# NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

**NCERT solutions for class 8 maths chapter 8 Comparing Quantities- **This chapter is a combination of several important topics, like ratios, percentages, discount, profit & loss, and simple and compound interest. To help the students, step by step NCERT solutions for class 8 maths chapter 8 comparing quantities are provided. The ratio of two quantities is known as comparing quantity. In this chapter, we are going to compare quantities like marks of two students, the height of two persons, a comparison of profit, sales value, etc. In the NCERT solutions for class 8 maths chapter 8 comparing quantities, there are 3 exercises with 26 questions. The first exercise of CBSE class 8 chapter comparing quantities is on percentage and ratio.

The second exercise is on finding the increase or decrease percent and finding discounts. Through these NCERT solutions for class 8 maths chapter 8 comparing quantities, students can cover all the problems related to ratios, percentages, discount, profit-loss, and simple-compound interest. The NCERT solutions from class 6 to 12 will help students to solve practice questions. Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.1 **

### ** (i) **

### In a primary school, the parents were asked the number of hours they spend per day in helping their children to do homework. There were 90 parents who helped hour to hours. The distribution of parents according to the time for which they said they helped is given in the adjoining figure; 20% helped for more than hours; 30% helped for hour to hours; 50% did not help at all.

Using this, answer the following :

(i) How many parents were surveyed?

### ** Answer: **

Let the number of parents surveyed be X.

30% of them helped for hour to hours.

In total, there were 90 such parents.

30% of X = 90

Therefore, there were a total of 300 parents who were surveyed.

** (ii) **

Using this, answer the following :

(ii) How many said that they did not help?

### ** Answer: **

from the previous question number of parents surveyed X = 300.

Given, 50% did not help.

Therefore, number of parents that did not help = 50% of 300 = 150.

** (iii) **

Using this, answer the following :

(iii) How many said that they helped for more than hours?

### ** Answer: **

from the first question number of parents surveyed X = 300

Given, 20% helped for more than hours.

Therefore, number of such parents = 20% of 300 = 2 x 30 = 60.

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.3 **

** 1(a) ** A shop gives 20% discount. What would the sale price of each of these be?

### ** Answer: **

The marked price of dress = Rs120

Discount rate = 20%

Selling price (SP) = Marked price - Total discount = Rs (120- 24) = Rs 96.

** 1(b) ** A shop gives 20% discount. What would the sale price of each of these be?

(b) A pair of shoes marked at Rs 750

### ** Answer: **

The marked price of shoes=Rs 750

Discount rate = 20%

Selling price= Marked price- Total discount =Rs (750- 150) = Rs 600.

** 1(c) ** A shop gives 20% discount. What would the sale price of each of these be?

### ** Answer: **

The marked price of bag = Rs 250

Discount rate=20%

Selling price = Marked price - Total discount = Rs (250- 50) = Rs 200.

### ** Answer: **

Marked price = Rs 15000

Selling Price = Rs 14400

We know, Discount amount = Marked price - Selling price

Discount amount = Rs 15000-Rs 14400 = Rs 600

Therefore, Discount percentage = 4%

** 3 ** An almirah is sold at Rs 5,225 after allowing a discount of 5%. Find its marked price.

### ** Answer: **

Selling price = Rs 5225

Discount percent = 5%

Therefore, the marked price = Rs 5500

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.4.1 **

** 1(a) ** Find selling price (SP) if a profit of 5% is made on

(a) A cycle of Rs 700 with Rs 50 as overhead charges.

### ** Answer: **

We know, Cost price = Buying price + Overhead expenses

Cost price = Rs 700 + Rs 50 = Rs 750

Profit percent = 5%

We know,

Profit = Rs 37.5

Now, Selling price (SP) = Cost price + Profit =Rs 750 + Rs 37.5 = Rs 787.50

Therefore, the selling price (SP) is Rs 787.50

** 1(b) ** Find selling price (SP) if a profit of 5% is made on

(b) a lawnmower bought at Rs 1150 with Rs 50 as transportation charges.

