{2} \\[1em] \Rightarrow x = 49\frac{1}{2}\end{aligned}$$

Hence, the required third proportional is $\mathbf{49\frac{1}{2}}$.

$3\frac{1}{2}, 8\frac{3}{4}$

We have : $3\frac{1}{2} = \frac{7}{2}$ and $8\frac{3}{4} = \frac{35}{4}$.

Let the third proportional to $\frac{7}{2}$ and $\frac{35}{4}$ be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{7}{2} : \frac{35}{4} :: \frac{35}{4} : x \\[1em] \Rightarrow \frac{7}{2} \times x = \frac{35}{4} \times \frac{35}{4} \\[1em] \Rightarrow \frac{7x}{2} = \frac{1225}{16} \\[1em] \Rightarrow x = \left(\frac{1225}{16} \times \frac{2}{7}\right) \\[1em] \Rightarrow x = \frac{175}{8} \\[1em] \Rightarrow x = 21\frac{7}{8}\end{aligned}$$

Hence, the required third proportional is $\mathbf{21\frac{7}{8}}$.

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

∴ Mean proportion between 8 and 18 = $\sqrt{8 \times 18} = \sqrt{144} = 12$

Hence, the mean proportion between 8 and 18 is 12.

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

0.3 and 2.7

Mean proportion between 0.3 and 2.7 = $\sqrt{0.3 \times 2.7} = \sqrt{0.81} = 0.9$

Hence, the mean proportion between 0.3 and 2.7 is 0.9.

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

$66\frac{2}{3}$ and 6

We have : $66\frac{2}{3} = \frac{200}{3}$.

Mean proportion = $\sqrt{\frac{200}{3} \times 6} = \sqrt{400} = 20$

Hence, the mean proportion between $\mathbf{66\frac{2}{3}}$ and 6 is 20.

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

1.25 and 0.45

Mean proportion between 1.25 and 0.45 = $\sqrt{1.25 \times 0.45} = \sqrt{0.5625} = 0.75$

Hence, the mean proportion between 1.25 and 0.45 is 0.75.

We know that the mean proportion between $a$ and $c$ is $\sqrt{ac}$.

$\frac{1}{7}$ and $\frac{4}{63}$

Mean proportion = $\sqrt{\frac{1}{7} \times \frac{4}{63}} = \sqrt{\frac{4}{441}} = \frac{2}{21}$

Hence, the mean proportion between $\mathbf{\frac{1}{7}}$ and $\mathbf{\frac{4}{63}}$ is $\mathbf{\frac{2}{21}}$.

Since 28 is the third proportional to 7 and $x$, we have :

$$\begin{aligned}\Rightarrow 7 : x = x : 28 \\[1em] \Rightarrow x \times x = 7 \times 28 \\[1em] \Rightarrow x^2 = 196 \\[1em] \Rightarrow x = \sqrt{196} = 14\end{aligned}$$

Hence, $\bm{x = 14}$.

Since 18, $x$, 50 are in continued proportion, we have :

$$\begin{aligned}\Rightarrow 18 : x = x : 50 \\[1em] \Rightarrow x \times x = 18 \times 50 \\[1em] \Rightarrow x^2 = 900 \\[1em] \Rightarrow x = \sqrt{900} = 30\end{aligned}$$

Hence, $\bm{x = 30}$.

Let the lengths of the two pieces be $7x$ cm and $5x$ cm.

Clearly, the smaller piece is of length $5x$ cm.

$$\begin{aligned}\Rightarrow 5x = 45.5 \\[1em] \Rightarrow x = \frac{45.5}{5} = 9.1\end{aligned}$$

∴ Length of the longer piece = (7 × 9.1) cm = 63.7 cm.

Hence, the length of the longer piece is 63.7 cm.

Let the areas of the two rectangular fields be $5x$ sq. metres and $9x$ sq. metres.

Clearly, the area of the larger field is $9x$ sq. metres.

$$\begin{aligned}\Rightarrow 9x = 2331 \\[1em] \Rightarrow x = \frac{2331}{9} = 259\end{aligned}$$

∴ Area of the smaller field = (5 × 259) sq. metres = 1295 sq. metres.

Hence, the area of the smaller field is 1295 sq. metres.

