Rs Aggarwal 2018 Solutions for Class 7 Math Chapter 10 Percentage are provided here with simple step-by-step explanations. These solutions for Percentage are extremely popular among Class 7 students for Math Percentage Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 7 Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 142:

#### Question 1:

Convert each of the following fraction into a percentage:

(i) $\frac{47}{100}$

(ii) $\frac{9}{20}$

(iii) $\frac{3}{8}$

(iv) $\frac{8}{125}$

(v) $\frac{19}{500}$

(vi) $\frac{4}{15}$

(vii) $\frac{2}{3}$

(viii) $1\frac{3}{5}$

#### Answer:

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

#### Page No 142:

#### Question 2:

Convert each of the following into a fraction:

(i) 32%

(ii) $6\frac{1}{4}\%$

(iii) $26\frac{2}{3}\%$

(iv) 120%

(v) 6.25%

(vi) 0.8%

(vii) 0.06%

(viii) 22.75%

#### Answer:

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

#### Page No 142:

#### Question 3:

Express each of the following as a ratio:

(i) 43%

(ii) 36%

(iii) 7.5%

(iv) 125%

#### Answer:

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

#### Page No 142:

#### Question 4:

Convert each of the following ratios into a percentage:

(i) 37 : 100

(ii) 16 : 25

(iii) 3 : 5

(iv) 5 : 4

#### Answer:

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

#### Page No 142:

#### Question 5:

Convert each of the following into decimal form:

(i) 45%

(ii) 127%

(iii) 3.6%

(iv) 0.23%

#### Answer:

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

#### Page No 142:

#### Question 6:

Convert each of the following decimals into a percentage:

(i) 0.6

(ii) 0.42

(iii) 0.07

(iv) 0.005

#### Answer:

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%

#### Page No 142:

#### Question 7:

Find:

(i) 32% of 425

(ii) $16\frac{2}{3}\%\mathrm{of}16$

(iii) 6.5% of 400

(iv) 136% of 70

(v) 2.8% of 35

(vi) 0.6% of 45

#### Answer:

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

#### Page No 142:

#### Question 8:

Find:

(i) 25% of Rs 76

(ii) 20% of Rs 132

(iii) 7.5% of 600 m

(iv) $3\frac{1}{3}\%\mathrm{of}90\mathrm{km}$

(v) 8.5% of 5 kg

(vi) 20% of 12 litres

#### Answer:

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 = $\left(5\times \frac{8.5}{100}\right)\mathrm{kg}=\left(5\times \frac{85}{1000}\right)\mathrm{kg}=0.425\mathrm{kg}=425\mathrm{g}$ [âˆµ1 kg = 1000 g]

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

#### Page No 142:

#### Question 9:

Find the number whose 13% is 65.

#### Answer:

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.

#### Page No 142:

#### Question 10:

Find the number whose $6\frac{1}{4}\%$ is 2.

#### Answer:

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.

#### Page No 143:

#### Question 11:

What amount is 10% more than Rs 90?

#### Answer:

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.

#### Page No 143:

#### Question 12:

What amount is 20% less than Rs 60?

#### Answer:

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.

#### Page No 143:

#### Question 13:

If 3% of *x* is 9, find the value of *x*.

#### Answer:

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.

#### Page No 143:

#### Question 14:

If 12.5% of *x* is 6, find the value of *x*.

#### Answer:

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.

#### Page No 143:

#### Question 15:

What per cent of 84 is 14?

#### Answer:

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.

#### Page No 143:

#### Question 16:

What percentage is

(i) Rs 15 of Rs 120?

(ii) 36 minutes of 2 hours?

(iii) 8 hours of 2 days

(iv) 160 metres of 4 km?

(v) 175 mL of 1 litre?

(vi) 25 paise of Rs 4?

#### Answer:

(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

⇒ 40*x* = 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

⇒ 10*x* = 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

⇒ 4*x* = 25

∴ *x* = $\left(\frac{25}{4}\right)\%$ = $6\frac{1}{4}\%$

Hence, $6\frac{1}{4}\%$ of Rs 4 is 25 paise.

