Exercise : Compound Interest - General Questions
✔ Compound Interest -
General Questions
11.
Find the Compound Interest on $12,000 for 2 years if the rate is 5% for the 1st year and 10% for the 2nd year.
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Answer: Option A
Explanation:
Explanation:
\(A = 12000 \times (1 + 5/100) \times (1 + 10/100) = 12000 \times 1.05 \times 1.10 = 13860\). \(CI = 13860 - 12000 = 1,860\).
12.
What is the difference between Compound Interest compounded annually and half-yearly on $10,000 for 1 year at 10%?
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Answer: Option A
Explanation:
Explanation:
Annual \(CI = 10% \text{ of } 10000 = 1000\). Half-yearly \(A = 10000(1.05)^2 = 11025\), so \(CI = 1025\). Difference = \(1025 - 1000 = 25\).
13.
The Compound Interest on a certain sum for 2 years at 10% is $420. Find the Simple Interest on the same sum for the same time and rate.
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Answer: Option A
Explanation:
Explanation:
\(P(1.1^2 - 1) = 420 \implies 0.21P = 420 \implies P = 2000\). \(SI = (2000 \times 10 \times 2) / 100 = 400\).
14.
Find the Compound Interest on $8,000 for 1 year 3 months at 10% per annum compounded annually.
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Answer: Option A
Explanation:
Explanation:
\(A = 8000(1.1)^1 \times (1 + \frac{1/4 \times 10}{100}) = 8800 \times 1.025 = 9020\). \(CI = 9020 - 8000 = 1,020\).
15.
A sum of $12,000 deposited at Compound Interest becomes double after 5 years. After 20 years, it will become:
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Answer: Option A
Explanation:
Explanation:
20 years consists of 4 cycles of 5 years. \(A = 12000 \times 2^4 = 12000 \times 16 = 192,000\).
16.
A man borrows $2,100 and agrees to pay it back in two equal annual installments at 10% Compound Interest. Find the value of each installment.
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Answer: Option A
Explanation:
Explanation:
Using the installment formula: \(P = \frac{x}{1+r} + \frac{x}{(1+r)^2}\). \(2100 = \frac{x}{1.1} + \frac{x}{1.21} \implies 2100 = \frac{1.1x + x}{1.21} \implies x = 1,210\).
17.
The difference between Compound Interest and Simple Interest for 3 years at 10% is $31. Find the sum.
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Answer: Option A
Explanation:
Explanation:
Formula for 3-year difference: \(D = P(R/100)^2 \times (300+R)/100\). \(31 = P(1/100) \times (310/100) \implies 31 = P \times 31/1000 \implies P = 1,000\).
18.
A sum of money is lent at 20% Compound Interest. At the end of 2 years, if the interest were compounded half-yearly instead of annually, how much more interest is earned on $10,000?
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Answer: Option A
Explanation:
Explanation:
Annual \(CI = 10000(1.2^2 - 1) = 4400\). Half-yearly \(CI = 10000(1.1^4 - 1) = 4641\). Extra interest = \(4641 - 4400 = 241\).
19.
Divide $13,010 between A and B such that A's share after 7 years is equal to B's share after 9 years at 4% Compound Interest. What is A's share?
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Answer: Option A
Explanation:
Explanation:
\(A(1.04)^7 = B(1.04)^9 \implies A/B = 1.04^2 = 1.0816 = 676/625\). A's share = \(\frac{676}{676+625} \times 13010 = 6,760\).
20.
If the Compound Interest on a sum for 3 years at 5% is $252.20, find the Simple Interest for the same period and rate.
View Answer
Answer: Option A
Explanation:
Explanation:
\(P[1.05^3 - 1] = 252.20 \implies P[0.157625] = 252.20 \implies P = 1600\). \(SI = 1600 \times 0.05 \times 3 = 240\).