NCERT Exemplar Solutions Class 8 Maths – Chapter 9 Comparing Quantities

# NCERT Exemplar Solutions Class 8 Maths – Chapter 9 Comparing Quantities

Subject specialists have designed NCERT Exemplar Solutions for Maths Class 8 Chapter 9 which includes thorough solutions for reference. These solutions are updated according to the latest CBSE syllabus for 2024-25 and are provided in easy language for understanding. Tips and tricks are also provided.

These solutions are provided so a student can clear his doubts and get help with a deep understanding of the concept. Also, you can refer them to make the chapter notes and revisions notes. PDF of this can also be downloaded from the website.

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The NCERT Exemplar Solutions of this chapter starts with basic concepts like terms, factors, coefficients, like and unlike terms, addition and subtraction of algebraic expressions, and multiplication of two or more polynomials. Students can go through the various algebraic expression identities and solve problems applying these identities. In NCERT Class 8, this chapter carries a total weightage of 8 to 10 marks in the final examination. Students can use them to solve exercise questions and prepare for their examinations.

### Access Answers to NCERT Exemplar Solutions for Class 8 Maths Chapter 9 Comparing Quantities

There are four options, out of which one is correct. Write the correct answer.

Question: Suppose for the principal P, rate R% and time T, the simple interest is S and compound interest is C. Consider the possibilities.
(i) C > S
(ii) C = S
(iii) C < S

Answer – (a) only (i) is correct.

Explanation: Let the principal P be Rs. 100, the rate R be 10% and the time T be 1 year.
Simple interest (SI) = (P × R × T) / 100 = (100 × 10 × 1) / 100 = Rs. 10
Amount = P(1 + R/100)T = 100(1 + 10/100)1 = 100(11/10) = Rs. 110
Compound interest (CI) = Amount – Principal = 110 – 100 = Rs. 10

As we can see, CI > SI, which is the first possibility.

Question: Suppose a certain sum doubles in 2 years at r % rate of simple interest per annum or at R% rate of interest per annum compounded annually. We have
(a) r < R
(b) R < r
(c) R = r
(d) can’t be decided

Answer – (b) R < r

Explanation: Let the sum be Rs. 100.

For simple interest,
Amount = P + SI = P + (P × r × T) / 100 = 100 + (100 × r × 2) / 100 = 100 + 2r

For compound interest,
Amount = P(1 + R/100)T = 100(1 + R/100)2 = 100(1 + R/50) = 100 + 2R

Since the sum doubles in 2 years, the amount becomes 200.

So, 200 = 100 + 2r and 200 = 100 + 2R

Solving these equations, we get r = 50 and R = 25.

Therefore, R < r.

Question: The compound interest on Rs 50,000 at 4% per annum for 2 years compounded annually is
(a) Rs 4,000
(b) Rs 4,080
(c) Rs 4,280
(d) Rs 4,050

Explanation: P = Rs. 50,000, R = 4%, T = 2 years

Amount = P(1 + R/100)T = 50000(1 + 4/100)2 = 50000(1 + 1/25)2

A = 50000(26/25)2 = 54080

Compound interest = A – P = 54080 – 50000 = Rs. 4080

Therefore, the compound interest is Rs. 4080.

Question: If the marked price of an article is Rs 1,200, and the discount is 12%, then the selling price of the article is
(a) Rs 1,056
(b) Rs 1,344
(c) Rs 1,212
(d) Rs 1,188

Explanation: Marked price = Rs. 1200
Discount = 12%

Discount price = 12% of 1200 = 12/100 × 1200 = 12 × 12 = 144

Selling price = Marked price – discount price = 1200 – 144 = Rs. 1056

Therefore, the selling price of the article is Rs. 1056.

Question: If 90% of x is 315 km, then the value of x is
(a) 325 km
(b) 350 km
(c) 350 m
(d) 325 m

Explanation: 90% of x is 315 km

90/100 × x = 315

X = 315 × 100/90 = 315 × 10/9 = 350

Therefore, the value of x is 350 km.

Question: To gain 25% after allowing a discount of 10%, the shopkeeper must mark the price of the article, which costs him Rs 360 as
(a) Rs 500
(b) Rs 450
(c) Rs 460
(d) Rs 486

Explanation: Let the marked price be x.

Cost price = Rs. 360

Discount = 10% of x = 10/100 × x

Selling price = x – (10/100) × x

Gain = 25% of 360 = 25/100 × 360 = 90

Selling price + Gain = Marked price

x – (10/100) × x + 90 = x

90 = (10/100) × x

x = 90 × 100/10 = 900

Therefore, the shopkeeper must mark the price of the article as Rs. 900.

