Ratio & Proportion

₹7.80 to 65 paise

Convert ₹ into paise: ₹7.80 = 7.80 × 100 paise = 780 paise.

Ratio = 780 : 65

H.C.F. of 780 and 65 is 65.

$\frac{780 \div 65}{65 \div 65} = \frac{12}{1} = 12 : 1$

Hence, the ratio is 12 : 1.

15 hours to 1 day

Convert day into hours: 1 day = 24 hours.

Ratio = 15 : 24

H.C.F. of 15 and 24 is 3.

$\frac{15 \div 3}{24 \div 3} = \frac{5}{8} = 5 : 8$

Hence, the ratio is 5 : 8.

750 gm to 2 kg

Convert kg into gm: 2 kg = 2 × 1000 gm = 2000 gm.

Ratio = 750 : 2000

H.C.F. of 750 and 2000 is 250.

$\frac{750 \div 250}{2000 \div 250} = \frac{3}{8} = 3 : 8$

Hence, the ratio is 3 : 8.

64 ml to 1 litre

Convert litre into ml: 1 litre = 1000 ml.

Ratio = 64 : 1000

H.C.F. of 64 and 1000 is 8.

$\frac{64 \div 8}{1000 \div 8} = \frac{8}{125} = 8 : 125$

Hence, the ratio is 8 : 125.

18 mm to 3 cm

Convert cm into mm: 3 cm = 3 × 10 mm = 30 mm.

Ratio = 18 : 30

H.C.F. of 18 and 30 is 6.

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

Hence, the ratio is 3 : 5.

80 kg to 2 quintals

Convert quintals into kg: 1 quintal = 100 kg, so 2 quintals = 200 kg.

Ratio = 80 : 200

H.C.F. of 80 and 200 is 40.

$\frac{80 \div 40}{200 \div 40} = \frac{2}{5} = 2 : 5$

Hence, the ratio is 2 : 5.

15 : 25

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

$\frac{15 \div 5}{25 \div 5} = \frac{3}{5} = 3 : 5$

Hence, the answer is 3 : 5.

49 : 35

H.C.F. of 49 and 35 is 7.

$\frac{49 \div 7}{35 \div 7} = \frac{7}{5} = 7 : 5$

Hence, the answer is 7 : 5.

90 : 72

H.C.F. of 90 and 72 is 18.

$\frac{90 \div 18}{72 \div 18} = \frac{5}{4} = 5 : 4$

Hence, the answer is 5 : 4.

146 : 365

H.C.F. of 146 and 365 is 73.

$\frac{146 \div 73}{365 \div 73} = \frac{2}{5} = 2 : 5$

Hence, the answer is 2 : 5.

111 : 259

H.C.F. of 111 and 259 is 37.

$\frac{111 \div 37}{259 \div 37} = \frac{3}{7} = 3 : 7$

Hence, the answer is 3 : 7.

$\frac{1}{6} : \frac{1}{8}$

Let us find the L.C.M. of 6 and 8:

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

L.C.M. = 2 × 2 × 2 × 3 = 24.

Multiplying each term by 24:

$\frac{1}{6} : \frac{1}{8} = \left(\frac{1}{6} \times 24\right) : \left(\frac{1}{8} \times 24\right)$

= 4 : 3

Hence, the answer is 4 : 3.

$\frac{3}{4} : \frac{4}{5} : \frac{5}{6}$

Let us find the L.C.M. of 4, 5 and 6:

$$\begin{array}{l|rrr} 2 & 4, & 5, & 6 \\ \hline 2 & 2, & 5, & 3 \\ \hline 3 & 1, & 5, & 3 \\ \hline 5 & 1, & 5, & 1 \\ \hline & 1, & 1, & 1 \end{array}$$

L.C.M. = 2 × 2 × 3 × 5 = 60.

Multiplying each term by 60:

$\frac{3}{4} : \frac{4}{5} : \frac{5}{6} = \left(\frac{3}{4} \times 60\right) : \left(\frac{4}{5} \times 60\right) : \left(\frac{5}{6} \times 60\right) = 45 : 48 : 50$

Since the H.C.F. of 45, 48 and 50 is 1, the ratio is already in simplest form.

Hence, the answer is 45 : 48 : 50.

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

Convert the mixed fractions into improper fractions:

$2\frac{1}{3} = \frac{7}{3}, \quad 3\frac{1}{4} = \frac{13}{4}, \quad 5\frac{1}{6} = \frac{31}{6}$

Let us find the 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. = 2 × 2 × 3 = 12.

