The NCERT solutions for Class 7 Maths Chapter 8 are easy to solve problems with the help of the solutions available. These solutions contain various problems related to the field of mathematics as well. The different ratios to be calculated is one example of the different problems that needs to be calculated. Some of the various types of solutions are for problems containing types of different problems such as simple calculation of different quantities. It is easy to find the NCERT solutions for class 7 maths, to download the PDF. Once you download the PDF you are able to download the entire content related to the different solutions provided. Some of the various problems that are available have their solutions within the PDF. With the help of NCERT Solutions for Class 7 one can easily figure out the NCERT Solutions for Class 7 maths chapter 8.

### NCERT Solutions For Class 7 Maths Chapter 8 Exercises

- NCERT Solutions For Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.1
- NCERT Solutions For Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.2
- NCERT Solutions For Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3

*EXERCISE- 8.1:*

*QUESTION 1:*

*Calculate the ratio:*

* (i) Rs10 to 50 paise (ii) 10kg to 500g*

* (iii) 60m to 20cm (iv) 20days to 50 hours*

*Solution; *

*(i) Rs.10 to 50 paise*

=> 10 x 100 = 1000 paise (Since , Rs.1 = 100 paise)

Thus, the ratio = 1000/50 = 20 : 1

*(ii) 10kg to 500g*

10 x 1000= 10000g (Since 1kg is 1000g)

Thus, the ratio = 10000/500 = 20 : 1

*(iii) 60m to 20cm*

60 x 100 = 6000cm (Since 1m = 100cm)

Thus, the ratio =6000/20 = 300 : 1

*(iv) 20days to 50 hours*

20 x 24 = 480 hours

Thus, the ratio = 480/50 = 48 : 5

*QUESTION 2:*

*A computer lab has 2 computers for every 4 students. For 20 students how many computers will be required?*

*Solution;*

Given,

4 students = 2 computers

1 student requires = 2/4 computers.

Therefore, 20 students require = (2/4) x 20 = 10 computers.

*QUESTION 3:*

*Mizoram has a population of 10 lakh and Sikkim has a population of 6 lakh. Area of Mizoram is 21,000 km ^{2} and the area of Sikkim is 7000 km^{2}.*

* (i) Find the number of people per km ^{2} in both the states.*

* (ii) Which state has a lesser population?*

*Solution;*

** (i)** People per km

^{2}= Population/ Area

In Mizoram = 1000000/21000 = 47.61 ≃48 people/km^{2}

In Sikkim = 600000/7000 = 85.71 ≃ 86 people/km^{2}

** (ii)** Sikkim has a lesser population.

* *

*EXERCISE- 8.2*

*QUESTION 1:*

*Convert the given fractional numbers to percent:*

*(a) 1/3 (b)6/5 (c)4/25 (d)3/7*

*Solution;*

** (a)** 1/3 = (1/3) x 100% = 100/3 = 33.33%

** (b)** 6/5 = (6/5) x 100% = 6 x 20 = 120%

** (c)** 4/25 = (4/25) x 100% = 16%

** (d)** 2/7 = (2/7) x 100% = 28.57%

* *

*QUESTION 2:*

*Express the given decimal numbers in percentages.*

*(a) 0.31 (b)0.03 (c)19.62 (d)0.007*

*Solutions;*

** (a)** 0.31 = (31/100) x 100% = 31%

** (b)** 0.03 = (3/100) x 100% = 3%

** (c)** 19.62 = (1962/100) x 100% = 1962%

** (d)** 0.007 = (7/1000) x 100% = 0.7%

* QUESTION 3:*

*Find what part of the following figures is colored; also find the percentage of the colored part.*

*Solutions;*

** (i)** Colored part = 1/4

Therefore, percentage of the colored portion = (1/4) x 100% = 25%

** (ii)** Colored portion = 3/5

Therefore, percentage of the colored portion = (3/5) x 100 =60%

** (iii)** Colored portion = (3/8) x 100% = 37.5%

*QUESTION 4:*

** Calculate** :

*(a) 15% of 450 (b) 2% of 2 hour*

* (c) 10% of ₹2600 (d) 50% of 5 kg*

*Solutions;*

** (a)** 15% of 450 = (15/100) x 450 = 67.5

** (b)** 2% of 2 hours= 2% of (2×3600) secs (Since 1 hour = 60 x 60 secs)

= (2/100) x 7200 =144secs.

