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S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7C

Exercise 7C

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Question 1

If 30 m of cloth cost Rs. 2400, what is the cost of 50 m cloth?

Sol :

Cost of 30m cloth is 2400

Cost of 50m cloth is $=\frac{2400 \times 50}{30}=4000$

Ans Cost of 50m cloth is 4000

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Question 2

If 8 ball pens cost Rs. 40, how much will 20 ball pens of the same kind will cost?

Sol :

Ball pens cost 40

20 ball pens cost $\frac{40\times 20}{8}$

=100

Ans : Cost of 20 ball pens is 100

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Question 3

Find the price of 744 articles at the rate of Rs. 60.25 a dozen.

Sol :

12 articles cost 60.25

744 articles cost $\frac{60.25\times 744}{12}$

=3735.5

Ans : Price of 744 articles is 3735.5

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Question 4

If the wages of 15 men for 8 days are Rs. 60,000, how much would they receive for 14 days?

Sol :

15 men's wage for 8 days are 60000

15 men's wage for 14 days are $\frac{60000 \times 14}{8}$

=105000

Ans : They receive 105000 in 14 days

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Question 5

If a car goes 45 m in 3 seconds, how far does it go in 5 seconds? How long does it take in going 120 m?

Sol :

In 3 sec the car goes 45m

In 5 sec the car goes $\frac{45 \times 5}{3}$

=75m

Ans : In 5 sec the car goes 75m

The car goes 45m in 3sec

The car goes 120m in $\frac{3\times 120}{45}$=8sec

Ans : It take 8sec for going 120m

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Question 6

If 48 men have ration for 16 days, how long will the food last, if a reinforcement of 16 men arrives?

Sol :

48 men have food for 16 days

48+16=64 men have food for $\frac{16\times 48}{64}$ days

=12 days

Ans : The food will last for 12 days

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Question 7

How long will 50 men take to do the same job, that 40 men do in 25 days?

Sol :

40 men can do a job in 25 days

40 men can do a job in 25×40 days

50 men can do a job in $\frac{25\times 40}{50}$ days

=20 days

Ans : 50 men can do it in 20 days

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Question 8

If 10 kg of mangoes cost Rs. 550, how much will 27 kg cost? How many kilograms of mangoes can be bought for Rs. 1925?

Sol :

10 kg of mangoes cost 550

1 kg of mangoes cost $\frac{550}{10}$

27 kg of mangoes cost $\frac{550 \times 27}{10}$

=1485

550 is cost of 10 kg mango

1 is cost of $\frac{10}{550}$ kg mangoes

1925 is cost of $\frac{10 \times 1925}{550}$=35kg

Ans : Cost of 27 kg mangoes is 1485

35 kg can be bought for 1925

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Question 9

A garrison of 1200 men has provisions for 60 days. How long will they last, if the garrison (a) is reduced to 600 men; (b) reinforced by 800 men?

Sol :

(a)

1250 men has provisions for 60 days

1 men has provisions for 60×1200 days

600 men has provisions for $\frac{60 \times 1200}{600}$=120 days

Ans : They last for 120 days

(b)

1200 men has provisions for 60 days

1 men has provisions for 60×1200 days

(1200+800)=2000 men has provisions for $\frac{60\times 1200}{2000}$=36 days

Ans : They last for 36 days

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Question 10

On a certain map, 2 cm represent 1 km. What area shows 20 sq cm on the map represent?

Sol :

20 sq cm is the area on the map

Let us consider length and breadth -2 and 10 respectively

2cm×10cm=20 sq cm

Converting in km

2cm $=\frac{2}{2}$km

10$=\frac{10}{2}$ km

∴Area$=\frac{2}{2} \times \frac{10}{2}=5$ sq km

Ans : 5 sq km does the map represent

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Question 11

The greatest safe load for a lift is 14 men, each weighing 80 kg. How many people, each weighing 70 kg, can it take safely?

Sol :

Weight of 80 kg each a safe lift 14 men

Weight of 1 kg each a safe lift 14×80 men

Weight of 80 kg each a safe lift $\frac{14 \times 80}{70}$=17 men

Ans : It can take 16 men safely

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Question 12

Five turns of the winder of my watch, keeps it going for 15 hours, What is the effect of 9 turns?

