RS Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 10 - Percentage

Question 1:

Question 2:

We have the following:

(i) $\frac{47}{100}=\left(\frac{47}{100}\times 100\right)\% = 47\%$

(ii) $\frac{9}{20}=\left(\frac{9}{20}\times 100\right)\% =\left(9\times 5\right)\%= 45\%$

(iii) $\frac{3}{8}=\left(\frac{3}{8}\times 100\right)\% =\left(\frac{3\times 25}{2}\right)\%=\left(\frac{75}{2}\right)\%=37\frac{1}{2}\%$

(iv) $\frac{8}{125}=\left(\frac{8}{125}\times 100\right)\%=\left(\frac{8\times 4}{5}\right)\%=\left(\frac{32}{5}\right)\%=6.4\%$

(v) $\frac{19}{500}=\left(\frac{19}{500}\times 100\right)\%=\left(\frac{19}{5}\right)\%=3.8\%$

(vi) $\frac{4}{15}=\left(\frac{4}{15}\times 100\right)\%=\left(\frac{4\times 20}{3}\right)\%=\left(\frac{80}{3}\right)\%=26\frac{2}{3}\%$

(vii) $\frac{2}{3}=\left(\frac{2}{3}\times 100\right)\%=\left(\frac{200}{3}\right)\%=66\frac{2}{3}\%$

(viii) $1\frac{3}{5}=\frac{8}{5}=\left(\frac{8}{5}\times 100\right)\%=(8\times 20)\%=160\%$

Question 3:

We have the following:

(i) $32\%=\left(\frac{32}{100}\right)=\frac{8}{25}$

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

(iii) $26\frac{2}{3}\%=\left(\frac{80}{3}\right)\%=\left(\frac{80}{3}\times \frac{1}{100}\right)=\left(\frac{4\times 1}{3\times 5}\right)=\frac{4}{15}$

(iv) $120\%=\left(\frac{120}{100}\right)=\frac{6}{5}=1\frac{1}{5}$

(v) $6.25\%=\left(\frac{6.25}{100}\right)=\left(\frac{625}{100\times 100}\right)=\left(\frac{25}{400}\right)=\frac{1}{16}$

(vi) $0.8\%=\left(\frac{0.8}{100}\right)=\left(\frac{8}{10\times 100}\right)=\left(\frac{8}{1000}\right)=\frac{1}{125}$

(vii) $0.06\%=\left(\frac{0.06}{100}\right)=\left(\frac{6}{100\times 100}\right)=\left(\frac{6}{10000}\right)=\frac{3}{5000}$

(viii)  $22.75\%=\left(\frac{22.75}{100}\right)=\left(\frac{2275}{100\times 100}\right)=\frac{91}{400}$

Question 4:

We have:

(i) $43\%=\frac{43}{100}=43 : 100$

(ii) $36\%=\frac{36}{100}=\frac{9}{25}=9 : 25$

(iii) $7.5\%=\left(\frac{7.5}{100}\right)=\left(\frac{75}{10\times 100}\right)=\frac{3}{40}=3 : 40$

(iv) $125\%=\frac{125}{100}=\frac{5}{4}=5 : 4$

Question 5:

We have the following:

(i) 37 : 100 = $\frac{37}{100}=\left(\frac{37}{100}\times 100\right)\%=37\%$

(ii) 16 : 25 = $\frac{16}{25}=\left(\frac{16}{25}\times 100\right)\%=\left(16\times 4\right)\%=64\%$

(iii) 3 : 5 =$\frac{3}{5}=\left(\frac{3}{5}\times 100\right)\%=\left(3\times 20\right)\%=60\%$

(iv) 5 : 4 = $\frac{5}{4}=\left(\frac{5}{4}\times 100\right)\%=\left(5\times 25\right)\%=125\%$

Question 6:

We have the following:

(i) 45% = $\left(\frac{45}{100}\right)=0.45$

(ii) 127% = $\left(\frac{127}{100}\right)=1.27$

(iii) 3.6% =$\left(\frac{3.6}{100}\right)=\left(\frac{36}{10\times 100}\right)=\frac{36}{1000}=0.036$

