Ratio & Proportion

The given ratio is 48 : 54.

H.C.F. of 48 and 54 is 6.

∴ 48 : 54 = $\frac{48 \div 6}{54 \div 6} = \frac{8}{9}$ = 8 : 9

Hence, 48 : 54 = 8 : 9.

Both quantities must be in the same unit.

5 kg = 5 × 1000 g = 5000 g

∴ 5 kg : 800 g = 5000 g : 800 g = 5000 : 800

H.C.F. of 5000 and 800 is 200.

$\frac{5000 \div 200}{800 \div 200} = \frac{25}{4}$ = 25 : 4

Hence, 5 kg : 800 g = 25 : 4.

Both quantities must be in the same unit.

3 m = 3 × 100 cm = 300 cm

∴ 3 m : 90 cm = 300 cm : 90 cm = 300 : 90

H.C.F. of 300 and 90 is 30.

$\frac{300 \div 30}{90 \div 30} = \frac{10}{3}$ = 10 : 3

Hence, 3 m : 90 cm = 10 : 3.

Both quantities must be in the same unit.

2 years = 2 × 12 months = 24 months

∴ 2 years : 9 months = 24 months : 9 months = 24 : 9

H.C.F. of 24 and 9 is 3.

$\frac{24 \div 3}{9 \div 3} = \frac{8}{3}$ = 8 : 3

Hence, 2 years : 9 months = 8 : 3.

Converting the mixed fractions into improper fractions :

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

Dividing the first term by the second term :

$= \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$ = 3 : 5

Hence, $1\frac{1}{2} : 2\frac{1}{2}$ = 3 : 5.

Converting the mixed fraction into an improper fraction :

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

Dividing the first term by the second term :

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

Hence, $3\frac{1}{2} : 7$ = 1 : 2.

Multiplying both terms by 10 to remove the decimals :

2.5 : 1.5 = 25 : 15

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

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

Hence, 2.5 : 1.5 = 5 : 3.

Given,

$x : (3x − 4)$ = 2 : 5

$\Rightarrow \frac{x}{3x − 4} = \frac{2}{5}$

By cross multiplication :

$$\begin{aligned}\Rightarrow 5x = 2(3x - 4) \\[1em] \Rightarrow 5x = 6x - 8 \\[1em] \Rightarrow 8 = 6x - 5x \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, the value of $\bm{x = 8}$.

Given,

$(2x − 5) : (x + 5)$ = 7 : 5

$\Rightarrow \frac{2x − 5}{x + 5} = \frac{7}{5}$

By cross multiplication :

$$\begin{aligned}\Rightarrow 5(2x − 5) = 7(x + 5) \\[1em] \Rightarrow 10x − 25 = 7x + 35 \\[1em] \Rightarrow 10x − 7x = 35 + 25 \\[1em] \Rightarrow 3x = 60 \\[1em] \Rightarrow x = \frac{60}{3} = 20\end{aligned}$$

Substituting value of $x$, we get :

$2x - 5 = 2 \times 20$ - 5 = 40 - 5 = 35.

$x + 5$ = 20 + 5 = 25.

(i) Hence, $\bm{x = 20}$.

(ii) Hence, $\bm{2x - 5 = 35}$.

(iii) Hence, $\bm{x + 5 = 25}$.

Solving,

$$\begin{aligned}\Rightarrow m : n = 4 : 3 \\[1em] \Rightarrow \frac{m}{n} = \frac{4}{3} \\[1em] \Rightarrow \frac{3m}{4n} = \frac{3}{4} \times \frac{4}{3} \\[1em] \Rightarrow \frac{3m}{4n} = \frac{1}{1}\end{aligned}$$

Hence, $\bm{3m : 4n = 1 : 1}$.

Solving,

$$\begin{aligned}\Rightarrow p : q = 5 : 4 \\[1em] \Rightarrow \frac{p}{q} = \frac{5}{4} \\[1em] \Rightarrow \frac{6p}{5q} = \frac{6}{5} \times \frac{p}{q} \\[1em] \Rightarrow \frac{6p}{5q} = \frac{6}{5} \times \frac{5}{4} \\[1em] \Rightarrow \frac{6p}{5q} = \frac{6}{4} = \frac{3}{2}\end{aligned}$$

Hence, $\bm{6p : 5q = 3 : 2}$.

