8.4 Compound Interest -- Practice MCQs

Instructions: Choose the best answer. Try to solve each problem in under 90 seconds. Answers with detailed explanations are at the end of each question.

Basic Level (Q1-Q15)

Q1.

Find the compound interest on Rs. 8000 at 10% per annum for 2 years, compounded annually.

(a) Rs. 1600 (b) Rs. 1680 (c) Rs. 1700 (d) Rs. 1760

A = 8000 (1.1)^2 = 8000 x 1.21 = 9680
CI = 9680 - 8000 = Rs. 1680

Q2.

What will Rs. 15,000 amount to at 20% per annum compound interest in 2 years?

(a) Rs. 21,000 (b) Rs. 21,200 (c) Rs. 21,600 (d) Rs. 22,000

A = 15000 (1.2)^2 = 15000 x 1.44 = Rs. 21,600

Q3.

The compound interest on Rs. 5000 for 1 year at 8% per annum is:

(a) Rs. 400 (b) Rs. 416 (c) Rs. 440 (d) Rs. 408

For 1 year, CI = SI = 5000 x 8 / 100 = Rs. 400

Q4.

Rs. 6000 becomes Rs. 7260 in 2 years at compound interest. The rate is:

(a) 8% (b) 9% (c) 10% (d) 11%

(1 + R/100)^2 = 7260/6000 = 1.21
1 + R/100 = 1.1
R = 10%

Q5.

The difference between CI and SI on Rs. 10,000 at 5% for 2 years is:

(a) Rs. 20 (b) Rs. 25 (c) Rs. 30 (d) Rs. 50

CI - SI = P(R/100)^2 = 10000 x (5/100)^2 = 10000 x 1/400 = Rs. 25

Q6.

The population of a town is 50,000. It increases at 10% per annum. The population after 2 years is:

(a) 55,000 (b) 60,000 (c) 60,500 (d) 61,000

Population = 50000 (1.1)^2 = 50000 x 1.21 = 60,500

Q7.

A machine worth Rs. 40,000 depreciates at 10% per annum. Its value after 1 year is:

(a) Rs. 36,000 (b) Rs. 38,000 (c) Rs. 32,400 (d) Rs. 34,000

Value = 40000 (1 - 10/100)^1 = 40000 x 0.9 = Rs. 36,000

Q8.

The compound interest on Rs. 1000 at 10% for 3 years is:

(a) Rs. 300 (b) Rs. 310 (c) Rs. 330 (d) Rs. 331

A = 1000 (1.1)^3 = 1000 x 1.331 = 1331
CI = 1331 - 1000 = Rs. 331

Q9.

Rs. 1600 invested at 25% per annum compound interest will give what amount after 2 years?

(a) Rs. 2400 (b) Rs. 2500 (c) Rs. 2600 (d) Rs. 2000

A = 1600 (1.25)^2 = 1600 x 1.5625 = Rs. 2500

Q10.

At what rate of compound interest will Rs. 10,000 become Rs. 12,100 in 2 years?

(a) 8% (b) 10% (c) 12% (d) 15%

(1 + R/100)^2 = 12100/10000 = 1.21
1 + R/100 = 1.1
R = 10%

Q11.

In how many years will Rs. 8000 become Rs. 9261 at 5% per annum CI?

(a) 2 years (b) 3 years (c) 4 years (d) 5 years

(1.05)^T = 9261/8000 = 1.157625
(1.05)^3 = 1.157625
T = 3 years

Q12.

Find the CI on Rs. 16,000 at 15% per annum for 2 years.

(a) Rs. 4800 (b) Rs. 5160 (c) Rs. 5000 (d) Rs. 5016

A = 16000 (1.15)^2 = 16000 x 1.3225 = 21160
CI = 21160 - 16000 = Rs. 5160

Q13.

If CI on a sum is Rs. 410 and SI is Rs. 400 for 2 years, the rate of interest is:

(a) 4% (b) 5% (c) 10% (d) 8%

CI - SI = 410 - 400 = 10
SI for 1 year = 400/2 = 200

Rate = (CI - SI) / (SI for 1 year) x 100
     = 10/200 x 100 = 5%

Q14.

