IBPS PO General Awareness: Compound Interest questions with solutions

Q1. A sum amounts to ₹2410 in 3 years and ₹2651 in 4 years at compound interest. Find the principal.

1. ₹1216
2. ₹1500
3. ₹1900
4. ₹1600

Answer: ₹1600

For compound interest, the amount after one more year is multiplied by the same factor. So, \frac{2651}{2410} = 1.1, which means the rate is 10%. Then the principal is \frac{2410}{1.1^3} = 1600. Hence, the principal is ₹1600.

Q2. What is the compound interest on ₹20,000 at 12% per annum for 3 years?

1. ₹7200
2. ₹7500
3. ₹8000
4. ₹8098.56

Answer: ₹8098.56

Compound amount = $20000(1+0.12)^3 = 20000 \times 1.404928 = ₹28,098.56$. Compound interest = amount - principal = ₹28,098.56 - ₹20,000 = ₹8,098.56.

Q3. A sum of money lent at compound interest for 2 years at 20% per annum would fetch ₹96.4 more interest if compounded half-yearly. Find the sum.

1. ₹4,000
2. ₹2,500
3. ₹3,750
4. ₹4,250

Answer: ₹4,000

Let the principal be $P$. Annual compounding for 2 years gives interest $P(1.2^2-1)=0.44P$. Half-yearly compounding gives interest $P(1.1^4-1)=0.4641P$. Their difference is $0.0241P=96.4$, so $P=4000$.

Q4. A man invested ₹(P+500) at 10% p.a. compound interest for two years, and he received interest of ₹1785. If he invested ₹(P-1000) at the same rate for the same period, then find the compound interest received by him (in ₹).

1. 1470
2. 1520
3. 2190
4. 1860

Answer: 1470

For 2 years at 10%, compound interest is 21% of the principal. So the first principal is 1785/0.21 = 8500, giving P + 500 = 8500 and P = 8000. The second principal is 7000, and 21% of 7000 is 1470.

Q5. Interest at 20% p.a. compounded annually for 2 years is ₹4,600 more than interest at 20% p.a. compounded half-yearly for 1 year. Find the principal.

1. ₹20,000
2. ₹32,000
3. ₹50,000
4. ₹47,000

Answer: ₹20,000

Annual CI for 2y: P(1.2²-1)=0.44P. Half-yearly CI for 1y (10% per half): P(1.1²-1)=0.21P. Difference: 0.44P-0.21P=0.23P=4600. P=₹20,000.

Q6. Compound interest at 10% p.a. for 2 years = ₹945. Find principal.

1. 3750
2. 5000
3. 4000
4. 4500

Answer: 4500

CI = P[(1+10/100)^2 - 1] = P[1.21-1] = P×0.21. 945 = P×0.21 → P = 945/0.21 = 4500.