RS Aggarwal 2019,2020 solution class 7 chapter 10 Percentage Test Paper 10

Test Paper 10

Page-151

Question 1:

Convert:

(i) $\frac{4}{5}$ into a percentage
(ii) $\frac{7}{4}$ into a percentage
(iii) 45% into a percentage
(iv) 105% into a percentage
(v) 15% into a percentage
(vi) 12 : 25 into a percentage

Answer 1:

We have:

(i) $\frac{4}{5}$= $\left(\frac{4}{5}\times 100\right)\%=\left(4\times 20\right)\%=80\%$

(ii) $\frac{7}{4}$=$\left(\frac{7}{4}\times 100\right)\%=\left(7\times 25\right)\%=175\%$

(iii) 45% = $\left(\frac{45}{100}\right)=\left(\frac{9}{20}\right)$

(iv) 105% =$\left(\frac{105}{100}\right)=\left(\frac{21}{20}\right)$

(v) 15% =$\frac{15}{100}=\frac{3}{20}$=  3 : 20

(vi) 12 : 25 = $\frac{12}{25}=\left(\frac{12}{25}\times 100\right)\%=\left(12\times 4\right)\%=48\%$

Question 2:

(i) What per cent of 1 kg is 125 g?
(ii) What per cent of 80 m is 24 m?

Answer 2:

(i) Let x% of 1 kg be 125g.
Then, $\left(\frac{x}{100}\times 1\times 1000\right) \mathrm{g}=125 \mathrm{g}$
     ⇒ 10x = 125
     ⇒ x =$\left(\frac{125}{10}\right)\%=12\frac{1}{2}\%$
Hence, $12\frac{1}{2}\%$ of 1 kg is 125 g.

(ii) Let x% of 80 m be 24 m.
Then, $\left(\frac{x}{100}\times 80\right) \mathrm{m}=24 \mathrm{m}$
     ⇒ $\left(\frac{4x}{5}\right)$ = 24
     ⇒ x =$\left(24\times \frac{5}{4}\right)\%=30\%$
Hence, $30\%$ of 80 m is 24 m.

Question 3:

(i) Find $16\frac{2}{3}\%$ of 30.
(ii) Find 15% of Rs 140

Answer 3:

(i) $16\frac{2}{3}\%$ of 30 = $\frac{50}{3}\%$ of 30
                          = $\left(\frac{50}{3\times 100}\times 30\right)$
                          = 5

(ii) 15% of Rs 140 = Rs $\left(\frac{15}{100}\times 140\right)$
                              = Rs (3 $\times$ 7)
                              = Rs 21

Question 4:

(i) Find the number whose $6\frac{1}{4}\%$ is 5.
(ii) Find 0.8% of 45.

Answer 4:

(i) Let x be the required number.
Then, $6\frac{1}{4}\%$ of x = 5
     ⇒ $\frac{25}{4}\%$ of x = 5
     ⇒ $\left(\frac{25}{4\times 100}\times x\right)=25$
     ⇒$\left(\frac{x}{16}\right)=5$
∴ x = (5 $\times$ 16) = 80
Hence, the required number is 80.

(i) 0.8% of 45 =$\left(\frac{0.8}{100}\times 45\right)$
                         =$\left(\frac{8}{10\times 100}\times 45\right)$
                         = $\left(\frac{72}{200}\right)=\left(\frac{36}{100}\right)=0.36$
Hence, 0.8% of 45 is 0.36.

Question 5:

A number is increased by 10% and the increased number is decreased by 10%. Show that the net decrease is 1%.

Answer 5:

Let x be the number.
The number is increased by 10%.
∴ Increased number = 110% of x = $\left(x\times \frac{110}{100}\right)=\left(\frac{11x}{10}\right)$
The number is, then, decreased by 10%.

∴ Decreased number = 90% of $\left(\frac{11x}{10}\right)$ = $\left(\frac{11x}{10}\times \frac{90}{100}\right)=\left(\frac{99x}{100}\right)$

Net decrease = $\left(x-\frac{99x}{100}\right)=\frac{\left(100x-99x\right)}{100}=\frac{x}{100}$
Net decrease percentage = $\left(\frac{x}{100}\times \frac{1}{x}\times 100\right)=1$

Question 6:

The value of a machine depriciates at the rate of 10% per annum. If its present value is Rs 10000, what will be its value after 2 years?

Answer 6:

The present value of the machine = Rs 10000
The decrease in its value after the 1^st year = 10% of Rs 10000
                                                            = Rs $\left(\frac{10}{100}\times 10000\right)$ = Rs 1000
The depreciated value of the machine after the 1^st year = Rs (10000 − 1000) =Rs 9000
The decrease in its value after the 2^nd year = 10% of Rs 9000
                                                               = Rs $\left(\frac{10}{100}\times 9000\right)$ = Rs 900
The depreciated value of the machine after the 2^nd year = Rs (9000 − 900) = Rs 8100

Hence, the value of the machine after two years will be Rs 8100.

Question 7:

The population of a town increases at 5% per annum. Its present population is 16000. What will be its population after 2 years?

Answer 7:

The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
                                                          = $\left(\frac{5}{100}\times 16000\right)$ = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
                                                           = $\left(\frac{5}{100}\times 16800\right)$ = 840
Increased population after two years = 16800 + 840 = 17640

Hence, the population of the town after two years will be 17,640.

Question 8:

The price of a teaset is increased by 5%. If the increased price is Rs 441, what is its original price?