Question: If a % is the discount per cent on a marked price x, then discount is
(a) (x/a) × 100
(b) (a/x) × 100
(c) x × (a/100)
(d) 100/(x × a)

Answer – (c) x × (a/100)

Explanation: Discount = Discount% on Marked price

Discount = a/100 × x

Therefore, discount is x × (a/100).

Question: Ashima took a loan of Rs 1,00,000 at 12% p.a. compounded half-yearly. She paid Rs. 1,12,360. If (1.06)2 is equal to 1.1236, then the period for which she took the loan is:
(a) 2 years
(b) 1 year
(c) 6 months
(d) 1(1/2) years

Explanation: P = Rs. 100000, R = 12% per annum compounded half-yearly, Amount = Rs. 112360

A = P(1 + R/100)T

112360 = 100000(1 + 12/100)T

112360/100000 = (1 + 12/100)T

(1.1236)1 = (1.06)T

T = 1 year

Therefore, Ashima took the loan for 1 year.

Question: For the calculation of interest compounded half yearly, keeping the principal same, which one of the following is true?
(a) Double the given annual rate and half the given number of years.
(b) Double the given annual rate as well as the given number of years.
(c) Half the given annual rate as well as the given number of years.
(d) Half the given annual rate and double the given number of years.

Answer – (d) Half the given annual rate and double the given number of years.

Explanation: If the interest is compounded half yearly, then the rate of interest is divided by 2 and the number of times the interest is charged is doubled.

Therefore, the correct option is (d) Half the given annual rate and double the given number of years.

Question: Shyama purchases a scooter costing Rs 36,450, and the rate of sales tax is 9%. Then the total amount paid by her is:
(a) Rs 36,490.50
(b) Rs 39,730.50
(c) Rs 36,454.50
(d) Rs 33,169.50

Explanation: Cost of scooter = Rs. 36450
Rate of sales tax = 9%

Total amount paid = Cost of scooter + (Cost of scooter × Sales tax) / 100

Total amount paid = 36450 + (36450 × 9) / 100 = 36450 + 3280.50 = Rs. 39730.50

Therefore, the total amount paid by Shyama is Rs. 39730.50.

Question: The marked price of an article is Rs 80, and it is sold at Rs 76. Then the discount rate is:
(a) 5%
(b) 95%
(c) 10%
(d) appx. 11%

Explanation: Marked price = Rs. 80
Selling price = Rs. 76

Discount = Marked price – Selling price = Rs. 80 – Rs. 76 = Rs. 4

Discount% = (Discount / Marked price) × 100 = (4 / 80) × 100 = 5%

Therefore, the discount rate is 5%.

Question: A bought a tape recorder for Rs 8,000 and sold it to B. B in turn sold it to C, each earning a profit of 20%. Which of the following is true:
(a) A and B earn the same profit.
(b) A earns more profit than B.
(c) A earns less profit than B.
(d) Cannot be decided.

Answer – (c) A earns less profit than B

Explanation: A bought a tape recorder for Rs 8,000 and sold it to B at a profit of 20%.

Cost price for B = 8000 × 100 / (100 + 20) = 8000 × 100 / 120 = Rs. 6666.67

B sold the tape recorder to C at a profit of 20%.

Cost price for C = 6666.67 × 100 / (100 + 20) = 6666.67 × 100 / 120 = Rs. 5555.56

Profit for A = 8000 – 6666.67 = Rs. 1333.33

Profit for B = 6666.67 – 5555.56 = Rs. 1111.11

Therefore, A earns more profit than B.

Question: Latika bought a teapot for Rs 120 and a set of cups for Rs 400. She sold the teapot at a profit of 5% and the cups at a loss of 5%. The amount received by her is:
(a) Rs 494
(b) Rs 546
(c) Rs 506
(d) Rs 534

Explanation: Cost price of teapot = Rs. 120
Selling price of teapot = 120 + (120 × 5) / 100 = 120 + 6 = Rs. 126

Cost price of cups = Rs. 400
Selling price of cups = 400 – (400 × 5) / 100 = 400 – 20 = Rs. 380

Amount received by Latika = Selling price of teapot + Selling price of cups = 126 + 380 = Rs. 506

Therefore, the amount received by Latika is Rs. 506.

Question: A jacket was sold for Rs 1,120 after allowing a discount of 20%. The marked price of the jacket is:
(a) Rs 1440
(b) Rs 1400
(c) Rs 960
(d) Rs 866.66

Explanation: Let the marked price be x.

