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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Q1 | Ex-7A |Class 7 |S.Chand | New Learning Composite maths |Ratio and Proportion and Unitary Method |Ch-7 |myhelper

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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 :

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

$=\frac{8}{2}$

=4 

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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Q2 | Ex-7A |Class 7 |S.Chand | New Learning Composite maths |Ratio and Proportion and Unitary Method |Ch-7 |myhelper

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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

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

or 

1 : 3

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Q3 | Ex-7A |Class 7 |S.Chand | New Learning Composite maths |Ratio and Proportion and Unitary Method |Ch-7 |myhelper

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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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Q12 | Ex-7A |Class 7 |S.Chand | New Learning Composite maths |Ratio and Proportion and Unitary Method |Ch-7 |myhelper

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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