myhelper: RS Aggarwal 2019,2020 solution class 7 chapter 10 Percentage Exercise 10C

RS Aggarwal 2019,2020 solution class 7 chapter 10 Percentage Exercise 10C

Exercise 10C

Page-148

Question 1:

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$\frac{3}{4}$ as rate per cent is

(a) 7.5%
(b) 75%
(c) 0.75%
(d) none of these

Answer 1:

(b) 75%

$\frac{3}{4}$ =

$(\frac{3}{4} \times 100) %$ = 75%

Question 2:

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The ratio 2 : 5 as rate per cent is

(a) 4%
(b) 0.4%
(c) 40%
(d) 14%

Answer 2:

$\frac{2}{5}$ =

$(\frac{2}{5} \times 100) %$ = 40%

Question 3:

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$8 \frac{1}{3} %$ expressed as a fraction, is

(a)

$\frac{25}{3}$
(b)

$\frac{3}{25}$
(c)

$\frac{1}{12}$
(d)

$\frac{1}{4}$

Answer 3:

(c)

$\frac{1}{12}$

$8 \frac{1}{3} %$ =

$\frac{25}{3} %$ =

$(\frac{25}{3} \times \frac{1}{100}) = (\frac{1}{3 \times 4}) = \frac{1}{12}$

Page-149

Question 4:

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If x% of 75 = 9, then the value of x is

(a) 16
(b) 14
(c) 12
(d) 8

Answer 4:

(c) 12
We have x% of 75 = 9
⇒ $(\frac{x}{100} \times 75) = 9$
∴ x = $(\frac{9 \times 100}{75}) = 12$
Hence, the value of x is 12

Question 5:

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What per cent of $\frac{2}{7}$ is $\frac{1}{35}$ ?

(a) 25%
(b) 20%
(c) 15%
(d) 10%

Answer 5:

(d) 10%

Let x be the required percent.
Then, x % of $\frac{2}{7}$ = $\frac{1}{35}$
⇒ $(\frac{x}{100} \times \frac{2}{7}) = \frac{1}{35}$
∴ x = $(\frac{100 \times 7}{35 \times 2})$ = 10
Hence, 10% of $\frac{2}{7}$ is $\frac{1}{35}$

Question 6:

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What per cent of 1 day is 36 minutes?

(a) 25%
(b) 2.5%
(c) 3.6%
(d) 0.25%

Answer 6:

(b) 2.5%

Let x % of 1 day be 36 min.
Then, $(\frac{x}{100} \times 1 \times 24 \times 60)$ min = 36 min
∴ x = $(\frac{36 \times 100}{24 \times 60})$ = $(\frac{3 \times 5}{2 \times 3}) % = (\frac{5}{2}) % = 2.5 %$

Hence, 2.5% of 1 day is 36 min.

Question 7:

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A number increased by 20% gives 42. The number is

(a) 35
(b) 28
(c) 36
(d) 30

Answer 7:

(a) 35
Let the required number be x.
Then, x + 20% of x = 42
⇒ $(x + \frac{20 x}{100}) = 42$
⇒ $(x + \frac{x}{5}) = 42$
⇒ $(\frac{5 x + x}{5}) = 42$ [∵ LCM of 1 and 5 = 5]
⇒ $(\frac{6 x}{5}) = 42$
∴ x = $(\frac{42 \times 5}{6}) = 35$
Hence, the required number is 35.

Question 8:

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A number decreased by 8% gives 69. The number is

(a) 80
(b) 75
(c) 85
(d) none of these

Answer 8:

(b) 75
Let the required number be x.
Then, x − 8% of x = 69
⇒ $(x - \frac{8 x}{100})$ = 69
⇒ $(x - \frac{2 x}{25})$ = 69
⇒ $(\frac{25 x - 2 x}{25})$ = 69 [Since L.C.M. of 1 and 25 = 25]
⇒ $(\frac{23 x}{25}) = 69$
∴ x = $(\frac{69 \times 25}{23})$ = 75
Hence, the required number is 75

Question 9:

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An ore contains 5% copper. How much ore is required to obtain 400 g of copper?