Sol :

5 turns of the winder , keep it going for 15 h

1 turns of the wimder , keep it going for $\frac{15}{5}$ h

9 turns of the winder , keep it going for $\frac{15\times 9}{5}$=27 h

Ans : In 9 turns its goes 27 h

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Question 13

24 men can repair a road in 15 days. How long will it take 20 men to do so, if all, work at the same rate?

Sol :

24 men can repair a road in 15 days

1 men can repair a road in 15×24 days

20 men can repair a road in $\frac{15 \times 24}{20}$=18 days

Ans : 20 men can do it in 18 days

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Question 14

I have enough money to take a 24 days holiday, If i spend Rs. 125 a day. How long a holiday can I take, if I spend Rs. 150 a day?

Sol :

If I spend 125 a day then I can take 24 holidays

If I spend 1 a day then I can take 24×125 holidays

If I spend 150 a day then I can take $\frac{24 \times 125}{150}$=20 holidays

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Question 15

A man walks 110 m in a minute. How long will he take to walk 26.4 km?

Sol :

26.4 km=26400 m

A man walks 100m in a 1m

A man walks 1m in a $\frac{1}{110}$m

A man walks 26400m in a $\frac{26400}{110}$=240 m or 4 h

Ans : He takes 4 h to walk 26.4 km

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7B

Exercise 7B

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Question 1

Find out whether the following groups of numbers are in proportion or not.

(a) 18, 20, 9, 10

Sol :

Product of extremes=18×10=180

Product of means=20×9=180

So, product of extremes=product of means

∴They are proportional

(b) 21, 6, 35, 10

Sol :

Product of extremes=21×10=210

Product of means=6×35=210

So, product of extremes=product of means

∴They are proportional

(c) 12, 18, 28, 12

Sol :

Product of extremes=12×12=144

Product of means=18×28=504

So, product of extremes≠product of means

∴They are not proportional

(d) 5.2, 3.9, 3, 4

Sol :

Product of extremes=5.2×4=20.8

Product of means=3.9×3=11.7

So, product of extremes≠product of means

∴They are not proportional

(e) 16, $3 \frac{1}{3}$, 18, $3\frac{3}{4}$

Sol :

Product of extremes$=16\times 3\frac{3}{4}=16\times \frac{15}{4}$

=60

Product of means$=3\frac{1}{3} \times 18=\frac{10}{3} \times 18$

=60

So, product of extremes=product of means

∴They are proportional

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Question 2

Find x in the following proportions.

(a) x, 6, 55, 11

Sol :

$\frac{x}{6}=\frac{55}{11}$

11x=6×55

$x=\frac{6\times 55}{11}=30$

(b) 16, 18, x, 108

Sol :

$\frac{16}{18}=\frac{x}{108}$

$x=\frac{16 \times 108}{18}=96$

(c) 4, x, x, 16

Sol :

$x=\sqrt{4\times 16}$

x=8

(d) 7.5, 15, 5, x

Sol :

$\frac{7.5}{15} =\frac{5}{x}$

$x=\frac{15\times 5 \times 10}{75}$

∴x=10

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Question 3

Find the fourth proportional to

(a) 32, 64, 6

Sol :

$\frac{32}{64}=\frac{6}{x}$

$x=\frac{6\times 64}{32}=12$

∴Fourth proportional is 12

(b) 8, 9, 24

Sol :

$\frac{8}{9}=\frac{24}{x}$

$x=\frac{9\times 24}{8}$

∴x=27

∴Fourth proportional is 27

(c) $\frac{1}{5}, \frac{1}{9}, \frac{1}{10}$

Sol :

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

$x=\frac{9\times 10}{5}=18$

∴Fourth proportional is $\frac{1}{18}$

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Question 4

Find the mean proportional between

(a) 16 and 25

Sol :

$\frac{16}{x}=\frac{x}{25}$

$x^2=16\times 25$

$x=\sqrt{16 \times 25}$

∴x=4×5=20

∴Mean proportional is 20

(b) 9 and $\frac{1}{36}$

Sol :

$\frac{9}{x}=x\times \frac{36}{1}$

$x^2=\frac{9}{36}$

$x=\sqrt{\frac{1}{4}}$

∴$x=\frac{1}{2}$

∴Mean proportional is $\frac{1}{2}$

(c) 0.01 and 0.81

Sol :