### ** Answer: **

If a profit of 5% is made on a lawnmower bought at Rs 1150 with Rs 50 as transportation charges then selling price is:

** 1 (c) ** Find selling price (SP) if a profit of 5% is made on

(c) a fan bought at Rs 560 and expenses of Rs 40 made on its repairs.

### ** Answer: **

We know, Cost price = Buying price + Overhead expenses

Cost price = Rs 560 + Rs 40 = Rs 600

Profit percent = 5%

Profit = Rs 30

Now, Selling price (SP) = Cost price + Profit =Rs 600 + Rs 30 = Rs 630

Therefore, the selling price (SP) is Rs 630

### ** Answer: **

Net profit = Rs 1000 - Rs 1000 = 0

He neither made an overall profit or loss since the profit made on the first tv equals the loss suffered on the second one.

** Q 1 ** Two times a number is a 100% increase in the number. If we take half the number what would be the decrease in percent?

### ** Answer: **

Let the number be x.

Half the number =

** = 50 % **

Therefore, ** ** If we take half the number, the decrease in percent is 50%

** Q: 2 ** By what percent is Rs 2,000 less than Rs 2,400? Is it the same as the percent by which Rs 2,400 is more than Rs 2,000?

### ** Answer: **

Difference between Rs 2400 and Rs 2000 = Rs 2400- Rs 2000 = Rs 400

The per cent by which Rs 2,000 is less than Rs 2,400 is

( ** Less with respect to 2400. Hence, 2400 will be in the denominator! ** )

Rs 2000 is less than Rs 2400 by ** 16.66% **

Now,

The per cent by which Rs 2,400 is more than Rs 2,000 is

( ** More with respect to 2000. Hence, 2000 will be in the denominator! ** )

Rs 2400 is more than Rs 2000 by ** 20% **

Therefore, they are not the same.

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.6 **

** Q ** Find interest and amount to be paid on Rs 15000 at 5% per annum after 2 years.

### ** Answer: **

=Rs1500

Amount to be paid = Interest + Principal Amount

=Rs 15000 + Rs 1500

=Rs 16500

Find the time period and rate for each.

1. A sum taken for years at 8% per annum is compounded half yearly.

### ** Answer: **

Since the sum taken is compounded half yearly:

Time period = 3 half years

Rate = 4% per half year

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.8 **

Find the time period and rate for each.

A sum taken for 2 years at 4% per annum compounded half yearly.

### ** Answer: **

Since the sum taken is compounded half yearly:

Time period = 4 half years.

Rate = 2% per half year

Find the amount to be paid

1. At the end of 2 years on Rs. 2,400 at 5% per annum compounded annually.

### ** Answer: **

The amount to be paid at the end of 2 years is Rs 2646

At the end of 1 year on Rs. 1800 at 8% per annum compounded quarterly.

### ** Answer: **

The Amount to be paid at the end of 4 years is Rs 1948.37

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities topic 8.9 **

** 1 ** A machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year.

### ** Answer: **

Price of machinery = Rs. 10,500

Value after one year = Rs 10,500- Rs 525 = Rs 9975

### ** Answer: **

The current population of city

= 12 lakh

1200000

Population after two years =

Thus, the population after two years is 1297920.

** NCERT solutions for class 8 maths chapter 8 comparing quantities exercise 8.1 **

1 (a) Find the ratio of the following.

Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

### ** Answer: **

The ratio of the speed of cycle to the speed of scooter =

= 1:2

1(b) Find the ratio of the following.

### ** Answer: **

To find the ratio, we need to make both quantities of the same unit.

We know, 1 km = 1000 m.

10 km = 10x1000 m = 10000 m

Therefore the required ratio =

= 1 : 2000

1(c) Find the ratio of the following.

### ** Answer: **

To find the ratio, we first make the quantities of the same unit.

We know, Rs. 1 = 100 paisa. Rs. 5 = 5x100 = 500 paisa

Therefore, the required ratio =

= 1:10

** 2 (a) ** Convert the following ratios into percentages.

### ** Answer: **

To convert a ratio to a percentage, we multiply the ratio by 100%.

