CBSE Class 10 Mathematics Simple and Compound Interest MCQs Set 17

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Simple and Compound Interest

Access targeted multiple-choice questions for Simple and Compound Interest designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Simple and Compound Interest MCQs for Class 10 Mathematics

Access the complete set of multiple-choice questions for Simple and Compound Interest below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6 1/4% p.a. for 2 years. Find his gain in the transaction per year.
(a) Rs. 112.50
(b) Rs. 125
(c) Rs. 150
(d) Rs. 167.50
Show Answer & Explanation

Answer: (a) Rs. 112.50

Explanation:
1. Interest paid per year = \( 5000 \times 4/100 = 200 \)
2. Interest received per year = \( 5000 \times 6.25/100 = 312.50 \)
3. Gain per year = Rs. 112.50.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 2: Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
(a) 3.6
(b) 6
(c) 18
(d) None of these
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. Let rate = time = \( r \): \( 1200 \times r \times r/100 = 432 \)
2. \( r^{2} = 36 \), so \( r = 6 \).

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 3: What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
(a) Rs. 112
(b) Rs. 118.80
(c) Rs. 120
(d) Rs. 122
Show Answer & Explanation

Answer: (c) Rs. 120

Explanation:
1. Present worth = \( 132/(1 + 5 \times 2/100) = 132/1.1 \) = Rs. 120.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 4: The rate at which a sum becomes four times itself in 15 years at S.I. will be
(a) 15%
(b) 17 1/2%
(c) 20%
(d) 25%
Show Answer & Explanation

Answer: (c) 20%

Explanation:
1. SI = \( 3P \) in 15 years, so \( R = 300/15 \) = 20%.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 5: A sum of money trebles itself in 15 years 6 months. In how many years would it double itself?
(a) 6 years 3 months
(b) 7 years 9 months
(c) 8 years 3 months
(d) 9 years 6 months
Show Answer & Explanation

Answer: (b) 7 years 9 months

Explanation:
1. Interest of \( 2P \) needs 15.5 years, so interest \( P \) needs 7.75 years = 7 years 9 months.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 6: Rs. 1000 is invested at 5% simple interest. If the interest is added to the principal every 10 years, the amount will become Rs. 2000 after
(a) 15 years
(b) 18 years
(c) 20 years
(d) 16 2/3 years
Show Answer & Explanation

Answer: (d) 16 2/3 years

Explanation:
1. After 10 years the amount = 1500
2. Interest on 1500 at 5% is Rs. \( 75 a \) year, so the extra Rs. 500 takes \( 6 2/3 \) years
3. Total = \( 16 2/3 \) years.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 7: The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is
(a) 2
(b) 2 1/2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. Amount = 34347
2. \( 34347/30000 = 1.1449 = 1.07^{2} \), so the period is 2 years.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Question 8: At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?
(a) 6%
(b) 6.5%
(c) 7%
(d) 7.5%
Show Answer & Explanation

Answer: (a) 6%

Explanation:
1. \( 1348.32/1200 = 1.1236 = 1.06^{2} \)
2. The rate is 6%.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 9: A sum of money invested at compound interest amounts to Rs. 800 in 3 years and to Rs. 840 in 4 years. The rate of interest per annum is
(a) 2 1/2%
(b) 4%
(c) 5%
(d) 6 2/3%
Show Answer & Explanation

Answer: (c) 5%

Explanation:
1. Growth in the 4th year = 40 on 800
2. \( R = 40/800 \times 100 \) = 5%.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 10: A sum of money invested at compound interest amounts to Rs. 4624 in 2 years and to Rs. 4913 in 3 years. The sum of money is
(a) Rs. 4096
(b) Rs. 4260
(c) Rs. 4335
(d) Rs. 4360
Show Answer & Explanation

Answer: (a) Rs. 4096

Explanation:
1. Rate = \( 289/4624 \times 100 \) = 6.25%
2. \( P = 4624/(17/16)^{2} \) = Rs. 4096.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 11: If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 27 times in
(a) 8 years
(b) 12 years
(c) 24 years
(d) 36 years
Show Answer & Explanation

Answer: (b) 12 years

Explanation:
1. \( 27 = 3^{3} \), so the time = \( 3 \times 4 = 12 \) years.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 12: The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. \( 1.2^{3} = 1.728 \lt 2 \) and \( 1.2^{4} = 2.0736 \gt 2 \)
2. The least number of years is 4.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Question 13: A sum of money is borrowed and paid back in two annual instalments of Rs. 882 each allowing 5% compound interest. The sum borrowed was
(a) Rs. 1620
(b) Rs. 1640
(c) Rs. 1680
(d) Rs. 1700
Show Answer & Explanation

Answer: (b) Rs. 1640

Explanation:
1. Sum = \( 882/1.05 + 882/1.05^{2} = 840 + 800 \) = Rs. 1640.

Teacher's Note:
1. Read which sum is the principal and which is the amount.
2. Break the problem into simple one-step calculations.