(iv) 0.23% =$\left(\frac{0.23}{100}\right)=\left(\frac{23}{100\times 100}\right)=\frac{23}{10000}=0.0023$

Question 7:

We have:

(i) 0.6 = (0.6 $\times$ 100)% = 60%
(ii) 0.42 = (0.42 $\times$ 100)% = 42%
(iii) 0.07 = (0.07 $\times$ 100)% = 7%
(iv) 0.005 = (0.005 $\times$ 100)% = 0.5%

Question 8:

We have:
 (i) 32% of 425 = $\left(\frac{32}{100}\times 425\right)=\left(\frac{32\times 17}{4}\right)=\left(8\times 17\right)=136$

(ii) $16\frac{2}{3}\%$ of 16 = $\frac{50}{3}\%$ of 16 = $\left(\frac{50}{3\times 100}\times 16\right)=\left(\frac{1}{6}\times 16\right)=\frac{8}{3}=2\frac{2}{3}$

(iii) 6.5% of 400 = $\left(\frac{6.5}{100}\times 400\right)=\left(\frac{65}{10\times 100}\times 400\right)=\left(\frac{65\times 4}{10}\right)=\frac{260}{10}=26$

(iv) 136% of 70 = $\left(\frac{136}{100}\times 70\right)=\left(\frac{136\times 7}{10}\right)=\left(\frac{952}{10}\right)=95.2$

(v) 2.8% of 35 = $\left(\frac{2.8}{100}\times 35\right)=\left(\frac{28}{10\times 100}\times 35\right)=\left(\frac{14\times 7}{100}\right)=\frac{98}{100}=0.98$

(vi) 0.6% of 45  = $\left(\frac{0.6}{100}\times 45\right)=\left(\frac{6}{10\times 100}\times 45\right)=\left(\frac{3\times 45}{5\times 100}\right)=\left(\frac{3\times 9}{100}\right)=\frac{27}{100}=0.27$

Question 9:

We have the following:

(i) 25% of Rs 76 = Rs $\left(76\times \frac{25}{100}\right)$ = Rs $\left(76\times \frac{1}{4}\right)$ = Rs 19

(ii) 20% of Rs 132 = Rs $\left(132\times \frac{20}{100}\right)$= Rs $\left(132\times \frac{1}{5}\right)$ = Rs 26.4

(iii) 7.5% of 600 m = $\left(600\times \frac{7.5}{100}\right) \mathrm{m}=\left(6\times 7.5\right) \mathrm{m}=45 \mathrm{m}$

(iv) $3\frac{1}{3}\%$ of 90 km = $\frac{10}{3}\%$of 90 km =$\left(90\times \frac{10}{3\times 100}\right) \mathrm{km}=\left(90\times \frac{1}{30}\right) \mathrm{km}=3 \mathrm{km}$

(v) 8.5% of 5 kg = $\left(5\times \frac{8.5}{100}\right) \mathrm{kg}=\left(5\times \frac{85}{1000}\right) \mathrm{kg}=0.425 \mathrm{kg}=425 \mathrm{g}$         [∵1 kg = 1000 g]

(vi) 20% of 12 L = $\left(12\times \frac{20}{100}\right) \mathrm{L}=\left(12\times \frac{1}{5}\right) \mathrm{L}=2.4 \mathrm{L}$

Question 10:

Let x be the required number.

Then, 13% of x = 65  
⇒ $\left(\frac{13}{100}\times x\right)=65$
⇒ x = $\left(65\times \frac{100}{13}\right)=500$
Hence, the required number is 500.

Question 11:

Let x be the required number.
Then, $6\frac{1}{4}\%$ of x = 2
⇒ $\left(6\frac{1}{4}\%\times x\right)=2$
⇒ $\left(\frac{25}{400}\times x\right)=2$
⇒  x = $\left(2\times \frac{400}{25}\right)=32$
Hence, the required number is 32.

Question 12:

10% of Rs 90 = Rs $\left(\frac{10}{100}\times 90\right)$ = Rs 9
∴ Amount that is 10% more than Rs 90 = Rs (90 + 9) = Rs 99

Hence, the required amount is Rs 99.