The compound interest on a certain sum for 2 years at 10% is Rs. 525. The sum is:

(a) Rs. 2000 (b) Rs. 2500 (c) Rs. 3000 (d) Rs. 3500

CI = P[(1 + R/100)^2 - 1]
525 = P[(1.1)^2 - 1]
525 = P[1.21 - 1]
525 = P x 0.21
P = 525/0.21 = Rs. 2500

Q15.

A sum amounts to Rs. 13,230 in 2 years at 5% per annum CI. The sum is:

(a) Rs. 10,000 (b) Rs. 11,000 (c) Rs. 12,000 (d) Rs. 12,500

P = A / (1 + R/100)^T
P = 13230 / (1.05)^2
P = 13230 / 1.1025
P = Rs. 12,000

Medium Level (Q16-Q30)

Q16.

Find the CI on Rs. 10,000 at 12% per annum compounded half-yearly for 1 year.

(a) Rs. 1200 (b) Rs. 1236 (c) Rs. 1255 (d) Rs. 1260

Half-yearly rate = 6%, periods = 2
A = 10000 (1.06)^2 = 10000 x 1.1236 = 11236
CI = 11236 - 10000 = Rs. 1236

Q17.

The difference between CI and SI on Rs. 5000 at 4% for 3 years is:

(a) Rs. 24.00 (b) Rs. 24.32 (c) Rs. 25.00 (d) Rs. 26.40

CI - SI = P(R/100)^2(3 + R/100)
        = 5000 x (4/100)^2 x (3 + 4/100)
        = 5000 x 0.0016 x 3.04
        = 5000 x 0.004864
        = Rs. 24.32

Q18.

A sum of Rs. 12,000 is deposited at 10% CI. The interest for the 3rd year is:

(a) Rs. 1200 (b) Rs. 1320 (c) Rs. 1452 (d) Rs. 1440

Interest for the 3rd year = Amount at end of Year 3 - Amount at end of Year 2

Amount after 2 years = 12000 (1.1)^2 = 12000 x 1.21 = 14520
Amount after 3 years = 12000 (1.1)^3 = 12000 x 1.331 = 15972

Interest for 3rd year = 15972 - 14520 = Rs. 1452

Alternatively: Interest for 3rd year = 10% of 14520 = Rs. 1452

Q19.

The CI on a sum for the 2nd year is Rs. 1320 and for the 3rd year is Rs. 1452. The rate is:

(a) 8% (b) 10% (c) 12% (d) 15%

Rate = [(CI_3rd - CI_2nd) / CI_2nd] x 100
     = [(1452 - 1320) / 1320] x 100
     = (132/1320) x 100
     = 10%

Q20.

A car depreciates at 20% per annum. After 3 years its value is Rs. 2,56,000. The original price was:

(a) Rs. 4,00,000 (b) Rs. 4,50,000 (c) Rs. 5,00,000 (d) Rs. 5,50,000

256000 = V (0.8)^3
256000 = V x 0.512
V = 256000/0.512 = Rs. 5,00,000

Q21.

The effective rate of 20% compounded semi-annually is:

(a) 20% (b) 20.5% (c) 21% (d) 21.5%

E = (1 + 20/200)^2 - 1
  = (1.1)^2 - 1
  = 1.21 - 1
  = 0.21 = 21%

Q22.

The population of a village was 10,000 two years ago. It has increased by 20% per annum. The present population is:

(a) 12,000 (b) 14,000 (c) 14,400 (d) 12,400

Population = 10000 (1.2)^2 = 10000 x 1.44 = 14,400

Q23.

Rs. 20,000 is invested at 10% per annum compounded quarterly. The amount after 6 months is:

(a) Rs. 20,500.00 (b) Rs. 21,000.00 (c) Rs. 21,012.50 (d) Rs. 20,050.00

Quarterly rate = 10/4 = 2.5%
Number of quarters in 6 months = 2

A = 20000 (1 + 2.5/100)^2
A = 20000 (1.025)^2
A = 20000 x 1.050625
A = Rs. 21,012.50

Q24.

