Rd Sharma 2018 Solutions for Class 6 Math Chapter 1 Knowing Our Numbers are provided here with simple step-by-step explanations. These solutions for Knowing Our Numbers are extremely popular among Class 6 students for Math Knowing Our Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 6 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 6 Math are prepared by experts and are 100% accurate.

#### Page No 9.14:

#### Question 1:

Which of the following statements are true?

(i) 16 : 24 = 20 : 30

(ii) 21 : 6 = 35 : 10

(iii) 12 : 18 = 28 : 12

(iv) 51 : 58 = 85 : 102

(v) 40 men : 200 men = Rs 5 : Rs 25

(vi) 99 kg : 45 kg = Rs 44 : Rs 20

#### Answer:

(i) 16 : 24 = 20 : 30

$\frac{16}{24}=\frac{2}{3}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}8)\phantom{\rule{0ex}{0ex}}\frac{20}{30}=\frac{2}{3}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}10)\phantom{\rule{0ex}{0ex}}\therefore \frac{16}{24}=\frac{20}{30}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}$

(ii) 21 : 6 = 35 : 10

$\frac{21}{6}=\frac{7}{2}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}3)\phantom{\rule{0ex}{0ex}}\frac{35}{10}=\frac{7}{2}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denomenator}\mathrm{by}5)\phantom{\rule{0ex}{0ex}}\therefore \frac{21}{6}=\frac{35}{10}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}$

(iii) 12 : 18 = 28 : 12

$\frac{12}{18}=\frac{2}{3}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}6)\phantom{\rule{0ex}{0ex}}\frac{28}{12}=\frac{7}{3}(\mathrm{Divi}\mathrm{ding}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}4)\phantom{\rule{0ex}{0ex}}\therefore \frac{16}{24}\mathrm{and}\frac{20}{30}\mathrm{are}\mathrm{not}\mathrm{equal}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{not}\mathrm{true}.\phantom{\rule{0ex}{0ex}}$

(iv) 51 : 58 = 85 : 102

$\frac{51}{58}\phantom{\rule{0ex}{0ex}}\mathrm{And}\frac{85}{102}=\frac{5}{6}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}17)\phantom{\rule{0ex}{0ex}}\therefore \frac{51}{58}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\frac{85}{102}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{not}\mathrm{true}.$

(v) 40 men : 200 men = Rs 5 : Rs 25

$\frac{40}{200}=\frac{1}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}40)\phantom{\rule{0ex}{0ex}}\frac{5}{25}=\frac{1}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}5)\phantom{\rule{0ex}{0ex}}\therefore \frac{40}{200}=\frac{5}{25}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{true}.$

(vi) 99 kg : 45 kg = Rs 44 : Rs 20

$\frac{99}{45}=\frac{11}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}9)\phantom{\rule{0ex}{0ex}}\frac{44}{20}=\frac{11}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}4)\phantom{\rule{0ex}{0ex}}\therefore \frac{99}{45}=\frac{11}{5}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{true}.$

#### Page No 9.14:

#### Question 2:

Find which of the following are in proportion:

(i) 8, 16, 6, 12

(ii) 6, 2, 4, 3

(iii) 150, 250, 200, 300

#### Answer:

(i) Consider $\frac{8}{16}=\frac{1}{2}$

And $\frac{6}{12}=\frac{1}{2}$

$\because $ 8 : 16 = 6 : 12

$\therefore $ 8, 16, 6, 12 are in proportion.

(ii) Consider $\frac{6}{2}=\frac{3}{1}$

And $\frac{4}{3}$

$\because $ 6 : 2 $\ne $ 4 : 3

$\therefore $ 6, 2, 4, 3 are not in proportion.

(iii) Consider $\frac{150}{250}=\frac{3}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}50)$

And $\frac{200}{300}=\frac{2}{3}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}100)$

$\because $ 150 : 250 $\ne $ 200 : 300

$\therefore $ 150, 250, 200, 300 are not in proportion.

#### Page No 9.14:

#### Question 3:

Find *x* in the following proportions:

(i) *x* : 6 = 55 : 11

(ii) 18 : *x* = 27 : 3

(iii) 7 : 14 = 15 :* x*

(iv) 16 : 18 = *x* : 96

#### Answer:

(i) *x* : 6 = 55 : 11

$\Rightarrow \frac{x}{6}=\frac{55}{11}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{6}=\frac{5}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow x=5\times 6=30$

(ii) 18 : *x* = 27 : 3

$\Rightarrow \frac{18}{x}=\frac{27}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{18}{x}=\frac{9}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{18}{9}=2$

(iii) 7 : 14 = 15 : *x*

$\Rightarrow \frac{7}{14}=\frac{15}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{15}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=15\times 2=30$

(iv) 16 : 18 = *x* : 96

$\Rightarrow \frac{16}{18}=\frac{x}{96}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8}{9}=\frac{x}{96}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8}{9}\times 96=\frac{256}{3}$

#### Page No 9.14:

#### Question 4:

Set up all proportions from the numbers 9, 150, 105, 1750.

#### Answer:

All proportions are:

9 : 150 = 3 : 50

9 : 105 = 3 : 35

9 : 1750

150 : 105 = 10 : 7

150 : 1750 = 3 : 35

105 : 1750 = 3 : 50

Thus, all proportions that can be formed are:

3 : 50, 3 : 35, 10 : 7, 9 : 1750, 1750 : 9, 7 : 10, 35 : 3 and 50 : 3

#### Page No 9.14:

#### Question 5:

Find the other three proportions involving terms of each of the following:

(i) 45 : 30 = 24 : 16

(ii) 12 : 18 = 14 : 21

#### Answer:

(i) 45 : 30 = 24 : 16 (3 : 2 is its simplest form)

Other three proportions are:

45 : 24 = 30 : 16 (15 : 8 is its simplest form)

30 : 45 = 16 : 24 (2 : 3 is its simplest form)

16 : 30 = 24 : 45 (8 : 15 is its simplest form)

(ii) 12 : 18 = 14 : 21 (2 : 3 is its simplest form)

Other three proportions are:

12 : 14 = 18 : 21 (6 : 7 is its simplest form)

21 : 18 = 14 : 12 (7 : 6 is its simplest form)

18 : 12 = 21 : 14 (3 : 2 is its simplest form)

#### Page No 9.14:

#### Question 6:

If 4, *x*, 9 are in continued proportion, find the value *x*.