(a) 2 kg
(b) 4 kg
(c) 6 kg
(d) 8 kg

Answer 9:

(d) 8 kg
Let x kg be the required amount of ore.
Then, 5% of x kg = 400 g = 0.4 kg [∵ 1 kg = 1000 g]
⇒ $(\frac{5}{100} \times x) = 0.4$
⇒ x = $(\frac{0.4 \times 100}{5})$ = 8
Hence, 8 kg of ore is required to obtain 400 g of copper.

Question 10:

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After deducting a commission of 10% a TV costs Rs 18000. What is its gross value?

(a) Rs 18800
(b) Rs 20000
(c) Rs 19800
(d) none of these

Answer 10:

(b) Rs. 20000
Suppose that the gross value of the TV is Rs x.
Commission on the TV = 10%
Price of the TV after deducting the commission = Rs (x − 10% of x)
= Rs $(x - \frac{10}{100} x)$ = Rs $(\frac{100 x - 10 x}{100})$ = Rs $(\frac{9 x}{10})$
However, price of the TV after deducting the commission = Rs 18000
Then, Rs $(\frac{9 x}{10})$ = Rs 18000
∴ x = $(\frac{18000 \times 10}{9})$ = Rs (2000 $\times$ 10) = Rs 20000
Hence, the gross value of the TV is Rs 20,000

Question 11:

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On increasing the salary of a man by 25%, it becomes Rs 20000. What was his original salary?

(a) Rs 15000
(b) Rs 16000
(c) Rs 18000
(d) Rs 25000

Answer 11:

(b) Rs. 16000
Let us assume that the original salary of the man is Rs x.
Increase in it = 25%
Value increased in the salary = 25% of Rs. x
= Rs $(\frac{25}{100} \times x)$ = Rs $(\frac{x}{4})$
Salary after increment= Rs $(x + \frac{x}{4})$ = Rs $(\frac{5 x}{4})$
However, increased salary = Rs 20000
Then, Rs $(\frac{5 x}{4})$ = Rs 20000
∴ x = Rs $(\frac{20000 \times 4}{5})$ = Rs 16000
Hence, the original salary of the man is Rs 16,000

Question 12:

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In an examination, 95% of the total examinees passed. If the number of failures is 28, how many examinees were there?

(a) 600
(b) 480
(c) 560
(d) 840

Answer 12:

(c) 560
Suppose that the number of examinees is 100.
Number of passed examinees = 95
Number of failed examinees = (100 − 95) = 5

Total number of examinees if 5 of them failed = 100
Total number of examinees if 28 of them failed = $(\frac{100}{5} \times 28) = (20 \times 28) = 560$
Hence, there were 560 examinees.

Question 13:

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A fruit-seller had some apples. He sells 40% of them and still has 420 apples. How many apples had he in all?

(a) 588
(b) 600
(c) 700
(d) 725

Answer 13:

(c) 700
Suppose that the fruit seller initially had 100 apples.
Number of apples sold = 40
∴ Number of remaining apples = (100 − 40) = 60

Initial number of apples if 60 of them are remaining = 100
Initial number of apples if 420 of them are remaining = $(\frac{100}{60} \times 420)$ = 700
Hence, the fruit seller originally had 700 apples with him.

Question 14:

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The value of a machine depreciated 10% annually. If its present value is Rs 25000, what will be its value after 1 year?

(a) Rs 27500
(b) Rs 22500
(c) Rs 25250
(d) none of these

Answer 14:

(c) Rs. 25250

Present value of the machine = Rs 25000
Decrease in its value after 1 year = 10% of Rs 25000
= Rs $(\frac{10}{100} \times 25000)$ = Rs 2500
Depreciated value after 1 year = Rs (25000 − 2500) = Rs 22500

Hence, the value of the machine after 1 year will be Rs 22500

Question 15:

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8% of a number is 6. What is the number?