The SI on a certain sum for 2 years is Rs. 2400 and CI is Rs. 2544. The rate is:

(a) 10% (b) 12% (c) 15% (d) 8%

CI - SI = 2544 - 2400 = 144
SI for 1 year = 2400/2 = 1200

Rate = (144 / 1200) x 100 = 12%

Q25.

Rs. 800 at 5% per annum CI. The amount after 3 years is:

(a) Rs. 920 (b) Rs. 926.10 (c) Rs. 930 (d) Rs. 940

A = 800 (1.05)^3
  = 800 x 1.157625
  = Rs. 926.10

Q26.

The difference between CI and SI on a sum at 10% for 2 years is Rs. 25. The sum is:

(a) Rs. 2000 (b) Rs. 2500 (c) Rs. 3000 (d) Rs. 5000

CI - SI = P(R/100)^2
25 = P x (10/100)^2
25 = P x 0.01
P = Rs. 2500

Q27.

A sum is invested at CI. It amounts to Rs. 2420 in 2 years and Rs. 2662 in 3 years. The rate is:

(a) 8% (b) 10% (c) 12% (d) 15%

Interest for 3rd year = 2662 - 2420 = 242
This is R% of 2420:
R = (242/2420) x 100 = 10%

Q28.

The CI on Rs. 6400 at 12.5% for 2 years is:

(a) Rs. 1600 (b) Rs. 1700 (c) Rs. 1800 (d) Rs. 1500

12.5% = 1/8, multiplier = 9/8

Year 1: 6400 x 9/8 = 7200
Year 2: 7200 x 9/8 = 8100

CI = 8100 - 6400 = Rs. 1700

Q29.

The population of a city increases by 10% in the 1st year and decreases by 10% in the 2nd year. If the present population is 99,000, what was it 2 years ago?

(a) 98,000 (b) 99,000 (c) 1,00,000 (d) 1,01,000

Let original population = P
P x (1.1) x (0.9) = 99000
P x 0.99 = 99000
P = 99000/0.99 = 1,00,000

Q30.

Rs. 5000 is invested at 8% CI compounded half-yearly. The CI for 1 year is:

(a) Rs. 400 (b) Rs. 408 (c) Rs. 416 (d) Rs. 412

Half-yearly rate = 4%, periods = 2
A = 5000 (1.04)^2 = 5000 x 1.0816 = 5408
CI = 5408 - 5000 = Rs. 408

Advanced Level (Q31-Q42)

Q31.

The CI on a certain sum at 10% per annum for 2 years (compounded annually) is Rs. 6,300. The sum is:

(a) Rs. 25,000 (b) Rs. 28,000 (c) Rs. 30,000 (d) Rs. 32,000

CI = P[(1 + R/100)^2 - 1]
6300 = P[(1.1)^2 - 1]
6300 = P[1.21 - 1]
6300 = P x 0.21
P = 6300 / 0.21
P = Rs. 30,000

Q32.

A certain sum becomes 8 times in 3 years at compound interest. The rate is:

(a) 50% (b) 100% (c) 200% (d) 150%

(1 + R/100)^3 = 8
1 + R/100 = 8^(1/3) = 2
R/100 = 1
R = 100%

Q33.

If a sum doubles in 5 years at CI, in how many years will it become 16 times?

(a) 15 years (b) 20 years (c) 25 years (d) 30 years

If sum doubles in 5 years: (1 + R/100)^5 = 2

16 = 2^4

We need (1 + R/100)^T = 16 = 2^4

Since (1 + R/100)^5 = 2, raising both sides to the power of 4:
  [(1 + R/100)^5]^4 = 2^4
  (1 + R/100)^20 = 16

T = 20 years

Q34.