#### Answer:

It is given that 4, *x*, 9 are in continued proportion; therefore, we have:

4 : *x* : : *x* : 9

$\Rightarrow \frac{4}{x}=\frac{x}{9}=4\times 9={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=36\Rightarrow x=6$

#### Page No 9.14:

#### Question 7:

If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third term.

#### Answer:

In a proportion, the first, second and fourth terms are 32, 112 and 217, respectively.

Let the third term is be *x. *

Then, we have:

32 : 112 : : *x *: 217

$\Rightarrow \frac{32}{112}=\frac{x}{217}\Rightarrow \frac{32}{112}\times 217=x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=62$

#### Page No 9.14:

#### Question 8:

Show that the following numbers are in continued proportion:

(i) 36, 90, 225

(ii) 48, 60, 75

(iii) 16, 84, 441

#### Answer:

(i) 36, 90, 225

Consider $\frac{36}{90}=\frac{2}{5}$ (Dividing numerator and denominator by 18)

$\frac{90}{225}=\frac{2}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}45)$

$\Rightarrow $ 36 : 90 : : 90 : 225

(ii) 48, 60, 75

Consider $\frac{48}{60}=\frac{4}{5}$ (Dividing numerator and denominator by 12)

$\frac{60}{75}=\frac{4}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}15)$

$\Rightarrow $ 48 : 60 : : 60 : 75

(iii) 16, 84, 441

Consider $\frac{16}{84}=\frac{4}{21}$ (Dividing numerator and denominator by 4)

$\frac{84}{441}=\frac{4}{21}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}21)$

$\Rightarrow $16 : 84 : : 84 : 441

#### Page No 9.14:

#### Question 9:

The ratio of the length of a school ground to its width is 5 : 2. Find its length if the width is 40 metres.

#### Answer:

Ratio of the length to the width of a school ground = 5 : 2

$\because $ Width = 40 m

$\therefore $ Length = $\frac{5}{2}\times 40=5\times 20=100$ m

#### Page No 9.14:

#### Question 10:

The ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop was 2 : 9. If the total sale of eggs in the same week was Rs 360, find the sale of eggs on Sunday.

#### Answer:

Ratio of the sale of eggs on Sunday to that of the whole week = 2 : 9

When total eggs of Rs. 9 is sold in a week, the sale of eggs on Sunday = Rs. 2

When total eggs of Rs. 1 sold in a week, th sale of eggs on Sunday = Rs. $\frac{2}{9}$

When total eggs of Rs. 360 sold in a week, the sale of eggs on Sunday = $\frac{2}{9}\times 360=2\times 40=\mathrm{Rs}.80$

#### Page No 9.14:

#### Question 11:

The ratio of copper and zinc in an alloy is 9 : 7. If the weight of zinc in the alloy is 9.8 kg, find the weight of copper in the alloy.

#### Answer:

The ratio of copper and zinc in an alloy is 9 : 7.

When the weight of zinc is 7 kg, the weight of copper = 9 kg

When the weight of zinc is 1 kg, the weight of copper = $\frac{9}{7}$ kg

When the weight of zinc is 9.8 kg, the weight of copper = $\frac{9}{7}\times 9.8=12.6$ kg

#### Page No 9.14:

#### Question 12:

The ratio of the income to the expenditure of a family is 7 : 6. Find the saving if the income is Rs 1400.

#### Answer:

The ratio of the income to the expenditure of a family is 7 : 6.

$\therefore $ Ratio of saving to the income = (7 $-$ 6) : 7 = 1 : 7 (Saving = Total income $-$ Expenditure )

$\because $ Income of the family = Rs. 1400

$\therefore $ Saving = $1400\times \frac{1}{7}=\mathrm{Rs}.200$

#### Page No 9.14:

#### Question 13:

The ratio of story books in a library to other books is 1 : 7. The total number of story books is 800. Find the total number of books in the library.

#### Answer:

The ratio of story books in a library to other books is 1 : 7.

Out of (1 + 7) = 8 books, 1 book is a story book

Therefore,

When number of story books is 1, the total number of books = 8

When number of story books is 800, the total number of books = 8 $\times $ 800

= 6400

#### Page No 9.18:

#### Question 1:

The price of 3 metres of cloth is Rs 79.50. Find the price of 15 metres of such cloth.

#### Answer:

Price of 3 metres of cloth = Rs. 79.50

Price of 1 metre of cloth = $\frac{79.50}{3}=\mathrm{Rs}.26.5$

Price of 15 metres of cloth = $26.5\times 15=\mathrm{Rs}.397.50$

#### Page No 9.18:

#### Question 2:

The cost of 17 chairs is Rs 9605. Find the number of chairs that can be purchased in Rs 56500.

#### Answer:

Number of chairs purchased for Rs. 9,605 = 17

Number of chairs purchased for Rs. 1 = $\frac{17}{9605}$

Number of chairs purchased for Rs. 56,500 = $\frac{17}{9605}\times 56500=100$

#### Page No 9.18:

#### Question 3:

Three ferryloads are needed to carry 150 people across a river. How many people will be carried on 4 ferryloads?

#### Answer:

Number of people needed to carry 3 ferryloads across a river = 150

Number of people needed to carry 1 ferryload across a river = $\frac{150}{3}=50$

Number of people needed to carry 4 ferryloads across a river = $50\times 4=200$

#### Page No 9.18:

#### Question 4:

If 9 kg of rice costs Rs 120.60, What will 50 kg of such a quality of rice cost?

#### Answer:

Cost of 9 kg rice = Rs. 120.60

Cost of 1 kg rice = $\frac{120.60}{9}=\mathrm{Rs}.13.4$

Cost of 50 kg rice = $13.4\times 50=\mathrm{Rs}.670$

#### Page No 9.18:

#### Question 5:

A train runs 200 kilometres in 5 hours. How many kilometres does it run in 7 hours?

#### Answer:

Distance covered by the train in 5 hours = 200 km

Distance covered by the train in 1 hour = $\frac{200}{5}=40$ km

Distance covered by the train in 7 hours = $40\times 7=280$ km

#### Page No 9.18:

#### Question 6:

10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?

#### Answer:

It is given that 10 boys can dig a pitch in 12 hours.

Therefore, time taken by one boy to dig the pitch = $10\times 12=120$ hours

Thus, time taken by 8 boys to dig the same pitch = $\frac{120}{8}=15$ hours

#### Page No 9.18:

#### Question 7:

A man can work 8 hours daily and finishes a work in 12 days. If he works 6 hrs daily, in how many days will the same work be finished?

#### Answer:

It is given that a man can work for 8 hours daily and finishes a work in 12 days.