Given,

Length of the field = 80 m

Width of the field = 60 m

Ratio of width to length = 60 : 80

H.C.F. of 60 and 80 is 20.

$\frac{60 \div 20}{80 \div 20} = \frac{3}{4} = 3 : 4$

Hence, the ratio of width to length is 3 : 4.

Given,

5 dozens = 5 × 12 = 60

2 scores = 2 × 20 = 40

Ratio = 60 : 40

H.C.F. of 60 and 40 is 20.

$\frac{60 \div 20}{40 \div 20} = \frac{3}{2} = 3 : 2$

Hence, the required ratio is 3 : 2.

Given,

Monthly income = ₹ 12,000

Monthly expenditure = ₹ 8,500

Monthly savings = Income − Expenditure = ₹ 12,000 − ₹ 8,500 = ₹ 3,500

(i) Ratio of income to expenditure = 12000 : 8500

H.C.F. of 12000 and 8500 is 500.

$\frac{12000 \div 500}{8500 \div 500} = \frac{24}{17} = 24 : 17$

Hence, the ratio of income to expenditure is 24 : 17.

(ii) Ratio of expenditure to savings = 8500 : 3500

H.C.F. of 8500 and 3500 is 500.

$\frac{8500 \div 500}{3500 \div 500} = \frac{17}{7} = 17 : 7$

Hence, the ratio of expenditure to savings is 17 : 7.

(iii) Ratio of savings to income = 3500 : 12000

H.C.F. of 3500 and 12000 is 500.

$\frac{3500 \div 500}{12000 \div 500} = \frac{7}{24} = 7 : 24$

Hence, the ratio of savings to income is 7 : 24.

Given,

Original expenses = ₹ 1,500

Increased (new) expenses = ₹ 2,250

Increase in expenses = ₹ 2,250 − ₹ 1,500 = ₹ 750

(i) Ratio of increase in expenses to original expenses = 750 : 1500

H.C.F. of 750 and 1500 is 750.

$\frac{750 \div 750}{1500 \div 750} = \frac{1}{2} = 1 : 2$

Hence, the required ratio is 1 : 2.

(ii) Ratio of original expenses to increased expenses = 1500 : 2250

H.C.F. of 1500 and 2250 is 750.

$\frac{1500 \div 750}{2250 \div 750} = \frac{2}{3} = 2 : 3$

Hence, the required ratio is 2 : 3.

(iii) Ratio of increased expenses to increase in expenses = 2250 : 750

H.C.F. of 2250 and 750 is 750.

$\frac{2250 \div 750}{750 \div 750} = \frac{3}{1} = 3 : 1$

Hence, the required ratio is 3 : 1.

Given,

Total members = 360

Carrom = 40, Table tennis = 96, Badminton = 144

Volley ball = 360 − (40 + 96 + 144) = 360 − 280 = 80

(i) Ratio of carrom to badminton = 40 : 144

H.C.F. of 40 and 144 is 8.

$\frac{40 \div 8}{144 \div 8} = \frac{5}{18} = 5 : 18$

Hence, the required ratio is 5 : 18.

(ii) Ratio of badminton to table tennis = 144 : 96

H.C.F. of 144 and 96 is 48.

$\frac{144 \div 48}{96 \div 48} = \frac{3}{2} = 3 : 2$

Hence, the required ratio is 3 : 2.

(iii) Ratio of table tennis to volley ball = 96 : 80

H.C.F. of 96 and 80 is 16.

$\frac{96 \div 16}{80 \div 16} = \frac{6}{5} = 6 : 5$

Hence, the required ratio is 6 : 5.

(iv) Members who play other games (carrom, table tennis and badminton) = 40 + 96 + 144 = 280

Ratio of volley ball to other games = 80 : 280

H.C.F. of 80 and 280 is 40.

$\frac{80 \div 40}{280 \div 40} = \frac{2}{7} = 2 : 7$

Hence, the required ratio is 2 : 7.