Let $x$ be the required number. Then,

$$\begin{aligned}\Rightarrow \frac{41 - x}{55 - x} = \frac{36 - x}{48 - x} \\[1em] \Rightarrow (41 - x)(48 - x) = (55 - x)(36 - x) \\[1em] \Rightarrow 1968 - 41x - 48x + x^2 = 1980 - 55x - 36x + x^2 \\[1em] \Rightarrow 1968 - 89x = 1980 - 91x \\[1em] \Rightarrow 91x - 89x = 1980 - 1968 \\[1em] \Rightarrow 2x = 12 \\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, the required number is 6.

Let the required quantity of zinc be $x$ kg.

We have : $2\frac{2}{5} = \frac{12}{5}$.

Then,

$$\begin{aligned}\Rightarrow 9 : 4 = \frac{12}{5} : x \\[1em] \Rightarrow 9 \times x = 4 \times \frac{12}{5} \\[1em] \Rightarrow 9x = \frac{48}{5} \\[1em] \Rightarrow x = \left(\frac{48}{5} \times \frac{1}{9}\right) \\[1em] \Rightarrow x = \frac{16}{15} \\[1em] \Rightarrow x = 1\frac{1}{15}\end{aligned}$$

Hence, the required quantity of zinc is $\mathbf{1\frac{1}{15}}$ kg.

Let the consequent be $x$. Then,

$$\begin{aligned}\Rightarrow 5 : 8 = 40 : x \\[1em] \Rightarrow \frac{5}{8} = \frac{40}{x} \\[1em] \Rightarrow 5x = 8 \times 40 \\[1em] \Rightarrow x = \frac{320}{5} = 64\end{aligned}$$

Hence, option 3 is the correct option.

Let the two numbers be $x$ and $y$. Then,

$$\begin{aligned}\Rightarrow 0.5 \times x = 0.07 \times y \\[1em] \Rightarrow \frac{x}{y} = \frac{0.07}{0.5} = \frac{7}{50}\end{aligned}$$

∴ $x : y$ = 7 : 50.

Hence, option 4 is the correct option.

Let $2x = 3y = 4z = k$. Then,

$x = \frac{k}{2}, \quad y = \frac{k}{3} \quad \text{and} \quad z = \frac{k}{4}$

Calculating L.C.M. of 2, 3 and 4:

$$\begin{array}{l|rrr} 2 & 2, & 3, & 4 \\ \hline 2 & 1, & 3, & 2 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 2, 3 and 4 = 2 × 2 × 3 = 12.

$$\begin{aligned}\Rightarrow x : y : z \\[1em] \Rightarrow \frac{k}{2} : \frac{k}{3} : \frac{k}{4} \\[1em] \Rightarrow \left(\frac{k}{2} \times \frac{12}{k}\right) : \left(\frac{k}{3} \times \frac{12}{k}\right) : \left(\frac{k}{4} \times \frac{12}{k}\right) \\[1em] \Rightarrow 6 : 4 : 3\end{aligned}$$

Hence, option 4 is the correct option.

Calculating L.C.M. of two values of $y$, namely 9 and 5:

$$\begin{array}{l|rr} 3 & 9, & 5 \\ \hline 3 & 3, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 9 and 5 = 3 × 3 × 5 = 45.

$$\begin{aligned}\Rightarrow x : y = 7 : 9 = (7 \times 5) : (9 \times 5) = 35 : 45 \\[1em] \Rightarrow y : z = 5 : 4 = (5 \times 9) : (4 \times 9) = 45 : 36\end{aligned}$$

∴ $x : y : z$ = 35 : 45 : 36.

Hence, option 2 is the correct option.

Let the fourth proportional to 3, 5 and 21 be $x$. Then,

$$\begin{aligned}\Rightarrow 3 : 5 :: 21 : x \\[1em] \Rightarrow 3 \times x = 5 \times 21 \\[1em] \Rightarrow 3x = 105 \\[1em] \Rightarrow x = \frac{105}{3} = 35\end{aligned}$$

Hence, option 1 is the correct option.

Mean proportional between 7 and 28 = $\sqrt{7 \times 28} = \sqrt{196} = 14$

Hence, option 3 is the correct option.

Let the third proportional to 9 and 12 be $x$. Then,

$$\begin{aligned}\Rightarrow 9 : 12 :: 12 : x \\[1em] \Rightarrow 9 \times x = 12 \times 12 \\[1em] \Rightarrow 9x = 144 \\[1em] \Rightarrow x = \frac{144}{9} = 16\end{aligned}$$

Hence, option 3 is the correct option.