Therefore, time taken by the man to finish the total work = $8\times 12=96$ hours

If he works for 6 hours daily, the number of days required to finish the total work = $\frac{96}{6}=16$ days

#### Page No 9.18:

#### Question 8:

Fifteen post cards cost Rs 2.25. What will be the cost of 36 post cards? How many postcards can we buy in Rs 45?

#### Answer:

Cost of 15 postcards = Rs. 2.25

Cost of 1 postcard = Rs. $\frac{2.25}{15}$

Cost of 36 postcards = $\frac{2.25}{15}\times 36=\frac{225}{15\times 100}\times 36=\mathrm{Rs}.5.40$

Number of postcards purchased in Rs. 1 = $\frac{15}{2.25}$

Number of postcards purchased in Rs. 45 = $\frac{15}{2.25}\times 45=\frac{15\times 100}{225}\times 45=300$

#### Page No 9.18:

#### Question 9:

A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?

#### Answer:

Cost of a rail journey of 75 km = Rs. 215

Cost of a rail journey of 1 km = Rs. $\frac{215}{75}$

Cost of a rail journey of 120 km = $\frac{215}{75}\times 120=\mathrm{Rs}.344$

#### Page No 9.18:

#### Question 10:

If the sales tax on a purchase worth Rs 60 is Rs 4.20. What wil be the sales tax on the purchase worth Rs 150?

#### Answer:

Sales tax on a purchase worth Rs. 60 = Rs. 4.20

Sales tax on a purchase worth Rs. 1 = $\mathrm{Rs}.\frac{4.20}{60}$

Sales tax on a purchase worth Rs. 150 = $\frac{4.20}{60}\times 150=\frac{420}{60\times 100}\times 150=\mathrm{Rs}.10.50$

#### Page No 9.19:

#### Question 11:

The cost of 17 chairs is Rs 19210. Find the number of such chairs that can be purchased in Rs 113000?

#### Answer:

Number of chairs purchased in Rs. 19,210 = 17

Number of chairs purchased in Rs. 1 = $\frac{17}{19210}$

Number of chairs purchased in Rs. 1,13,000 = $\frac{17}{19210}\times 113000=100$

#### Page No 9.19:

#### Question 12:

A car travels 165 km in 3 hours.

(i) How long will it take to travel 440 km?

(ii) How far will it travel in 7 hours?

#### Answer:

A car travels 165 km in 3 hours.

$\therefore $ Speed of the car = $\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{165}{3}=55\mathrm{km}/\mathrm{h}$

(i) Time taken by the car to travel 440 km = $\frac{440}{55}=8\mathrm{hours}$

(ii) Distance travel by the car in 7 hours = $55\times 7=385\mathrm{km}$

#### Page No 9.19:

#### Question 13:

2 dozens of oranges cost Rs 60. Find the cost of 120 similar oranges?

#### Answer:

Cost of 2 dozens or 24 oranges = Rs. 60

Cost of 1 orange = Rs. $\frac{60}{24}$

Cost of 120 oranges = $\frac{60}{24}\times 120=\mathrm{Rs}.300$

#### Page No 9.19:

#### Question 14:

A family of 4 members consumes 6 kg of sugar in a month. What will be the monthly consumption of sugar, if the number off family members becomes 6?

#### Answer:

Amount of sugar consumed by a family of 4 members = 6 kg

Amount of sugar consumed by a family of 1 member = $\frac{6}{4}\mathrm{kg}$

Amount of sugar consumed by a family of 6 members = $\frac{6}{4}\times 6=9\mathrm{kg}$

#### Page No 9.19:

#### Question 15:

The weight of 45 folding chairs is 18 kg.How many such chairs can be loaded on a truck having a capacity of carrying 4000 kg load?

#### Answer:

Number of folding chairs weighing 18 kg = 45

Number of folding chairs weighing 1 kg = $\frac{45}{18}$

Number of folding chairs weighing 4,000 kg = $\frac{45}{18}\times 4000=10000$

#### Page No 9.19:

#### Question 1:

A ratio equivalent of 2 : 3 is

(a) 4 : 3

(b) 2 : 6

(c) 6 : 9

(d) 10 : 9

#### Answer:

(c) 6 : 9

2 : 3 is equivalent to 6 : 9. (Dividing by 3)

#### Page No 9.19:

#### Question 2:

The angles of a triangle are in the ratio 1 : 2 : 3. The measure of the largest angle is

(a) 30°

(b) 60°

(c) 90°

(d) 120°

#### Answer:

(c) 90^{o}

Sum of all the angles of a triangle = 180^{o}

Largest angle = $\frac{3}{(1+2+3)}\times 180=\frac{3}{6}\times 180={90}^{o}$

#### Page No 9.19:

#### Question 3:

The sides of a triangle are in the ratio 2 : 3 : 5. If its perimeter is 100 cm, the length of its smallest side is

(a) 2 cm

(b) 20 cm

(c) 3 cm

(d) 5 cm

#### Answer:

(b) 20 cm

Length of the smallest side = $100\times \frac{2}{(2+3+5)}=100\times \frac{2}{10}=20cm$

*Disclaimer: In actual terms, the sides of a triangle cannot be in the ratio 2 : 3 : 5.
Because, to form a triangle, the sum of two sides must be greater than the third side.
Here, the sum of two sides of the triangle is equal to the third side.*

#### Page No 9.19:

#### Question 4:

Two numbers are in the ratio 7 : 9. If the sum of the numbers is 112, then the larger number is

(a) 63

(b) 42

(c) 49

(d) 72

#### Answer:

(a) 63

Let the larger number be *x.*

Then

$\frac{9}{(7+9)}=\frac{x}{112}\phantom{\rule{0ex}{0ex}}x=\frac{9}{16}\times 112=63$

#### Page No 9.19:

#### Question 5:

Two ratio 384 : 480 in its simplest form is

(a) 3 : 5

(b) 5 : 4

(c) 4 : 5

(d) 2 : 5

#### Answer:

(c) 4 : 5

$\frac{384}{480}=\frac{4}{5}(\mathrm{Dividing}\mathrm{the}\mathrm{denominator}\mathrm{and}\mathrm{numerator}\mathrm{by}96)$

#### Page No 9.19:

#### Question 6:

If *A*, *B*, *C*,divide Rs 1200 in the ratio 2 : 3 : 5, then B's share is

(a) Rs 240

(b) Rs 600

(c) Rs 380

(d) Rs 360

#### Answer:

(d) Rs. 360

B's share =$1200\times \frac{3}{(2+3+5)}=1200\times \frac{3}{10}=\mathrm{Rs}.360$

#### Page No 9.19:

#### Question 7:

If a bus travels 126 km in 3 hours and a train travels 315 km in 5 hours, then the ratio of their speeds is

(a) 2 : 5

(b) 2 : 3

(c) 5 : 2

(d) 25 : 6

#### Answer:

(b) 2 : 3

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Speed of the bus = $\frac{126}{3}=42\mathrm{km}/\mathrm{h}$

Speed of the train = $\frac{315}{5}=63\mathrm{km}/\mathrm{h}$

Ratio of their speeds = 42 : 63 = 2 : 3

#### Page No 9.19:

#### Question 8:

The ratio of male and female employees in a multinational company is 5 : 3. If there are 115 male employees in the company, then the number off female employees is

(a) 96

(b) 52

(c) 69

(d) 66

#### Answer:

(c) 69

If the number of male employees is 115, let the number of female employees be *x*.