$\frac{0.01}{x}=\frac{x}{0.81}$

$x^2=0.01\times 0.81$

$x=\sqrt{0.01 \times 0.81}$

∴x=0.09

∴Mean proportional is 0.09

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Question 5

Find the third proportional to

(a) 25, 40

Sol :

25×x=40×40

$x=\frac{40\times 40}{25}=64$

∴x=64

∴Third proportional is 64

(b) 18, 30

Sol :

18×x=30×30

$x=\frac{30\times 30}{18}$

∴x=50

∴Third proportional is 50

(c) 3.9, 11.7

Sol :

3.9×x=11.7×11.7

$x=\frac{11.7\times 11.7}{3.9}$

∴x=35.1

∴Third proportional is 35.1

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Question 6

Show that 5, 25, 125 are in continued proportions.

Sol :

Since 5×5=25

and 25×5=125

∴It can be said that they are in proportion

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Question 7

If the cost of 8 m of cloth is Rs. 600, find the cost of 15 m of cloth.

Sol :

The cost of 8m of cloth is 600

The cost of 15m of cloth is $=\frac{600\times 15}{8}$

=1125

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Question 8

When a man spends $\frac{5}{9}$ of his monthly income, he saves Rs. 3600 a month, find his income.

Sol :

Let , his monthly income be x

ATQ,

$x-\frac{5x}{9}=3600$

4x=3600×9

$x=\frac{3600 \times 9}{4}$

∴x=8100

His monthly income is 8100

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Question 9

After selling 85 copies of a book for every 100 copies printed, a publisher finds that 450 copies are left unsold. How many copies were printed?

Sol :

(100-85)=15 copies left

15 copies left for every 100 copies printed

450 copies left for every $\frac{100 \times 450}{15}= 3000$ or copies printed

Ans 3000 copies were printed

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Question 10

A truck requires 108 L of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Sol :

For covering 594 km distance diesel required 108L

For covering 1650 km distance diesel required $\frac{108\times 165}{594}$

=300 L

Ans 300L diesel will be required

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7A

Exercise 7A

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Question 1

Express each of the following ratios in its simplest form.

(a) A length of 6 cm to a length of 8 cm.

Sol :

$\frac{6}{8}=\frac{3}{4}$

Ans : 3 : 4

(b) An area of 20 cm square to an area of 45 cm square.

Sol :

$\frac{20}{45}=\frac{4}{9}$

Ans 4 : 9

(c) A volume of 26 litres to a volume of 65 litres.

Sol :

$\frac{26}{65}$

Ans 26 : 65

(d) A population of 50 thousand to a population of 1.5 lakh.

Sol :

$\frac{0.5}{1.5}=\frac{1}{3}$

Ans 1 : 3

(e) $\frac{1}{4} : 2$

Sol :

$=\dfrac{\frac{1}{4}}{2}$

$=\frac{1}{8}$

Ans 1 : 8

(f) $\frac{1}{2}: \frac{1}{8}$

Sol :

$=\frac{1}{2} \times \frac{8}{1}$

$=\frac{8}{2}$

Ans 4 : 1

(g) 0.75 : 1

Sol :

$=\frac{0.75}{100}$

Ans 3 : 4

(h) 14 cm to 1 m

Sol :

$=\frac{14}{100}=\frac{7}{50}$

Ans 7 : 50

(i) 25 cm to 50 mm

Sol :

$=\frac{250}{50}=\frac{5}{1}$

Ans 5 : 1

(j) 15 m to 1 km

Sol :

$=\frac{15}{1000}=\frac{3}{200}$

Ans 3 : 200

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Question 2

Ravi earns Rs. 84,000 a year and spends Rs. 63,000 a year. Find the simplest form the ratio of

(a) Ravi’s income to his savings

Sol :

Ravi's savings

$=(84000-63000)$

=21000

ATQ,

84000 : 21000

$\frac{84}{21}=\frac{28}{7}$

or 28 : 7

(b) Money that Ravi saves to the money he spends.