Multiplying each term by 12:

$\frac{7}{3} : \frac{13}{4} : \frac{31}{6} = \left(\frac{7}{3} \times 12\right) : \left(\frac{13}{4} \times 12\right) : \left(\frac{31}{6} \times 12\right)$

= 28 : 39 : 62

Since the H.C.F. of 28, 39 and 62 is 1, the ratio is already in simplest form.

Hence, the answer is 28 : 39 : 62.

Given:

Number of boys = 420

Number of girls = 180

Total number of students = 420 + 180 = 600

(i) Ratio of girls to boys = 180 : 420

H.C.F. of 180 and 420 is 60.

$\frac{180 \div 60}{420 \div 60} = \frac{3}{7} = 3 : 7$

Hence, the ratio of girls to boys is 3 : 7.

(ii) Ratio of girls to total number of students = 180 : 600

H.C.F. of 180 and 600 is 60.

$\frac{180 \div 60}{600 \div 60} = \frac{3}{10} = 3 : 10$

Hence, the ratio of girls to total number of students is 3 : 10.

(iii) Ratio of boys to total number of students = 420 : 600

H.C.F. of 420 and 600 is 60.

$\frac{420 \div 60}{600 \div 60} = \frac{7}{10} = 7 : 10$

Hence, the ratio of boys to total number of students is 7 : 10.

Given:

Total population = 3540

Number of males = 2065

Number of females = Total population − Number of males

Number of females = 3540 − 2065 = 1475

Ratio of males to females = 2065 : 1475

H.C.F. of 2065 and 1475 is 295.

$\frac{2065 \div 295}{1475 \div 295} = \frac{7}{5} = 7 : 5$

Hence, the ratio of males to females is 7 : 5.

Given:

Cost of coffee = ₹48 per 100 gm

Cost of tea = ₹280 per kg

To compare the two prices, express the cost of coffee per kg (1 kg = 1000 gm).

Cost of coffee per kg = ₹48 × 10 = ₹480 per kg

Ratio of price of coffee to that of tea = 480 : 280

H.C.F. of 480 and 280 is 40.

$\frac{480 \div 40}{280 \div 40} = \frac{12}{7} = 12 : 7$

Hence, the ratio of the price of coffee to that of tea is 12 : 7.

Given:

Total weight of alloy = 21 gm

Ratio (Tin : Zinc) = 3 : 4

Sum of ratio terms = 3 + 4 = 7.

(i) Quantity of tin = $21 \times \frac{3}{7}$ = 3 × 3 gm = 9 gms.

Hence, the quantity of tin is 9 gms.

(ii) Quantity of zinc = $21 \times \frac{4}{7}$ = 4 × 3 gm = 12 gms.

Hence, the quantity of zinc is 12 gms.

Given:

Total money = ₹1,155

Ratio (Amit : Sanya) = 7 : 4

Sum of ratio terms = 7 + 4 = 11.

Amit's share = $1{,}155 \times \frac{7}{11}$ = 7 × ₹105 = ₹735.

Sanya's share = $1{,}155 \times \frac{4}{11}$ = 4 × ₹105 = ₹420.

Hence, Amit gets ₹735 and Sanya gets ₹420.

Given:

Total money = ₹3,500

Ratio (A : B : C) = 3 : 4 : 7

Sum of ratio terms = 3 + 4 + 7 = 14.

A's share = $3{,}500 \times \frac{3}{14}$ = 3 × ₹250 = ₹750.

B's share = $3{,}500 \times \frac{4}{14}$ = 4 × ₹250 = ₹1,000.

C's share = $3{,}500 \times \frac{7}{14}$ = 7 × ₹250 = ₹1,750.

Hence, A gets ₹750, B gets ₹1,000 and C gets ₹1,750.

Given:

Total money = ₹3,010

Suppose C gets ₹1. Then B gets double of C = ₹2, and A gets double of B = ₹4.

So, Ratio (A : B : C) = 4 : 2 : 1

Sum of ratio terms = 4 + 2 + 1 = 7.

A's share = $3{,}010 \times \frac{4}{7}$ = 4 × ₹430 = ₹1,720.

B's share = $3{,}010 \times \frac{2}{7}$ = 2 × ₹430 = ₹860.

C's share = $3{,}010 \times \frac{1}{7}$ = 1 × ₹430 = ₹430.

Hence, A gets ₹1,720, B gets ₹860 and C gets ₹430.

Given:

Total = 182

Ratio = $\frac{1}{10} : \frac{1}{15} : \frac{1}{20}$

Let us find the L.C.M. of 10, 15 and 20:

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

L.C.M. = 2 × 2 × 3 × 5 = 60.

Multiplying each term by 60:

$$\begin{aligned}\Rightarrow \left(\frac{1}{10} \times 60\right) : \left(\frac{1}{15} \times 60\right) : \left(\frac{1}{20} \times 60\right) \\[1em] \Rightarrow 6 : 4 : 3\end{aligned}$$

Sum of ratio terms = 6 + 4 + 3 = 13.