According to the question:

$\frac{5}{3}=\frac{115}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{115}{5}\times 3=69$

#### Page No 9.19:

#### Question 9:

Length and width of a field are in the ratio 5 : 3. If the width of the field is 42 m, then its length is

(a) 50 m

(b) 70 m

(c) 80 m

(d) 100 m

#### Answer:

(b) 70 m

Ratio of length and width of the field = 5 : 3

Let the length be *x *m.

$\because $ Width = 42 m

$\therefore $ Length = $\frac{5}{3}=\frac{x}{42}\Rightarrow x=\frac{5}{3}\times 42=70\mathrm{m}$

#### Page No 9.19:

#### Question 10:

If 57 : *x* = 51 : 85, then the value of *x* is

(a) 95

(b) 76

(c) 114

(d) None of these

#### Answer:

(a) 95

Consider

$\frac{57}{x}=\frac{51}{85}\phantom{\rule{0ex}{0ex}}\Rightarrow 57\times \frac{85}{51}=x\phantom{\rule{0ex}{0ex}}\Rightarrow x=95$

#### Page No 9.19:

#### Question 11:

The ratio of boys and girls in a school is 12 : 5. If there are 840 girls in the school, then the number of boys is

(a) 1190

(b) 2380

(c) 2856

(d) 2142

#### Answer:

None of the given options are correct.

If the number of girls in the school is 840, let the number of boys be *x.*

It is given that the ratio of boys and girls is 12 : 5. Therefore, we get:

$\therefore \frac{12}{5}=\frac{x}{840}\phantom{\rule{0ex}{0ex}}x=\frac{12}{5}\times 840=2016$

#### Page No 9.19:

#### Question 12:

If 4, *a*, *a*, 36 are in proportion, then *a* =

(a) 24

(b) 12

(c) 3

(d) 24

#### Answer:

(b) 12

4, *a*, *a*, 36 are in proportion; therefore, we get:

4 : a : : a : 36

$\Rightarrow \frac{4}{a}=\frac{a}{36}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\times 36=a\times a\phantom{\rule{0ex}{0ex}}\Rightarrow 144={a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow a=12$

#### Page No 9.19:

#### Question 13:

If 5 : 4 : : 30 : *x*, then the value of *x* is

(a) 24

(b) 12

(c) $\frac{3}{2}$

(d) 6

#### Answer:

(a) 24

5 : 4 : : 30 : *x
$\Rightarrow \frac{5}{4}=\frac{30}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=30\times \frac{4}{5}=24$*

#### Page No 9.19:

#### Question 14:

If *a*, *b*, *c*, *d* are in proportion, then

(a) *ab = cd*

(b) *ac = bd*

(c) *ad = bc*

(d) None of these

#### Answer:

(c) *ad* =* bc*

$\because $ *a*, *b*, *c **and **d* are in proportion.

$\therefore $ *a* : *b* = *c* : *d*

$\Rightarrow \frac{a}{b}=\frac{c}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow ad=bc$

#### Page No 9.20:

#### Question 15:

If a, b, c, are in proportion, then

(a) *a*^{2} = *bc*

(b) *b*^{2}^{ }= *ac*

(c) *c*^{2} = *ab*

(d) None of these

#### Answer:

(b)* **b*^{2} = *ac*

$\because $ *a* , *b* and *c* are in proportion.

$\therefore $ *a* : *b* :: *b* : *c*

$\Rightarrow \frac{a}{b}=\frac{b}{c}\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}=ac\phantom{\rule{0ex}{0ex}}$

#### Page No 9.20:

#### Question 16:

If the cost of 5 bars of a soap is Rs. 30, then the cost of one dozen bars is

(a) Rs 60

(b) Rs 120

(c) Rs 72

(d) Rs 140

#### Answer:

(c) Rs. 72

Let the cost of one dozen bars be Rs. *x.*

$\therefore \frac{30}{5}=\frac{x}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{30}{5}\times 12=\mathrm{Rs}.72$

Cost of one dozen (12) bars = Rs. 72

#### Page No 9.20:

#### Question 17:

12 men can finish a piece of work in 25 days. The number of days in which the same piece of work can be done by 20 men, is

(a) 10 days

(b) 12 days

(c) 15 days

(d) 14 days

#### Answer:

(c) 15 days

It is given that 12 men can finish a piece of work in 25 days.

Let the number of days required to do the same piece of work by 20 men be *x *days.

We get:

$\frac{20}{12}=\frac{25}{x}\phantom{\rule{0ex}{0ex}}x=12\times \frac{20}{25}=15\mathrm{Days}$

#### Page No 9.20:

#### Question 18:

If the cost of 25 packets of 12 pencils each is Rs 750, then the cost of 30 packets of 8 pencils each is

#### Answer:

(a) Rs. 600

It is given that the cost of 25 packets of 12 pencils each is Rs 750; therefore, we have:

Cost of (25$\times $12) pencils = 300 pencils = Rs. 750

Let the cost of (30$\times $8) or 240 pencils be Rs. *x.*

$\because $ 750 : 300 : : *x *: 240

$\therefore $ Cost of (30$\times $8) = 240 pencils = $\frac{750}{300}\times 240=\mathrm{Rs}.600$

#### Page No 9.20:

#### Question 19:

If a, b, c are in proportion, then

(a) a : b : : b : c

(b) a : b : : c : a

(c) a : b : : c : b

(d) a : c : : b : c

#### Answer:

(a) a : b :: b : c

$\Rightarrow $ ac = b^{2}

#### Page No 9.20:

#### Question 20:

The first, second and fourth terms of a proportion are 16, 24 and 54 respectively. The third term is

(a) 32

(b) 48

(c) 28

(d) 36

#### Answer:

(d) 36

The first, second and fourth term of a proportion are 16, 24 and 54, respectively.

