Q: The compound interest on a sum of money for 2 years is ₹615 and the simple interest for the same period is ₹600. Find the principal.

₹6000. For 2 years, $\text{CI} - \text{SI} = P\left(\frac{r}{100}\right)^2$. Here the difference is 15, so $P(r/100)^2 = 15$. Also, $\text{SI} = \frac{Pr}{50} = 600$, which gives the rate and principal; solving gives $P = 6000$.

Q: The difference between compound interest and simple interest on a certain sum for 3 years at 5% per annum is ₹122. The sum is:

₹ 16000. For 3 years, $\text{CI} - \text{SI} = P\left[3\left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3\right]$. Substituting $r=5\%$ and difference $=122$ gives $P=16000$.

Q: The difference between simple interest and compound interest (compounded annually) on a sum of money for 2 years at 10% per annum is ₹65. The sum is:

₹ 6500. For 2 years, the difference between CI and SI is $P\left(\frac{r}{100}\right)^2$. With $r=10\%$, we get $65 = P\times 0.01$, so $P=6500$.

Q: A sum of money becomes 1.331 times in 3 years under compound interest. The rate of interest is:

10%. Since the amount becomes 1.331 times in 3 years, $(1+r/100)^3 = 1.331$. Now $1.331 = 1.1^3$, so $1+r/100 = 1.1$, giving $r=10\%$.

Question 1

The compound interest on a sum of money for 2 years is ₹615 and the simple interest for the same period is ₹600. Find the principal.

Accepted Answer

₹6000. For 2 years, $\text{CI} - \text{SI} = P\left(\frac{r}{100}\right)^2$. Here the difference is 15, so $P(r/100)^2 = 15$. Also, $\text{SI} = \frac{Pr}{50} = 600$, which gives the rate and principal; solving gives $P = 6000$.

Question 2

The difference between the simple interest and compound interest on a certain sum of money for 2 years at 4% per annum is ₹4. The sum is:

Accepted Answer

₹ 2500. For 2 years, the difference between compound interest and simple interest is $P\left(\frac{r}{100}\right)^2$. Here, $4 = P\left(\frac{4}{100}\right)^2$. Solving gives $P = 2500$.

Question 3

The difference between compound interest and simple interest on a certain sum for 3 years at 5% per annum is ₹122. The sum is:

Accepted Answer

₹ 16000. For 3 years, $\text{CI} - \text{SI} = P\left[3\left(\frac{r}{100}\right)^2 + \left(\frac{r}{100}\right)^3\right]$. Substituting $r=5\%$ and difference $=122$ gives $P=16000$.

Question 4

The difference between simple interest and compound interest (compounded annually) on a sum of money for 2 years at 10% per annum is ₹65. The sum is:

Accepted Answer

₹ 6500. For 2 years, the difference between CI and SI is $P\left(\frac{r}{100}\right)^2$. With $r=10\%$, we get $65 = P\times 0.01$, so $P=6500$.

Question 5

A sum of money placed at compound interest doubles itself in 5 years. In how many years will it amount to eight times itself at the same rate of interest?

Accepted Answer

15 years. If the money doubles in 5 years, then one doubling takes 5 years. To become 8 times, it must double 3 times because $8=2^3$. Therefore, the required time is $3\times 5=15$ years.

Question 6

A sum of money becomes 1.331 times in 3 years under compound interest. The rate of interest is:

Accepted Answer

10%. Since the amount becomes 1.331 times in 3 years, $(1+r/100)^3 = 1.331$. Now $1.331 = 1.1^3$, so $1+r/100 = 1.1$, giving $r=10\%$.

Question 7

A sum of ₹210 was taken as a loan. It is to be paid back in two equal instalments. If the rate of interest is 10% compounded annually, then the value of each instalment is:

Accepted Answer

₹ 121. The present value of the two equal instalments must equal the loan amount. So $\frac{x}{1.1}+\frac{x}{1.21}=210$, which gives $x=121$.

Question 8

A sum of money is paid back in two annual instalments of ₹17,640 each, allowing 5% compound interest compounded annually. The sum borrowed was:

Accepted Answer

₹ 32800. The borrowed sum equals the present value of the two instalments. So it is $\frac{17640}{1.05}+\frac{17640}{1.05^2}$, which comes to ₹32800.

SSC CGL (Prelims) Maths: Simple and Compound Interest questions with solutions

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