The required percentage of the ratio 3:4 =

** 2 (b) ** Convert the following ratios to percentages.

** Answer: **

To convert a ratio to a percentage, we multiply the ratio by 100%.

The required percentage of ratio 2:3 =

** 3 ** 72% of 25 students are interested in mathematics. How many are not interested in mathematics?

### ** Answer: **

Given;

72% of 25 students are interested in mathematics

Percentage of students not interested in mathematics = (100 - 72)% = 28 %

The number of students not interested in mathematics = 28% of 25

Therefore, 7 students (out of 25) are not interested in mathematics.

### ** Answer: **

Given,

Win percentage of the team = 40%

This means that they won 40 matches out of 100 matches played.

They won 10 matches out of = 25 matches played.

Therefore, they played a total of 25 matches in all.

** 5 ** If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning?

### ** Answer: **

Let the amount of money Chameli had in the beginning = Rs. X

She spends 75% of the money.

Percentage of money left = (100 - 75)% = 25%

Since she has Rs. 600 left.

25% of X = 600

X =2400

Therefore, Chameli had Rs. 2400 with her in the beginning.

** 6. ** If 60% of people in the city like cricket, 30% football and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

### ** Answer: **

Given,

Total number of people = 50 lakhs

Percentage of people who like cricket = 60%

Percentage of people who like football = 30%

Since remaining people like other games,

Percentage of people who like other games = {100 - (60 + 30) } = (100 - 90) = 10%

** 10% of the people like other games. **

Now,

Number of people who like ** cricket ** = 60% of 50 lakhs

Number of people who like ** football ** = 30% of 50 lakhs

Number of people who like ** other games ** = 10% of 50 lakhs

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities exercise 8.2 **

** 1 ** A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his original salary.

### ** Answer: **

Given,

Percentage increase in the salary = 10%

Therefore, if the original salary was Rs. 100, the new salary is Rs. 110

If new salary is Rs 1,54,000, the original salary was

Therefore, the original salary was Rs. 14,00,00.

### ** Answer: **

Given,

Number of people who went to the zoo on Sunday = 845

Number of people who went to the zoo on Monday = 169

The decrease in the number of people visiting the zoo = (845 - 169) = 676

Percentage decrease

(Decrease from the original number)

The percent decrease in the people visiting the Zoo is 80%

### ** Answer: **

Given,

Cost price (CP) of the 80 articles = Rs. 2,400

Profit percentage = 16%

Profit amount on all 80 articles = 16% of 2400 = 16/100 x 2400 = Rs. 384

The selling price of the 80 articles = Rs. (2400 + 384) = Rs. 2784

Selling price (SP) of each item = Rs. 2784/80 = Rs. 34.8

Therefore, the selling price of one article is Rs.34.8

### ** Answer: **

Given,

Cost of the article = Rs. 15500

Cost of repair = Rs. 450

Cost price(CP) of the article = Rs. (15500 + 450) = Rs. 15950

Profit percentage = 15%

Profit amount = 15% of Rs.15950 = Rs. 2392.50

Selling Price = Rs.(15950 + 2392.50) = Rs. 18342.50

OR

Selling price (SP) of the article = CP + Profit = CP + (15% of CP)

= ** 115% of CP **

### ** Answer: **

Given,

The cost price of the VCR = Rs. 8000

The cost price of TV = Rs. 8000

Now, He made a loss of 4% on VCR.

Selling price (SP) of the VCR = (100-4)% of CP = 96% x Rs. 8000 = Rs. 7680

Again, He made a profit of 8% on TV.

Selling price (SP) of the TV = (100+8)% of CP = 108% x Rs. 8000 = Rs. 8640

Net Selling price = Rs. (7680 + 8640) = Rs. 16320

Net Cost price = Rs. (2x8000) = Rs. 16000

Since, SP > CP, he made a net profit (gain)

Now, Net Gain = CP - SP = Rs. (16320-16000) = Rs. 320

Gain % = = 2%

The shopkeeper made a gain of 2% on the whole transaction.

