NCERT SOLUTION FOR CLASS 8 MATHS CHAPTER 8 COMPARING QUANTITIES

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Your academic journey begins a new stage as you reach the eighth grade, and a new facet of learning awaits you. This is the year where pupils begin to acquire an interest in the topic at hand, and it is also the year in which you choose whether you will study business, humanity, or science in the following year (class 12). As a consequence of this, having a solid foundation of knowledge in fields such as mathematics and physics comes highly recommended. The NCERT textbook for class 8 is the greatest book now available on the market for enhancing one’s education in the areas of mathematics and science. Read up on the theory that is presented in the book for grade 8 and answer all of the questions that are posed in the practice section of the NCERT textbook.

NCERT SOLUTION FOR CLASS 8 MATHS CHAPTER 8 COMPARING QUANTITIES – Exercises

NCERT Solutions for Class 8 Maths Exercise 8.1

Question 1.

Find the ratio of the following:
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 50 m to 10 km
(c) 50 paise to Rs.5

Solution :

(a) Speed of cycle = 15 km/hr
Speed of scooter = 30 km/hr
Hence ratio of speed of cycle to that of scooter = 15 : 30 = 15 /30 = 1/2 = 1 : 2
(b) ∵ 1 km = 1000 m
10 km =10 × 1000 = 10000 m
Ratio = = = 1 : 2000
(c) `1 = 100 paise
∴`5 = 5 × 100= 500 paise
∴Hence Ratio = Comparing Quantities/image008.png = Comparing Quantities/image009.png= 1 : 10

Question 2.

Convert the following ratios to percentages:
(a) 3 : 4
(b) 2 : 3
Solution :

(a) Percentage of 3 : 4 =Comparing Quantities/image010.png
= 75%
(b) Percentage of 2 : 3 =Comparing Quantities/image011.png
=Comparing Quantities/image012.png

Question 3.

72% of 25 students are good in mathematics. How many are not good in mathematics?

Solution :

Total number of students = 25
Number of good students in mathematics = 72% of 25 =Comparing Quantities/image013.png = 18
Number of students not good in mathematics = 25 – 18 = 7
Hence percentage of students not good in mathematics = Comparing Quantities/image014.png = 28%

Question 4.

A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Solution : Let total number of matches be x,
According to question,

Comparing Quantities/image018.png

Hence total number of matches are 25.

Question 5.

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

Solution :

Let her money in the beginning be `x.
According to question,
Comparing Quantities/image020.png

Hence the money in the beginning was `2,400.

Question 6.

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

Solution :

Number of people who like cricket
= 60%
Number of people who like football
= 30%
Number of people who like other games
= 100% – (60% + 30%) = 10%
Now Number of people who like cricket
= 60% of 50,00,000
=Comparing Quantities/image027.png = 30,00,000
And Number of people who like football
= 30% of 50,00,000
= 15,00,000
Number of people who like other games = 10% of 50,00,000
Comparing Quantities/image028.png = 5,00,000
Hence, number of people who like other games are 5 lakh.

NCERT Solutions for Class 8 Maths Exercise 8.2

Question 1.

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

Solution :

Let original salary be Rs.100.
Therefore New salary i.e., 10% increase
= 100 + 10 = Rs.110
∵ New salary is Rs.110, when original salary = Rs.100
∴ New salary is Rs.1, when original salary =Comparing Quantities/image003.png
∴ New salary is Rs.1,54,000, when original salary = Comparing Quantities/image004.png= Rs.1,40,000
Hence original salary is Rs. 1,40,000.

Question 2.

On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?

Solution :

On Sunday, people went to the Zoo
= 845
On Monday, people went to the Zoo = 169
Number of decrease in the people
= 845 – 169 = 676
Decrease percent = Comparing Quantities/image005.png = 80%
Hence decrease in the people visiting the Zoo is 80%.

Question 3.

A shopkeeper buys 80 articles for Rs.2,400 and sells them for a profit of 16%. Find the selling price of one article.

Solution :

No. of articles = 80
∵Cost Price of articles = Rs. 2,400And Profit
= 16%
Cost price of articles is Rs.100, then selling price = 100 + 16 = Rs.116
Cost price of articles is Rs.1, then selling price =
Cost price of articles is Rs.2400, then selling price = Rs.2784
Hence, Selling Price of 80 articles = Rs.2784
Therefore Selling Price of 1 article
= Rs.34.80

Question 4.

The cost of an article was Rs.15,500, Rs.450 were spent on its repairs. If it sold for a profit of 15%, find the selling price of the article.