** (c)** 10% of Rs.2600 = (10/100) x 2600

=Rs.260

** (d)** 50% of 5kg = 50% of (5000g) (Since 1kg = 1000g)

= (50/100) x 5000

= 2500 g or 2.5kg

* *

* QUESTION 5:*

*Calculate the total quantity if:*

* (a) 5% of it is 100 *

*(b) 10% of it is ₹1060*

*(c) 20% of it is 800 km*

*(d) 90% of it is 18 minutes*

*(e) 9% of it is 60 liters*

*Solutions;*

let the total quantity be Y in all above cases,

** (a)**5% of Y = 100

=> (5/100) x Y =100

=> Y=(100×100)/5 =2000

** (b)**10% of Y = 1060

=> (10/100) x Y = 1060

=> Y = (1060 x 100)/10 = Rs.10600

** (c)**20% of Y = 800

=> (20/100) x Y = 800

=> Y = (800 x 100)/20 = 4000km

** (d)** 90% of Y = 18min

=> (90/100) x Y = 18

=> Y=(18 x 100)/90 =20mins.

** (e)** 9% of Y = 60

=> (9/100) x Y = 60

=> Y = (60 x 100)/9 =666.66liters.

*QUESTION 6:*

*Convert the following percentages to fractions , simplest fractions and decimal numbers:*

*(a)20% (b)25% (c)5% (d)120%*

*Solution;*

Percentage |
Fractions |
Simplest Fraction |
Decimal Numbers. |

20% | 20/100 | 1/5 | 0.2 |

25% | 25/100 | 1/4 | 0.25 |

5% | 5/100 | 1/20 | 0.05 |

120% | 120/100 | 6/5 | 1.2 |

*QUESTION 7: *

*12000 voters constituted a constituency, out of which 80% voted. What percentage and number of voters did not vote?*

*Solution;*

Given,

Total no. of voters = 12000

Percentage who voted = 80%

Thus, percentage who did not vote = 100 -80 = 20%

Therefore, the number of people who did not vote = 20% of 12000

= (20/100) x 12000 = 2400

*QUESTION 8: *

*Sanjay saves Rs.5000 every month, if this is 20% of his salary. What is his monthly salary?*

*Solution;*

Let Sanjay’s salary be Y

Now, 20% of Y = 5000

(20/100) x Y = 5000

Y = Rs.25000

Therefore,Sanjay’s salary is Rs.25000

*QUESTION 9: *

*A local**dota team played 10 matches in a season. If they won 40% of their matches, in how many matches do they win?*

*Solution;*

Given,

Number of matches played by the dota2 team =10

Percentage of matches won = 40%

Therefore, total matches they were victorious in = 40% of 10

= (40/100) x 10

=4.

*EXERCISE- 8.3*

** QUESTION 1**:

*What is the profit and theloss for the following transaction.Also find what is the profit percentage and the loss percentage in the cases given:*

*(i) Gardening shears were bought for Rs 250 and were sold for Rs 325*

*(ii) A refrigerator was bought for Rs 12,000 and sold for Rs 13,500*

*(iii) A cupboard was bought for Rs 2,500 and sold for Rs 3,000.*

*(iv) A shirt was bought for Rs 250 and sold for Rs 150*

*Solution;*

** (i)**the cost price of the gardening shears: Rs 250

The selling price of the gardening shears: Rs 325

Since, SP> CP therefore here it is a profit.

Therefore profit = SP – CP = Rs 325 – Rs 250

= Rs 75

Now profit % = \(\frac{profit}{C.P}\times 100\)

= \(\frac{75}{250}\times 100=\) %

Therefore, Profit = Rs 75 and Profit % = 30%

** (ii)** Cost prince of refrigerator = Rs 12,000

Selling price of Refrigerator = Rs 13,500

Since, S.P > C.P Therefore, it is a profit.

Therefore, Profit = SP – CP = Rs 13500 – Rs 12000 = Rs 1500

Now profit % = \(\frac{profit}{C.P}\times 100\)

= \(\frac{1500}{12000}\times 100\) = 12.5%

Therefore, Profit = Rs 1,500 and Profit % = 12.5 %

** (iii)** Cost price of cupboard = Rs 2500

Selling price of cupboard = Rs 3,000

Since, SP> CP, Therefore it is a profit.