First part = $182 \times \frac{6}{13}$ = 6 × 14 = 84.

Second part = $182 \times \frac{4}{13}$ = 4 × 14 = 56.

Third part = $182 \times \frac{3}{13}$ = 3 × 14 = 42.

Hence, the three parts are 84, 56 and 42.

Given:

Perimeter of the triangle = 54 cm

Ratio of sides = 2 : 3 : 4

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

First side = $54 \times \frac{2}{9}$ = 2 × 6 cm = 12 cm.

Second side = $54 \times \frac{3}{9}$ = 3 × 6 cm = 18 cm.

Third side = $54 \times \frac{4}{9}$ = 4 × 6 cm = 24 cm.

Hence, the lengths of the sides of the triangle are 12 cm, 18 cm and 24 cm.

Given:

Ratio of angles = 3 : 5 : 7

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

Sum of ratio terms = 3 + 5 + 7 = 15.

First angle = $180 \times \frac{3}{15}$ = 3 × 12° = 36°.

Second angle = $180 \times \frac{5}{15}$ = 5 × 12° = 60°.

Third angle = $180 \times \frac{7}{15}$ = 7 × 12° = 84°.

Hence, the angles of the triangle are 36°, 60° and 84°.

Given:

Ratio in simplest form = 11 : 15

Antecedent = 121

It is given that $\frac{11}{15} = \frac{121}{?}$.

We know that 121 = 11 × 11.

So, we multiply the numerator and denominator of $\frac{11}{15}$ by 11.

$\frac{11}{15} = \frac{11 \times 11}{15 \times 11} = \frac{121}{165}$

Hence, the required consequent is 165.

Given:

Ratio in simplest form = 7 : 12

Consequent = 144

It is given that $\frac{7}{12} = \frac{?}{144}$.

We know that 144 = 12 × 12.

So, we multiply the numerator and denominator of $\frac{7}{12}$ by 12.

$\frac{7}{12} = \frac{7 \times 12}{12 \times 12} = \frac{84}{144}$

Hence, the required antecedent is 84.

12 : 18 = 8 : 12

We express each of the given ratios in simplest form.

Consider 12 : 18.

$$\begin{array}{l} 12\overline{\smash{\big)}18 \smash{\big(}}1 \\ \phantom{111}{12} \\ \phantom{111}\overline{6\smash{\big)}12 \smash{\big(}2} \\ \phantom{11111}\underline{12\phantom{1}} \\ \phantom{111111}0\ \end{array}$$

H.C.F. of 12 and 18 = 6.

$\frac{12 \div 6}{18 \div 6} = \frac{2}{3} = 2 : 3$

Again, consider 8 : 12.

$$\begin{array}{l} 8\overline{\smash{\big)}12 \smash{\big(}}1 \\ \phantom{111}{8} \\ \phantom{111}\overline{4\smash{\big)}8 \smash{\big(}2} \\ \phantom{11111}\underline{8\phantom{1}} \\ \phantom{11111}0\ \end{array}$$

H.C.F. of 8 and 12 = 4.

$\frac{8 \div 4}{12 \div 4} = \frac{2}{3} = 2 : 3$

Thus, both the ratios reduce to 2 : 3, so they are equal.

Hence, 12 : 18 = 8 : 12 is verified.

81 kg : 45 kg = 18 men : 10 men

We express each of the given ratios in simplest form.

Consider 81 : 45.

$$\begin{array}{l} 45\overline{\smash{\big)}81 \smash{\big(}}1 \\ \phantom{111}{45} \\ \phantom{111}\overline{36\smash{\big)}45 \smash{\big(}1} \\ \phantom{111111}{36} \\ \phantom{111111}\overline{9\smash{\big)}36 \smash{\big(}4} \\ \phantom{11111111}\underline{36\phantom{1}} \\ \phantom{111111111}0\ \end{array}$$

H.C.F. of 81 and 45 = 9.

$\frac{81 \div 9}{45 \div 9} = \frac{9}{5} = 9 : 5$

Again, consider 18 : 10.

$$\begin{array}{l} 10\overline{\smash{\big)}18 \smash{\big(}}1 \\ \phantom{111}{10} \\ \phantom{111}\overline{8\smash{\big)}10 \smash{\big(}1} \\ \phantom{111111}{8} \\ \phantom{111111}\overline{2\smash{\big)}8 \smash{\big(}4} \\ \phantom{11111111}\underline{8\phantom{1}} \\ \phantom{11111111}0\ \end{array}$$

H.C.F. of 18 and 10 = 2.

