Practice Compound Interest questions covering principal, rate, time, annual and half-yearly compounding, and the difference between compound and simple interest for Quantitative Aptitude.
Question 1
Find the compound interest on Rs. 5,000 for 2 years at 10% per annum, interest being compounded annually.
A
Rs. 1,000✓✗
B
Rs. 1,050✓✗
C
Rs. 1,100✓✗
D
Rs. 1,200✓✗
Explanation
Amount = 5000 × 1.1 × 1.1 = 5000 × 1.21 = Rs. 6,050.
Compound interest = 6050 - 5000 = Rs. 1,050.
Shortcut Method
Ratio = 11 : 10
5000 →(×11/10)→ 5500 →(×11/10)→ 6050
∴ CI = 6050-5000 = Rs.1,050 ✓
Question 2
What amount will Rs. 8,000 become in 2 years if invested at 5% per annum compound interest?
A
Rs. 8,800✓✗
B
Rs. 8,850✓✗
C
Rs. 8,820✓✗
D
Rs. 8,720✓✗
Explanation
Compound interest for 2 years means the amount is multiplied by (1 + rate) once for every year.
Amount after year 1 = 8000 × 1.05 = Rs. 8,400.
Amount after year 2 = 8400 × 1.05 = Rs. 8,820.
Shortcut Method
Ratio = 21 : 20 (5%)
8000 →(×21/20)→ 8400 →(×21/20)→ 8820
∴ Amount = Rs.8,820 ✓
Question 3
A sum of Rs. 10,000 is invested at 10% per annum, compounded half-yearly. Find the compound interest earned in 1 year.
A
Rs. 1,000✓✗
B
Rs. 1,025✓✗
C
Rs. 1,050✓✗
D
Rs. 1,100✓✗
Explanation
Compounded half-yearly means the rate per half-year is 5%, applied twice in 1 year.
Amount = 10000 × 1.05 × 1.05 = 10000 × 1.1025 = Rs. 11,025.
Compound interest = 11025 - 10000 = Rs. 1,025.
Shortcut Method
Half-yearly → rate/period = 5%, periods = 2
Ratio = 21 : 20
10000 →(×21/20)→ 10500 →(×21/20)→ 11025
∴ CI = Rs.1,025 ✓
Question 4
Find the difference between the compound interest and the simple interest on Rs. 4,000 for 2 years at 5% per annum.
A
Rs. 8✓✗
B
Rs. 10✓✗
C
Rs. 15✓✗
D
Rs. 20✓✗
Explanation
Simple interest for 2 years = 4000 × 5 × 2 / 100 = Rs. 400.
Compound interest for 2 years = 4000 × (1.05 × 1.05 - 1) = 4000 × 0.1025 = Rs. 410.
Difference = 410 - 400 = Rs. 10.
Shortcut Method
2-year CI-SI diff = P(r/100)²
= 4000 × (5/100)² = 4000 × 1/400
∴ Difference = Rs.10 ✓
Question 5
The compound interest on a certain sum of money for 2 years at 10% per annum is Rs. 630. Find the sum.
A
Rs. 2,800✓✗
B
Rs. 3,000✓✗
C
Rs. 3,200✓✗
D
Rs. 3,500✓✗
Explanation
Let the sum be P. Amount factor for 2 years at 10% = 1.1 × 1.1 = 1.21.
Compound interest = P × (1.21 - 1) = 0.21P.
0.21P = 630, so P = 630 / 0.21 = Rs. 3,000.
Shortcut Method
CI = P[(1+r/100)²-1]
630 = P × 0.21
∴ P = 630/0.21 = Rs.3,000 ✓
Question 6
A sum of Rs. 2,000 amounts to Rs. 2,420 in 2 years, interest being compounded annually. Find the rate of interest per annum.
A
8%✓✗
B
10%✓✗
C
12%✓✗
D
15%✓✗
Explanation
Amount factor for 2 years = 2420 / 2000 = 1.21.
So (1 + r/100)2 = 1.21, giving 1 + r/100 = 1.1, so r = 10%.
Shortcut Method
Amount factor = 2420/2000 = 1.21 = (1.1)²
∴ Rate = 10% ✓
Question 7
The value of a machine depreciates at the rate of 10% per annum. If its present value is Rs. 8,000, find its value after 2 years.