### ** Answer: **

Since the 10% discount is on all the items, we can calculate the Selling price by totalling the Cost price of all item bought.

Now,

Total Cost price(CP) of the items he bought = CP of a pair of jeans + CP of two shirts = Rs.(1450 + 850 + 850) = Rs. 3150

The selling price of these items = (100- 10)% of Rs. 3150 = 90% x Rs.3150 = Rs. 2835

The customer has to pay Rs. 2835.

### ** Answer: **

Given,

The milkman sold two of his buffaloes for Rs 20,000 each. This is the Selling price of the buffaloes.

Let CP of one of the buffalo be Rs. X and the other be Rs. Y

Since he made a profit of 5% on one of them.

105% of X = Rs. 20000

X = Rs. 19047.6

Similarly, since he made a loss of 10% on the other.

90% of Y = Rs. 20000

Y = Rs. 22222.2

Net CP = Rs.(19047.6 + 22222.2) = Rs. 41269.8

And net SP = 2 x Rs.20000 = Rs. 40000

Since SP < CP, he made a net loss.

His overall loss = Rs. (41269.8 - 40000) = Rs. 1269.8 = Rs. 1270 (approx)

** 8 ** The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

### ** Answer: **

Given,

Cost price of the TV = Rs. 13000

Sales tax at the rate of 12%

Selling price = CP + Sales Tax = CP + 12% of CP

= 112% of CP =

Vinod will have to pay Rs. 14,560.

### ** Answer: **

The discount given was 20% which means if CP is Rs. 100 then the SP is Rs. 80

If SP is Rs. 80 then CP is Rs. 100

For SP of Rs. 1600, CP

The marked price is Rs. 2000.

** 10 ** I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

### ** Answer: **

Given,

VAT = 8%

Let the original price be Rs. 100

Original price + VAT = Rs. 100 + Rs.

Original price + VAT = Rs. 100 + Rs. 8 = Rs. 108

If the price after VAT is Rs.5400, then the price before VAT is

The price before VAT was added is Rs. 5000.

** 11 ** An article was purchased for Rs 1,239 including GST of 18%. Find the price of the article before GST was added?

### ** Answer: **

Given,

GST = 18%

Cost with GST included = Rs. 1239

Cost without GST = x Rs.

cost before GST+ GST = cost with GST

x = 1050

Price before GST = 1050 rupees

## ** NCERT solutions for class 8 maths chapter 8 comparing quantities exercise 8.3 **

** 1 (a) ** Calculate the amount and compound interest on

(a) Rs 10,800 for 3 years at % per annum compounded annually .

### ** Answer: **

Given,

Principal,P = Rs 10800

Compound Interese Rate,R = p.a.

Time period,n = 3 years.

We know,

Amount when interest is compounded annually, A =

Therefore, the required amount = = Rs. 15377.34

And Compound Interest, CI = Amount - Principal = Rs. (15377.34 - 10800) = Rs. 4577.34

** 1 (b) ** Calculate the amount and compound interest on

(b) Rs 18,000 for years at 10% per annum compounded annually .

### ** Answer: **

Given,

Principal,P =Rs.18000, Rate,R = 10% and time period,n = 2.5 years.

We know, Amount when interest is compounded annually =

Amount after 2 years at 10% , A = = Rs.21780

This acts as the principal amount for the next half year.

SI on next 1/2 year at = = Rs. 1089

Therefore, Total amount to be paid after 2.5 years = Rs. (21780+1089) = ** Rs.22869 **

Now, Compound Interest after 2 years = A - P = Rs.(21780-18000) = Rs. 3780

Therefore, Compound Interest after 2.5 years, CI = Rs. 3780 + SI = ** Rs.4869 **

** Q: 1 (c) ** Calculate the amount and compound interest on

(c) Rs 62,500 for years at 8% per annum compounded half yearly.

### ** Answer: **

Given,

Principal,P =Rs 62500,

Compound interest Rate,R = 8% compounded half yearly for 1.5 years.