Let the required fraction be $x$. Then,

$x : \frac{1}{27} = \frac{3}{7} : \frac{5}{9}$

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow x \times \frac{5}{9} = \frac{1}{27} \times \frac{3}{7} \\[1em] \Rightarrow \frac{5x}{9} = \frac{3}{189} = \frac{1}{63} \\[1em] \Rightarrow x = \left(\frac{1}{63} \times \frac{9}{5}\right) \\[1em] \Rightarrow x = \frac{9}{315} \\[1em] \Rightarrow x = \frac{1}{35}\end{aligned}$$

Hence, option 2 is the correct option.

Let the required number be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{49 + x}{68 + x} = \frac{3}{4} \\[1em] \Rightarrow 4(49 + x) = 3(68 + x) \\[1em] \Rightarrow 196 + 4x = 204 + 3x \\[1em] \Rightarrow 4x - 3x = 204 - 196 \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, option 1 is the correct option.

Let the required number be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{6 + x}{14 + x} = \frac{18 + x}{38 + x} \\[1em] \Rightarrow (6 + x)(38 + x) = (14 + x)(18 + x) \\[1em] \Rightarrow 228 + 6x + 38x + x^2 = 252 + 14x + 18x + x^2 \\[1em] \Rightarrow 228 + 44x = 252 + 32x \\[1em] \Rightarrow 44x - 32x = 252 - 228 \\[1em] \Rightarrow 12x = 24 \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, option 2 is the correct option.

Let the required number be $x$. Then,

$$\begin{aligned}\Rightarrow \frac{14 - x}{17 - x} = \frac{34 - x}{42 - x} \\[1em] \Rightarrow (14 - x)(42 - x) = (17 - x)(34 - x) \\[1em] \Rightarrow 588 - 14x - 42x + x^2 = 578 - 17x - 34x + x^2 \\[1em] \Rightarrow 588 - 56x = 578 - 51x \\[1em] \Rightarrow 588 - 578 = 56x - 51x \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, option 3 is the correct option.

Calculating L.C.M. of two values of B, namely 5 and 2:

$$\begin{array}{l|rr} 2 & 5, & 2 \\ \hline 5 & 5, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 5 and 2 = 2 × 5 = 10.

A : B = 4 : 5 = (4 × 2) : (5 × 2) = 8 : 10

B : C = 2 : 3 = (2 × 5) : (3 × 5) = 10 : 15

∴ A : B : C = 8 : 10 : 15.

Let A's money, B's money and C's money be ₹ $8x, ₹ 10x$ and ₹ $15x$ respectively.

But, A has ₹ 800.

$$\begin{aligned}\Rightarrow 8x = 800 \\[1em] \Rightarrow x = 100\end{aligned}$$

∴ C's money = ₹ (15 × 100) = ₹ 1,500.

Hence, option 3 is the correct option.

Calculating L.C.M. of two values of B, namely 3 and 4:

$$\begin{array}{l|rr} 2 & 3, & 4 \\ \hline 2 & 3, & 2 \\ \hline 3 & 3, & 1 \\ \hline & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 3 and 4 = 2 × 2 × 3 = 12.

A : B = 2 : 3 = (2 × 4) : (3 × 4) = 8 : 12

B : C = 4 : 5 = (4 × 3) : (5 × 3) = 12 : 15

∴ A : B : C = 8 : 12 : 15.

Sum of ratio terms = (8 + 12 + 15) = 35.

∴ C's share = ₹ $\left(7000 \times \frac{15}{35}\right)$ = ₹ 3,000

Hence, option 3 is the correct option.

Let C's share be ₹ $x$. Then,

B's share = ₹ ($x$ + 8) and A's share = ₹ ($x$ + 8 + 7) = ₹ ($x$ + 15).

$$\begin{aligned}\Rightarrow (x + 15) + (x + 8) + x = 53 \\[1em] \Rightarrow 3x + 23 = 53 \\[1em] \Rightarrow 3x = 30 \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ C's share = ₹ 10, B's share = ₹ 18 and A's share = ₹ 25.

∴ Ratio of their shares = 25 : 18 : 10.

Hence, option 2 is the correct option.