$\frac{18 \div 2}{10 \div 2} = \frac{9}{5} = 9 : 5$

Thus, both the ratios reduce to 9 : 5, so they are equal.

Hence, 81 kg : 45 kg = 18 men : 10 men is verified.

$3\frac{1}{3} : 2\frac{1}{2} = 12 : 9$

We express each of the given ratios in simplest form.

Convert the mixed numbers into improper fractions: $3\frac{1}{3} = \frac{10}{3}$ and $2\frac{1}{2} = \frac{5}{2}$.

Consider $\frac{10}{3} : \frac{5}{2}$.

L.C.M. of 3 and 2 is 6. Multiplying each term by 6:

$\frac{10}{3} : \frac{5}{2} = \left(\frac{10}{3} \times 6\right) : \left(\frac{5}{2} \times 6\right) = 20 : 15$

$$\begin{array}{l} 15\overline{\smash{\big)}20 \smash{\big(}}1 \\ \phantom{111}{15} \\ \phantom{111}\overline{5\smash{\big)}15 \smash{\big(}3} \\ \phantom{11111}\underline{15\phantom{1}} \\ \phantom{111111}0\ \end{array}$$

H.C.F. of 20 and 15 = 5.

$\frac{20 \div 5}{15 \div 5} = \frac{4}{3} = 4 : 3$

Again, consider 12 : 9.

$$\begin{array}{l} 9\overline{\smash{\big)}12 \smash{\big(}}1 \\ \phantom{111}{9} \\ \phantom{111}\overline{3\smash{\big)}9 \smash{\big(}3} \\ \phantom{11111}\underline{9\phantom{1}} \\ \phantom{11111}0\ \end{array}$$

H.C.F. of 12 and 9 = 3.

$\frac{12 \div 3}{9 \div 3} = \frac{4}{3} = 4 : 3$

Thus, both the ratios reduce to 4 : 3, so they are equal.

Hence, $3\frac{1}{3} : 2\frac{1}{2} = 12 : 9$ is verified.

6 : 45 ≠ 4.8 : 3.6

We express each of the given ratios in simplest form.

Consider 6 : 45.

$$\begin{array}{l} 6\overline{\smash{\big)}45 \smash{\big(}}7 \\ \phantom{11}{42} \\ \phantom{11}\overline{3\smash{\big)}6 \smash{\big(}2} \\ \phantom{1111}\underline{6\phantom{1}} \\ \phantom{1111}0\ \end{array}$$

H.C.F. of 6 and 45 = 3.

$\frac{6 \div 3}{45 \div 3} = \frac{2}{15} = 2 : 15$

Again, consider 4.8 : 3.6.

Multiplying each term by 10, we get 48 : 36.

$$\begin{array}{l} 36\overline{\smash{\big)}48 \smash{\big(}}1 \\ \phantom{111}{36} \\ \phantom{111}\overline{12\smash{\big)}36 \smash{\big(}3} \\ \phantom{111111}\underline{36\phantom{1}} \\ \phantom{1111111}0\ \end{array}$$

H.C.F. of 48 and 36 = 12.

$\frac{48 \div 12}{36 \div 12} = \frac{4}{3} = 4 : 3$

Thus, 2 : 15 ≠ 4 : 3, so the two ratios are not equal.

Hence, 6 : 45 ≠ 4.8 : 3.6 is verified.

Four numbers a, b, c, d are in proportion if and only if a × d = b × c (product of extremes = product of means).

(i) 30, 42, 5, 7

Product of extremes = 30 × 7 = 210

Product of means = 42 × 5 = 210

Since 210 = 210, the numbers are in proportion.

Hence, 30, 42, 5, 7 are in proportion.

(ii) 4, 11, 22, 33

Product of extremes = 4 × 33 = 132

Product of means = 11 × 22 = 242

Since 132 ≠ 242, the numbers are not in proportion.

Hence, 4, 11, 22, 33 are not in proportion.

(iii) 9, 13, 10, 14

Product of extremes = 9 × 14 = 126

Product of means = 13 × 10 = 130

Since 126 ≠ 130, the numbers are not in proportion.

Hence, 9, 13, 10, 14 are not in proportion.

(iv) 16, 18, 24, 27

Product of extremes = 16 × 27 = 432

Product of means = 18 × 24 = 432

Since 432 = 432, the numbers are in proportion.

Hence, 16, 18, 24, 27 are in proportion.

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

36 : 81 :: x : 63

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 36 \times 63 = 81 \times x\\[1em] \Rightarrow x = \frac{36 \times 63}{81} \\[1em] = \frac{2268}{81} \\[1em] = 28\end{aligned}$$

Hence, x = 28.

