Machine-translated by Google. Questions are authored in English.
Top

Compound Interest Questions and Answers

Important Instructions
121.

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

Answer: B

The money more than doubles once \( (1.2)^{n} > 2 \). Test successive powers:

\[ (1.2)^{1} = 1.2, \qquad (1.2)^{2} = 1.44 \]

\[ (1.2)^{3} = 1.728, \qquad (1.2)^{4} = 2.0736 \]

The third year falls short and the fourth passes 2, so the answer is 4 years.

The question asks for complete years, so the answer must be a whole number even though the exact doubling time is about 3.8 years.

Please log in or register to report this question.

122.

The difference between simple and compound interest on a certain sum of money for 2 years at 4 percent per annum is Re.1. The sum of money is

Answer: B

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ \text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^{2} \]

\[ 1 = P(0.04)^{2} = 0.0016P \]

\[ P = \frac{1}{0.0016} = \text{Rs. } 625 \]

Checking: SI = Rs. 50 and CI = \( 625(1.0816 - 1) = 51 \) - the Re. 1 gap is 4% of the first year's Rs. 25 of interest.

Please log in or register to report this question.

123.

Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:

Answer: C

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

Work out the compound interest first.

\[ \text{CI} = 4000\left[(1.1)^{2} - 1\right] = 4000(1.21 - 1) = \text{Rs. } 840 \]

The simple interest is half of that:

\[ \text{SI} = 420 \]

\[ 420 = \frac{P \times 8 \times 3}{100} = 0.24P \]

\[ P = \frac{420}{0.24} = \text{Rs. } 1750 \]

The two sums are unrelated; the only link is the numerical value of the interest, so the compound side must be finished before the simple side can begin.

Please log in or register to report this question.

124.

At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.20000?

Answer: A

\[ A = P\left(1 + \frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P, \qquad \text{SI} = \frac{PRn}{100} \]

\[ \text{CI} = 20000\left[(1.1)^{3} - 1\right] = 20000(1.331 - 1) \]

\[ = 20000 \times 0.331 = \text{Rs. } 6620 \]

Simple interest would be Rs. 6000; the extra Rs. 620 is the compounding. The multiplier \( 1.1^{3} = 1.331 \) is worth memorising, as it appears constantly in these questions.

Please log in or register to report this question.

Loading…
Tags: Compound Interest Questions and Answers || Compound Interest MCQ Questions and Answers || Compound Interest GK Questions and Answers || Compound Interest GK MCQ Questions || Compound Interest Multiple Choice Questions and Answers || Compound Interest GK || GK on Compound Interest || Quantitative Aptitude Questions and Answers || Quantitative Aptitude MCQ Questions and Answers || Quantitative Aptitude GK Questions and Answers || GK on Quantitative Aptitude