Let third term is be *x.*

According to the question, we have:

16 : 24 = *x* : 54

$\frac{16}{24}=\frac{x}{54}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16}{24}\times 54=x\phantom{\rule{0ex}{0ex}}\Rightarrow x=36$

#### Page No 9.20:

#### Question 1:

If *a* = 2*b*, then *a* : *b* =

(a) 2 : 1

(b) 1 : 2

(c) 3 : 4

(d) 4 : 3

#### Answer:

It is given that,

$a=2b\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{b}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow a:b=2:1$

Hence, the correct option is (a).

#### Page No 9.20:

#### Question 2:

If 4 : 3 = *x*^{2} : 12, then the value of *x* (*x* > 0).

(a) 16

(b) 4

(c) 9

(d) 3

#### Answer:

It is given that,

$\frac{4}{3}=\frac{{x}^{2}}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4\times 4}{3\times 4}=\frac{{x}^{2}}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16}{12}=\frac{{x}^{2}}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow 16={x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {4}^{2}={x}^{2}\left(\because x0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 4=x$

Hence, the correct option is (b).

#### Page No 9.20:

#### Question 3:

If* x* : *y* = 2 : 3 and *y* : *z* = 2 : 3, then *x* : *z* =

(a) 2 : 3

(b) 3 : 4

(c) 5 : 7

(d) 4 : 9

#### Answer:

It is given that,

$\frac{x}{y}=\frac{2}{3}....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\frac{y}{z}=\frac{2}{3}....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{x}{y}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}\times \frac{y}{z}=\frac{2}{3}\times \frac{y}{z}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}\times \frac{y}{z}=\frac{2}{3}\times \frac{2}{3}\left[\mathrm{From}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{z}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow x:z=4:9$

Hence, the correct option is (d).

#### Page No 9.20:

#### Question 4:

If 80 : 60 = *x* : 12, then *x* =

(a) 16

(b) 7

(c) 24

(d) 50

#### Answer:

It is given that,

$\frac{80}{60}=\frac{x}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{80\xf75}{60\xf75}=\frac{x}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16}{12}=\frac{x}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=16$

Hence, the correct option is (a).

#### Page No 9.20:

#### Question 5:

Two numbers are in the ratio 3 : 5 and their sum is 96. The larger number is

(a) 36

(b) 42

(c) 60

(d) 70

#### Answer:

Let the larger number be *x*.

Then, the smaller number be (96 − *x*).

According to the question,

$\frac{96-x}{x}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(96-x\right)\times x\times 5}{x}=\frac{3\times x\times 5}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\left(96-x\right)=3x\phantom{\rule{0ex}{0ex}}\Rightarrow 480-5x=3x\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+5x=480\phantom{\rule{0ex}{0ex}}\Rightarrow 8x=480\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8x}{8}=\frac{480}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=60$

Thus, the larger number is 60.

Hence, the correct option is (c).

#### Page No 9.20:

#### Question 6:

If *x* : *y* = 1 : 1, then $\frac{3x+4y}{5x+6y}=$

(a) $\frac{7}{11}$

(b) $\frac{17}{11}$

(c) $\frac{17}{23}$

(d) $\frac{4}{5}$

#### Answer:

It is given that,

$\frac{x}{y}=\frac{1}{1}=1$ ...(1)

Now,

$\frac{3x+4y}{5x+6y}=\frac{\left(3x+4y\right)\xf7y}{\left(5x+6y\right)\xf7y}\phantom{\rule{0ex}{0ex}}=\frac{3\left({\displaystyle \frac{x}{y}}\right)+4}{5\left({\displaystyle \frac{x}{y}}\right)+6}\phantom{\rule{0ex}{0ex}}=\frac{3\left(1\right)+4}{5\left(1\right)+6}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{3+4}{5+6}\phantom{\rule{0ex}{0ex}}=\frac{7}{11}$

Hence, the correct option is (a).

#### Page No 9.20:

#### Question 7:

If six men can do a piece of work in 6 days, then 3 men can do same work in

(a) 10 days

(b) 12 days

(c) 15 days

(d) 18 days

#### Answer:

Let the required number of days be *x*.

Number of days is inversaly proportional to the number of men.

According to the question,

$6:3::x:6\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6}{3}=\frac{x}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6\times 6}{3}=\frac{x\times 6}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{36}{3}=x\phantom{\rule{0ex}{0ex}}\Rightarrow x=12$

Thus, 3 men can do same work in 12 days.

Hence, the correct option is (b).

#### Page No 9.20:

#### Question 8:

If 4 : 5 : : *x* : 45, then *x* =

(a) 54

(b) 60

(c) 36

(d) 30

#### Answer:

$4:5::x:45\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{5}=\frac{x}{45}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4\times 45}{5}=\frac{x\times 45}{45}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4\times 9}{1}=x\phantom{\rule{0ex}{0ex}}\Rightarrow x=36$

Hence, the correct option is (c).

#### Page No 9.20:

#### Question 9:

If 343 is the third proportional of *a* and *b*, where *a* : *b* = 1 : 7, then the value of *a* + *b* is

(a) 14

(b) 24

(c) 56

(d) 63

#### Answer:

It is given that,

$\frac{a}{b}=\frac{1}{7}...\left(1\right)$

According to the question,

$a:b::b:343\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{b}=\frac{b}{343}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{7}=\frac{b}{343}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1\times 343}{7}=\frac{b\times 343}{343}\phantom{\rule{0ex}{0ex}}\Rightarrow 49=b$

Now,

$\frac{a}{b}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{49}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a\times 49}{49}=\frac{1\times 49}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow a=7$

Thus, *a* + *b* = 7 + 49 = 56

Hence, the correct option is (c).

#### Page No 9.20:

#### Question 10:

The first three terms of a proportion are 12, 21 and 8 respectively. Then 4^{th} term is

(a) 18

(b) 16

(c) 14

(d) 20

#### Answer:

Let the 4^{th} term be *x*.

According to the question,

$12:21::8:x\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Product}\mathrm{of}\mathrm{extreme}\mathrm{terms}=\mathrm{product}\mathrm{of}\mathrm{mean}\mathrm{terms}\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=8\times 21\phantom{\rule{0ex}{0ex}}\Rightarrow 12x=168\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{12x}{12}=\frac{168}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=14$

Hence, the correct option is (c).