Let the number of one-rupee, 50-paise and 25-paise coins be $3x, 4x$ and $12x$ respectively. Then,

$$\begin{aligned}\Rightarrow 3x + \frac{4x}{2} + \frac{12x}{4} = 600 \\[1em] \Rightarrow 3x + 2x + 3x = 600 \\[1em] \Rightarrow 8x = 600 \\[1em] \Rightarrow x = 75\end{aligned}$$

∴ Number of 25-paise coins = (12 × 75) = 900.

Hence, option 2 is the correct option.

Let the total amount be ₹ $x$.

Sum of ratio terms = (4 + 3) = 7.

∴ Q's share = ₹ $\left(x \times \frac{3}{7}\right) = ₹\ \frac{3x}{7}$

But, Q's share = ₹ 4,800.

$$\begin{aligned}\Rightarrow \frac{3x}{7} = 4800 \\[1em] \Rightarrow x = \left(4800 \times \frac{7}{3}\right) = 11200\end{aligned}$$

Hence, option 1 is the correct option.

Ratio of boys : girls = 7 : 5.

Sum of ratio terms = (7 + 5) = 12.

No. of boys = $\left(720 \times \frac{7}{12}\right) = 420$

No. of girls = $\left(720 \times \frac{5}{12}\right) = 300$

Let $x$ more girls be admitted to make the ratio 1 : 1.

$$\begin{aligned}\Rightarrow \frac{420}{300 + x} = \frac{1}{1} \\[1em] \Rightarrow 300 + x = 420 \\[1em] \Rightarrow x = 120\end{aligned}$$

Hence, option 2 is the correct option.

Let the prices of the scooter and the moped be ₹ $9x$ and ₹ $5x$ respectively. Then,

$$\begin{aligned}\Rightarrow 9x - 5x = 6800 \\[1em] \Rightarrow 4x = 6800 \\[1em] \Rightarrow x = 1700\end{aligned}$$

∴ Price of the scooter = ₹ (9 × 1700) = ₹ 15,300.

Hence, option 2 is the correct option.

Ratio of zinc : copper = 13 : 7.

Sum of ratio terms = (13 + 7) = 20.

∴ Quantity of zinc = $\left(100 \times \frac{13}{20}\right) \text{ kg} = 65 \text{ kg}$

Hence, option 4 is the correct option.

Let the numbers be $3x$ and $5x$. Then,

$$\begin{aligned}\Rightarrow \frac{3x - 9}{5x - 9} = \frac{12}{23} \\[1em] \Rightarrow 23(3x - 9) = 12(5x - 9) \\[1em] \Rightarrow 69x - 207 = 60x - 108 \\[1em] \Rightarrow 69x - 60x = 207 - 108 \\[1em] \Rightarrow 9x = 99 \\[1em] \Rightarrow x = 11\end{aligned}$$

∴ Second number = (5 × 11) = 55.

Hence, option 4 is the correct option.

For two whole numbers in the ratio $a : b$ with sum 64, the sum of ratio terms ($a + b$) must divide 64 exactly.

Option 1 : 5 : 3 ⇒ sum of terms = 8, and 64 ÷ 8 = 8. Numbers are 40 and 24.

Option 2 : 7 : 1 ⇒ sum of terms = 8, and 64 ÷ 8 = 8. Numbers are 56 and 8.

Option 3 : 3 : 4 ⇒ sum of terms = 7, and 64 ÷ 7 is not a whole number.

Option 4 : 9 : 7 ⇒ sum of terms = 16, and 64 ÷ 16 = 4. Numbers are 36 and 28.

So, the numbers cannot be in the ratio 3 : 4.

Hence, option 3 is the correct option.

Given ratios are $\frac{7}{20}, \frac{13}{25}, \frac{17}{30}$ and $\frac{11}{15}$.

Calculating L.C.M. of 20, 25, 30 and 15:

$$\begin{array}{l|rrrr} 2 & 20, & 25, & 30, & 15 \\ \hline 2 & 10, & 25, & 15, & 15 \\ \hline 3 & 5, & 25, & 15, & 15 \\ \hline 5 & 5, & 25, & 5, & 5 \\ \hline 5 & 1, & 5, & 1, & 1 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 20, 25, 30 and 15 = 2 × 2 × 3 × 5 × 5 = 300.

$$\begin{aligned}\Rightarrow \frac{7}{20} = \frac{7 \times 15}{20 \times 15} = \frac{105}{300};\\[1em] \Rightarrow \frac{13}{25} = \frac{13 \times 12}{25 \times 12} = \frac{156}{300}; \\[1em] \Rightarrow \frac{17}{30} = \frac{17 \times 10}{30 \times 10} = \frac{170}{300}; \\[1em] \Rightarrow \frac{11}{15} = \frac{11 \times 20}{15 \times 20} = \frac{220}{300}\end{aligned}$$

Clearly, $\frac{105}{300}$ is the smallest.