#### Page No 147:

#### Question 1:

Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him.

#### Answer:

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.

#### Page No 147:

#### Question 2:

The monthly salary of a typist is Rs 15625. If he gets an increase of 12%, find his new salary.

#### Answer:

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.

#### Page No 147:

#### Question 3:

The excise duty on a certain item has been reduced to Rs 760 from Rs 950. Find the reduction per cent in the excise duy on that item.

#### Answer:

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%.

#### Page No 147:

#### Question 4:

96% of the cost of a TV is Rs 10464. What is its total cost?

#### Answer:

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.

#### Page No 147:

#### Question 5:

70% of the students in a school are boys and the number of girls is 504. Find the number of boys in the school.

#### Answer:

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.

#### Page No 147:

#### Question 6:

An ore contains 12% copper. How many kilograms of the ore are required to get 69 kg of copper?

#### Answer:

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.

#### Page No 147:

#### Question 7:

36% of the maximum marks are required to pass a test. A student gets 123 marks and is declared failed by 39 marks. Find the maximum marks.

#### Answer:

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

#### Page No 147:

#### Question 8:

A fruit-seller had some apples. He sells 40% of them and still has 420 apples. Find the number of apples he had originally.

#### Answer:

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.

#### Page No 147:

#### Question 9:

In an examination, 72% of the total examinees passed. If the number of failures is 392, find the total number of examinees.

#### Answer:

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.

#### Page No 147:

#### Question 10:

After decuting a commission of 5%, a moped costs Rs 15200. What is its gross value?

#### Answer:

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.

#### Page No 147:

#### Question 11:

Gunpowder contains 75% of nitre and 10% of sulphur, and the rest of it is charcoal. Find the amount of charcoal in 8 kg of gunpowder.

#### Answer:

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.

#### Page No 147:

#### Question 12:

Chalk contains 3% of carbon, 10% of calcium and 12% of oxygen. Find the amount in grams of each of these substances in 1 kg of chalk.

#### Answer:

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

#### Page No 148:

#### Question 13:

Sonal went to school for 219 days in a full year. If her attendance is 75%, find the number of days on which the school was open.

#### Answer:

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.

#### Page No 148:

#### Question 14:

3% commission on the sale of a property amounts to Rs 42660. What is the total value of the property?

#### Answer:

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.

#### Page No 148:

#### Question 15:

In an election, there were two candidates A and B. The total number of voters in this constituency was 60000 and 80% of the total votes were polied. If 60% of the polied votes were cast in favour of A, how many votes were received by B?

#### Answer:

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.

#### Page No 148:

#### Question 16:

The price of a shirt is reduced by 12% in a discount sale. If its present price is Rs 1188, find its original price.

#### Answer:

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

⇒ 88*x* = (1188 $\times $ 100)

⇒ 88*x* = 118800

∴ *x* = $\left(\frac{118800}{88}\right)$ = 1350

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

#### Page No 148:

#### Question 17:

The price of a sweater is increased by 8%. If its increased price is Rs 1566, find the original price.

#### Answer:

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

#### Page No 148:

#### Question 18:

After spending 80% of his income and giving 10% of the remainder in a charity, a man has Rs 46260 left with him. Find his income.

#### Answer:

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.

#### Page No 148:

#### Question 19:

A number is increased by 20% and the increased number is decreased by 20%. Find the net increase of decrease per cent.

#### Answer:

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%.

#### Page No 148:

#### Question 20:

The salary of an officer is increased by 20%. By what percentage should the new salary be reduced to restore the original salary?

#### Answer:

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 $16\frac{2}{3}\%$.

#### Page No 148:

#### Question 21:

A property dealer charges commission at the rate of 2% on the first Rs 200000, 1% on the next Rs 200000 and 0.5% on the remaining price. Find his commission on the property that has been sold for Rs 540000.

#### Answer:

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.

#### Page No 148:

#### Question 22:

Nikhil's income is 20% less than that of Akhil. How much per cent is Akhil's income more than that of Nikhil's?

#### Answer:

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%.