Given,

Total money = ₹ 120

Ratio = 5 : 3

Total parts : 5 + 3 = 8 parts.

The value of 1 part : 120 ÷ 8 = ₹ 15.

Hari's share : 5 × ₹ 15 = ₹ 75

Gopi's share : 3 × ₹ 15 = ₹ 45

Hence, Hari gets ₹ 75 and Gopi gets ₹ 45.

Given,

Ratio = $2\frac{1}{2} : 1\frac{1}{2} = \frac{5}{2} : \frac{3}{2}$

Multiplying both terms by 2 :

$= \frac{5}{2} \times 2 : \frac{3}{2} \times 2 = 5 : 3$

Total parts : 5 + 3 = 8 parts.

The value of 1 part : 72 ÷ 8 = 9.

First part : 5 × 9 = 45

Second part : 3 × 9 = 27

Hence, 72 is divided into 45 and 27.

Given ratio = $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$

L.C.M. of the denominators 2, 3 and 4 is 12.

Multiplying each term by 12 :

$\frac{1}{2} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 6 : 4 : 3$

Let the shares of A, B and C be 6x, 4x and 3x respectively.

⇒ 6x + 4x + 3x = 10400

⇒ 13x = 10400

⇒ x = $\frac{10400}{13}$

⇒ x = ₹ 800

Shares are :

A's share = 6x = 6 × ₹ 800 = ₹ 4,800,

B's share = 4x = 4 × ₹ 800 = ₹ 3,200,

C's share = 3x = 3 × ₹ 800 = ₹ 2,400.

Hence, A gets ₹ 4,800, B gets ₹ 3,200 and C gets ₹ 2,400.

Given,

Ratio of the angles = 3 : 7 : 8

Let the angles be 3x, 7x and 8x.

The sum of the angles of a triangle = 180°

⇒ 3x + 7x + 8x = 180°

⇒ 18x = 180°

⇒ x = $\frac{180^{\circ}}{18}$

⇒ x = 10°

Angles are :

3x = 3 × 10° = 30°,

7x = 7 × 10° = 70°,

8x = 8 × 10° = 80°.

Hence, the greatest angle is 80° and the smallest angle is 30°.

Given,

Ratio of the sides = 3 : 2 : 4

Perimeter of the triangle = 27 cm

Let the sides be 3x, 2x and 4x.

⇒ 3x + 2x + 4x = 27

⇒ 9x = 27

⇒ x = $\frac{27}{9}$

⇒ x = 3 cm

Sides are :

3x = 3 × 3 cm = 9 cm,

2x = 2 × 3 cm = 6 cm,

4x = 4 × 3 cm = 12 cm.

Hence, the lengths of the sides are 9 cm, 6 cm and 12 cm.

(i) 8 : 16 and 12 : 15

Simplifying first ratio :

$\Rightarrow \frac{8}{16} = \frac{1}{2}$

Simplifying second ratio :

$\Rightarrow \frac{12}{15} = \frac{4}{5}$

∴ Ratio 8 : 16 ≠ 12 : 15.

Hence, 8 : 16 and 12 : 15 do not form a proportion.

(ii) 16 : 28 and 24 : 42

Simplifying first ratio :

$\Rightarrow \frac{16}{28} = \frac{4}{7}$

Simplifying second ratio :

$\Rightarrow \frac{24}{42} = \frac{4}{7}$

∴ Ratio 16 : 28 = 24 : 42.

Hence, 16 : 28 and 24 : 42 form a proportion.

(iii) 98 : 112 and 14 : 16

Simplifying first ratio :

$\Rightarrow \frac{98}{112} = \frac{7}{8}$

Simplifying second ratio :

$\Rightarrow \frac{14}{16} = \frac{7}{8}$

∴ Ratio 98 : 112 = 14 : 16.

Hence, 98 : 112 and 14 : 16 form a proportion.

Four numbers a, b, c, d are in proportion if a × d = b × c.

x, 6, 10 and 15

⇒ x × 15 = 6 × 10

⇒ 15x = 60

⇒ x = $\frac{60}{15}$ = 4

Hence, x = 4.

Four numbers a, b, c, d are in proportion if a × d = b × c.