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

27 : x :: 63 : 84

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 27 \times 84 = x \times 63\\[1em] \Rightarrow x = \frac{27 \times 84}{63} \\[1em] = \frac{2268}{63} \\[1em] = 36\end{aligned}$$

Hence, x = 36.

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

x : 92 :: 87 : 116

Product of extremes = product of means

$$\begin{aligned}\Rightarrow x \times 116 = 92 \times 87\\[1em] \Rightarrow x = \frac{92 \times 87}{116} \\[1em] = \frac{8004}{116} \\[1em] = 69\end{aligned}$$

Hence, x = 69.

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

45 : x :: 25 : 35

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 45 \times 35 = x \times 25\\[1em] \Rightarrow x = \frac{45 \times 35}{25} \\[1em] = \frac{1575}{25} \\[1em] = 63\end{aligned}$$

Hence, x = 63.

Given:

1st term = 32, 2nd term = 112, 4th term = 217

Let the 3rd term be x. Then, 32 : 112 :: x : 217.

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 32 \times 217 = 112 \times x\\[1em] \Rightarrow x = \frac{32 \times 217}{112} \\[1em] = \frac{6944}{112} \\[1em] = 62\end{aligned}$$

Hence, the 3rd term of the proportion is 62.

Given:

1st term = 51, 3rd term = 81, 4th term = 108

Let the 2nd term be x. Then, 51 : x :: 81 : 108.

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 51 \times 108 = x \times 81\\[1em] \Rightarrow x = \frac{51 \times 108}{81} \\[1em] = \frac{5508}{81} \\[1em] = 68\end{aligned}$$

Hence, the 2nd term of the proportion is 68.

Given:

Ratio (Ruchi : Rajan) = 4 : 7

Ruchi's income = ₹16,800

Let Rajan's income be ₹x. Then, 4 : 7 :: 16800 : x.

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 4 \times x = 7 \times 16{,}800\\[1em] \Rightarrow x = \frac{7 \times 16{,}800}{4} \\[1em] = \frac{1{,}17{,}600}{4} \\[1em] = 29{,}400\end{aligned}$$

Hence, Rajan earns ₹29,400 per month.

Given:

Shadow of the pole = 20 m

Height of the tree = 6 m and shadow of the tree = 8 m

At the same time, the ratio of height to shadow is the same for all objects.

Let the height of the pole be x metres. Then,

Height of pole : Shadow of pole = Height of tree : Shadow of tree

$x : 20 = 6 : 8$

Product of extremes = product of means

$$\begin{aligned}\Rightarrow x \times 8 = 20 \times 6\\[1em] \Rightarrow x = \frac{20 \times 6}{8} \\[1em] = \frac{120}{8} \\[1em] = 15\end{aligned}$$

Hence, the height of the pole is 15 metres.

Given:

Ratio (Copper : Zinc) = 9 : 5

Weight of zinc = 9.5 g

Let the weight of copper be x g. Then, 9 : 5 :: x : 9.5.

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 9 \times 9.5 = 5 \times x\\[1em] \Rightarrow x = \frac{9 \times 9.5}{5} \\[1em] = \frac{85.5}{5} \\[1em] = 17.1\end{aligned}$$

Hence, the weight of copper in the alloy is 17.1 g.

Given:

Ratio (Length : Breadth) = 9 : 5

Breadth = 60 m

Let the length be x metres. Then, 9 : 5 :: x : 60.

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 9 \times 60 = 5 \times x\\[1em] \Rightarrow x = \frac{9 \times 60}{5} \\[1em] = \frac{540}{5} \\[1em] = 108\end{aligned}$$

Hence, the length of the rectangular plot is 108 m.

H.C.F. of 108 and 135 is 27.

$\frac{108 \div 27}{135 \div 27} = \frac{4}{5} = 4 : 5$

Hence, option 4 is the correct option.

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

$$\begin{aligned}\Rightarrow 30 \times x = 42 \times 55\\[1em] \Rightarrow x = \frac{42 \times 55}{30} \\[1em] = \frac{2310}{30} \\[1em] = 77\end{aligned}$$

Hence, option 2 is the correct option.

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

$$\begin{aligned}\Rightarrow 30 \times x = 42 \times 55\\[1em] \Rightarrow x = \frac{42 \times 55}{30} \\[1em] = \frac{2310}{30} \\[1em] = 77\end{aligned}$$

Hence, option 1 is the correct option.

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

$$\begin{aligned}\Rightarrow 7 \times 63 = 9 \times x\\[1em] \Rightarrow x = \frac{7 \times 63}{9} \\[1em] = \frac{441}{9} \\[1em] = 49\end{aligned}$$

Hence, option 2 is the correct option.

If a : b :: c : d, then the product of extremes is equal to the product of means.

Here a and d are the extremes and b and c are the means.