#### Page No 148:

#### Question 23:

Jhon's income is 20% more than that of Mr Thomas. How much per cent is the income of Mr Thomas less than that of John?

#### Answer:

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 $16\frac{2}{3}\%$.

#### Page No 148:

#### Question 24:

The value of a machine depreciated 10% every year. If its present value is Rs 387000, what was its value 1 year ago?

#### Answer:

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.

#### Page No 148:

#### Question 25:

The value of a car decreases annually by 20%. If the present value of the car be Rs 450000, what will be its value after 2 years?

#### Answer:

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.

#### Page No 148:

#### Question 26:

The population of a town increases 10% annually. If its present population is 60000, what will be its population after 2 years?

#### Answer:

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.

#### Page No 148:

#### Question 27:

Due to an increase in the price of sugar by 25%, by how much per cent must a householder decrease the consumption of sugar so that there is no increase in the expenditure on sugar?

#### Answer:

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%

#### Page No 148:

#### Question 1:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 7.5%

(b) 75%

(c) 0.75%

(d) none of these

#### Answer:

(b) 75%

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

#### Page No 148:

#### Question 2:

**Mark ****(âœ“)**** against the correct answer**

The ratio 2 : 5 as rate per cent is

(a) 4%

(b) 0.4%

(c) 40%

(d) 14%

#### Answer:

(c) 40%

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

#### Page No 148:

#### Question 3:

**Mark ****(âœ“)**** against the correct answer**

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

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

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

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

(d) $\frac{1}{4}$

#### Answer:

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

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

#### Page No 149:

#### Question 4:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 16

(b) 14

(c) 12

(d) 8

#### Answer:

(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

#### Page No 149:

#### Question 5:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 25%

(b) 20%

(c) 15%

(d) 10%

#### Answer:

(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 $\frac{1}{35}$

#### Page No 149:

#### Question 6:

**Mark ****(âœ“)**** against the correct answer**

What per cent of 1 day is 36 minutes?

(a) 25%

(b) 2.5%

(c) 3.6%

(d) 0.25%

#### Answer:

(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.

#### Page No 149:

#### Question 7:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 35

(b) 28

(c) 36

(d) 30

#### Answer:

(a) 35

Let the required number be *x*.

Then, *x *+ 20% of *x* = 42

⇒ $\left(x+\frac{20x}{100}\right)=42$

⇒ $\left(x+\frac{x}{5}\right)=42$

⇒ $\left(\frac{5x+x}{5}\right)=42$ [âˆµ 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.

#### Page No 149:

#### Question 8:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 80

(b) 75

(c) 85

(d) none of these

#### Answer:

(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

#### Page No 149:

#### Question 9:

**Mark ****(âœ“)**** against the correct answer**

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:

(d) 8 kg

Let *x* kg be the required amount of ore.

Then, 5% of *x* kg = 400 g = 0.4 kg [âˆµ 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.

#### Page No 149:

#### Question 10:

**Mark ****(âœ“)**** against the correct answer**

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:

(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

#### Page No 149:

#### Question 11:

**Mark ****(âœ“)**** against the correct answer**

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:

(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

#### Page No 149:

#### Question 12:

**Mark ****(âœ“)**** against the correct answer**

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:

(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.

#### Page No 149:

#### Question 13:

**Mark ****(âœ“)**** against the correct answer**

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:

(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.

#### Page No 149:

#### Question 14:

**Mark ****(âœ“)**** against the correct answer**

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:

(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

#### Page No 149:

#### Question 15:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 48

(b) 96

(c) 75

(d) 60

#### Answer:

(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

#### Page No 149:

#### Question 16:

**Mark ****(âœ“)**** against the correct answer**

60% of 450 = ?

(a) 180

(b) 210

(c) 270

(d) none of these

#### Answer:

(c) 270

60% of 450 = $\left(\frac{60}{100}\times 450\right)$

= $\left(\frac{3}{5}\times 450\right)$ = (3 $\times $ 90) = 270

#### Page No 149:

#### Question 17:

**Mark ****(âœ“)**** against the correct answer**

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:

(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

#### Page No 149:

#### Question 18:

**Mark ****(âœ“)**** against the correct answer**

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:

(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.