2, x, 10 and 25

⇒ 2 × 25 = x × 10

⇒ 50 = 10x

⇒ x = $\frac{50}{10}$ = 5

Hence, x = 5.

Four numbers a, b, c, d are in proportion if a × d = b × c.

9, 12, x and 8

⇒ 9 × 8 = 12 × x

⇒ 72 = 12x

⇒ x = $\frac{72}{12}$ = 6

Hence, x = 6.

Four numbers a, b, c, d are in proportion if a × d = b × c.

6, 4, 18 and x

⇒ 6 × x = 4 × 18

⇒ 6x = 72

⇒ x = $\frac{72}{6}$ = 12

Hence, x = 12.

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

(x + 5) : 24 = 3 : 4

⇒ 4(x + 5) = 24 × 3

⇒ 4(x + 5) = 72

⇒ x + 5 = $\frac{72}{4}$

⇒ x + 5 = 18

⇒ x = 18 − 5 = 13

Hence, x = 13.

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

14 : (x − 1) = 7 : 9

⇒ 14 × 9 = 7(x − 1)

⇒ 126 = 7(x − 1)

⇒ x − 1 = $\frac{126}{7}$

⇒ x − 1 = 18

⇒ x = 18 + 1 = 19

Hence, x = 19.

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

2 : 3 = (x − 2) : 18

⇒ 2 × 18 = 3(x − 2)

⇒ 36 = 3(x − 2)

⇒ x − 2 = $\frac{36}{3}$

⇒ x − 2 = 12

⇒ x = 12 + 2 = 14

Hence, x = 14.

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

3 : 4 = 30 : (x + 10)

⇒ 3(x + 10) = 4 × 30

⇒ 3(x + 10) = 120

⇒ x + 10 = $\frac{120}{3}$

⇒ x + 10 = 40

⇒ x = 40 − 10 = 30

Hence, x = 30.

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

1 : 2 = (x − 1) : 26

⇒ 1 × 26 = 2(x − 1)

⇒ 26 = 2(x − 1)

⇒ x − 1 = $\frac{26}{2}$

⇒ x − 1 = 13

⇒ x = 13 + 1 = 14

Hence, x = 14.

Let the third term be x.

∴ 6, 18, x and 75 are in proportion, i.e. 6 : 18 : : x : 75

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

⇒ 6 × 75 = 18 × x

⇒ 450 = 18x

⇒ x = $\frac{450}{18}$ = 25

Hence, the third term is 25.

Given,

Ratio (Copper : Zinc) = 9 : 8

Weight of zinc = 9.6 kg

Since 8 parts represent the zinc, 8 parts = 9.6 kg.

1 part = $\frac{9.6}{8}$ = 1.2 kg.

Weight of copper = 9 parts = 9 × 1.2 = 10.8 kg.

Hence, the weight of copper in the alloy is 10.8 kg.

Given,

Ratio (Girls : Boys) = 2 : 5

Number of boys = 225

Since 5 parts represent the boys, 5 parts = 225.

1 part = $\frac{225}{5}$ = 45.

(i) Number of girls = 2 parts = 2 × 45 = 90.

Hence, the number of girls in the school is 90.

(ii) Total students = Girls + Boys = 90 + 225 = 315.

Hence, the total number of students in the school is 315.

Given,

$(3x - 4) : (x + 2)$ = 2 : 1

⇒ $\frac{3x - 4}{x + 2} = \frac{2}{1}$

By cross multiplication :

$$\begin{aligned}\Rightarrow 3x − 4 = 2(x + 2) \\[1em] \Rightarrow 3x − 4 = 2x + 4 \\[1em] \Rightarrow 3x − 2x = 4 + 4 \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, the value of $\bm{x = 8}$.

Given,

(x − 7) : 2 : : (x − 3) : 3

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

⇒ 3(x − 7) = 2(x − 3)

⇒ 3x − 21 = 2x − 6

⇒ 3x − 2x = 21 − 6

⇒ x = 15

Now, x − 5 = 15 − 5 = 10

Hence, the value of x − 5 is 10.

Given,

Out of every 5 students, 3 play cricket.

