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NCERT Exemplar · Class 8 · Maths

Chapter 9: Comparing Quantities

Practice each question, then open the answer or worked solution.

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Questions

50 questions
Q1Written answer

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 Then:

(a) only (i) is correct. (b) either (i) or (ii) is correct. (c) either (ii) or (iii) is correct. (d) only (iii) is correct.

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Worked solution

(a) only (i) is correct.

Explanation: Let Principal, P = Rs. 100, Rate = 10% and Time = 1 year

Simple interest (SI)= (P×R×T)/100 = (100×10×1)/100 = Rs.10

Since, Amount = P(1+R/100)T=100(1+10/100)1= 100(11/10) = Rs. 110

Compound interest (CI) = Amount – Principal = 110 -100 = 10

So, CI>SI

Q2Written answer

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

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Worked solution

(b) R < r

Q3Written answer

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

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Worked solution

(b) Rs 4,080

Explanation: P = Rs.50000, R = 4%, T = 2 years

A = 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

Q4Written answer

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

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Worked solution

(a) Rs 1,056

Explanation: Marked price = Rs.1200

Discount = 12%

Since, Discount = Discount% on Marked price

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

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

Q5Written answer

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

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Worked solution

(b) 350 km

Explanation: 90% of x is 315 km

90/100 × x = 315

X = 315 × 100/90 = 315 × 10/9 = 350

Q6Written answer

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

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Worked solution

(a) Rs 500

Explanation: Say, marked price = x

Cost price = Rs.360

As per the question;

x – [x×(10/100)] – [(25×360)/100] = 360

x – x/10 – 90 = 360

9x/10 = 360 + 90

9x = 4500

x = 500

Q7Written answer

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)

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Worked solution

(c) x × (a/100)

(Discount = Discount% on Marked Price)

Q8Written answer

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

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Worked solution

(b) 1 year

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

Amount = Rs.112360

Since we know,

A = P (1+R/100)T

112360 = 100000(1+12/100)T

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

(1.1236)1 = (1.12) T

If we compare the base terms, 1.1236 is approximately equal to 1.12

Hence, T = 1 year.

Q9Multiple choice

For the calculation of interest compounded half yearly, keeping the principal same, which one of the following is true?

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Correct option: D
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Worked solution

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

Q10Written answer

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

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Worked solution

(b) Rs 39,730.50

Explanation: The scooter costs Rs.36450 at the rate of sales tax = 9%.

Total cost of scooter paid by Shyama = 9% of 36450 + 36450

= (9/100 × 36450) + 36450

= 3280.5 + 36450

= 39730.5