Since it is compounded half yearly, R becomes half = 4%, and time period doubles, n = 3 years.

We know, Amount when interest is compounded annually, A =

Therefore, the required amount = = Rs.70304

And Compound Interest, CI = Amount - Principal = Rs. (70304 - 62500) = Rs. 7804

** 1 (d) ** Calculate the amount and compound interest on:

### ** Answer: **

Given,

Principal,P =Rs.8000, Rate, R = 9% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 9% = 4.5% half yearly.

Therefore, the required amount = = Rs. 8736.20

And Compound Interest, CI = Amount - Principal = Rs. (8736.20 - 8000) = Rs. 736.20

** 1 (e) ** Calculate the amount and compound interest on

Rs 10,000 for 1 year at 8% per annum compounded half yearly.

### ** Answer: **

Given,

Principal, P =Rs.10000, Rate, R = 8% per annum compounded half yearly for 1 year.

Now, there are two half years in a year. Therefore compounding has to be 2 times.

And rate = half of 10% = 5% half yearly.

Therefore, the required amount = = ** Rs. 10816 **

And Compound Interest, C.I. = Amount - Principal = Rs. (10816 - 10000) = ** Rs. 816 ** .

** 2 ** Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? ( ** Hint: ** Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for years )

### ** Answer: **

The amount borrowed from the bank = Principal amount, P = Rs 26400

Compound interest rate, R = 15% p.a.

Time period = 2 years 4 months =

We know, Amount when interest is compounded annually, A =

Therefore, for the first 2 years, amount, A = = Rs 34914

Now, this would act as principal for the next 1/3 year. We find the SI on Rs 34914 for 1/3 year.

SI = = Rs 1745.70

Therefore, Required amount at the end of 2 years and 4 months = A + SI = Rs (34914 + 1745.70) = ** Rs 36659.70 **

### ** Answer: **

For Fabina,

Principal,P =Rs 12500

Simple interest Rate,R = 12% p.a.

Time period,n = 3 years.

Simple Interest, SI at 12% for 3 years = = Rs 4500

For Radha,

Principal,P =Rs 12500

Compound interest Rate,R = 10% p.a.

Time period,n = 3 years.

We know, Amount when interest is compounded annually,

Fabina pays more interest and Rs (4500 - 4137.50) = Rs 362.50 more.

### ** Answer: **

Given,

Principal,P =Rs 12000

Simple interest Rate,R = 6% p.a.

Time period,n = 2 years.

Simple Interest, SI at 6% for 2 years =

If he would have borrowed it at a compound interest rate, R = 6% p.a.

We know, Amount when interest is compounded annually, A =

He would have to pay Rs (1483.20 - 1440) = Rs 43.20 extra.

### ** Answer: **

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half yearly

For a period of 6 months. Time period, n = 1 half year (As there is 1 half year in 6 months.)

We know, Amount when interest is compounded annually, A =

After 6 months, Vasudevan would get an amount Rs. 63600.

### ** Answer: **

Given,

Principal, P = Rs 60,000

Compound interest rate, R = 12% p.a

= 6 % half-yearly

For a period of 1 year. Time period, n = 2 half years (As there are 2 half years in a year.)

We know, Amount when interest is compounded annually, A =

After 1 year, Vasudevan would get an amount Rs. 67416.

### ** Answer ** :

(i) Given,

Principal amount, P = Rs 80000

Rate of interest, R = 10% p.a.

Time period = years.

We know, Amount when interest is compounded annually, A =

Now, For the first year, A=

For the next half-year, this will act as the principal amount.

Interest for 1/2 year at 10% p.a =

Required total amount = Rs (88000 + 4400) = ** Rs 92400 **

(ii) If it is compounded half-yearly, then there are 3 half years in years.

n = 3 half years.

And, Rate of interest = half of 10% p.a = 5% half yearly

The difference in the two amounts = Rs (92610 - 92400) = ** Rs 210 **

(i) The amount credited against her name at the end of the second year.

### ** Answer: **

Given,

Principal amount, P = Rs 8,000

Compound rate of interest, R = 5% p.a.