∴ Fraction of students who play cricket = $\frac{3}{5}$

Number of students who play cricket = $\frac{3}{5} \times 2400$

= 3 × 480

= 1440.

Hence, 1440 students play cricket.

Given,

Price of 25 articles = ₹ 1,750

Price of 1 article = $\frac{1750}{25}$ = ₹ 70.

Price of 13 articles = 13 × ₹ 70 = ₹ 910.

Hence, the price of 13 articles is ₹ 910.

Given,

Distance covered in 5 litres = 330 km

Distance covered in 1 litre = $\frac{330}{5}$ = 66 km.

Distance covered in 2.5 litres = 2.5 × 66 = 165 km.

Hence, the motorbike will cover 165 km consuming 2.5 litres of petrol.

Given,

Cost of 35 envelopes = ₹ 105

Number of envelopes bought for ₹ 105 = 35

Number of envelopes bought for ₹ 1 = $\frac{35}{105}$ envelopes.

Number of envelopes bought for ₹ 90 = $\frac{35}{105} \times 90$

= $\frac{1}{3} \times 90$

= 30 envelopes.

Hence, 30 envelopes can be bought for ₹ 90.

Given,

Cost of 10 kg of wheat = ₹ 180

Cost of 1 kg of wheat = $\frac{180}{10}$ = ₹ 18.

(i) Cost of 18 kg of wheat = 18 × ₹ 18 = ₹ 324.

Hence, 18 kg of wheat costs ₹ 324.

(ii) Quantity of wheat purchased for ₹ 18 = 1 kg

Quantity of wheat purchased for ₹ 432 = $\frac{432}{18}$ = 24 kg.

Hence, 24 kg of wheat can be purchased for ₹ 432.

Given,

Height of the first tree = 24 m, shadow of the first tree = 15 m

Shadow of the second tree = 6 m

For a shadow of 15 m, the height = 24 m

For a shadow of 1 m, the height = $\frac{24}{15}$ m.

For a shadow of 6 m, the height = $\frac{24}{15} \times 6$

= $\frac{144}{15}$

= 9.6 m

Hence, the height of the second tree is 9.6 m.

Given,

Distance travelled in 25 minutes = 18 km

2 hours = 2 × 60 minutes = 120 minutes

Distance travelled in 1 minute = $\frac{18}{25}$ km.

Distance travelled in 120 minutes = $\frac{18}{25} \times 120$

= $\frac{2160}{25}$

= 86.4 km.

Hence, the truck will travel 86.4 km in 2 hours.

Given,

Jai Singh's investment = ₹ 6,00,000

Vijay Singh's investment = ₹ 9,00,000

Total profit = ₹ 1,50,000

Ratio of investments = 600000 : 900000 = 6 : 9 = 2 : 3

Let the shares of Jai Singh and Vijay Singh be 2x and 3x respectively.

⇒ 2x + 3x = 150000

⇒ 5x = 150000

⇒ x = $\frac{150000}{5}$

⇒ x = ₹ 30,000

Vijay Singh's share = 3x = 3 × ₹ 30,000 = ₹ 90,000.

Hence, the share of Vijay Singh is ₹ 90,000.

Given,

Total students = 350

Number of girls = 150

Number of boys = 350 − 150 = 200

(i) Ratio of girls to total students = 150 : 350

H.C.F. of 150 and 350 is 50.

$\frac{150 \div 50}{350 \div 50} = \frac{3}{7} = 3 : 7$

Hence, the ratio of girls to the total number of students is 3 : 7.

(ii) Ratio of boys to girls = 200 : 150

H.C.F. of 200 and 150 is 50.

$\frac{200 \div 50}{150 \div 50} = \frac{4}{3} = 4 : 3$

Hence, the ratio of boys to girls is 4 : 3.

Given,

Cost of one dozen (12) articles = ₹ 166

36 articles = 3 dozens

Cost of 36 articles = 3 × ₹ 166 = ₹ 498.

Hence, the cost of 36 articles is ₹ 498.

Given,

Distance covered in 3 hours = 180 km

Speed of the car = $\frac{180}{3}$ = 60 km/h.

