Ratio and Proportion — Exercise 9(A)

Express the following ratio in simplest form:

18 : 30

Answer

18 : 30

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

∴ H.C.F. of 18 and 30 = 2 × 3 = 6.

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

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

Express the following ratio in simplest form:

7.5 : 9

Answer

7.5 : 9

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

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

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

∴ H.C.F. of 75 and 90 = 3 × 5 = 15.

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

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

Express the following ratio in simplest form:

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

Answer

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

We have :

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

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 :

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

We have : 40 = 2³ × 5 and 45 = 3² × 5

∴ H.C.F. of 40 and 45 = 5.

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

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

Express the following ratio in simplest form:

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

Answer

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

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 :

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

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

Express the following ratio in simplest form:

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

Answer

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

Multiplying each term by 2, we get :

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

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

Express the following ratio in simplest form:

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

Answer

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

We have :

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

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 :

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

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

∴ H.C.F. of 48, 80 and 100 = 2² = 4.

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

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

Express the following ratio in simplest form:

75 paise : 4 rupees

Answer

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

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

Express the following ratio in simplest form:

1 m 8 cm : 72 cm

Answer

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

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

Express the following ratio in simplest form:

1 hour 15 minutes : 45 minutes

Answer

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

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

Express the following ratio in simplest form:

2 kg 750 g : 3 kg

Answer

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

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

Express the following ratio in simplest form:

1 year 9 months : 2 years 4 months

Answer

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

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

Which ratio is greater?

(4 : 9) or (3 : 7)

Answer

(4 : 9) or (3 : 7)

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

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.

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

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

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

Which ratio is greater?

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

Answer

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

We have :

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

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

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

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.

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

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

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

Which ratio is greater?

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

Answer

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

We have :

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

And,

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

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

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.

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

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

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

Which ratio is greater?

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

Answer

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

We have :

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

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

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

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.

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

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

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

Arrange the following ratios in ascending order of magnitude:

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

Answer

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

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

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,

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

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

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

Arrange the following ratios in ascending order of magnitude:

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

Answer

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

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

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,

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

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

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

Divide ₹ 142.20 between Gagan and Mukesh in the ratio $\dfrac{1}{4} : \dfrac{1}{5}$.

Answer

Given,

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

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:

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

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

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

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

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

Divide ₹ 3,726 among A, B, C in the ratio $\dfrac{1}{3} : \dfrac{1}{4} : \dfrac{1}{6}$.

Answer

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

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:

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

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

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

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

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

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

Divide ₹ 810 among A, B, C in the ratio $\dfrac{1}{4} : \dfrac{2}{5} : 1\dfrac{3}{8}$.

Answer

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

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:

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

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

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

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

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

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

Divide ₹ 1,050 between Geeta and Renu in the ratio $2\dfrac{2}{3} : 6\dfrac{2}{3}$.

Answer

Given,

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

Multiplying both the terms by 3, we get :

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

We have : 8 = 2³ and 20 = 2² × 5

∴ H.C.F. of 8 and 20 = 2² = 4.

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

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

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

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

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

Divide ₹ 747 among A, B, C such that 4A = 5B = 7C.

Answer

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

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

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.

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

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

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

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

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

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

A bag contains one rupee, 50 p and 25 p coins in the ratio 5 : 6 : 8 amounting to ₹ 210. Find the number of coins of each type.

Answer

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 = ₹ $\left(6x \times \dfrac{50}{100}\right) = ₹\ \dfrac{6x}{2}$.

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

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

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.

Find A : B : C when:

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

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

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

Answer

(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 = $\dfrac{1}{2} : \dfrac{1}{3}$ and B : C = $\dfrac{1}{4} : \dfrac{1}{5}$

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 = $\dfrac{1}{2} : \dfrac{1}{3} = \left(\dfrac{1}{2} \times 6\right) : \left(\dfrac{1}{3} \times 6\right) = 3 : 2$

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

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.

If A : B = 4 : 9 and A : C = 2 : 3, find B : C and A : B : C.

Answer

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

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

∴ 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.

If A : C = 5 : 8 and B : C = 5 : 6, find A : B and A : B : C.

Answer

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

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

∴ 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.

Two numbers are in the ratio 6 : 11. On adding 2 to the first and 7 to the second, their ratio becomes 8 : 15. Find the numbers.

Answer

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

Given,

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

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

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

Hence, the required numbers are 78 and 143.

A sum of money is divided between Mohan and Sohan in the ratio 5 : 7. If Sohan's share is ₹ 665, find the total amount.

Answer

Let the total amount be ₹ $x$.

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

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

But, Sohan's share = ₹ 665.

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

Hence, the total amount is ₹ 1,140.

Two numbers are in the ratio 7 : 4. If their difference is 72, find the numbers.

Answer

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

$\Rightarrow 7x - 4x = 72 \\[1em] \Rightarrow 3x = 72 \\[1em] \Rightarrow x = \dfrac{72}{3} = 24$

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

Hence, the required numbers are 168 and 96.

A certain sum of money is divided among A, B, C in the ratio 5 : 6 : 7. If A's share is ₹ 175, find the total amount and the share of each one of B and C.

Answer

Let the total amount be ₹ $x$.

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

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

But, A's share = ₹ 175.

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

∴ Total amount = ₹ 630.

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

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

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

The ratio of number of boys to the number of girls in a school of 1440 students is 7 : 5. If 40 new boys are admitted, find how many new girls may be admitted to make this ratio 4 : 3.

Answer

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

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

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

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

Let $x$ new girls be admitted.

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

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

Hence, 60 new girls may be admitted.

The sum of three numbers is 212. If the ratio of the first to the second is 13 : 16 and that of the second to the third is 2 : 3, then find the numbers.

Answer

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

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

∴ $a : b : c$ = 13 : 16 : 24.

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

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

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

Find the number which when added to each term of the ratio 27 : 35, changes the ratio to 4 : 5.

Answer

Let the required number be $x$. Then,

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

Hence, the required number is 5.

The present ages of Mr. Sen and his son are in the ratio 17 : 9. If the ratio of their ages, 9 years ago was 7 : 3, then find their present ages.

Answer

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.

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

∴ 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.

A sample of vinegar was prepared by dissolving 705 ml of acetic acid in 4.7 litres of water. Find the ratio by volume of acetic acid to water in this sample.

Answer

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.

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

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

The salaries of A, B and C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?

Answer

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

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

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

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

∴ New ratio of their salaries

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

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

Anuj got 58 marks out of 75 in Physics and 97 marks out of 120 in Maths. In which of the two subjects did he perform better?

Answer

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

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

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.

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

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

Hence, Anuj performed better in Maths.

A mixture consists of only two components A and B. In 60 litres of this mixture, the components A and B are present in the ratio 2 : 1. What quantity of component B has to be added to this mixture so that the new ratio is 1 : 2?

Answer

Ratio of A : B = 2 : 1.

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

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

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

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

$\Rightarrow \dfrac{40}{20 + x} = \dfrac{1}{2} \\[1em] \Rightarrow 20 + x = 40 \times 2 \\[1em] \Rightarrow 20 + x = 80 \\[1em] \Rightarrow x = 60$

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