So, ad = bc.

Hence, option 3 is the correct option.

Sum of ratio terms = 5 + 7 = 12.

Larger number = $504 \times \frac{7}{12}$ = 7 × 42 = 294.

Hence, option 2 is the correct option.

Sum of ratio terms = 8 + 11 = 19.

A's share = $1{,}520 \times \frac{8}{19}$ = 8 × ₹80 = ₹640.

Hence, option 3 is the correct option.

Sum of ratio terms = 13 + 8 = 21.

So, the girls form $\frac{8}{21}$ of the total strength.

Total strength $\times \frac{8}{21}$ = 840

Total strength = $840 \times \frac{21}{8}$ = 105 × 21 = 2205.

Hence, option 3 is the correct option.

Sum of ratio terms = 1 + 3 + 5 = 9.

Largest side = $90 \times \frac{5}{9}$ = 5 × 10 cm = 50 cm.

Hence, option 3 is the correct option.

A ratio a : b is in its simplest form when the H.C.F. of a and b is 1. To reduce it, we divide each of its terms by the H.C.F. of a and b.

Hence, option 1 is the correct option.

(i) Simplest form of 78 : 91

H.C.F. of 78 and 91 is 13.

$\frac{78 \div 13}{91 \div 13} = \frac{6}{7} = 6 : 7$

Hence, the simplest form of 78 : 91 is 6 : 7.

(ii) A ratio exists only between two quantities of the same kind. Since ₹12 (money) and 24 cm (length) are quantities of different kinds, their ratio cannot be found.

Hence, no, it is not possible.

(iii) For 3, 6, 9, 12 to be in proportion, product of extremes = product of means.

Product of extremes = 3 × 12 = 36

Product of means = 6 × 9 = 54

Since 36 ≠ 54, they are not in proportion.

Hence, no, 3, 6, 9, 12 are not in proportion.

(iv) x : 36 :: 33 : 108

Product of extremes = product of means

$$\begin{aligned}\Rightarrow x \times 108 = 36 \times 33\\[1em] \Rightarrow x = \frac{36 \times 33}{108} \\[1em] = \frac{1188}{108} \\[1em] = 11\end{aligned}$$

Hence, x = 11.

(v) 4 : 9 :: 9 : x

Product of extremes = product of means

$$\begin{aligned}\Rightarrow 4 \times x = 9 \times 9\\[1em] \Rightarrow x = \frac{81}{4}\end{aligned}$$

Hence, $x = \frac{81}{4}$.

(i) The simplest form of $2\frac{1}{5} : 1\frac{2}{5}$ is 11 : 7.

Convert the mixed numbers into improper fractions: $2\frac{1}{5} = \frac{11}{5}$ and $1\frac{2}{5} = \frac{7}{5}$.

$\frac{11}{5} : \frac{7}{5} = 11 : 7$

Hence, the statement is True.

(ii) The ratio 3 : 4 is the same as 51 : 68.

H.C.F. of 51 and 68 is 17.

$\frac{51 \div 17}{68 \div 17} = \frac{3}{4} = 3 : 4$

Hence, the statement is True.

(iii) The ratio 70 g : 1 kg is not defined.

Convert kg into g: 1 kg = 1000 g. Both quantities are of the same kind (weight), so the ratio is defined: 70 : 1000 = 7 : 100.

Hence, the statement is False.

(iv) 70 P : ₹7 = 1 : 10.

Convert ₹ into paise: ₹7 = 700 paise.

$70 : 700 = \frac{70}{700} = \frac{1}{10} = 1 : 10$

Hence, the statement is True.

(v) If ₹120 is divided in the ratio 5 : 7, the larger part is ₹$\left(120 \times \frac{7}{5}\right)$.

The larger part must be found by multiplying the total by the larger ratio term divided by the sum of the ratio terms, i.e. ₹$\left(120 \times \frac{7}{12}\right)$ = ₹70, not ₹$\left(120 \times \frac{7}{5}\right)$.

Hence, the statement is False.

1. Given,

Total money = ₹21,600

Ratio (A : B : C) = 5 : 6 : 7

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

A's share = $21{,}600 \times \frac{5}{18}$ = 5 × ₹1,200 = ₹6,000.

Hence, option (b) is the correct option.

2. Sum of ratio terms = 5 + 6 + 7 = 18.

B's share = $21{,}600 \times \frac{6}{18}$ = 6 × ₹1,200 = ₹7,200.

Hence, option (a) is the correct option.

3. Sum of ratio terms = 5 + 6 + 7 = 18.

C's share = $21{,}600 \times \frac{7}{18}$ = 7 × ₹1,200 = ₹8,400.

Hence, option (c) is the correct option.