(i) Time taken to cover 240 km = $\frac{240}{60}$ = 4 hours.

Hence, the car will cover 240 km in 4 hours.

(ii) Distance covered in 2 hours = 60 × 2 = 120 km.

Hence, the car will cover 120 km in 2 hours.

Ratio of 15 kg and 35 kg = 15 : 35

H.C.F. of 15 and 35 is 5.

$\frac{15 \div 5}{35 \div 5} = \frac{3}{7} = 3 : 7$

Ratio of 6 years and 14 years = 6 : 14

H.C.F. of 6 and 14 is 2.

$\frac{6 \div 2}{14 \div 2} = \frac{3}{7} = 3 : 7$

Since both ratios are equal to 3 : 7, they are the same.

Hence, yes, the two ratios are the same.

Given,

Speed of bus = $\frac{180}{3}$ = 60 km/h

Speed of train = $\frac{450}{5}$ = 90 km/h

Ratio of speed of train to speed of bus = 90 : 60

H.C.F. of 90 and 60 is 30.

$\frac{90 \div 30}{60 \div 30} = \frac{3}{2} = 3 : 2$

Hence, the ratio of the speed of the train to the speed of the bus is 3 : 2.

Given,

School opens at 10 a.m. and closes at 3.30 p.m.

Total time the school is open = 5 hours 30 minutes = 5 × 60 + 30 = 330 minutes

Lunch interval = 30 minutes

Total time of the class periods = 330 − 30 = 300 minutes

Ratio of lunch interval to total time of the class periods = 30 : 300

H.C.F. of 30 and 300 is 30.

$\frac{30 \div 30}{300 \div 30} = \frac{1}{10} = 1 : 10$

Hence, the ratio of lunch interval to total time of the class periods is 1 : 10.

Given,

Speed of Rohit = 60 km/h

Speed of Manoj = 40 km/h

Since both cover the same distance, time is inversely proportional to speed.

∴ Ratio of time taken by Rohit to Manoj = $\frac{1}{60} : \frac{1}{40}$

Multiplying both terms by the L.C.M. of 60 and 40, i.e. 120 :

$= \frac{1}{60} \times 120 : \frac{1}{40} \times 120 = 2 : 3$

Hence, the ratio between the time taken by Rohit and Manoj is 2 : 3.

Given,

Total weight of the alloy = $12\frac{1}{2}$ kg = $\frac{25}{2}$ kg = 12.5 kg

Ratio (Zinc : Copper) = 1 : 4

Let the weights of zinc and copper be x and 4x respectively.

⇒ x + 4x = 12.5

⇒ 5x = 12.5

⇒ x = $\frac{12.5}{5}$

⇒ x = 2.5 kg

Weight of copper = 4x = 4 × 2.5 = 10 kg.

Hence, the weight of copper in the alloy is 10 kg.

Let the fourth term be x.

∴ 18, 27, 32 and x are in proportion, i.e. 18 : 27 : : 32 : x

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

⇒ 18 × x = 27 × 32

⇒ 18x = 864

⇒ x = $\frac{864}{18}$ = 48

Hence, the fourth term is 48.

Given,

Ratio (Length : Width) = 5 : 2

Width = 40 m

Since 2 parts represent the width, 2 parts = 40 m.

1 part = $\frac{40}{2}$ = 20 m.

Length = 5 parts = 5 × 20 = 100 m.

Hence, the length of the school ground is 100 m.

Given,

Ratio of the two numbers = 9 : 2

The smaller number corresponds to 2 parts.

Since 2 parts represent the smaller number, 2 parts = 320.

1 part = $\frac{320}{2}$ = 160.

Larger number = 9 parts = 9 × 160 = 1440.

Hence, the larger number is 1440.

Given,

Ratio of the distance of the school from A's house to that from B's house = 2 : 1.

∴ Distance from A's house = 2 × (Distance from B's house), and Distance from B's house = $\frac{1}{2}$ × (Distance from A's house).

Completing each column using this relation :

When distance from A's house = 4 km, distance from B's house = $\frac{1}{2} \times 4 = 2$ km.

When distance from B's house = 9 km, distance from A's house = 2 × 9 = 18 km.