Discount = 20% of x = 20/100 × x

Selling price = x – (20/100) × x = x – x/5 = 4x/5

Selling price = 1120

4x/5 = 1120

x = 1120 × 5/4 = 1400

Therefore, the marked price of the jacket is Rs. 1400.

Question: A sum is taken for two years at 16% p.a. If interest is compounded after every three months, the number of times for which interest is charged in 2 years is:
(a) 8
(b) 4
(c) 6
(d) 9

Explanation: If interest is compounded after every three months, then the number of times interest is charged in 2 years is 2 × 4 = 8 times.

Therefore, the correct option is (a) 8.

Question: The original price of a washing machine which was bought for Rs 13,500 inclusive of 8% VAT, is:
(a) Rs 14,420
(b) Rs 14,580
(c) Rs 12,500
(d) Rs 13,492

Explanation: Let the original price of the washing machine be x.

Cost price of washing machine including VAT = x + (8/100) × x = x + 0.08x = 1.08x = 13500

x = 13500 / 1.08 = 12500

Therefore, the original price of the washing machine is Rs. 12500.

Question: Avinash bought an electric iron for Rs 900 and sold it at a gain of 10%. He sold another electric iron at 5% loss which was bought for Rs 1200. On the transaction, he has a:
(a) Profit of Rs 75
(b) Loss of Rs 75
(c) Profit of Rs 30
(d) Loss of Rs 30

Answer – (c) Profit of Rs 30

Explanation: Cost price of first electric iron = Rs. 900
Selling price of first electric iron = 900 + (900 × 10) / 100 = 900 + 90 = Rs. 990

Cost price of second electric iron = Rs. 1200
Selling price of second electric iron = 1200 – (1200 × 5) / 100 = 1200 – 60 = Rs. 1140

Total cost price = 900 + 1200 = Rs. 2100

Total selling price = 990 + 1140 = Rs. 2130

Profit = Total selling price – Total cost price = 2130 – 2100 = Rs. 30

Therefore, Avinash has a profit of Rs. 30.

Question: A TV set was bought for Rs 26,250, including 5% VAT. The original price of the TV set is:
(a) Rs 27,562.50
(b) Rs 25,000
(c) Rs 24,937.50
(d) Rs 26,245

Explanation: Let the original price of the TV set be x.

Cost price of TV set including VAT = x + (5/100) × x = x + 0.05x = 1.05x = 26250

x = 26250 / 1.05 = 25000

Therefore, the original price of the TV set is Rs. 25000.

Question – To calculate the growth of a bacteria if the rate of growth is known, the formula for calculation of the amount in compound interest can be used.

Question – Discount is a reduction given on the cost price of an article.

Question – Compound interest is the interest calculated on the previous year’s amount.

Question – C.P. = M.P. – Discount.

## NCERT Exemplar Solutions Class 8 Maths Solutions Chapter 9 Comparing Quantities

### Key Features of NCERT Exemplar Solutions for this chapter

1. NCERT Exemplar Solutions helps in providing fully resolved step-by-step solutions to all textbook questions.
2. Set of solutions that come with a list of all important formulas on algebraic identities.
3. These solutions are made according to the latest syllabus.
4. Solutions are prepared by subject matter experts.
5. NCERT Exemplar Solutions provides help for the preparation of competitive exams.

## Chapter 9 Comparing Quantities FAQ’s

### How is NCERT Exemplar Solutions from this chapter helpful for board exams?

NCERT Exemplar Solutions of this chapter comes with detailed descriptions as per the term limit particularised by the Board for self-evaluation. Solving these solutions can help in providing good marks in exams. Students can improve their skills on siting on exam dah

### Is it necessary to practice all the exercises present in Chapter 9 of NCERT Exemplar Solutions for Class 8 Maths?

Yes. As these questions seem to be important for exams. These questions are solved by subject matter experts for helping students to crack these exercises easily. These solutions give students knowledge about data handling. Solutions can be downloaded in PDF format on the Infinity Learn website.

### Mention the topics that are covered in these NCERT Exemplar Solutions?

The topics included in NCERT Exemplar Solutions of this chapter are 9.1 – introduction to expressions 9.2 – terms, factors, and coefficients 9.3 – monomials, binomials, and polynomials 9.4 – like and unlike terms 9.5 – addition and subtraction of algebraic expressions 9.6 – multiplication of algebraic expressions: introduction 9.7 – multiplying a monomial by a monomial 9.8 – multiplying a monomial by a polynomial 9.9 – multiplying a polynomial by a polynomial 9.10 – definition and meaning of identity 9.11 – standard identities 9.12 – applying identities.

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