Profit = SP – CP = Rs 3,000 – Rs 2,500 = Rs 500

\(\frac{1500}{12000}\times 100\)

= \(\frac{500}{2500}\times 100\) =20%

Therefore, Profit = 500 and profit % = 20%

** (iv)** Cost price of shirt = Rs 250

Selling price of shirt = Rs 150

Since CP > SP, therefore here it is a loss.

Therefore Loss = CP – SP = Rs 250 – Rs 150

= Rs 100

Now loss % = \(\frac{Loss}{C.P}\times 100\)

= \(\frac{100}{250}\times 100\) = 40%

Therefore , Loss = Rs 100 and Loss% = 40%

*QUESTION 2: *

*Convert the following ratio to its percentages:*

*(i) 3 : 1*

*(ii) 2 : 3 : 5*

*(iii) 1 : 4*

*(iv) 1 : 2 : 5*

*Solution;*

** (i)**3 : 1

Total part = 3 + 1 = 4

Therefore, Fractional part = \(\frac{3}{4}:\frac{1}{4}\)

=> Percentage of parts = \(\frac{3}{4}\times 100:\frac{1}{4}\times 100\)

=> Percentage of parts = 75 % : 25 %

** (ii)**2 : 3 : 5

Total part = 2 + 3 + 5 = 10

Therefore, Fractional part = \(\frac{2}{10}:\frac{3}{10}:\frac{5}{10}\)

=> Percentage of parts = \(\frac{2}{10}\times 100:\frac{3}{10}\times 100:\frac{5}{10}\times 100\)

=> Percentage of parts = 20 % : 30 % : 50 %

** (iii)**1 : 4

Total part = 1 + 4 = 5

Therefore, Fractional part = \(\frac{1}{5}:\frac{4}{5}\)

=> Percentage of parts = \(\frac{1}{5}\times 100:\frac{4}{5}\times 100\)

=> Percentage of parts = 20% : 80%

** (iv)**1 : 2 : 5

Total part = 1 + 2 + 5 = 8

Therefore, Fractional part = \(\frac{1}{8}:\frac{2}{8}:\frac{5}{8}\)

=> Percentage of parts = \(\frac{1}{8}\times 100:\frac{2}{8}\times 100:\frac{5}{8}\times 100\)

=> Percentage of parts = 12.5 % : 25 % : 62.5 %

*QUESTION 3: *

*The population of a city decreased from 25,000 to 24,500. Find the decrease in percentage.*

*Solution;*

The decrease in population of a city is from 25,000 to 24,500.

Population Decreased = 25000 – 24500 = 500

Decreased percentage = \(\frac{Population Decreased}{Original Population}\times 100\)

= \(\frac{500}{25000}\times 100\) = 2 %

Hence, the percentage of decrease in the population is 2 %.

** QUESTION 4**:

*Reah and Reuben** bought a caravan for Rs. 3,50,000. The next year, the price raised to Rs 3,70,000. What is the percentage of price increased?*

*Solution;*

Increased in price of the caravan from Rs 3, 50,000 to 3, 70,000

Amount change = Rs 3, 70,000 – Rs 3, 50,000 = Rs 20,000

Therefore, Increased percentage = \(\frac{Amount of change}{Original Amount}\times 100\)

\(\frac{20000}{350000}\times 100 = 5\frac{5}{7}\) %

Hence, The percentage of price increased is \(5\frac{5}{7}\) %

*QUESTION 5: *

*Harismitha** bought a T.V for Rs 10,000 and sold it at a profit of 20%. How much money did she get for it?*

*Solution;*

The cost price of T.V = Rs 10000

Profit percent = 20%

Now, Profit = Profit % of CP

= \(\frac{20}{100}\times 10000\)

= Rs 2000

Selling price = CP + Profit = Rs 10000 + Rs 2000 = Rs 12000

She gets Rs 12000 on selling the T.V

*QUESTION 6: *

*Nancy sold a washing machine for Rs 13500. She Lost 20% in the bargain. What is the price that she bought the washing machine for?*

*Solution;*

Selling price of washing machine= Rs 13,500

Loss percent = 20%

Let the cost price of washing machine be Rs x

Since, Loss = Loss % of CP

=> Loss = 20 % of Rs x = \(\frac{20}{100}\times x=\frac{x}{5}\)

Therefore, SP = CP – Loss

\(\\=>13500=x-\frac{x}{5}\\=>13500=\frac{4x}{5}\\=>x=\frac{13500\times 5}{4}=Rs16,875\)

Hence, The cost price of washing machine is Rs 16,875.