Hence, option 1 is the correct option.

Let the consequent be $x$. Then,

$$\begin{aligned}\Rightarrow 5 : 8 = 40 : x \\[1em] \Rightarrow \frac{5}{8} = \frac{40}{x} \\[1em] \Rightarrow 5x = 8 \times 40 \\[1em] \Rightarrow x = \frac{320}{5} = 64\end{aligned}$$

Hence, option 3 is the correct option.

(i) In $a : b, a$ is called the antecedent while $b$ is called the consequent.

(ii) $a : b \gt c : d$ if $ad$ > $bc$.

(Reason : $\frac{a}{b} \gt \frac{c}{d} \Leftrightarrow ad \gt bc$.)

(iii) The simplest ratio of 3 hours : 1 day is 1 : 8.

(Reason : 1 day = 24 hours. So, 3 : 24 = $\frac{3 \div 3}{24 \div 3} = \frac{1}{8}$ = 1 : 8.)

(iv) If 10% of $x$ is equal to 20% of $y$, then $x : y$ is equal to 2 : 1.

(Reason : $\frac{10}{100} \times x = \frac{20}{100} \times y \Rightarrow \frac{x}{y} = \frac{20}{10} = \frac{2}{1}$.)

(v) The mean proportion between 49 and 81 is 63.

(Reason : Mean proportion = $\sqrt{49 \times 81} = \sqrt{3969} = 63$.)

(vi) The ratio $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$ is the same as 6 : 4 : 3.

(Reason : Calculating L.C.M. of 2, 3 and 4:

$$\begin{array}{l|rrr} 2 & 2, & 3, & 4 \\ \hline 2 & 1, & 3, & 2 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 2, 3 and 4 = 2 × 2 × 3 = 12.

So, $\left(\frac{1}{2} \times 12\right) : \left(\frac{1}{3} \times 12\right) : \left(\frac{1}{4} \times 12\right) = 6 : 4 : 3$.)

(vii) If a sum of ₹ 455 is divided between A and B in the ratio 4 : 3, then A gets ₹ 260.

(Reason : Sum of ratio terms = (4 + 3) = 7. So, A's share = ₹ $\left(455 \times \frac{4}{7}\right)$ = ₹ 260.)

(i) False. If $a : b :: c : d$, then product of extremes = product of means, i.e., $ad = bc$ and not $ac = db$.

(ii) True. $\frac{1}{x} : \frac{1}{y} = \frac{1}{x} \times \frac{y}{1} = \frac{y}{x} = y : x$.

(iii) False. Three numbers $a, b, c$ are in continued proportion if $a : b = b : c$, i.e., if $\frac{a}{b} = \frac{b}{c}$ and not $\frac{a}{b} = \frac{c}{b}$.

(iv) True. If $a : b :: b : c$, then $b^2 = ac \Rightarrow b = \sqrt{ac}$, and $b$ is called the mean proportion between $a$ and $c$.

1. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's savings = 10% of $5x = \left(\frac{10}{100} \times 5x\right) = \frac{5x}{10}$

Q's savings = 15% of $6x = \left(\frac{15}{100} \times 6x\right) = \frac{9x}{10}$

R's savings = 20% of $8x = \left(\frac{20}{100} \times 8x\right) = \frac{16x}{10}$

∴ Ratio of their savings

$= \frac{5x}{10} : \frac{9x}{10} : \frac{16x}{10} = 5 : 9 : 16$

Hence, option (d) is the correct option.

2. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P saves $\frac{1}{2}$ of his salary, so he spends $\left(1 - \frac{1}{2}\right) = \frac{1}{2}$ of his salary.

Q saves $\frac{1}{3}$ of his salary, so he spends $\left(1 - \frac{1}{3}\right) = \frac{2}{3}$ of his salary.

R saves $\frac{1}{4}$ of his salary, so he spends $\left(1 - \frac{1}{4}\right) = \frac{3}{4}$ of his salary.