Solution :

Here, C.P. = Rs.15,500 and Repair cost = Rs.450
Therefore Total Cost Price = 15500 + 450 = Rs.15,950
Let C.P be Rs.100, then S.P. = 100 + 15
= Rs.115
When C.P. is Rs.100, then S.P. = Rs.115
When C.P. is Rs.1, then S.P. = 
When C.P. is Rs.15950, then S.P.
Comparing Quantities/image010.png15950 = Rs.18,342.50

Question 5.

A VCR and TV were bought for Rs.8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Solution :

Cost price of VCR = Rs.8000 and Cost price of TV = Rs.8000
Total Cost Price of both articles
= Rs.8000 + Rs.8000 = Rs. 16,000
Now VCR is sold at 4% loss.
Let C.P. of each article be Rs.100, then S.P. of VCR = 100 – 4 = Rs.96
∵ When C.P. is Rs.100, then S.P. = Rs.96
∴ When C.P. is Rs.1, then S.P. =
∴ When C.P. is Rs.8000, then S.P.
= Rs.7,680
And TV is sold at 8% profit, then S.P. of TV = 100 + 8 = Rs.108
When C.P. is Rs.100, then S.P. = Rs.108
∴When C.P. is Rs.1, then S.P. =Comparing Quantities/image011.png
∴When C.P. is Rs.8000, then S.P.
Comparing Quantities/image012.png= Rs.8,640
Then, Total S.P.
= Rs.7,680 + Rs.8,640 = Rs. 16,320
Since S.P. > C.P.,
Therefore Profit = S.P. – C.P.
= 16320 – 16000 = Rs.320
And Profit% =Comparing Quantities/image015.png
=Comparing Quantities/image016.png= 2%

Question 6.

During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs.1450and two shirts marked at Rs.850 each?

Solution :

Rate of discount on all items = 10%
Marked Price of a pair of jeans = Rs.1450 and Marked Price of a shirt = Rs.850
Discount on a pair of jeans
Comparing Quantities/image017.png = Rs.145
∴ S.P. of a pair of jeans = Rs.1450 – Rs.145
= Rs.1305
Marked Price of two shirts = 2 × 850
= Rs.1700Comparing Quantities/image019.png
Discount on two shirts =
= Rs.170
∴ S.P. of two shirts = Rs.1700 – Rs.170
= Rs.1530
Therefore the customer had to pay
= 1305 + 1530
= Discount on a pair of jeans
=Comparing Quantities/image017.png
= Rs.145
∴ S.P. of a pair of jeans
= Rs.1450 – Rs.145 = Rs.2,835

Question 7.

A milkman sold two of his buffaloes for Rs.20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

Solution :

S.P. of each buffalo = Rs.20,000
S.P. of two buffaloes = 20 000 × 2
= Rs.40,000
One buffalo is sold at 5% gain.
Let C.P. be Rs.100, then S.P. = 100 + 5
= Rs.105
∵When S.P. is Rs.105, then C.P. = Rs.100
∴ When S.P. is Rs.1, then C.P. = Comparing Quantities/image021.png
∴ When S.P. is Rs.20,000, then C.P.
Comparing Quantities/image022.png = Rs.19,047.62
Another buffalo is sold at 10% loss.
Let C.P. be Rs.100, then S.P. = 100 – 10
= Rs.90
∵ When S.P. is Rs.90, then C.P. = Rs.100
∴ When S.P. is Rs.1, then C.P. =
∴When S.P. is Rs.20,000, then C.P.
=Comparing Quantities/image024.png= Rs.22,222.22
Total C.P. = Rs.19,047.62 + Rs.22,222.22
= Rs.41,269.84
Since C.P. >S.P.
Therefore here it is loss.
Loss = C.P. – S.P.
= Rs.41,269.84 – Rs. 40,000.00 = Rs.1,269.84

Question 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.

Solution :

C.P. = Rs.13,000 and S.T. rate = 12%
Let C.P. be Rs.100, then S.P. for purchaser
= 100 + 12 = Rs.112
∵ When C.P. is Rs.100, then S.P. = Rs.112
∴ When C.P. is Rs.1, then S.P. = Comparing Quantities/image025.png
∴ When C.P. is Rs.13,000, then S.P.
=Comparing Quantities/image026.png= Rs.14,560

Question 9.

Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs.1,600, find the marked price.

Solution :

S.P. = Rs.1,600 and Rate of discount
= 20%
Let M.P. be Rs.100, then S.P. for customer
= 100 – 20 = Rs.80
∵ When S.P. is Rs.80, then M.P. = Rs.100
∴ When S.P. is Rs.1, then M.P. =Comparing Quantities/image027.png
∴ When S.P. is Rs.1600, then M.P.
Comparing Quantities/image028.png = Rs.2,000

Question 10.