Question 14: At what rate of simple interest per annum does the interest on Rs. 1200 in 2 years equal the interest on Rs. 600 at 7/2% p.a. for 4 years?
(a) 3/4%
(b) 7/2%
(c) 4/3%
(d) 7/8%
Show Answer & Explanation

Answer: (b) 7/2%

Explanation:
1. Interest on \( 600 = 600 \times 3.5 \times 4/100 \) = Rs. 84
2. \( 1200 \times R \times 2/100 = 84 \), so \( R \) = 3.5% = 7/2%.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 15: Raju invested a sum of Rs. 5832 at a rate of interest n% per annum, compounded annually. Find the value of n if he received Rs. 13824 after 3 years.
(a) 33 1/3
(b) 33 2/3
(c) 33 4/3
(d) 33 5/3
Show Answer & Explanation

Answer: (a) 33 1/3

Explanation:
1. \( 13824/5832 = 2.3704 = (4/3)^{3} \)
2. The rate is \( 1/3 = 33 \) 1/3%.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 16: A boy's height is increasing at the rate of 2% over that of the previous year. If his present height is 156.06 cm, what was his height two years ago?
(a) 148 cm
(b) 150 cm
(c) 152 cm
(d) 151 cm
Show Answer & Explanation

Answer: (b) 150 cm

Explanation:
1. Height two years ago = \( 156.06/1.02^{2} = 150 \) cm.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Question 17: Ramu invested a sum of Rs. 12500 at 12% per annum compound interest. He received an amount of Rs. 15680 after x years. Then the value of x is
(a) 1
(b) 4
(c) 3
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. \( 15680/12500 = 1.2544 = 1.12^{2} \)
2. \( x = 2 \).

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Question 18: The rate of interest for a sum that becomes 729/576 times itself in 2 years, when compounded annually, is
(a) 32/5%
(b) 12/5%
(c) 25/2%
(d) 12/7%
Show Answer & Explanation

Answer: (c) 25/2%

Explanation:
1. \( 729/576 = (27/24)^{2} = (9/8)^{2} \)
2. Rate = 1/8 = 12.5% = 25/2%.

Teacher's Note:
1. Interest rate questions are solved from the ratio of amounts.
2. Spot perfect squares or cubes in the ratio.

Question 19: A sum amounts to Rs. 3600 at 2% p.a. under simple interest and Rs. 4800 at 4% p.a. under simple interest. The time taken is
(a) 2.5 years
(b) 3 years
(c) 30 years
(d) 25 years
Show Answer & Explanation

Answer: (d) 25 years

Explanation:
1. \( P(1 + 0.02t) = 3600 \) and \( P(1 + 0.04t) = 4800 \)
2. \( (1 + 0.04t)/(1 + 0.02t) = 4/3 \), so \( 3 + 0.12t = 4 + 0.08t \) and \( t = 25 \) years.

Teacher's Note:
1. Simple interest = \( P \times R \times T/100 \) and amount = \( P \) + SI.
2. Keep time in years and the rate per annum.

Question 20: Kailash set up a factory by investing Rs. 1000000. During the first two years, his profits were 10% and 15% respectively. If he reinvested the profit of each year at the beginning of the next year, his total profit (in Rs.) is
(a) 265000
(b) 25000
(c) 275000
(d) 27060
Show Answer & Explanation

Answer: (a) 265000

Explanation:
1. After year 1 the capital = 1100000
2. After year \( 2 = 1100000 \times 1.15 = 1265000 \)
3. Total profit = Rs. 265000.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

Question 21: Suresh and Naresh borrowed Rs. 62500 and Rs. 60000 respectively for a period of 2 years. Suresh paid simple interest at the rate of 4% per annum, while Naresh paid compound interest at the same rate compounded annually. Who paid more interest and by how much?
(a) Naresh paid more by Rs. 104
(b) Suresh paid more by Rs. 104
(c) Naresh paid more by Rs. 94
(d) Both paid the same interest
Show Answer & Explanation

Answer: (b) Suresh paid more by Rs. 104

Explanation:
1. Suresh's interest = \( 62500 \times 0.04 \times 2 = 5000 \)
2. Naresh's interest = \( 60000 \times (1.04^{2} - 1) = 4896 \)
3. Suresh paid Rs. 104 more.

Teacher's Note:
1. The difference between CI and SI for 2 years is \( P(R/100)^{2} \).
2. The difference for 3 years is \( P(R/100)^{2}(3 + R/100) \).

Question 22: A doctor wants to divide Rs. 145000 between his son and daughter who are 12 years and 14 years respectively, in such a way that the sum invested at 12 1/2% p.a. compounded annually will give the same amount to each when they attain 16 years. How should he divide the sum?
(a) Rs. 81000 to son and Rs. 64000 to daughter
(b) Rs. 64000 to son and Rs. 81000 to daughter
(c) Rs. 45000 to son and Rs. 100000 to daughter
(d) Rs. 100000 to son and Rs. 45000 to daughter
Show Answer & Explanation

Answer: (b) Rs. 64000 to son and Rs. 81000 to daughter

Explanation:
1. The son's money grows for 4 years and the daughter's for 2 years
2. \( S \times (9/8)^{4} = D \times (9/8)^{2} \), so \( S = D \times 64/81 \)
3. Dividing 145000 in the ratio 64 : 81 gives son Rs. 64000 and daughter Rs. 81000.

Teacher's Note:
1. Compound interest uses amount = \( P(1 + R/100)^n \) and CI = amount \( - P \).
2. Compounding at a different frequency changes both rate and number of periods.

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Multiple Choice Questions (MCQs) for Class 10 Mathematics Simple and Compound Interest

Class 10 Mathematics Simple and Compound Interest Objective Test Questions

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