4. Sum of ratio terms = 5 + 6 + 7 = 18.

A's share = $21{,}600 \times \frac{5}{18}$ = 5 × ₹1,200 = ₹6,000.

B's share = $21{,}600 \times \frac{6}{18}$ = 6 × ₹1,200 = ₹7,200.

C's share = $21{,}600 \times \frac{7}{18}$ = 7 × ₹1,200 = ₹8,400.

After adding ₹600 to each: A = ₹6,600, B = ₹7,800, C = ₹9,000.

New ratio = 6600 : 7800 : 9000

Dividing each term by 600:

$\frac{6600}{600} : \frac{7800}{600} : \frac{9000}{600} = 11 : 13 : 15$

Hence, option (b) is the correct option.

1. Given,

Total number of students = 725

Ratio (P : Q : R) = $\frac{1}{6} : \frac{1}{8} : \frac{1}{9}$

Let us find the L.C.M. of 6, 8 and 9:

$$\begin{array}{l|rrr} 2 & 6, & 8, & 9 \\ \hline 2 & 3, & 4, & 9 \\ \hline 2 & 3, & 2, & 9 \\ \hline 3 & 3, & 1, & 9 \\ \hline 3 & 1, & 1, & 3 \\ \hline & 1, & 1, & 1 \end{array}$$

L.C.M. = 2 × 2 × 2 × 3 × 3 = 72.

Multiplying each term by 72:

$\frac{1}{6} : \frac{1}{8} : \frac{1}{9} = \left(\frac{1}{6} \times 72\right) : \left(\frac{1}{8} \times 72\right) : \left(\frac{1}{9} \times 72\right) = 12 : 9 : 8$

Sum of ratio terms = 12 + 9 + 8 = 29.

Number of students in school P = $725 \times \frac{12}{29}$ = 12 × 25 = 300.

Hence, option (b) is the correct option.

2. Sum of ratio terms = 12 + 9 + 8 = 29.

Number of students in school Q = $725 \times \frac{9}{29}$ = 9 × 25 = 225.

Hence, option (d) is the correct option.

3. Sum of ratio terms = 12 + 9 + 8 = 29.

Number of students in school R = $725 \times \frac{8}{29}$ = 8 × 25 = 200.

Hence, option (a) is the correct option.

4. Sum of ratio terms = 12 + 9 + 8 = 29.

Number of students in school P = $725 \times \frac{12}{29}$ = 12 × 25 = 300.

Number of students in school Q = $725 \times \frac{9}{29}$ = 9 × 25 = 225.

Number of students in school R = $725 \times \frac{8}{29}$ = 8 × 25 = 200.

New number of students in P = 300 + 16% of 300 = $300 + \frac{16}{100} \times 300$ = 300 + 48 = 348.

New number of students in Q = 225 + 12% of 225 = $225 + \frac{12}{100} \times 225$ = 225 + 27 = 252.

New number of students in R = 200 + 20% of 200 = $200 + \frac{20}{100} \times 200$ = 200 + 40 = 240.

New ratio = 348 : 252 : 240

Dividing each term by 12:

$\frac{348}{12} : \frac{252}{12} : \frac{240}{12} = 29 : 21 : 20$

Hence, option (b) is the correct option.

Convert 2 hours into minutes: 2 hours = 2 × 60 = 120 minutes.

Ratio of 30 minutes to 2 hours = 30 : 120 = $\frac{30}{120} = \frac{1}{4}$ = 1 : 4.

So, Assertion (A) is true.

Ratio exists between two quantities of the same kind and in the same units. This is a correct statement, and it is exactly why the two times were first converted into the same unit (minutes) before finding the ratio.

So, Reason (R) is true and it correctly explains the Assertion (A).

Hence, option 1 is the correct option.

The numbers a, b, c, d are in proportion if and only if the product of extremes equals the product of means, i.e. ad = bc.

So, Assertion (A) is true.

The order of terms in a proportion does matter. For example, 2 : 3 :: 4 : 6 is a proportion, but changing the order to 3 : 2 :: 4 : 6 is not. So the Reason (R) is a false statement.

So, Reason (R) is false.

Hence, option 3 is the correct option.

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

So, the flour forms $\frac{3}{9}$ of the total quantity.

Total cups $\times \frac{3}{9}$ = 12

Total cups = $12 \times \frac{9}{3}$ = 4 × 9 = 36 cups.

Hence, option 4 is the correct option.

A ratio compares two fixed given quantities. The antecedent (first term) and the consequent (second term) are determined by these quantities, so they cannot be changed unless the given statements themselves change.

Hence, option 2 is the correct option.

In the given figure, each shape is divided so that the shaded region and the unshaded region are equal in area. Therefore, the shaded area and the unshaded area are in the ratio 1 : 1.