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

Solution :

C.P. = Rs.5,400 and Rate of VAT = 8%
Let C.P. without VAT is Rs. 100, then price including VAT = 100 + 8 = Rs.108
When price including VAT is Rs.108, then original price = Rs.100
∴ When price including VAT is Rs.1, then original price =Comparing Quantities/image029.png
∴ When price including VAT is Rs.5400, then original price = Comparing Quantities/image030.png = Rs.5000

NCERT Solutions for Class 8 Maths Exercise 8.3

Question 1.

Calculate the amount and compound interest on:
(a) Rs.10,800 for 3 years at Comparing Quantities/image001.png per annum compounded annually.
(b) Rs.18,000 forComparing Quantities/image002.pngyears at 10% per annum compounded annually.
(c) Rs.62,500 for Comparing Quantities/image003.png years at 8% per annum compounded annually.
(d) Rs.8,000 forComparing Quantities/image005.png years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).
(e) Rs.10,000 for years at 8% per annum compounded half yearly.

Solution :

(a) Here, Principal (P) = Rs. 10800, Time = 3 years,
Rate of interest (R) =Comparing Quantities/image006.png
Comparing Quantities/image007.png

= Rs. 15,377.34
Compound Interest (C.I.) = A – P
= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34
(b) Here, Principal (P) = Rs. 18,000, Time Comparing Quantities/image005.png=Comparing Quantities/image002.png years, Rate of interest (R)
= 10% p.a.
Amount (A) =Comparing Quantities/image007.png
Comparing Quantities/image013.png

Interest for Comparing Quantities/image017.png years on Rs. 21,780 at rate of 10% = Comparing Quantities/image018.png = Rs. 1,089
Total amount for Comparing Quantities/image002.png years
= Rs. 21,780 + Rs. 1089 = Rs. 22,869
Compound Interest (C.I.) = A – P
= Rs. 22869 – Rs. 18000 = Rs. 4,869
(c) Here, Principal (P) = Rs. 62500, Time  Comparing Quantities/image019.pngyears = 3 years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Comparing Quantities/image007.png

= Rs. 70,304
Compound Interest (C.I.) = A – P
= Rs. 70304 – Rs. 62500 = Rs. 7,804
(d) Here, Principal (P) = Rs. 8000, Time (n)= 1 years = 2 years(compounded half yearly)
Rate of interest (R) = 9% =Comparing Quantities/image024.png (compounded half yearly)
Amount (A) =Comparing Quantities/image007.png
Comparing Quantities/image025.png

= Rs. 8,736.20
Compound Interest (C.I.) = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20
(e) Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) =Comparing Quantities/image007.png
Comparing Quantities/image029.png

= Rs. 10,816
Compound Interest (C.I.) = A – P
= Rs. 10,816 – Rs. 10,000 = Rs. 816

Question 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 4/12 years).

Solution :

Here, Principal (P) = Rs. 26,400, Time (n) = 2 years 4 months, Rate of interest (R) = 15% p.a.
Amount for 2 years (A) =Comparing Quantities/image007.png
Comparing Quantities/image034.png

Interest for 4 months =Comparing Quantities/image038.png years at the rate of 15% = Comparing Quantities/image039.png
= Rs. 1745.70
Total amount = Rs. 34,914 + Rs. 1,745.70
= Rs. 36,659.70

Question 3.

Fabina borrows Rs.12,500 per annum for 3 years at simple interest and Radhaborrows the same amount for the same time period at 10% per annnum, compounded annually. Who pays more interest and by how much?

Solution :

Here, Principal (P) = Rs.12,500, Time (T) = 3 years, Rate of interest (R)
= 12% p.a.
Simple Interest for Fabina =Comparing Quantities/image042.png

Comparing Quantities/image007.png

= Rs. 16,637.50
C.I. for Radha = A – P
= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50
Here, Fabina pays more interest
= Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

Question 4.

I borrowsRs.12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Solution :

Here, Principal (P) = Rs.12,000, Time(T) = 2 years, Rate of interest (R) = 6% p.a.
Simple Interest =Comparing Quantities/image042.png
Comparing Quantities/image049.png

= Rs. 13,483.20 – Rs. 12,000
= Rs. 1,483.20
Difference in both interests
= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

Question 5.

Vasudevan invested Rs.60,000 at an interest rate of 12% per ann^um compounded half yearly. What amount would he get:
(i) after 6 months?
(ii) after 1 year?