A
Rs. 6,400✓✗
B
Rs. 6,480✓✗
C
Rs. 6,500✓✗
D
Rs. 7,200✓✗
Explanation
Depreciation reduces the value by 10% each year, so the value is multiplied by 0.9 each year.
Value after 2 years = 8000 × 0.9 × 0.9 = 8000 × 0.81 = Rs. 6,480.
Shortcut Method
Ratio = 9 : 10 (10% fall)
8000 →(×9/10)→ 7200 →(×9/10)→ 6480
∴ Value = Rs.6,480 ✓
Question 8
The population of a town is 20,000. If it increases at the rate of 5% per annum, find the population after 2 years.
A
21,500✓✗
B
22,000✓✗
C
22,050✓✗
D
22,100✓✗
Explanation
The population grows by 5% each year, so it is multiplied by 1.05 once for every year.
Population after year 1 = 20000 × 1.05 = 21,000.
Population after year 2 = 21000 × 1.05 = 22,050.
Shortcut Method
Ratio = 21 : 20 (5%)
20000 →(×21/20)→ 21000 →(×21/20)→ 22050
∴ Population = 22,050 ✓
Question 9
Calculate the compound interest on Rs. 6,000 for 3 years at 10% per annum, interest compounded annually.
A
Rs. 1,866✓✗
B
Rs. 1,890✓✗
C
Rs. 1,986✓✗
D
Rs. 2,000✓✗
Explanation
Amount = 6000 × 1.1 × 1.1 × 1.1 = 6000 × 1.331 = Rs. 7,986.
Compound interest = 7986 - 6000 = Rs. 1,986.
Shortcut Method
Ratio = 11 : 10
6000 →(×11/10)→ 6600 →(×11/10)→ 7260 →(×11/10)→ 7986
∴ CI = 7986-6000 = Rs.1,986 ✓
Question 10
Find the difference between the compound interest and the simple interest on Rs. 5,000 for 3 years at 10% per annum.
A
Rs. 140✓✗
B
Rs. 150✓✗
C
Rs. 155✓✗
D
Rs. 165✓✗
Explanation
Simple interest for 3 years = 5000 × 10 × 3 / 100 = Rs. 1,500.
Amount at compound interest = 5000 × 1.13 = 5000 × 1.331 = Rs. 6,655, so compound interest = Rs. 1,655.
Difference = 1655 - 1500 = Rs. 155.
Shortcut Method
3-year CI-SI diff = P(r/100)²(3+r/100)
= 5000 × 0.01 × 3.1
∴ Difference = Rs.155 ✓
Question 11
A sum of Rs. 10,000 is invested for 2 years at compound interest, with the rate being 10% for the first year and 20% for the second year. Find the amount at the end of 2 years.
A
Rs. 12,800✓✗
B
Rs. 13,000✓✗
C
Rs. 13,200✓✗
D
Rs. 13,500✓✗
Explanation
Amount after year 1 = 10000 × 1.1 = Rs. 11,000.
Amount after year 2 = 11000 × 1.2 = Rs. 13,200.
Shortcut Method
10000 →(×1.1)→ 11000 →(×1.2)→ 13200
∴ Amount = Rs.13,200 ✓
Question 12
Find the compound interest on Rs. 10,000 for 6 months at 8% per annum, interest being compounded quarterly.
A
Rs. 400✓✗
B
Rs. 404✓✗
C
Rs. 410✓✗
D
Rs. 420✓✗
Explanation
Compounded quarterly at 8% per annum means the rate per quarter is 2%. 6 months = 2 quarters.
Amount = 10000 × 1.02 × 1.02 = 10000 × 1.0404 = Rs. 10,404.
Compound interest = 10404 - 10000 = Rs. 404.
Shortcut Method
Quarterly → rate/period = 2%, periods = 2
Ratio = 51 : 50
10000 →(×51/50)→ 10200 →(×51/50)→ 10404
∴ CI = Rs.404 ✓
Question 13
In how many years will a sum of Rs. 8,000 amount to Rs. 27,000 at 50% per annum compound interest?
A
2 years✓✗
B
3 years✓✗
C
4 years✓✗
D
5 years✓✗
Explanation
Required growth factor = 27000 / 8000 = 3.375.
At 50% per annum, the growth factor per year is 1.5, and 1.53 = 3.375.
So the sum takes 3 years to reach Rs. 27,000.
Shortcut Method
27000/8000 = 3.375
1.5² = 2.25, 1.5³ = 3.375 ✓ match
∴ Time = 3 years ✓
Question 14
The population of a village was 40,000. Due to migration, it decreases at the rate of 5% per annum. Find the population after 2 years.