Sol :

21000 : 63000

$\frac{21}{63}=\frac{7}{21}=\frac{1}{3}$

1 : 3

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Question 3

There are 12,000 students in a university, out of which 5,600 are girls. Find the simplest form, the ratio of

(a) number of girls to the number of students

Sol :

5600 : 12000

$\frac{56}{120}=\frac{28}{60}=\frac{7}{15}$

(b) number of boys to the number of girls

Sol :

Number of boys=12000-5600

=6400

ATQ,

6400 : 5600

$\frac{64}{56}=\frac{8}{7}$

(c) number of boys to the number of students

Sol :

6400 : 12000

$\frac{64}{120}=\frac{16}{30}=\frac{8}{15}$

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Question 4

Which is a greater ratio of the following pairs?

(a) 3:2 or 4:3

Sol :

$\frac{3}{2}=\frac{3\times 3}{2\times 3}=\frac{9}{6}$

$\frac{4}{3}=\frac{4\times 2}{3\times 2}=\frac{8}{6}$

Ans 3 : 2

(b) 7:10 or 8:11

Sol :

$\frac{7}{10}=\frac{7\times 11}{10\times 11}=\frac{77}{110}$

$\frac{8}{11}=\frac{8\times 10}{11\times 10}=\frac{80}{110}$

Ans 8 : 11

(c) 3² : 2² or 3 : 2

Sol :

$\frac{3^2}{2^2}=\frac{9}{4}$

$\frac{3}{2}=\frac{3\times 2}{2\times 2}=\frac{6}{4}$
Ans 3² : 2²

(d) 3 : 5 or 0.66 : 1

Sol :

$\frac{3}{5}=\frac{3\times 10}{5\times 10}=\frac{30}{50}$

$\frac{0.66}{100}=\frac{33}{50}$

Ans 0.66 : 1

(e) $\frac{1}{5} : \frac{1}{7}$ or $\frac{1}{4} : \frac{1}{9}$

Sol :

$\frac{1}{5} : \frac{1}{7}=\dfrac{\frac{1}{5}}{\frac{7}{1}}$

$=\frac{7\times 4}{5\times 4}=\frac{28}{20}$

$\frac{1}{4} : \frac{1}{9}=\dfrac{\frac{1}{4}}{\frac{1}{9}}$

$=\frac{9\times 5}{4\times 5}=\frac{45}{20}$

Ans $\frac{1}{4} : \frac{1}{9}$

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Question 5

Write the ratios – 4:9, 5:8 and 3:7 in order, from the least to the greatest.

Sol :

4 : 9

$=\frac{4}{9}$

$=\frac{4\times 56}{9\times 56}$

$=\frac{224}{504}$

5 : 8

$=\frac{5}{8}$

$=\frac{5\times 63}{8\times 63}$

$=\frac{315}{504}$

3 : 7

$=\frac{3}{7}$

$=\frac{3\times 72}{7\times 72}$

$=\frac{216}{504}$

Ans 3:7 , 4:9, 5:8

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Question 6

If a:b = 4:7 and b:c = 5:3, then find a:b:c

Sol :

a:b=4:7

=(4×5):(7×5)

=20:35

b:c=5:3

=(5×7):(3×7)

=35 : 21

∴a : b : c= 20 : 35 : 21

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Question 7

If a:b = 5:6 and b:c = 9:4, then find a:c

Sol :

a : b=5 : 6

=(5×3):(6×3)

=15 : 18

b : c=9 : 4

=(9×2):(4×2)

∴a:c=15:8

=15 : 8

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Question 8

The sides of a triangle are in the ratios, 1:1.5:2 and its perimeter is 18 cm. Find the length of each side.

Sol :

Let, the sides of triangle =x, 15x, 2x

ATQ,

x+1.5x+2x=18

4.5x=18

$x=\frac{180}{4.5}$

∴x=4

∴The sides of triangle=4,(1.5×4),(2×4.5)

=4 cm,6 cm ,8 cm

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Question 9

In an examination, 100 marks are distributed among 3 questions so that they are very proportional to 8,5,7. Find the marks of each question.

Sol :

100 marks distributed in 8 : 5 : 7

ATQ,

8x+5x+7x=100

20x=100

$x=\frac{100}{20}$

∴x=5

Marks for 1st question=(8×5)=40

Marks for 2nd question=(5×5)=25

Marks for 3rd question=(5×7)=35

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Question 10

The length of three ribbons are in the ratio 7:5:9. If the sum of the lengths of the three ribbons is 42 cm, find the length of the smallest ribbon.