Answer 8:

Let us assume that the original price of the tea set is Rs. x
Increase in it = 5%
So, value increased on the tea set = 5% of Rs. x
                                                      = Rs. $\left(\frac{5}{100}\times x\right)$ = Rs. $\left(\frac{x}{20}\right)$
Then, increased price of the tea set = Rs. $\left(x+\frac{x}{20}\right)$
                                                          = Rs. $\left(\frac{20x+x}{20}\right)$ = Rs. $\left(\frac{21x}{20}\right)$
However, increased price = Rs. 441
Then, Rs. $\left(\frac{21x}{20}\right)$ = Rs. 441
∴  x = $\left(\frac{441\times 20}{21}\right)$ = 420
Hence, the original price of the tea set is Rs 420

Question 9:

Mark (✓) against the correct answer
$6\frac{1}{4}\%$ expressed as a fraction is

(a) $\frac{1}{8}$
(b) $\frac{1}{16}$
(c) $\frac{4}{25}$
(d) $\frac{1}{25}$

Answer 9:

(b) $\frac{1}{16}$

$6\frac{1}{4}\%$ = $\frac{25}{4}\%$ =$\left(\frac{25}{4\times 100}\right)=\frac{1}{16}$

Question 10:

Mark (✓) against the correct answer
If x% of 75 = 12, then the value of x is

(a) 8
(b) 10
(c) 12
(d) 16

Answer 10:

(c) 12

Given that x% of 75 = 12
Then, $\left(\frac{x}{100}\times 75\right)=12$
⇒ x = $\left(\frac{12\times 100}{75}\right)$ =16
Hence, the value of x is 16

Question 11:

Mark (✓) against the correct answer
A number increased by 20% gives 30. The number is

(a) 150
(b) 6
(c) 25
(d) 60

Answer 11:

(c) 25

Let the number be x. Then, we have:
120% of x  = increased number
⇒ 30 = $\left(x\times \frac{120}{100}\right)$
⇒ 30 = $\left(\frac{6x}{5}\right)$
⇒ x = $\left(30\times \frac{5}{6}\right)=25$
Hence, the required number is 25

Question 12:

Mark (✓) against the correct answer
5% of a number is 9. The number is

(a) 120
(b) 140
(c) 160
(d) 180

Answer 12:

(d) 180

Let the required number be x. Then, we have:
5% of x = 9
⇒ $\left(\frac{5}{100}\times x\right)=9$
⇒ x = $\left(9\times \frac{100}{5}\right)=\left(9\times 20\right)=180$

Question 13:

Mark (✓) against the correct answer
If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 70

Answer 13:

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

Question 14:

Mark (✓) against the correct answer
In an examination it is required to get 36% to pass. A student gets 160 marks and fails by 20 marks. The maximum marks are

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

Answer 14:

(c) 500
Let x be the maximum marks.
Pass marks = (160 + 20) = 180
∴ 36% of x = 180
⇒ $\left(\frac{36}{100}\times x\right)=180$
⇒ x = $\left(\frac{180\times 100}{36}\right)=\left(5\times 100\right)=500$
Hence, maximum marks = 500

Question 15:

Fill in the blanks.

(i) 3 : 4 = (......)%
(ii) 0.75 = (......)%
(iii) 6% = ...... (express in decimals)
(iv) If x decreased by 40% gives 135, then x = ...... .
(v) (11% of x) − (7% of x) = 18 ⇨ x = ......

Answer 15:

We have the following:

(i) 3 : 4 = (75)%
     Explanation: 3 : 4 = $\frac{3}{4}$ = $\left(\frac{3}{4}\times 100\right)\%=\left(3\times 25\right)\%=75\%$

(ii) 0.75 = (75)%
      Explanation: ( 0.75 $\times$ 100)% = 75%


(iii) 6% = 0.06 (expressed in decimals)
       Explanation: 6% = $\frac{6}{100}=0.06$

(iv) If x decreased by 40% gives 135, then x = 225
       Explanation:
       Let the number be x.
       According to question, we have:
       x − 40% of x = 135
       ⇒ $\left(x-\frac{40x}{100}\right)=135$
       ⇒ $\left(\frac{100x-40x}{100}\right)=135$
       ⇒ $\left(\frac{60x}{100}\right)=135$
        ⇒ x = $\left(\frac{135\times 100}{60}\right)$ = 225

(v) (11% of x) − (7% of x) = 18
      ⇒ x = 450
      Explanation:
       (11% of x) − (7% of x) = 18
        ⇒ $\left(\frac{11x}{100}-\frac{7x}{100}\right)=18$
        ⇒ $\frac{4x}{100}=18$
         ∴ x = $\left(\frac{18\times 100}{4}\right)=\left(18\times 25\right)=450$

Question 16:

Write 'T' for true and 'F' for false

(i) $\frac{3}{4}$ as rate per cent is 75%
(ii) $12\frac{1}{2}\%$ expressed as a fraction is $\frac{1}{8}.$
(iii) 2 : 5 = 25%
(iv) 80 % of 450 = 360.
(v) 20% of 1 litre = 200 mL.

Answer 16:

(i) True (T)
     Justification: $\left(\frac{3}{4}\times 100\right)\%$ = 75%

(ii) True (T)
     Justification: $12\frac{1}{2}\%=\frac{25}{2}\%$ = $\left(\frac{25}{2\times 100}\right)=\frac{1}{8}$

(iii) False (F)
       Justification: $\frac{2}{5}$ = $\left(\frac{2}{5}\times 100\right)$% = $\left(2\times 20\right)\%$ = 40%

(iv) True (T)
       Justification: 80% of 450 = $\left(\frac{80}{100}\times 450\right)=\left(\frac{80\times 9}{2}\right)=\left(40\times 9\right)=360$

(v) True (T)
      Justification: 20% of 1 L = 20% of 1000 mL
                                                     = $\left(\frac{20}{100}\times 1000\right)$ mL = 200 mL