A
35,000✓✗
B
36,000✓✗
C
36,100✓✗
D
37,000✓✗
Explanation
The population falls by 5% each year, so it is multiplied by 0.95 once for every year.
Population after year 1 = 40000 × 0.95 = 38,000.
Population after year 2 = 38000 × 0.95 = 36,100.
Shortcut Method
Ratio = 19 : 20 (5% fall)
40000 →(×19/20)→ 38000 →(×19/20)→ 36100
∴ Population = 36,100 ✓
Question 15
A sum of money, invested at compound interest, amounts to Rs. 4,410 in 2 years and Rs. 4,630.50 in 3 years. Find the rate of interest per annum.
A
4%✓✗
B
5%✓✗
C
6%✓✗
D
8%✓✗
Explanation
The amount for the 3rd year is obtained from the amount for the 2nd year by multiplying by (1 + r/100) once more.
So (1 + r/100) = 4630.50 / 4410 = 1.05, giving r = 5%.
Shortcut Method
Year3/Year2 = 4630.50/4410 = 1.05
∴ Rate = 5% ✓
Question 16
The difference between the compound interest and the simple interest on a certain sum for 2 years at 8% per annum is Rs. 48. Find the sum.
A
Rs. 7,000✓✗
B
Rs. 7,200✓✗
C
Rs. 7,500✓✗
D
Rs. 8,000✓✗
Explanation
For 2 years, the difference between CI and SI equals P × (r/100)2.
So P × (0.08)2 = 48, giving P × 0.0064 = 48, so P = 48 / 0.0064 = Rs. 7,500.
Shortcut Method
2-year CI-SI diff = P(r/100)²
48 = P × (0.08)² = P × 0.0064
∴ P = 48/0.0064 = Rs.7,500 ✓
Question 17
The simple interest on a sum for 2 years at a certain rate of interest is Rs. 1,000. The difference between the compound interest and the simple interest for the same sum, period and rate is Rs. 25. Find the rate of interest per annum.
A
4%✓✗
B
5%✓✗
C
6%✓✗
D
10%✓✗
Explanation
Simple interest for 2 years = 2Pr/100 = 1000, so Pr = 50,000.
The CI-SI difference for 2 years = P(r/100)2 = 25, so Pr2 = 250,000.
Dividing, r = Pr2 / Pr = 250000 / 50000 = 5, so the rate is 5%.
Shortcut Method
SI(2yr) = 2Pr/100 = 1000 → Pr = 50,000
diff = P(r/100)² = 25 → Pr² = 250,000
r = Pr²/Pr = 250000/50000
∴ Rate = 5% ✓
Question 18
Find the compound interest on Rs. 6,400 for 2 years at 1212% per annum.
A
Rs. 1,600✓✗
B
Rs. 1,700✓✗
C
Rs. 1,750✓✗
D
Rs. 1,800✓✗
Explanation
12½% = 18, so the amount factor for 1 year is 9/8.
Amount = 6400 × 98 × 98 = 6400 × 8164 = Rs. 8,100.
Compound interest = 8100 - 6400 = Rs. 1,700.
Shortcut Method
12½% = 1/8 → ratio = 9 : 8
6400 →(×9/8)→ 7200 →(×9/8)→ 8100
∴ CI = 8100-6400 = Rs.1,700 ✓
Question 19
Find the compound interest on Rs. 8,000 for 2 years 4 months at 15% per annum, interest being compounded annually.
A
Rs. 3,000✓✗
B
Rs. 3,100✓✗
C
Rs. 3,109✓✗
D
Rs. 3,200✓✗
Explanation
For the first 2 years, amount = 8000 × 1.15 × 1.15 = 8000 × 1.3225 = Rs. 10,580.
For the remaining 4 months (one-third of a year), simple interest is applied on this amount: extra interest = 10580 × 15 × (13) / 100 = Rs. 529.
Total amount = 10580 + 529 = Rs. 11,109.
Compound interest = 11109 - 8000 = Rs. 3,109.
Shortcut Method
2 years: 8000 →(×1.15)→ 9200 →(×1.15)→ 10580
4 months = 1/3 yr → extra SI = 10580×15×(1/3)/100 = 529
Total = 10580+529 = 11109
∴ CI = 11109-8000 = Rs.3,109 ✓
Question 20
The difference between the compound interest and the simple interest on a sum for 3 years at 10% per annum is Rs. 93. Find the sum.