Time period, n = 2 years

We know, Amount when interest is compounded annually,

Therefore, the amount credited against her name at the end of the second year is ** Rs 8,820 **

(ii) The interest for the 3rd year.

### ** Answer: **

Now, the amount after 2nd year will become the principal amount for the 3rd year

Principal amount, P = Rs 8,820

Compound rate of interest, R = 5% p.a.

Time period, n' = 1 year

Therefore, the interest for the 3rd year is ** Rs 441. **

** 8. ** Find the amount and the compound interest on Rs 10,000 for years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

### ** Answer: **

Principal = Rs.10,000

Time = years

Rate = 10% per annum

CASE 1 Interest on compounded half yearly.

Rate = 10% per annum = 5 % per half yearly

= Amount

CI = Amount - principal

CI =

CI = 1576.25

CASE 2 Interest on compounded anually

Rate = 10% per annum

= Amount

CI = Amount - principal

CI =

CI = 1000

Interest for half years on 11000 =

= 550

Total interest =

= RS 1550

Since 1576.25 1000

Thus, interest would be more in CASE 1 i.e. compounded half yearly

### ** Answer: **

Given,

Principal amount, P = Rs 4,096

Rate of interest, R

Time period, n = 18 months = (12 + 6) months = 1.5 years = 3 half years

(There are 3 half years in 1.5 years)

We know,

Amount when interest is compounded annually, (A)

Therefore, the required amount

Ram will get Rs 4,913 after 18 months.

** 10 ** The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

### ** Answer: **

Let the population in 2001 be P

Compound rate of increase = 5% p.a.

The population in 2003 will be more than in 2001

Time period, n = 2 years (2001 to 2003)

Therefore, the population in 2001 was 48980 (approx)

** 10 ** ** ** The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(ii) what would be its population in 2005?

### ** Answer: **

Let the population in 2001 be P'

Compound rate of increase = 5% p.a.

The population in 2005 will be more than in 2003

Time period, n = 2 years (2003 to 2005)

Therefore, the population in 2005 will be 59535 (approx)

### ** Answer: **

Given,

Initial count of bacteria, P = 5, 06,000 (Principal Amount)

Rate of increase, R = 2.5% per hour.

Time period, n = 2 hours

(This question is done in a similar manner as compound interest)

Number of bacteria after 2 hours =

Therefore, the number of bacteria at the end of 2 hours will be 531616 (approx)

### ** Answer: **

Given,

Principal = Rs 42,000

Rate of depreciation = 8% p.a

Reduction = 8% of Rs 42000 per year

Value at the end of 1 year = Rs (42000 – 3360) = ** Rs 38,640 **

** NCERT solutions for class 8 maths: Chapter-wise **

** NCERT solutions for class 8: Subject-wise **

** Importance of NCERT solutions for class 8 maths chapter 8 comparing quantities: **

- To familiarise with the concept studied in the chapter NCERT solutions for class 8 maths chapter 8 comparing quantities are helpful.
- Solving homework exercise becomes easy with the NCERT solutions for class 8 maths chapter 8 comparing quantities in hand.
- Practicing NCERT solutions for class 8 maths chapter 8 comparing quantities helps to score well in the exam.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities

**Question: **What are the important topics of chapter Comparing Quantities ?

**Answer: **

Increase or decrease percentage, discounts, profit and loss, simple and compound interest are the important topics of this chapter.

**Question: **Does CBSE class 8 maths is tough ?

**Answer: **

CBSE class 8 maths is damn simple and basic. Most of the chapters are related to the previous classes.

**Question: **How does the NCERT solutions are helpful ?

**Answer: **

NCERT solutions are helpful for the students if they are not able to NCERT problems on their own. These solutions are provided in a very detailed manner which will give them conceptual clarity.

**Question: **How many chapters are there in the CBSE class 8 maths ?

**Answer: **

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

**Question: **Where can I find the complete solutions of NCERT for class 8 ?

**Answer: **

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

**Question: **Where can I find the complete solutions of NCERT for class 8 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

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