Question 13:

20% of Rs 60 = Rs $\left(60\times \frac{20}{100}\right)$= Rs 12
∴ Amount that is 20% less than Rs 60 = Rs (60 − 12) = Rs 48

Hence, the required amount is Rs 48.

Question 14:

3% of x = 9
⇒ $\left(\frac{3}{100}\times x\right)=9$
⇒ x = $\left(9\times \frac{100}{3}\right)=300$
Hence, the value of x is 300.

Question 15:

12.5% of x = 6
⇒ $\left(\frac{12.5}{100}\times x\right)=6$
⇒ x = $\left(6\times \frac{100}{12.5}\right)=\left(6\times 8\right)=48$
Hence, the value of x is 48.

Question 16:

Let x% of 84 be 14.
Then, $\left(\frac{x}{100}\times 84\right)=14$
⇒ $\frac{21x}{25}=14$
⇒ x = $\left(14\times \frac{25}{21}\right)=\left(\frac{2\times 25}{3}\right)=\frac{50}{3}=16\frac{2}{3}\%$
Hence, $16\frac{2}{3}\%$ of 84 is 14.

Question 1:

(i) Let x% of Rs 120 be Rs 15.
Then, Rs $\left(\frac{x}{100}\times 120\right)$ = Rs 15  
    ⇒ $\left(\frac{6x}{5}\right)$ = 15
    ∴ x = $\left(\frac{15\times 5}{6}\right)\%$ = $\left(\frac{25}{2}\right)\%$ = 12.5%
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min.
Then, $\left(\frac{x}{100}\times 2\times 60\right)$ min = 36 min
      ⇒ $\left(\frac{120x}{100}\right)$ = 36
     ∴ x = $\left(\frac{36\times 100}{120}\right)\%$ = 30%
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h.
Then, $\left(\frac{x}{100}\times 2\times 24\right)$ h = 8 h
     ⇒ $\left(\frac{48x}{100}\right)$ = 8
     ∴ x = $\left(\frac{8\times 100}{48}\right)\%$ = $16\frac{2}{3}\%$
     Hence, $16\frac{2}{3}\%$ of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m.
Then, $\left(\frac{x}{100}\times 4\times 1000\right)$ m = 160 m
     ⇒ 40x = 160
     ∴ x = $\left(\frac{160}{40}\right)\%$ = 4%
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL.
Then, $\left(\frac{x}{100}\times 1\times 1000\right)$ mL = 175 mL
     ⇒ 10x  = 175
     ∴ x = $\left(\frac{175}{10}\right)\%$ = 17.5%
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise.
Then, $\left(\frac{x}{100}\times 4\times 100\right)$ paise = 25 paise
     ⇒ 4x  = 25
     ∴ x = $\left(\frac{25}{4}\right)\%$ = $6\frac{1}{4}\%$
     Hence, $6\frac{1}{4}\%$ of Rs 4 is 25 paise.

Question 2:

Maximum marks of the examination = 750
Marks secured by Rupesh = 495
Percentage of marks secured = $\left(\frac{495}{750}\times 100\right)\%$ = 66%

Hence, Rupesh scored 66% in the examination.

Question 3:

Total monthly salary = Rs 15625
Increase percentage = 12%
∴ Amount increase = 12% of Rs 15625
                                 = Rs $\left(15625\times \frac{12}{100}\right)$ = Rs 1875
∴ New salary = Rs 15625 + Rs 1875
                       = Rs 17500
Hence, the new salary of the typist is Rs 17,500.

Question 4:

Original excise duty on the item = Rs 950
Amount reduced on excise duty = Rs (950 − 760) = Rs 190
∴ Reduction percent = $\left(\frac{\mathrm{Reduction} \mathrm{amount}}{\mathrm{Original} \mathrm{value}}\times 100\right)$
                                  = $\left(\frac{190}{950}\times 100\right)$ = 20
Hence, the excise duty on that item is reduced by 20%.

Question 5:

Let Rs x be the total cost of the TV set.