A
Rs. 2,800✓✗
B
Rs. 3,000✓✗
C
Rs. 3,200✓✗
D
Rs. 3,500✓✗
Explanation
For 3 years, the CI-SI difference equals P × (r/100)2 × (3 + r/100).
With r = 10%: difference = P × 0.01 × 3.1 = 0.031P.
0.031P = 93, so P = 93 / 0.031 = Rs. 3,000.
Shortcut Method
3-year CI-SI diff = P(r/100)²(3+r/100)
93 = P × 0.01 × 3.1 = 0.031P
∴ P = 93/0.031 = Rs.3,000 ✓
Question 21
Rs. 10,000 is invested for 1 year at 10% per annum. Find the difference between the compound interest earned when it is compounded annually and when it is compounded half-yearly.
A
Rs. 20✓✗
B
Rs. 25✓✗
C
Rs. 30✓✗
D
Rs. 50✓✗
Explanation
Compounded annually: amount = 10000 × 1.1 = Rs. 11,000, so CI = Rs. 1,000.
Compounded half-yearly (5% per half-year, 2 periods): amount = 10000 × 1.05 × 1.05 = 10000 × 1.1025 = Rs. 11,025, so CI = Rs. 1,025.
Difference = 1025 - 1000 = Rs. 25.
Shortcut Method
Annual: 10000×1.1 = 11000 (CI=1000)
Half-yearly: 10000×1.05×1.05 = 11025 (CI=1025)
∴ Difference = Rs.25 ✓
Question 22
The value of a share increases by 20% in the first year and decreases by 20% in the second year. Find the net percentage change in its value over the two years.
A
No change✓✗
B
4% increase✓✗
C
4% decrease✓✗
D
5% decrease✓✗
Explanation
Let the original value be 100. After a 20% increase, value = 100 × 1.2 = 120.
After a 20% decrease, value = 120 × 0.8 = 96.
Net change = 96 - 100 = -4, that is, a 4% decrease. A rise and an equal fall never cancel out exactly, because the fall is applied to a larger base.
Shortcut Method
100 →(×1.2)→ 120 →(×0.8)→ 96
∴ Net change = 4% decrease ✓
Question 23
A sum of Rs. 4,000 amounts to Rs. 4,410 in 1 year, interest being compounded half-yearly. Find the rate of interest per annum.
A
5%✓✗
B
8%✓✗
C
10%✓✗
D
20%✓✗
Explanation
With half-yearly compounding over 1 year (2 periods), the growth factor is (1 + r/200)2, where r is the annual rate.
(1 + r/200)2 = 4410 / 4000 = 1.1025, so 1 + r/200 = 1.05, giving r/200 = 0.05, so r = 10%.
Shortcut Method
(1+r/200)² = 4410/4000 = 1.1025
1+r/200 = 1.05 → r/200 = 0.05
∴ Rate = 10% ✓
Question 24
A sum of money, invested at compound interest, amounts to Rs. 1,600 in 2 years and Rs. 2,500 in 4 years. Find the rate of interest per annum.
A
15%✓✗
B
20%✓✗
C
25%✓✗
D
30%✓✗
Explanation
Going from year 2 to year 4 is 2 more years of growth, so (1 + r/100)2 = 2500 / 1600 = 1.5625.
Taking the square root: 1 + r/100 = 1.25, so r = 25%.
Shortcut Method
Gap of 2 years: (1+r/100)² = 2500/1600 = 1.5625
√1.5625 = 1.25
∴ Rate = 25% ✓
Question 25
For a certain sum of money, the simple interest for 3 years exceeds the compound interest for 2 years, both at 10% per annum, by Rs. 900. Find the sum.
A
Rs. 9,000✓✗
B
Rs. 10,000✓✗
C
Rs. 11,000✓✗
D
Rs. 12,000✓✗
Explanation
Simple interest for 3 years at 10% = 3Pr/100 = 0.3P.
Compound interest for 2 years at 10% = P × (1.21 - 1) = 0.21P.
Given: 0.3P - 0.21P = 900, so 0.09P = 900, giving P = Rs. 10,000.
Shortcut Method
SI(3yr) = 0.3P, CI(2yr) = 0.21P
0.3P - 0.21P = 900 → 0.09P = 900
∴ P = Rs.10,000 ✓
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