The CI on Rs. 30,000 at 7% per annum is Rs. 4347. The period is:

(a) 1 year (b) 2 years (c) 3 years (d) 2.5 years

A = 30000 + 4347 = 34347
(1 + 7/100)^T = 34347/30000 = 1.1449

(1.07)^2 = 1.1449
T = 2 years

Q35.

The difference between CI and SI on Rs. 40,000 at 10% for 3 years is:

(a) Rs. 1200 (b) Rs. 1220 (c) Rs. 1240 (d) Rs. 1210

CI - SI = P(R/100)^2(3 + R/100)
        = 40000 x (0.1)^2 x 3.1
        = 40000 x 0.01 x 3.1
        = Rs. 1240

Q36.

Rs. 25,000 at 12% CI compounded quarterly. The CI for 9 months is:

(a) Rs. 2250.00 (b) Rs. 2292.03 (c) Rs. 2300.00 (d) Rs. 2318.18

Quarterly rate = 12/4 = 3%
Number of quarters in 9 months = 3

A = 25000 (1.03)^3

(1.03)^2 = 1.0609
(1.03)^3 = 1.0609 x 1.03 = 1.092727

A = 25000 x 1.092727 = 27318.18
CI = 27318.18 - 25000 = Rs. 2318.18

Q37.

A bacteria culture doubles every 3 hours. If the initial count is 500, what will be the count after 15 hours?

(a) 8,000 (b) 12,000 (c) 16,000 (d) 32,000

Number of doublings in 15 hours = 15/3 = 5

Count = 500 x 2^5 = 500 x 32 = 16,000

Q38.

A machine depreciates at 10% per annum. In how many years will its value become 72.9% of its original value?

(a) 2 years (b) 3 years (c) 4 years (d) 5 years

(1 - 10/100)^T = 72.9/100 = 0.729
(0.9)^T = 0.729
(0.9)^3 = 0.729

T = 3 years

Q39.

If the CI on a sum for 2 years at 20% p.a. is Rs. 8800, the sum is:

(a) Rs. 18,000 (b) Rs. 20,000 (c) Rs. 22,000 (d) Rs. 25,000

CI = P[(1 + R/100)^2 - 1]
8800 = P[(1.2)^2 - 1]
8800 = P[1.44 - 1]
8800 = 0.44P
P = 8800/0.44 = Rs. 20,000

Q40.

Rs. 10,000 is lent at 10% CI compounded annually. After 1 year, Rs. 3000 is returned. The amount outstanding at the end of 2 years is:

(a) Rs. 8800 (b) Rs. 8700 (c) Rs. 8000 (d) Rs. 8580

After 1 year: 10000 x 1.1 = 11000
After repayment of 3000: Outstanding = 11000 - 3000 = 8000
After 2nd year: 8000 x 1.1 = Rs. 8800

Q41.

The CI on Rs. 50,000 for 3 years if the rate is 5% for the first year, 10% for the second year, and 20% for the third year is:

(a) Rs. 19,300 (b) Rs. 18,900 (c) Rs. 19,100 (d) Rs. 19,000

A = 50000 x (1.05) x (1.10) x (1.20)

Step by step:
  50000 x 1.05 = 52500
  52500 x 1.10 = 57750
  57750 x 1.20 = 69300

CI = 69300 - 50000 = Rs. 19,300

Q42.

A sum of Rs. 10,000 is borrowed at 10% CI and is paid back in 2 equal annual installments. The value of each installment is approximately:

(a) Rs. 5500 (b) Rs. 5762 (c) Rs. 6000 (d) Rs. 6100

Let each installment = X

10000 = X/(1.1) + X/(1.1)^2
10000 = X/1.1 + X/1.21

LCM approach:
10000 = (1.21X + 1.1X) / (1.1 x 1.21)
10000 = 2.31X / 1.331
10000 x 1.331 = 2.31X
13310 = 2.31X
X = 13310/2.31
X = Rs. 5762.34 (approx.)

Each installment ≈ Rs. 5762

Answer Key (Quick Reference)

Difficulty Distribution