Now, 96% of the total cost of TV = Rs 10464
⇒ 96% of Rs x = Rs 10464
⇒ $\left(\frac{96}{100}\times x\right)$ = 10464
∴ x = $\left(\frac{10464\times 100}{96}\right)$ =  10900
Hence, the total cost of the TV set is Rs 10900.

Question 6:

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

Now, total number of students when the number of girls is 30 = 100
Then, total number of students when the number of girls is 504 = $\left(\frac{100}{30}\times 504\right)$ = 1680
∴ Number of boys = (1680 − 504) = 1176

Hence, there are 1176 boys in the school.

Question 7:

Let x kg be the amount of the required ore.

Then, 12% of x kg = 69 kg
⇒ $\left(\frac{12}{100}\times x\right)$ kg = 69 kg
⇒ x = $\left(\frac{69\times 100}{12}\right)$ kg = 575 kg

Hence, 575 kg of ore is required to get 69 kg of copper.

Question 8:

Let x be the maximum marks.
Pass marks = (123 + 39) = 162
Then, 36% of x = 162
⇒ $\left(\frac{36}{100}\times x\right)=162$
⇒ x = $\left(\frac{162\times 100}{36}\right)$ = 450
∴ Maximum marks = 450

Question 9:

Suppose that the fruit seller initially had 100 apples.
Apples sold = 40
∴ Remaining apples = (100 − 40) = 60

Initial amount of apples if 60 of them are remaining = 100
Initial amount of apples if 1 of them is remaining = $\left(\frac{100}{60}\right)$
Initial amount of apples if 420 of them are remaining = $\left(\frac{100}{60}\times 420\right)$ = 700
Hence, the fruit seller originally had 700 apples.

Question 10:

Suppose that 100 candidates took the examination.
Number of passed candidates = 72
Number of failed candidates = (100 − 72) = 28

Total number of candidates if 28 of them failed = 100
Total number of candidates if 392 of them failed = $\left(\frac{100}{28}\times 392\right)$ = 1400
Hence, the total number of examinees is 1400.

Question 11:

Suppose that the gross value of the moped is Rs x.
Commission on the moped = 5%
Price of moped after deducting the commission = Rs ( x − 5% of x)
                                                                              = Rs $\left(x-\frac{5x}{100}\right)$ = Rs $\left(\frac{100x-5x}{100}\right)$ = Rs $\left(\frac{95x}{100}\right)$
Now, price of the moped after deducting the commission = Rs 15200
Then, Rs $\left(\frac{95x}{100}\right)$= Rs 15200
∴ x  = Rs $\left(\frac{15200\times 100}{95}\right)$ = Rs (160 $\times$ 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

Question 12:

Total quantity of gunpowder = 8 kg = 8000 g                         (1 kg = 1000 g)
Quantity of nitre in it = 75% of 8000 g
                                   = $\left(\frac{75}{100}\times 8000\right)$ g = 6000 g = 6 kg

Quantity of sulphur in it = 10% of 8000 g
                                         = $\left(\frac{10}{100}\times 8000\right)$ g = 800 g = 0.8 kg
∴ Quantity of charcoal in it = {8000 − (6000 + 800)} g
                                              = (8000 − 6800) g
                                              = 1200 g = 1.2 kg

Hence, the amount of charcoal in 8 kg of gunpowder is 1.2 kg.

Question 13:

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

Quantity of carbon in it = 3% of 1000 g
                                       =$\left(\frac{3}{100}\times 1000\right)$ = 30 g
Quantity of calcium in it = 10% of 1000 g
                                          = $\left(\frac{10}{100}\times 1000\right)$ g = 100 g
Quantity of oxygen in it = 12% of 1000 g
                                        = $\left(\frac{12}{100}\times 1000\right)$ g = 120 g

Question 14:

Let x be the total number of days on which the school was open.
Number of days when Sonal went to school = 219
Percentage of attendance = 75

Thus, 75% of x = 219
⇒ $\left(\frac{75}{100}\times x\right)=219$
∴ x = $\left(\frac{219\times 100}{75}\right)=292$ days
Hence, the school was open for a total of 292 days.