When distance from A's house = 8 km, distance from B's house = $\frac{1}{2} \times 8 = 4$ km.

When distance from B's house = 8 km, distance from A's house = 2 × 8 = 16 km.

When distance from A's house = 6 km, distance from B's house = $\frac{1}{2} \times 6 = 3$ km.

The completed table is :

Distance (in km) from A's house to school	Distance (in km) from B's house to the same school
4	2
18	9
8	4
16	8
6	3

Given,

Total money = ₹ 81,000

Let Geeta's share = 1 part.

Mohit gets 2.5 times what Geeta gets, so Mohit's share = 2.5 parts.

Ashok gets four times what Mohit gets, so Ashok's share = 4 × 2.5 = 10 parts.

∴ Ratio (Ashok : Mohit : Geeta) = 10 : 2.5 : 1

Multiplying each term by 2 to remove the decimal :

= 20 : 5 : 2

Let the shares of Ashok, Mohit and Geeta be 20x, 5x and 2x respectively.

⇒ 20x + 5x + 2x = 81000

⇒ 27x = 81000

⇒ x = $\frac{81000}{27}$

⇒ x = ₹ 3,000

Shares are :

Ashok's share = 20x = 20 × ₹ 3,000 = ₹ 60,000,

Mohit's share = 5x = 5 × ₹ 3,000 = ₹ 15,000,

Geeta's share = 2x = 2 × ₹ 3,000 = ₹ 6,000.

Hence, Ashok gets ₹ 60,000, Mohit gets ₹ 15,000 and Geeta gets ₹ 6,000.

(i) Given,

$(4x + 3y) : (3x + 5y)$ = 6 : 7

⇒ $\frac{4x + 3y}{3x + 5y} = \frac{6}{7}$

By cross multiplication :

$$\begin{aligned}\Rightarrow 7(4x + 3y) = 6(3x + 5y) \\[1em] \Rightarrow 28x + 21y = 18x + 30y \\[1em] \Rightarrow 28x − 18x = 30y − 21y \\[1em] \Rightarrow 10x = 9y \\[1em] \Rightarrow \frac{x}{y} = \frac{9}{10}\end{aligned}$$

Hence, $\bm{x : y = 9 : 10}$.

(ii) Since, $\frac{x}{y} = \frac{9}{10}$, we have $x = \frac{9}{10} y$.

$x = \frac{9}{10} \times 10$ = 9

Hence, $\bm{x = 9}$ when $\bm{y = 10}$.

(iii) Since, $\frac{x}{y} = \frac{9}{10}$, we have $y = \frac{10}{9} x$.

$y = \frac{10}{9} \times 27$ = 10 × 3 = 30

Hence, $\bm{y = 30}$ when $\bm{x = 27}$.

(i) Given,

$\frac{2y + 5x}{3y − 5x} = 2\frac{1}{2} = \frac{5}{2}$

By cross multiplication :

$$\begin{aligned}\Rightarrow 2(2y + 5x) = 5(3y − 5x) \\[1em] \Rightarrow 4y + 10x = 15y − 25x \\[1em] \Rightarrow 10x + 25x = 15y − 4y \\[1em] \Rightarrow 35x = 11y \\[1em] \Rightarrow \frac{x}{y} = \frac{11}{35}\end{aligned}$$

Hence, $\bm{x : y = 11 : 35}$.

(ii) Since, $\frac{x}{y} = \frac{11}{35}$, we have $x = \frac{11}{35} y$.

$x = \frac{11}{35} \times 70$ = 11 × 2 = 22

Hence, $\bm{x = 22}$ when $\bm{y = 70}$.

(iii) Since, $\frac{x}{y} = \frac{11}{35}$, we have $y = \frac{35}{11} x$.

$y = \frac{35}{11} \times 33$ = 35 × 3 = 105

Hence, $\bm{y = 105}$ when $\bm{x = 33}$.

Both quantities must be in the same unit.

1 km = 1000 m

Ratio = 350 : 1000

H.C.F. of 350 and 1000 is 50.

$\frac{350 \div 50}{1000 \div 50} = \frac{7}{20} = 7 : 20$

Hence, option 2 is the correct option.