(a) 48
(b) 96
(c) 75
(d) 60

Answer 15:

(c) 75

Let the required number be x. Then, we have:
8% of x = 6
⇒ $(\frac{8}{100} \times x) = 6$
∴ x = $(\frac{6 \times 100}{8}) = 75$
Hence, the required number is 75

Question 16:

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60% of 450 = ?

(a) 180
(b) 210
(c) 270
(d) none of these

Answer 16:

Question 17:

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On reducing the value of a chair by 6% it becomes Rs 658. The original value of the chair is

(a) Rs 750
(b) Rs 720
(c) Rs 500
(d) Rs 700

Answer 17:

(d) Rs. 700
Let us assume that the original price of the chair is Rs x.
Reduce percentage on the chair = 6%
So, value of reduction on the chair = 6% of Rs. x
= Rs $(\frac{6}{100} \times x)$ = Rs $(\frac{3 x}{50})$
Reduced price of the chair = Rs $(x - \frac{3 x}{50})$
= Rs $(\frac{50 x - 3 x}{50})$ = Rs $(\frac{47 x}{50})$
However, present price of the chair = Rs 658
Then, Rs $(\frac{47 x}{50})$ = Rs 658
⇒ Rs $(\frac{47 x}{50})$ = Rs 658
⇒ x = Rs $(\frac{658 \times 50}{47})$ = Rs $(14 \times 50) = 700$
Hence, the original price of the chair is Rs 700

Question 18:

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70% of students in a school are boys. If the number of girls is 240, how many boys are there in the school?

(a) 420
(b) 560
(c) 630
(d) 480

Answer 18:

(b) 560

Let the total number of students be 100.
Then, number of boys = 70
∴ Number of girls = (100 − 70) = 30

Now, total number of students if there are 30 girls = 100
Total number of students if there are 240 girls = $(\frac{100}{30} \times 240) = 800$
∴ Number of boys = (800 − 240) = 560

Hence, there are 560 boys in the school.

Question 19:

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If 11% of a number exceeds 7% of the number by 18, the number is

(a) 72
(b) 360
(c) 450
(d) 720

Answer 19:

(c) 450

Let x be the number.
(11% of x) − (7% of x) = 18
⇒ $(\frac{11 x}{100} - \frac{7 x}{100}) = 18$
⇒ $\frac{4 x}{100} = 18$
∴ x = $(\frac{18 \times 100}{4}) = (18 \times 25) = 450$
Hence, the required number is 450

Question 20:

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If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 105

Answer 20:

(a) 60
Let x be the number.
According to question, we have:
(35% of x ) + 39 = x
⇒ $(\frac{35}{100} \times x) + 39 = x$
⇒ $(\frac{7 x}{20}) + 39 = x$
⇒ $(x - \frac{7 x}{20}) = 39$
⇒ $(\frac{20 x - 7 x}{20}) = 39$
⇒ $(\frac{13 x}{20}) = 39$
∴ x = $(\frac{39 \times 20}{13})$ = 60
Hence, the required number is 60

Question 21:

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In an examination it is required to get 36% to pass. A student gets 145 marks and fails by 35 marks. The maximum marks are

(a) 400
(b) 450
(c) 500
(d) 600

Answer 21:

(c) 500
Let x be the maximum marks.
Pass marks = (145 + 35) = 180
∴ 36% of x = 180
⇒ $(\frac{36}{100} \times x) = 180$
⇒ x = $(\frac{180 \times 100}{36}) = (5 \times 100) = 500$
Hence, maximum marks = 500

Page-150

Question 22:

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A number decreased by 40% gives 135. The number is

(a) 175
(b) 200
(c) 250
(d) 225

Answer 22:

(d) 225
Let x be the number.
According to question, we have:
x − 40% of x = 135
⇒ $(x - \frac{40 x}{100}) = 135$
⇒ $(\frac{100 x - 40 x}{100}) = 135$
⇒ $(\frac{60 x}{100}) = 135$
⇒ x = $(\frac{135 \times 100}{60})$ = 225
Hence, the required number is 225