Question 15:

Let the total value of the property be Rs x.
Percentage of commission = 3
Amount of commission  = Rs 42660
Thus, 3% of Rs x = Rs 42660
⇒ $\left(\frac{3}{100}\times x\right)$ = 42660
∴ x = $\left(\frac{42660\times 100}{3}\right)=1422000$
Hence, the total value of the property is Rs 14,22,000.

Question 16:

Total number of eligible voters = 60000
Number of voters who gave their votes = 80% of 60000
                                                                 = $\left(\frac{80}{100}\times 60000\right)$ = 48000
Number of votes in favour of candidate A = 60% of 48000
                                                                    = $\left(\frac{60}{100}\times 48000\right)$ = 28800
∴ Number of votes received by candidate B = (48000 − 28800) = 19200

Hence, candidate B recieved 19,200 votes.

Question 17:

Let us assume that the original price of the shirt is Rs x.
Discount on the shirt = 12%
So, value of discount on the shirt = 12% of Rs x
                                                      = Rs $\left(\frac{12}{100}\times x\right)$ = Rs $\left(\frac{12x}{100}\right)$
Value of the shirt after discount = Rs $\left(x-\frac{12x}{100}\right)$
                                                      = Rs $\left(\frac{100x-12x}{100}\right)$ = Rs $\left(\frac{88x}{100}\right)$
Present price of the shirt = Rs 1188
Then, Rs $\left(\frac{88x}{100}\right)$ = Rs 1188
     ⇒ 88x = (1188 $\times$ 100)
     ⇒ 88x = 118800
∴  x = $\left(\frac{118800}{88}\right)$ = 1350

Hence, the original price of the shirt is Rs 1350.

Question 18:

Let us assume that the original price of the sweater is Rs. x
Increased percentage = 8%
So, value of increase on the sweater = 8% of Rs x
                                                             = Rs $\left(\frac{8}{100}\times x\right)$ = Rs $\left(\frac{2x}{25}\right)$
Increased price of the sweater = Rs $\left(x+\frac{2x}{25}\right)$
                                                  = Rs $\left(\frac{25x+2x}{25}\right)$ = Rs $\left(\frac{27x}{25}\right)$
However, increased price of the sweater = Rs 1566
Then, Rs $\left(\frac{27x}{25}\right)$ = Rs 1566
∴  x =  $\left(\frac{1566\times 25}{27}\right)$ = 1450
Hence, the original price of the sweater is Rs 1450

Question 19:

Let the income of the man be Rs x.
Then, income spent = 80% of Rs. x
                                = Rs $\left(\frac{80}{100}\times x\right)$ = Rs $\left(\frac{80x}{100}\right)$ = Rs $\left(\frac{4x}{5}\right)$
Amount left after all the expenditure = Rs $\left(x-\frac{4x}{5}\right)$ = Rs $\left(\frac{5x-4x}{5}\right)$ = Rs $\left(\frac{x}{5}\right)$
Amount given to the charity = 10% of Rs $\left(\frac{x}{5}\right)$
                                       = Rs $\left(\frac{10}{100}\times \frac{x}{5}\right)$ = Rs $\left(\frac{10x}{500}\right)$= Rs $\left(\frac{x}{50}\right)$
Amount left after the charity = Rs $\left(\frac{x}{5}-\frac{x}{50}\right)$
                                                = Rs $\left(\frac{10x-x}{50}\right)$ = Rs $\left(\frac{9x}{50}\right)$
Now, we have:
Rs $\left(\frac{9x}{50}\right)$ = Rs 46260
∴ x = Rs $\left(\frac{46260\times 50}{9}\right)$ = Rs 257000
Hence, the income of the man is Rs 2,57,000.

Question 20:

Let the number be 100.
Increase in the number = 20%
Increased number = (100 + 20) =120
Now, decrease in the number = (20% of 120)
                                                = $\left(\frac{20}{100}\times 120\right)=24$
New number = (120 − 24) = 96
Net decrease = (100 − 96) = 4
Net decrease percentage = $\left(\frac{4}{100}\times 100\right)$ = 4
Hence, the net decrease is 4%.