Given ratio = $\frac{7}{3} : 3 : \frac{5}{4}$

L.C.M. of the denominators 3, 1 and 4 is 12.

Multiplying each term by 12 :

$\frac{7}{3} \times 12 : 3 \times 12 : \frac{5}{4} \times 12 = 28 : 36 : 15$

Hence, option 3 is the correct option.

If p, q, r and s are in proportion, i.e. p : q : : r : s, then

product of extremes = product of means

⇒ p × s = q × r

Hence, option 3 is the correct option.

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

Product of extremes = 18 × 15 = 270

Let the other mean term be x.

⇒ 30 × x = 270

⇒ x = $\frac{270}{30}$ = 9.

Hence, option 1 is the correct option.

The sum of the angles of a triangle = 180°

Let the angles be 3x, 7x and 8x.

⇒ 3x + 7x + 8x = 180°

⇒ 18x = 180°

⇒ x = $\frac{180^{\circ}}{18}$

⇒ x = 10°

The angles are 3 × 10° = 30°, 7 × 10° = 70° and 8 × 10° = 80°.

Smallest angle = 30° and largest angle = 80°.

Sum of the smallest and the largest angle = 30° + 80° = 110°.

Hence, option 3 is the correct option.

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

⇒ 5 × x = 7 × 9

⇒ 5x = 63

⇒ x = $\frac{63}{5}$ = 12.6

Hence, option 1 is the correct option.

Ratio of the two numbers = 7 : 5

Let the two numbers be 7x and 5x.

⇒ 7x + 5x = 126

⇒ 12x = 126

⇒ x = $\frac{126}{12}$

The larger number = 7x = 7 × $\frac{126}{12} = \frac{126}{12} × 7$.

Hence, option 4 is the correct option.

Ratio (Zinc : Copper) = 3 : 7

Since 7 parts represent the copper, 7 parts = 70 g.

1 part = $\frac{70}{7}$ = 10 g.

Weight of zinc = 3 parts = 3 × 10 = 30 g.

Hence, option 1 is the correct option.

Cost of 20 articles = ₹ 600

Cost of 1 article = $\frac{600}{20}$ = ₹ 30.

Cost of 30 articles = 30 × ₹ 30 = ₹ 900.

Hence, option 2 is the correct option.

First ratio : 40 cm to 2 m

2 m = 200 cm

40 : 200 = $\frac{40}{200} = \frac{1}{5}$ = 1 : 5

Second ratio : ₹ 30 to ₹ 150

30 : 150 = $\frac{30}{150} = \frac{1}{5}$ = 1 : 5

Since both ratios are equal to 1 : 5, they form a proportion.

Hence, option 1 is the correct option.

Given,

$A$ and $B$ are in the ratio 3 : 5.

∴ $\frac{A}{B} = \frac{3}{5}$

$\Rightarrow A = \frac{3}{5} × B$.

So, Statement 1 is true.

From $A : B$ = 3 : 5, we get $\frac{A}{B} = \frac{3}{5}$, which on cross multiplication gives $5A = 3B$, i.e. $\frac{A}{3} = \frac{B}{5}$.

So, Statement 2 is true.

Hence, option 1 is the correct option.

Given,

$m : n$ = 5 : 4, i.e. $\frac{m}{n} = \frac{5}{4}$.

Solving,

$$\begin{aligned}\Rightarrow 5m : 4n = \frac{5m}{4n} \\[1em] = \frac{5}{4} \times \frac{m}{n} \\[1em] = \frac{5}{4} \times \frac{5}{4} \\[1em] = \frac{25}{16} \\[1em] = 25 : 16\end{aligned}$$

So, Statement 2 is true.

From the working above, $5m : 4n$ = 25 : 16.

So, Statement 1 is true.

Hence, option 1 is the correct option.

We know that,

Average speed = $\frac{\text{Distance}}{\text{Time}} = \frac{140}{2} = 70$ km/h.

So, Assertion (A) is true.

For a fixed distance, speed and time are inversely proportional, so the greater the speed, the less the time taken.

So, Reason (R) is true.

Hence, option 3 is the correct option.