Sol :

Let, the length of three ribbons=7x, 5x, 9x

∴7x+5x+9x=42

or 21x=42

$x=\frac{42}{21}$

x=2

Length of smallest ribbon=(5×2)cm

=10cm

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Question 11

Shyam lost his weight in the ratio 9:5. His original weight was 90 kg. What is his new weight?

Sol :

Shyam lost his weight in the ratio 9 : 5

∴9x=90

x=10

∴Shyam's new weight=(5×10)=50kg

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Question 12

A, B and C contributed Rs. 25,000, Rs, 50,000, Rs. 75,000 respectively in a business and their share of his profits are proportional to the capital they contributed. If the profits are Rs. 48,000, what is the share of each of them?

Sol :

A,B,C's capital in a ratio=25000 : 50000 : 75000

=25 : 50 : 75

=5 : 10 : 15

=1 : 2 : 3

Total profit 48000

A's share $=48000 \times \frac{1}{6}$

=8000

B's share $=48000 \times \frac{2}{6}$

=16000

C's share $=48000 \times \frac{3}{6}$

=24000

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Question 13

A legacy of Rs. 1,08,000 is to be divided among 3 sons in the ratio $\frac{3}{2} : \frac{9}{4} : 3$. How much does each of them receive?

Sol :

Ratio of the share of 3 son $=1\frac{1}{2} : 2\frac{1}{4} : 3$

$=\frac{3}{2} : \frac{9}{4} : 3$

=6 : 9 : 12

=2 : 3 : 4

Share of 1st son$=108000 \times \frac{2}{9}$

=24000

Share of 2nd son$=108000 \times \frac{3}{9}$

=36000

Share of 3rd son$=108000 \times \frac{4}{9}$

=48000

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Question 14

A map drawn to a scale of 1 cm to 10 km, measures 24 cm by 18 cm. If the areas of land and water represented are in the ratio of 7:2, find in square km, the land area represented.

Sol :

Given : 1 cm to 10 km

∴24 cm=24×10=240 km

∴18 cm=18×10=180 km

Area of map=(240×180)sq km

=43200 sq km

Sum of terms in ratio=7+2=9

Area of land$=43200 \times \frac{7}{9}$

=33600 sq km

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Question 15

Solve:

(a) Increase 180 in the ratio 6:9.

Sol :

6x=180

x=30

Ans (9×30)=270

(b) Decrease Rs. 900 in the ratio 5:3

Sol :

5x=900

$x=\frac{900}{5}$

x=180

=(3×180)

=540

(c) Find the multiplying factor which decreases 120 kg to 84 kg.

Sol :

120 : 84

40 : 28

10 : 7

(d) Two distances are in the ratio 15:8, the larger is 60 km, what is the smaller?

Sol :

15x=60

∴$x=\frac{60}{15}=4$

∴Smaller distance=(8×4)=32km

(e) A boy worked 8 hours a day. In what ratio , did his earnings change when, the pay was raised from Rs. 50 per hours Rs. 500 a day.

Sol :

Number of hours the boy worked in a day=8

His wage=50 per hour

So, per day wage becomes =(50×8)

=400

Now his wage per day=500

∴Ratio 400 : 500

=4 : 5

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Question 16

A photo measuring 7.5 cm by 5 cm is enlarged, so that the larger side becomes 18 cm. What does the shorter side become? In what ratio is the area increased?

Sol :

Area of photo=(7.5×5)=37.5sq cm

∴7.5x=18

x=2.4

∴5x=5×5.4=12

∴New area=(18×12)sq cm

=216 sq cm

∴Ratio=216 : 37.5

=5.76 : 1

The shorter side becomes 12 cm

The ratio is 5.76 : 1

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Question 17

At the beginning of a war, the numbers of war planes possessed by two powers of 1.6 :1. the weaker power having 400. In a general engagement, each power loses the same number of planes but, the ratio is changed to 2:1. How many planes does each lose?

Sol :

Let, 1st power have planes=1.6x

2nd power have planes=x

ATQ,

Less power have planes=400

∴x=400

1.6x=(1.6×400)

1.6x=640

New ratio= 2 : 1

∴640-y : 400-y= 2 : 1

$\frac{640-y}{400-y}=\frac{2}{1}$

640y=800-2y

2y-y=800-640

y=160

∴160 planes less by both to get new ratio 2 : 1

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7C

Exercise 7C

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7B

Exercise 7B

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7A

Exercise 7A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

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