P's expenditure = $\left(\frac{1}{2} \times 5x\right) = \frac{5x}{2}$

Q's expenditure = $\left(\frac{2}{3} \times 6x\right) = 4x$

R's expenditure = $\left(\frac{3}{4} \times 8x\right) = 6x$

∴ Ratio of their expenditures

$= \frac{5x}{2} : 4x : 6x = 5 : 8 : 12 \quad \text{[multiplying each term by 2]}$

Hence, option (d) is the correct option.

3. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's new salary = 120% of $5x = \left(\frac{120}{100} \times 5x\right) = 6x$

Q's new salary = 110% of $6x = \left(\frac{110}{100} \times 6x\right) = \frac{33x}{5}$

R's new salary = 115% of $8x = \left(\frac{115}{100} \times 8x\right) = \frac{46x}{5}$

∴ Ratio of their new salaries

$= 6x : \frac{33x}{5} : \frac{46x}{5} = 30 : 33 : 46 \quad \text{[multiplying each term by 5]}$

Hence, option (c) is the correct option.

4. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's new salary = 90% of $5x = \left(\frac{90}{100} \times 5x\right) = \frac{45x}{10}$

Q's new salary = 120% of $6x = \left(\frac{120}{100} \times 6x\right) = \frac{72x}{10}$

R's new salary = 70% of $8x = \left(\frac{70}{100} \times 8x\right) = \frac{56x}{10}$

∴ Ratio of their new salaries

$= \frac{45x}{10} : \frac{72x}{10} : \frac{56x}{10} = 45 : 72 : 56$

Hence, option (a) is the correct option.

Let the fourth proportional to 7, 12 and 28 be $x$. Then,

$$\begin{aligned}\Rightarrow 7 : 12 :: 28 : x \\[1em] \Rightarrow 7 \times x = 12 \times 28 \\[1em] \Rightarrow 7x = 336 \\[1em] \Rightarrow x = \frac{336}{7} = 48\end{aligned}$$

So, Assertion (A) is true.

If $a, b, c, d$ are in proportion, i.e., $a : b :: c : d$, then product of extremes = product of means, i.e., $ad = bc$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

Mean proportion between 8 and 98 $= \sqrt{8 \times 98} = \sqrt{784} = 28$

So, Assertion (A) is true.

By definition, three numbers $a, b, c$ are said to be in continued proportion if $a : b :: b : c$. So, Reason (R) is true.

Reason (R) only defines continued proportion; it does not by itself explain how the mean proportion $\sqrt{ac}$ is calculated. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

On adding 3 to each of the numbers 3, 7, 18 and 32, the resulting numbers are 6, 10, 21 and 35.

Product of extremes = 6 × 35 = 210.

Product of means = 10 × 21 = 210.

Since product of extremes = product of means, the numbers 6, 10, 21, 35 are in proportion. So, Assertion (A) is true.

The numbers $a, b, c, d$ are in proportion if $a : b :: c : d$, i.e., if $ad = bc$, and not if "$ab : cd$". So, Reason (R) is false.

Hence, option 3 is the correct option.

Let the number of men and women be $12x$ and $13x$ respectively.

Statement (i): The number of women is $13x$, which need not be 13 as $x$ can be any positive integer. So, statement (i) is false.

Statement (ii): Taking $x$ = 2, number of men = (12 × 2) = 24 and number of women = (13 × 2) = 26. So, statement (ii) is true.

Hence, option 2 is the correct option.

Let the numbers be $2x, 3x$ and $5x$. Then,

$$\begin{aligned}\Rightarrow (2x)^2 + (3x)^2 + (5x)^2 = 608 \\[1em] \Rightarrow 4x^2 + 9x^2 + 25x^2 = 608 \\[1em] \Rightarrow 38x^2 = 608 \\[1em] \Rightarrow x^2 = \frac{608}{38} = 16 \\[1em] \Rightarrow x = \sqrt{16} = 4\end{aligned}$$

∴ The numbers are (2 × 4), (3 × 4) and (5 × 4), i.e., 8, 12 and 20.

Hence, option 1 is the correct option.

Let the two numbers be $x$ and $y$. Then,

$$\begin{aligned}\Rightarrow 0.6 \times x = 0.09 \times y \\[1em] \Rightarrow \frac{x}{y} = \frac{0.09}{0.6} = \frac{9}{60} = \frac{3}{20}\end{aligned}$$

∴ $x : y$ = 3 : 20.

Hence, option 3 is the correct option.