Solution :

(i) Here, Principal (P) = Rs. 60,000,
Time (n)= 6 months = 1 year(compounded half yearly)
Rate of interest (R) = 12% = 6% (compounded half yearly)

Comparing Quantities/image055.png

= Rs. 63,600
After 6 months Vasudevan would get amount Rs. 63,600.
(ii) Here, Principal (P) = Rs. 60,000,
Time (n) = 1 year = 2 year(compounded half yearly)
Rate of interest (R) = 12% = 6% (compounded half yearly)
Amount (A) =Comparing Quantities/image007.png
Comparing Quantities/image059.png

= Rs. 67,416
After 1 year Vasudevan would get amount Rs. 67,416.

Question 6.

Arif took a loan of Rs.80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after

Solution : Comparing Quantities/image002.png years if the interest is:
(i) compounded annually.
(ii) compounded half yearly.

Solution :

(i) Here, Principal (P) = Rs. 80,000, Time (n)=Comparing Quantities/image002.png years, Rate of interest (R) = 10%
Amount for 1 year (A)Comparing Quantities/image007.png
Comparing Quantities/image063.png

= Rs. 88,000
Interest forComparing Quantities/image066.png
= Rs. 4,400
Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400
(ii) Here, Principal (P) = Rs.80,000,
Time (n)  =Comparing Quantities/image003.png year = 3year (compounded half yearly)
Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) = Comparing Quantities/image007.png
Comparing Quantities/image007.png

= Rs. 92,610
Difference in amounts
= Rs. 92,610 – Rs. 92,400 = Rs. 210

Question 7.

Maria invested Rs.8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the third year.

Solution :

(i) Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time (n) = 2 years
Amount (A)Comparing Quantities/image007.png
Comparing Quantities/image071.png

= Rs. 8,820
(ii) Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time (n) = 3 years
Amount (A)Comparing Quantities/image007.png
= id=”docs-internal-guid-2c688946-7fff-2a9f-114e-3f51e432463a” > Comparing Quantities/image075.png
Comparing Quantities/image076.png

= Rs. 9,261
Interest for 3rd year = A – P
= Rs. 9,261 – Rs. 8,820 = Rs. 441

Question 8.

Find the amount and the compound interest on Rs.10,000 for Comparing Quantities/image003.png years at 10% per annum, compounded half yearly.
Would this interest be more than the interest he would get if it was compounded annually?

Solution :

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)
Time Comparing Quantities/image005.png =Comparing Quantities/image003.png= 3 years (compounded half yearly)
Amount (A)Comparing Quantities/image007.png
Comparing Quantities/image079.png

= Rs. 11,576.25
Compound Interest (C.I.) = A – P
= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25
If it is compounded annually, then
Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n)  = Comparing Quantities/image003.png years
Amount (A) for 1 yearComparing Quantities/image007.png
Comparing Quantities/image083.png

Comparing Quantities/image040.png Total amount = Rs. 11,000 + Rs. 550
= Rs. 11,550
Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000
= Rs. 1,550
Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

Question 9.

Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months atComparing Quantities/image001.png per annum, interest being compounded half yearly.

Solution :

Here, Principal (P) = Rs. 4096,
Rate of Interest (R) =Comparing Quantities/image088.png
Comparing Quantities/image089.png= (compounded half yearly)
Time (n) = 18 months Comparing Quantities/image003.png = years = 3 years (compounded half yearly)
Comparing Quantities/image040.pngAmount (A) Comparing Quantities/image007.png
Comparing Quantities/image090.png

= Rs. 4,913

Question 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.
(ii) What would be its population in 2005?

Solution :

(i) Here, A2003 = Rs. 54,000, R = 5%, = 2 years
Population would be less in 2001 than 2003 in two years.
Here population is increasing.
Comparing Quantities/image094.png

(ii) According to question, population is increasing. Therefore population in 2005,

Comparing Quantities/image102.png

= 59,535
Hence population in 2005 would be 59,535.

Question 11.

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Solution :

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours
After 2 hours, number of bacteria,
Comparing Quantities/image007.png

= 5,31,616.25
Hence, number of bacteria after two hours are 531616 (approx.).

Question 12.

A scooter was bought at Rs.42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution :

Here, Principal (P) = Rs. 42,000, Rate of Interest (R) = 8%, Time (n)= 1 years
Comparing Quantities/image111.png

= Rs. 38,640
Hence, the value of scooter after one year is Rs. 38,640.

Conclusions for NCERT Solutions For Class 8 Maths Chapter 8

The academic team of SWC has provided NCERT answers for class 8 mathematics chapter-8. We’ve given solutions for all of Chapter 8’s exercises. These are step-by-step answers to all of the problems in Chapter 8 of the NCERT textbook. Read the first theoretical chapter. Make sure you’ve gone through the theoretical section of chapter 8 in the NCERT textbook and that you know the formula for the chapter.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 8 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

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