#### Page No 9.21:

#### Question 11:

If *a* : *b* = 2 : 5, find the value of $\frac{3a+2b}{4a+b}$.

#### Answer:

It is given that,

$\frac{a}{b}=\frac{2}{5}$ ...(1)

Now,

$\frac{3a+2b}{4a+b}=\frac{\left(3a+2b\right)\xf7b}{\left(4a+b\right)\xf7b}\phantom{\rule{0ex}{0ex}}=\frac{3\left(a\xf7b\right)+2}{4\left(a\xf7b\right)+1}\phantom{\rule{0ex}{0ex}}=\frac{3\left({\displaystyle \frac{a}{b}}\right)+2}{4\left({\displaystyle \frac{a}{b}}\right)+1}\phantom{\rule{0ex}{0ex}}=\frac{3\left({\displaystyle \frac{2}{5}}\right)+2}{4\left({\displaystyle \frac{2}{5}}\right)+1}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{6+2\times 5}{5}}}{{\displaystyle \frac{8+1\times 5}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{6+10}{8+5}\phantom{\rule{0ex}{0ex}}=\frac{16}{13}$

Hence, $\frac{3a+2b}{4a+b}=\frac{16}{13}$.

#### Page No 9.21:

#### Question 12:

The mean proportional of *a* and *b* is 10 and the value of *a* is four times the value of *b*. If *a* > 0, *b* > 0, find the value of *a *+ *b*?

#### Answer:

It is given that,

$a:10::10:b\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{10}=\frac{10}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow a\times b=10\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow a\times b=100....\left(1\right)$

and $a=4b....\left(2\right)$

From (1) and (2),

$\left(4b\right)b=100\phantom{\rule{0ex}{0ex}}\Rightarrow b\times b=\frac{100}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow b\times b=25\phantom{\rule{0ex}{0ex}}\Rightarrow b\times b=5\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow b=5\phantom{\rule{0ex}{0ex}}\Rightarrow a=4\times 5\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow a=20$

$\therefore a+b=20+5\phantom{\rule{0ex}{0ex}}=25$

Hence, *a *+ *b* = 25.

#### Page No 9.21:

#### Question 13:

A given quantity of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?

#### Answer:

A given quantity of rice is sufficient for 60 persons for 3 days.

A given quantity of rice is sufficient for 1 person for 3 × 60 days.

A given quantity of rice is sufficient for 18 persons for $\frac{3\times 60}{18}$ = 10 days.

Hence, the rice last for 10 days for 18 persons.

#### Page No 9.21:

#### Question 14:

12 men can reap a field in 25 days. In how many days can 20 men reap the same field?

#### Answer:

12 men can reap a field in 25 days.

1 man can reap a field in 25 × 12 days.

20 men can reap a field in $\frac{25\times 12}{20}=15$ days.

Hence, 20 men can reap the same field in 15 days.

#### Page No 9.21:

#### Question 15:

Shikha and Aarushi are sisters. The ratio of their ages is 3 : 4. Their parents give them ₹1400 to share in the ratio of their ages. How much does each of them recieve?

#### Answer:

Let the age of Shikha be 3*x* and the age of Aarushi be 4*x*.

According to the question,

3*x* + 4*x* = 1400

⇒ 7*x* = 1400

$\Rightarrow x=\frac{1400}{7}=200$

Hence, Shikha recieves 3 × 200 = 600 rupees

and Aarushi recieves 4 × 200 = 800 rupees.

#### Page No 9.21:

#### Question 16:

The ratio of copper and zinc in an alloy is 5 : 3. If the weight of copper in the alloy is 30.5 g, find the weight of zinc in it.

#### Answer:

Let the weight of copper be 5*x* and the weight of zinc be 3*x*.

According to the question,

5*x* = 30.5 g

$\Rightarrow x=\frac{30.5}{5}=6.1\mathrm{g}$

Hence, the weight of zinc in the alloy is $3\times 6.1=18.3\mathrm{g}$.

#### Page No 9.21:

#### Question 17:

Divide ₹3450 among A, B and C in the ratio 3 : 5 : 7.

#### Answer:

Let the value of A be 3*x*, the value of B be 5*x* and the value of C be 7*x*.

According to the question,

$3x+5x+7x=3450\phantom{\rule{0ex}{0ex}}\Rightarrow 15x=3450\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{3450}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow x=230$

Hence, the value of A is 3 × 230 = ₹690,

the value of B is 5 × 230 = ₹1150,

and the value of C is 7 × 230 = ₹1610.

#### Page No 9.21:

#### Question 18:

40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?

#### Answer:

26 days are required to finish a piece of work by 40 men.

1 day is required to finish a piece of work by 40 × 26 men.

16 days are required to finish a piece of work by $\frac{40\times 26}{16}=65$ men.

Hence, 65 men will be needed to finish a piece of work in 16 days.

#### Page No 9.21:

#### Question 19:

Tropical fruit juice is mixed from mango, papaya and orange juice in the ratio 3 : 1 : 4. Aarushi has 400 m*l *of orange juice. If she uses it all:

(i) How many mango and papaya will he need?

(ii) How much tropical fruit juice will he make altogether?

#### Answer:

Let the quantity of mango juice be 3*x*, the quantity of papaya juice be *x* and the quantity of orange juice be 4*x*.

According to the question,

$4x=400\mathrm{m}l\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{400}{4}\mathrm{m}l\phantom{\rule{0ex}{0ex}}\Rightarrow x=100\mathrm{m}l$

(i) Quantity of mango juice = $3\times 100=300$ m*l*

and Quantity of papaya juice = 100 m*l*

(ii) Total quantity of tropical juice = 400 + 300 + 100 = 800 m*l.*

#### Page No 9.21:

#### Question 20:

If 48 boxes contain 6000 pens, how many such boxes will be needed for 1875 pens?

#### Answer:

48 boxes contain 6000 pens.

Number of pens contained in 1 box = $\frac{6000}{48}=125$pens.

Therefore, number of boxes needed to contain 1875 pens = $\frac{1875}{125}=15$ boxes.

Hence, 15 boxes will be needed for 1875 pens.

#### Page No 9.21:

#### Question 21:

Fill in the blanks:

If 1, 2, 3 and *x* are in proportion, then *x* = .............

#### Answer:

1 : 2 : : 3 : *x*

$\Rightarrow \frac{1}{2}=\frac{3}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1\times 2\times x}{2}=\frac{3\times 2\times x}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=6$

Hence, *x* = __ 6 __.