Question 21:

Let the original salary be Rs 100.
Increase in it = 20%
Salary after increment = Rs (100 + 20) = Rs 120
To restore the original salary, reduction required = Rs (120 − 100) = Rs 20
Reduction on Rs 120 = Rs 20
∴ Reduction percentage = $\left(\frac{20}{120}\times 100\right)$ = $\left(\frac{100}{6}\right)$ = $16\frac{2}{3}$
Hence, the required reduction on the new salary is $16\frac{2}{3}\%$.

Question 22:

Total cost of the property = Rs 540000
Commission on the first Rs 200000 = 2% of Rs 200000
                                                            = $\left(\frac{2}{100}\times 200000\right)$ = Rs 4000

Commission on the next Rs 200000 = 1% of Rs 200000
                                                           = $\left(\frac{1}{100}\times 200000\right)$ = Rs 2000
Remaining amount = Rs (540000 − 400000) = Rs 140000
∴ Commission on Rs 140000 = 0.5% of Rs 140000
                                                  = Rs $\left(\frac{0.5}{100}\times 140000\right)$
                                                  = Rs $\left(\frac{5}{1000}\times 140000\right)$ = Rs 700
Thus, total commission on the property worth Rs 540000 = Rs (4000 + 2000 + 700)
                                                                                             = Rs 6700
Hence, the commission of the property dealer on the property that has been sold for Rs 540000 is Rs 6700.

Question 23:

Let Akhil's income be Rs 100.
∴ Nikhil's income = Rs 80
Akhil's income when Nikhil's income is Rs 80 = Rs 100
Akhil's income when Nikhil's income is Rs 100 = Rs $\left(\frac{100}{80}\times 100\right)$ = Rs 125
i.e., if Nikhil's income is Rs.100, then Akhil's income is Rs 125.
Hence, Akhil's income is more than that of Nikhil's by 25%.

Question 24:

Let Rs 100 be the income of Mr. Thomas.
∴ John's income = Rs 120
Mr. Thomas' income when John's income is Rs 120 = Rs 100
Mr. Thomas' income when John's income is Rs 100 = Rs $\left(\frac{100}{120}\times 100\right)$ = Rs $83\frac{1}{3}$
Hence, Mr Thomas' income is less than that of John's by $16\frac{2}{3}\%$.

Question 25:

Let Rs x be the value of the machine one year ago.

Then, its present value = 90% of Rs x
                                     = Rs $\left(\frac{90}{100}\times x\right)$ = Rs $\left(\frac{9x}{10}\right)$
It is given that present value of the machine = Rs 387000
⇒ x = Rs$\left(\frac{387000\times 10}{9}\right)$= Rs $\left(43000\times 10\right)$= Rs 430000

Hence, the value of the machine a year ago was Rs 430000.

Question 26:

The present value of  the car = Rs 450000
The decrease in its value after the first year = 20% of Rs 450000
                                                                         = Rs $\left(\frac{20}{100}\times 450000\right)$= Rs 90000
The depreciated value of the car after the first year = Rs (450000 − 90000) = Rs 360000
The decrease in its value after the second year = 20% of Rs 360000
                                                                            = Rs $\left(\frac{20}{100}\times 360000\right)$ = Rs 72000
The depreciated value of the car after the second year = Rs (360000 − 72000) = Rs 288000

Hence, the value of the car after two years will be Rs 288000.

Question 27:

Present population of the town = 60000
Increase in population of the town after the 1 year = 10% of 60000
                                                                              = $\left(\frac{10}{100}\times 60000\right)$ = 6000
Thus, population of the town after 1 year = 60000 + 6000 = 66000
Increase in population after 2 years = 10% of 66000
                                                           = $\left(\frac{10}{100}\times 66000\right)$ = 6600
Thus, population after the second year = 66000 + 6600 = 72600
Hence, the population of the town after 2 years will be 72600.