Hence, option 4 is the correct option.

Number of sides of a square = 4.

Number of edges of a cube = 12.

Ratio = 4 : 12

H.C.F. of 4 and 12 is 4.

$\frac{4 \div 4}{12 \div 4} = \frac{1}{3} = 1 : 3$

Hence, option 2 is the correct option.

Sum of ratio terms = 2 + 3 = 5.

So, the green-cover books form $\frac{2}{5}$ of the total number of books.

Total books $\times \frac{2}{5}$ = 18

Total books = $18 \times \frac{5}{2}$ = 9 × 5 = 45.

Number of brown-cover books = $45 \times \frac{3}{5}$ = 3 × 9 = 27.

Hence, option 3 is the correct option.

Sum of ratio terms = 5 + 2 = 7.

So, the incorrect answers form $\frac{2}{7}$ of the total number of answers.

Total answers $\times \frac{2}{7}$ = 16

Total answers = $16 \times \frac{7}{2}$ = 8 × 7 = 56.

Number of correct answers = $56 \times \frac{5}{7}$ = 5 × 8 = 40.

Hence, option 2 is the correct option.

Let the volume of water in bottle I be $3x$ ml and in bottle II be $4x$ ml.

After Rajeev drinks 40 ml from bottle I, the ratio becomes:

$$\begin{aligned}\Rightarrow \frac{3x - 40}{4x} = \frac{13}{20}\\[1em] \Rightarrow 20(3x - 40) = 13 \times 4x\\[1em] \Rightarrow 60x - 800 = 52x\\[1em] \Rightarrow 60x - 52x = 800\\[1em] \Rightarrow 8x = 800\\[1em] \Rightarrow x = 100\end{aligned}$$

Volume of water in bottle II at first = 4x = 4 × 100 = 400 ml.

Hence, option 4 is the correct option.

Four numbers are in proportion if product of extremes = product of means.

Option 1: 7 × 1380 = 9660 and 12 × 805 = 9660. Equal, so in proportion.

Option 2: 16 × 7 = 112 and 28 × 4 = 112. Equal, so in proportion.

Option 3: 150 × 90 = 13500 and 100 × 135 = 13500. Equal, so in proportion.

Option 4: 51 × 98 = 4998 and 85 × 57 = 4845. Since 4998 ≠ 4845, they are not in proportion.

Hence, option 4 is the correct option.

Sum of ratio terms = 2 + 3 + 1 = 6.

X's share = $3{,}000 \times \frac{2}{6}$ = 2 × ₹500 = ₹1,000.

Y's share = $3{,}000 \times \frac{3}{6}$ = 3 × ₹500 = ₹1,500.

Difference = ₹1,500 − ₹1,000 = ₹500.

Hence, option 3 is the correct option.

Given that each share is $\frac{4}{5}$ of the next one, so:

$P = \frac{4}{5}Q$, $Q = \frac{4}{5}R$ and $R = \frac{4}{5}S$.

Let us express each share in terms of S. Substituting one relation into the next:

$$\begin{aligned}R = \frac{4}{5}S\\[1em] Q = \frac{4}{5}R = \frac{4}{5} \times \frac{4}{5}S = \frac{16}{25}S\\[1em] P = \frac{4}{5}Q = \frac{4}{5} \times \frac{16}{25}S = \frac{64}{125}S\end{aligned}$$

So, $P : Q : R : S = \frac{64}{125}S : \frac{16}{25}S : \frac{4}{5}S : S$.

The denominators here are 125, 25, 5 and 1, and their L.C.M. is 125. So S must be a multiple of 125 for all four shares to come out as whole numbers.

Therefore we take S = 125k, where k is a positive number. Putting S = 125k in the above relations:

$$\begin{aligned}R = \frac{4}{5} \times 125k = 100k\\[1em] Q = \frac{16}{25} \times 125k = 80k\\[1em] P = \frac{64}{125} \times 125k = 64k\end{aligned}$$

So, P : Q : R : S = 64k : 80k : 100k : 125k = 64 : 80 : 100 : 125.

Sum of ratio terms = 64 + 80 + 100 + 125 = 369.

Q's share = $3{,}690 \times \frac{80}{369}$ = 80 × ₹10 = ₹800.

Hence, option 4 is the correct option.

Number of cupcakes had by 6 guests = 6 × 2 = 12.

Number of cupcakes had by 3 guests = 3 × 1 = 3.

Total cupcakes had by the guests = 12 + 3 = 15.

Number of cupcakes left = 21 − 15 = 6.

Ratio of cupcakes left to cupcakes had = 6 : 15

H.C.F. of 6 and 15 is 3.

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

Hence, option 1 is the correct option.