#### Page No 9.21:

#### Question 22:

Fill in the blanks:

Ratio of ₹2 and 50 paise and ₹3 and 75 paise is ............

#### Answer:

₹2 and 50 paise = (2 × 100 + 50) paise

= 250 paise

₹3 and 75 paise = (3 × 100 + 75) paise

= 375 paise

Now,

$\frac{250}{375}=\frac{250\xf725}{375\xf725}\phantom{\rule{0ex}{0ex}}=\frac{10}{15}\phantom{\rule{0ex}{0ex}}=\frac{10\xf75}{15\xf75}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$

Hence. Ratio of ₹2 and 50 paise and ₹3 and 75 paise is __2 : 3__.

#### Page No 9.21:

#### Question 23:

Fill in the blanks:

Mean proportion of 2 and 8 is ..........

#### Answer:

Let the mean proportion of 2 and 8 be *x*.

Then,

$2:x::x:8\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{x}=\frac{x}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times 8=x\times x\phantom{\rule{0ex}{0ex}}\Rightarrow 16=x\times x\phantom{\rule{0ex}{0ex}}\Rightarrow 4\times 4=x\times x\phantom{\rule{0ex}{0ex}}\Rightarrow 4=x$

Hence, mean proportion of 2 and 8 is __4__.

#### Page No 9.21:

#### Question 24:

If *a* : *b* = 4 : 5 and *b* : *c* = 2 : 3, then *a* : *c* = ...........

#### Answer:

It is given that,

$\frac{a}{b}=\frac{4}{5}....\left(1\right)$

and

$\frac{b}{c}=\frac{2}{3}....\left(2\right)$

Now,

$\frac{a}{c}=\frac{a\xf7b}{c\xf7b}\phantom{\rule{0ex}{0ex}}=\frac{a}{b}\times \frac{b}{c}\phantom{\rule{0ex}{0ex}}=\frac{4}{5}\times \frac{2}{3}\left[\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{8}{15}$

Hence, *a* : *c* = __8 : 15__.

#### Page No 9.21:

#### Question 25:

Fill in the blanks:

If 2*x* = 5*y*, then *x* : *y* = ........

#### Answer:

It is given that,

2*x* = 5*y * ......(1)

Now,

$2x=5y\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2x}{2y}=\frac{5y}{2y}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}=\frac{5}{2}$

Hence, *x* : *y* = __5 : 2__.

#### Page No 9.5:

#### Question 1:

Express each of the following in the language of ratios:

(i) In a class, the number of girls in the merit list of the board examination is two times that of boys.

(ii) The number of student passing mathematics test is $\frac{2}{3}$ of the number that appeared.

#### Answer:

(i) The ratio of the number of girls in the merit list of the board examination to the number of boys in that list is 2 : 1.

(ii) The ratio of the number of students passing mathematics test to the total number of students appeared is 2 : 3.

#### Page No 9.6:

#### Question 2:

Express the following ratios in language of daily life:

(i) The ratio of the number of bad pencils to that good pencils produced in a factory is 1:9.

(ii) In India, the ratio of the number of villages to that of cities is about 2000:1.

#### Answer:

(i) The ratio of the number of bad pencils to that of good pencils produced in a factory is 1 : 9 means out of 10 or (9 + 1) pencils, 9 pencils are good and 1 is bad.

(ii) In India, the number of villages is 2,000 times that of the cities.

#### Page No 9.6:

#### Question 3:

Express each of the of the following rations in its simplest from:

(i) 60:72

(ii) 324:144

(iii) 85:391

(iv) 186:403

#### Answer:

(i) 60 : 72 = 5 : 6 (dividing by 12)

(ii) 324 : 144 = 9 : 4 (dividing by 36)

(iii) 85 : 391 = 5 : 23 (dividing by 17)

(iv) 186 : 403 = 6 : 13 (dividing by 31)

#### Page No 9.6:

#### Question 4:

Find the ratio of the following in the simplest form:

(i) 75 paisa to Rs 3

(ii) 35 minutes to 45 minutes

(iii) 8 kg to 400 gm

(iv) 48 minutes to 1 hour

(v) 2 metres to 35 cm

(vi) 35 minutes to 45 seconds

(vii) 2 dozen to 3 scores

(viii) 3 weeks to 3 days

(ix) 48 min to 2 hour 40 min

(x) 3 m 5 cm to 35 cm.

#### Answer:

$\left(\mathrm{i}\right)\because \mathrm{Rs}.1=100\mathrm{paise}\phantom{\rule{0ex}{0ex}}\therefore 75:300=1:4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)35:45=7:9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\because 1\mathrm{kg}=1000\mathrm{gm}\phantom{\rule{0ex}{0ex}}\therefore 8000:400=20:1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\because 1\mathrm{Hour}=60\mathrm{minute}\phantom{\rule{0ex}{0ex}}\therefore 48:60=4:5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{v}\right)\because 1\mathrm{meter}=100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore 200:35=40:7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{vi}\right)\because 1\mathrm{min}=60\mathrm{sec}.\phantom{\rule{0ex}{0ex}}\therefore 35\times 60:45=140:\hspace{0.17em}3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{vii}\right)\because 1\mathrm{dozen}=12\phantom{\rule{0ex}{0ex}}\therefore 1\mathrm{score}=20\phantom{\rule{0ex}{0ex}}\mathrm{or}3\mathrm{score}=3\times 20=60\phantom{\rule{0ex}{0ex}}\therefore 24:60=2:5.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{viii}\right)\because 1\mathrm{week}=7\mathrm{days}\phantom{\rule{0ex}{0ex}}\therefore 21:3=7:1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ix}\right)\because 1\mathrm{hour}=60\mathrm{minute}\phantom{\rule{0ex}{0ex}}\therefore 2\mathrm{hour}40\mathrm{min}=2\times 60+40=160\phantom{\rule{0ex}{0ex}}\therefore 48:160=3:10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{x}\right)\because 1\mathrm{meter}=100\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore 3\mathrm{m}5\mathrm{cm}=3\times 100+5=305\mathrm{cm}\phantom{\rule{0ex}{0ex}}\therefore 305:35=61:7$

#### Page No 9.6:

#### Question 5:

Find the ratio of

(i) 3.2 metres to 56 metres

(ii) 10 metres to 25 cm.