Question 1:

Let the consumption of sugar originally be 1 unit and let its cost be Rs 100
New cost of 1 unit of sugar = Rs 125
Now, Rs 125 yield 1 unit of sugar.
∴ Rs 100 will yield $\left(\frac{1}{125}\times 100\right)$ unit = $\left(\frac{4}{5}\right)$ unit of sugar.
Reduction in consumption = $\left(1-\frac{4}{5}\right)$ = $\left(\frac{1}{5}\right)$ unit
∴ Reduction percent in consumption = $\left(\frac{1}{5}\times \frac{1}{1}\times 100\right)$ %= $\left(\frac{100}{5}\right)$ %= 20%

Question 2:

(b) 75%

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

Question 3:

(c) 40%

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

Question 4:

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

$8\frac{1}{3}\%$ = $\frac{25}{3}\%$ = $\left(\frac{25}{3}\times \frac{1}{100}\right)=\left(\frac{1}{3\times 4}\right)=\frac{1}{12}$

Question 5:

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

Question 6:

(d) 10%

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

Question 7:

(b) 2.5%

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

Hence, 2.5% of 1 day is 36 min.

Question 8:

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

Question 9:

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

Question 10:

(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]
⇒ $\left(\frac{5}{100}\times x\right)=0.4$
⇒ x = $\left(\frac{0.4\times 100}{5}\right)$ = 8
Hence, 8 kg of ore is required to obtain 400 g of copper.

Question 11:

(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 $\left(x-\frac{10}{100}x\right)$ = Rs $\left(\frac{100x-10x}{100}\right)$ = Rs $\left(\frac{9x}{10}\right)$
However, price of the TV after deducting the commission = Rs 18000
Then, Rs $\left(\frac{9x}{10}\right)$ = Rs 18000
∴ x = $\left(\frac{18000\times 10}{9}\right)$ = Rs (2000 $\times$ 10) = Rs 20000
Hence, the gross value of the TV is Rs 20,000

Question 12:

(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 $\left(\frac{25}{100}\times x\right)$ = Rs $\left(\frac{x}{4}\right)$
Salary after increment= Rs $\left(x+\frac{x}{4}\right)$ = Rs $\left(\frac{5x}{4}\right)$
However, increased salary = Rs 20000
Then, Rs $\left(\frac{5x}{4}\right)$ = Rs 20000
∴  x = Rs $\left(\frac{20000\times 4}{5}\right)$ = Rs 16000
Hence, the original salary of the man is Rs 16,000

Question 13:

(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 = $\left(\frac{100}{5}\times 28\right)=\left(20\times 28\right)=560$
Hence, there were 560 examinees.

Question 14:

(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 = $\left(\frac{100}{60}\times 420\right)$ = 700
Hence, the fruit seller originally had 700 apples with him.

Question 15:

(c) Rs. 25250

Present value of the machine = Rs 25000
Decrease in its value after 1 year = 10% of Rs 25000
                                                      = Rs $\left(\frac{10}{100}\times 25000\right)$ = 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 16:

(c) 75

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

Question 17:

(c) 270
60% of 450 = $\left(\frac{60}{100}\times 450\right)$
                    = $\left(\frac{3}{5}\times 450\right)$ = (3 $\times$ 90) = 270

Question 18:

(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 $\left(\frac{6}{100}\times x\right)$ = Rs $\left(\frac{3x}{50}\right)$
Reduced price of the chair = Rs $\left(x-\frac{3x}{50}\right)$
                                            = Rs $\left(\frac{50x-3x}{50}\right)$ = Rs $\left(\frac{47x}{50}\right)$
However, present price of the chair = Rs 658
Then, Rs $\left(\frac{47x}{50}\right)$ = Rs 658
⇒ Rs $\left(\frac{47x}{50}\right)$ = Rs 658
⇒ x = Rs $\left(\frac{658\times 50}{47}\right)$ = Rs $\left(14\times 50\right)=700$
Hence, the original price of the chair is Rs 700

Question 19:

(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 = $\left(\frac{100}{30}\times 240\right)=800$
∴ Number of boys = (800 − 240) = 560

Hence, there are 560 boys in the school.

Question 20:

(c) 450

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

Question 21:

(a) 60
Let x be the number.
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 22:

(c) 500
Let x be the maximum marks.
Pass marks = (145 + 35) = 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 1:

(d) 225
Let x be the number.
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
Hence, the required number is 225

Question 2:

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 3:

(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 4:

(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