#### Page No 149:

#### Question 19:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 72

(b) 360

(c) 450

(d) 720

#### Answer:

(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

#### Page No 149:

#### Question 20:

**Mark ****(âœ“)**** 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) 105

#### Answer:

(a) 60

Let *x* be the number.

According to question, we have:

(35% of *x *) + 39 = *x*

⇒ $\left(\frac{35}{100}\times x\right)+39=x$

⇒ $\left(\frac{7x}{20}\right)+39=x$

⇒ $\left(x-\frac{7x}{20}\right)=39$

⇒ $\left(\frac{20x-7x}{20}\right)=39$

⇒ $\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$

∴ *x* = $\left(\frac{39\times 20}{13}\right)$ = 60

Hence, the required number is 60

#### Page No 149:

#### Question 21:

**Mark ****(âœ“)**** against the correct answer**

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:

(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

#### Page No 150:

#### Question 22:

**Mark ****(âœ“)**** against the correct answer**

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

(a) 175

(b) 200

(c) 250

(d) 225

#### Answer:

(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

#### Page No 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:

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

#### Page No 151:

#### Question 2:

(i) What per cent of 1 kg is 125 g?

(ii) What per cent of 80 m is 24 m?

#### Answer:

(i) Let *x*% of 1 kg be 125g.

Then, $\left(\frac{x}{100}\times 1\times 1000\right)\mathrm{g}=125\mathrm{g}$

⇒ 10*x* = 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.

#### Page No 151:

#### Question 3:

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

(ii) Find 15% of Rs 140

#### Answer:

(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

#### Page No 151:

#### Question 4:

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

(ii) Find 0.8% of 45.

#### Answer:

(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.

#### Page No 151:

#### Question 5:

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

#### Answer:

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 = $\left(\frac{x}{100}\times \frac{1}{x}\times 100\right)=1$

#### Page No 151:

#### 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:

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.

#### Page No 151:

#### 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:

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.

#### Page No 151:

#### Question 8:

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

#### Answer:

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

#### Page No 151:

#### Question 9:

**Mark ****(âœ“)**** 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:

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

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

#### Page No 151:

#### Question 10:

**Mark ****(âœ“)**** against the correct answer**

If *x*% of 75 = 12, then the value of *x* is

(a) 8

(b) 10

(c) 12

(d) 16

#### Answer:

(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

#### Page No 151:

#### Question 11:

**Mark ****(âœ“)**** against the correct answer**

A number increased by 20% gives 30. The number is

(a) 150

(b) 6

(c) 25

(d) 60

#### Answer:

(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

#### Page No 151:

#### Question 12:

**Mark ****(âœ“)**** against the correct answer**

5% of a number is 9. The number is

(a) 120

(b) 140

(c) 160

(d) 180

#### Answer:

(d) 180

Let the required number be *x.* Then, we have:

5% of *x* = 9

⇒ $\left(\frac{5}{100}\times x\right)=9$

⇒ *x* = $\left(9\times \frac{100}{5}\right)=\left(9\times 20\right)=180$

#### Page No 151:

#### Question 13:

**Mark ****(âœ“)**** 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:

(a) 60

Let the number be *x*.

According to question, we have:

(35% of *x *) + 39 = *x*

⇒ $\left(\frac{35}{100}\times x\right)+39=x$

⇒ $\left(\frac{7x}{20}\right)+39=x$

⇒ $\left(x-\frac{7x}{20}\right)=39$

⇒ $\left(\frac{20x-7x}{20}\right)=39$

⇒ $\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$

∴ *x* = $\left(\frac{39\times 20}{13}\right)$ = 60

Hence, the required number is 60.

#### Page No 151:

#### Question 14:

**Mark ****(âœ“)**** 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:

(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

#### Page No 151:

#### 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 â‡¨ *x* = ......

#### Answer:

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 = $\left(\frac{18\times 100}{4}\right)=\left(18\times 25\right)=450$*

#### Page No 151:

#### 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:

(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

View NCERT Solutions for all chapters of Class 7