(iii) 25 paisa to Rs 60

(iv) 10 litres to 0.25 litre

#### Answer:

(i) The ratio of 3.2 metres to 56 metres = 3.2 : 56 = 0.4 : 7 = 2 : 35

(ii) The ratio of 10 metres to 25 cm = 1000 : 25 = 40 : 1

(iii) The ratio of 25 paisa to Rs 60 = 25 : 6000 = 1 : 240

(iv) The ratio of 10 litres to 0.25 litre = 10 : 0.25 = 40 : 1

#### Page No 9.6:

#### Question 6:

The number of boys and girls in a school are 1168 and 1095 respectively. Express the ratio of the number of boys to that of the girls in the simplest form.

#### Answer:

Ratio of the number of boys to that of the girls is 1168 : 1095 = 16 : 15 (dividing by 73)

#### Page No 9.6:

#### Question 7:

Avinash works as a lecturer and earns Rs 12000 per month. His wife who is a doctor earns Rs 15000 per month. Find the following ratios:

(i) Avinash's income to the income of his wife

(ii) Avinash's income to their total income.

#### Answer:

Avinash's income = Rs. 12000

Avinash's wife income = Rs. 15000

Total income = 12000 + 15000 = Rs. 27000

(i) Ratio of Avinash's income to the income of his wife = 12000 : 15000 = 4 : 5

(ii) Ratio of Avinash's income to their total income = 12000 : 27000 = 4 : 9

#### Page No 9.6:

#### Question 8:

Of the 72 persons working in an office, 28 are men and the remaining are women. Find the ratio of the number of:

(i) men to that of women

(ii) men to the total number persons

(iii) persons to that of women.

#### Answer:

Total number of persons = 72

Number of men = 28

Number of women = 72 $-$ 28 = 44

(i) Ratio of men to that of women = 28 : 44 = 7 : 11

(ii) Ratio of men to the total number of persons = 28 : 72 = 7 : 18

(iii) Ratio of persons to that of women = 72 : 44 = 18 : 11

#### Page No 9.6:

#### Question 9:

The length of a steel tape for measurements of buildings is 10 m and its width is 2.4 cm. What is the ratio of its length to width?

#### Answer:

Length of the steel tape = 10 m

= $10\times 100\mathrm{cm}$ = 1000 cm

Breadth of the steel tape = 2.4 cm

Ratio of its length to width = $\frac{1000}{2.4}=\frac{10000}{24}=\frac{1250}{3}\phantom{\rule{0ex}{0ex}}=1250:3.$

#### Page No 9.6:

#### Question 10:

An office opens at 9 a.m. and closed at 5 p.m. with a lunch interval of 30 minutes. What is the ratio of lunch interval to the total period in office?

#### Answer:

Total period in office = 9 a.m. to 12 p.m. and 12 p.m. to 5 p.m.

= 3 hours + 5 hours = 8 hours

$\because $ 1 hour = 60 minutes

$\therefore $ 8 hours = 8 $\times $ 60 = 480 minutes

Ratio of lunch interval to total period = 30 : 480 = 1 : 16 (dividing by 30)

#### Page No 9.6:

#### Question 11:

A bullock-cart travels 24 km in 3 hours and a train travels 120 km in 2 hours. Find the ratio fo their speeds.

#### Answer:

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}\mathrm{taken}}$

Distance covered by bullock-cart = 24 km

Time taken = 3 hours

Speed of bullock-cart = $\frac{24}{3}=8\mathrm{km}/\mathrm{h}$

Distance covered by train = 120 km

Time taken = 2 hours

Speed of train = $\frac{120}{2}=60\mathrm{km}/\mathrm{h}$

Ratio for their speeds = Speed of bullock-cart : Speed of train = 8 : 60

= 2 : 15

#### Page No 9.6:

#### Question 12:

Margarette works in a factory and earns Rs 955 per month. She saves Rs 185 per month from her earnings. Find the ratio of:

(i) her savings to her income

(ii) her income to her expenditure

(iii) her savings to her expenditure.

#### Answer:

Income of Margarette per month = Rs. 955

Saving of Margarette per month = Rs. 185

Expenditure of Margarette per month = 955$-$185 = Rs. 770

(i) Ratio of her saving to her income = 185 : 955 = 37 : 191

(ii) Ratio of her income to her expenditure = 955 : 770 = 191 : 154

(iii) Ratio of her saving to her expenditure = 185 : 770 = 37 : 154

#### Page No 9.9:

#### Question 1:

Which ratio is larger in the following pairs?

(i) 3 : 4 or 9 : 16

(ii) 15 : 16 or 24 : 25

(iii) 4 : 7 or 5 : 8

(iv) 9 : 20 or 8 : 13

(v) 1 : 2 or 13 : 27

#### Answer:

(i) 3 : 4

3 : 4 or 9 : 16

$\because $ 3 : 4 = 12 : 16 (After multiplying by 4)

And 12 > 9

$\therefore $ 3 : 4 > 9 : 16

(ii) 24 : 25

15 : 16 or 24 : 25

$\because $ 15 : 16 : : 375 : 400 (After multiplying by 25)

And, 24 : 25 : : 384 : 400 (After multiplying by 16)

And 375 < 384

$\therefore $ 15 : 16 < 24 : 25

(iii) 5 : 8

4 : 7 or 5 : 8

$\because $ 4 : 7 : : 32 : 56 (After multiplying by 8)

And, 5 : 8 : : 35 : 56 (After multiplying by 7)

And 32 < 35

$\therefore $ 4 : 7 < 5 : 8

(iv) 8 : 13

9 : 20 or 8 : 13

$\because $ 9 : 20 : : 117 : 260 (After multiplying by 13)

And, 8 : 13 : : 160 : 260 (After multiplying by 20)

And 117 < 160

$\therefore $ 9 : 20 < 8 : 13

(v) 1 : 2

1 : 2 or 13 : 27

$\because $ 1 : 2 : : 27 : 54 (After multiplying by 27)

And, 13 : 27 : : 26 : 54 (After multiplying by 16)

And 27 > 26

$\therefore $ 1 : 2 > 13 : 27

#### Page No 9.9:

#### Question 2:

- Give two equivalent ratios of 6 : 8.

#### Answer:

Two equivalent ratios of 6 : 8 are 3 : 4 and 9 : 12.

#### Page No 9.9:

#### Question 3:

Fill in the following blanks: $\frac{12}{20}=\frac{\overline{)}}{5}=\frac{9}{\overline{)}}$

#### Answer:

$\frac{12}{20}=\frac{3}{5}(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}4)\phantom{\rule{0ex}{0ex}}\frac{3}{5}=\frac{9}{15}(\mathrm{Multiplying}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}3)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \frac{12}{20}=\frac{3}{5}=\frac{9}{15}$

View NCERT Solutions for all chapters of Class 6