We have : 2 : A = 1 : 2.

$\Rightarrow \frac{2}{\text{A}} = \frac{1}{2}$

⇒ A = 4

Also, A : B = 2 : 3.

$$\begin{aligned}\Rightarrow \frac{4}{\text{B}} = \frac{2}{3} \\[1em] \Rightarrow 2 \times \text{B} = 12 \\[1em] \Rightarrow \text{B} = 6\end{aligned}$$

Hence, option 1 is the correct option.

Let the required fraction be $x$. Then,

$x : \frac{1}{27} = \frac{3}{7} : \frac{5}{9}$

Product of extremes = product of means.

$$\begin{aligned}\Rightarrow x \times \frac{5}{9} = \frac{1}{27} \times \frac{3}{7} \\[1em] \Rightarrow \frac{5x}{9} = \frac{3}{189} = \frac{1}{63} \\[1em] \Rightarrow x = \left(\frac{1}{63} \times \frac{9}{5}\right) \\[1em] \Rightarrow x = \frac{9}{315} \\[1em] \Rightarrow x = \frac{1}{35}\end{aligned}$$

Hence, option 4 is the correct option.

Let the total strength of the class be $x$ and let the number of boys and girls be $b$ and $g$ respectively. Then,

$b + g = x$ .........(1)

$b - g$ = 12% of $x$

$b - g = \frac{12x}{100}$

$b - g = \frac{3x}{25}$ .....(2)

Adding equations (1) and (2), we get :

$$\begin{aligned}\Rightarrow 2b = x + \frac{3x}{25} \\[1em] \Rightarrow 2b = \frac{28x}{25} \\[1em] \Rightarrow b = \frac{14x}{25}\end{aligned}$$

Subtracting equation (2) from (1), we get :

$$\begin{aligned}\Rightarrow 2g = x - \frac{3x}{25} \\[1em] \Rightarrow 2g = \frac{22x}{25} \\[1em] \Rightarrow g = \frac{11x}{25}\end{aligned}$$

$\Rightarrow b : g = \frac{14x}{25} : \frac{11x}{25} = 14 : 11$

Hence, option 3 is the correct option.

We know that, $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

For the same distance, time taken is inversely proportional to the speed.

∴ Ratio of times taken $= \frac{1}{3} : \frac{1}{4} : \frac{1}{5}$

Calculating L.C.M. of 3, 4 and 5:

$$\begin{array}{l|rrr} 2 & 3, & 4, & 5 \\ \hline 2 & 3, & 2, & 5 \\ \hline 3 & 3, & 1, & 5 \\ \hline 5 & 1, & 1, & 5 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 3, 4 and 5 = 2 × 2 × 3 × 5 = 60.

$$\begin{aligned}\Rightarrow \frac{1}{3} : \frac{1}{4} : \frac{1}{5} \\[1em] \Rightarrow \left(\frac{1}{3} \times 60\right) : \left(\frac{1}{4} \times 60\right) : \left(\frac{1}{5} \times 60\right) \\[1em] \Rightarrow 20 : 15 : 12\end{aligned}$$

Hence, option 2 is the correct option.

Let the two numbers be $3x$ and $4x$, where $x$ is their H.C.F.

Then, L.C.M. of the two numbers = $3 \times 4 \times x = 12x$.

$$\begin{aligned}\Rightarrow 12x = 180 \\[1em] \Rightarrow x = \frac{180}{12} = 15\end{aligned}$$

∴ First number = (3 × 15) = 45.

Hence, option 3 is the correct option.

Given, $x : y = y : z$.

$$\begin{aligned}\Rightarrow \frac{x}{y} = \frac{y}{z} \\[1em] \Rightarrow y^2 = xz \\[1em] \Rightarrow \frac{x^2}{y^2} = \frac{x^2}{xz} = \frac{x}{z}\end{aligned}$$

∴ $x^2 : y^2 = x : z$.

Hence, option 3 is the correct option.

Let the first, second and third numbers be $a, b$ and $c$ respectively. Then,

$b : c$ = 9 : 16 and $a : c = 1 : 4 = (1 \times 4) : (4 \times 4) = 4 : 16$

∴ $a : b : c$ = 4 : 9 : 16.

Sum of ratio terms = (4 + 9 + 16) = 29.

∴ Second number = $\left(174 \times \frac{9}{29}\right) = 54$

Hence, option 1 is the correct option.