*QUESTION 7: *

*(a) Chalk contains Calcium, Carbon and Oxygen in the ratio 10: 3 : 12. Find the percentage of carbon in chalk.*

*(b) If in a stick of chalk, Carbon is 3g, what is the weight of the chalk stick?*

*Solution;*

** (a)** Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of carbon = \(\frac{3}{25}\)

Percentage of Carbon part in Chalk = \(\frac{3}{25}\) x 100 = 12 %

** (b)** Quality of Carbon in chalk stick = 3g

Let the weight of chalk be x g.

Then, 12 % of x = 3

=> \(\frac{12}{100}\) x X = 3

=> x = \(\frac{3\times 100}{12}\) = 25g

Hence, the weight of chalk stick is 25g.

*QUESTION 8: *

*Amulya buys a book for Rs 275 and sells it at a loss of 15% . How much does she sell it for?*

*Solution;*

The cost of a book = Rs 275

Loss percentage = 15 %

Loss = Loss % of CP = 15 % of 275

= \(\frac{15}{100}\times 275\) = Rs 41.25

Therefore , SP = CP – Loss = 275 – 41.25 = Rs 233.75

Hence, Amulya sold the book for Rs 233.75

*QUESTION 9: *

*Find the amount to be paid at the end of 3 years in the following cases:*

*(i ) Principal = Rs 1200 at 12 % p. a*

*(ii) Principal = Rs 7500 at 5 % p. a*

*Solution;*

** (i)** Here, Principal (p) = 1200

Rate ( r ) = 12 % p a

Time ( T ) = 3 years

\(Simple interest = \frac{P \times R\times T}{100} = \frac{1200\times 12\times 3}{100}\)

=432

Now, Amount = Principal + Simple Interest

= Rs 1200 + Rs 432

= Rs 1632

** (ii)** Here, Principal (p) = 7500

Rate ( r ) = 5 % p a

Time ( T ) = 3 years

\(Simple interest = \frac{P \times R\times T}{100} = \frac{7500\times 5\times 3}{100}\)

= 1125

Now, Amount = Principal + Simple Interest

= Rs 7500 + Rs 1125

= Rs 8625

*QUESTION 10: *

*What rate gives Rs 280 as interest on a sum onRs 56,000 in 2 years?*

*Solution;*

Here, Principal (P ) = Rs 56000

Simple Interest (SI ) = Rs 280

Time ( T) = 2 yrs

Simple interest = \(\frac{P \times R\times T}{100}\)

=> 280 = \(\frac{56000\times R\times 2}{100}\)

=> R = \(\frac{280\times 100}{56000\times 2}\)

=> Hence, the rate of interest on sum is 0.25%

*QUESTION 11: *

*If Deepak gives an interest of Rs 45 for one year at 9% rate per annum. What is the sum he has borrowed from Nancy?*

*Solution;*

Simple Interest = Rs 45

Rate (R ) = 9%

Time ( T) = 1 year

Simple interest = \(\frac{P \times R\times T}{100}\)

=> 45 = \(\frac{P\times 9\times 1}{100}\)

=> P = \(\frac{45\times 100}{9}\)

=> P = Rs 500

Hence, she had borrowed Rs 500.

The NCERT solutions for class 7 maths chapter 8 are available below. The main topic discussed in the field of mathematics along with the examples, exercises and solutions. With the help of NCERT books for class 7 math, we can easily study for the major portions of the exam. With the help of PDF’s one can easily download the solutions to the problems available. These problems also gives a much better explanation of conversion of fractions into percentages. It is important to remember the conversion of various decimal numbers into percentages, as this is the kind of mathematical calculations that a student should be prepared for. Understanding the differences between Profit and Loss is also important as a student learns all the differences between cost price, selling price, increases and decreases in percentage. Thus, this chapter teaches us the difference between percentages and amounts as well. A student can simply refer to the various NCERT textbooks along with the solutions to achieve the maximum amount of marks possible.