Class 10: Compound Interest – Exercise 2(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: If the interest is compounded half-yearly, calculate the amount when the principal is Rs. $\displaystyle 7400$ ; the rate of interest is $\displaystyle 5\%$ per annum and the duration is one year. [2005]

Answer:

$\displaystyle P=7400 \text{ Rs.; } r=5\% \text{ ; Compounded half yearly } n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{n \times 2} = 12000 \Big(1+ \frac{5}{2 \times 100} \Big)^{1 \times 2} = 7774.63 \text{ Rs. }$

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Question 2: Find the difference between the compound interest compounded yearly and half-yearly on Rs. $\displaystyle 10000$ for $\displaystyle 18$ months at $\displaystyle 10\%$ per annum.

Answer:

Compounded Yearly

$\displaystyle P=10000 \text{ Rs.; } r=10\% \text{ ; Compounded yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{1 \times 100} \Big)^{1}. \Big(1+ \frac{r}{2 \times 100} \Big)^{\frac{1}{2} \times 2}$

$\displaystyle A=10000 \Big(1+ \frac{10}{1 \times 100} \Big)^{1}. \Big(1+ \frac{10}{2 \times 100} \Big)^{\frac{1}{2} \times 2} = 11550 \text{ Rs. }$

Compounded Half Yearly

$\displaystyle P=10000 \text{ Rs.; } r=10\% \text{ ; Compounded half yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = 10000 \Big(1+ \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = 11576.25 \text{ Rs. }$

Difference $\displaystyle 11576.25-11550 = 26.50 \text{ Rs. }$

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Question 3: A man borrowed Rs. $\displaystyle 16000$ for $\displaystyle 3 \text{ years }$ under the following terms:

$\displaystyle 20\%$ simple interest for the first $\displaystyle 2 \text{ years }$ ;
$\displaystyle 20\%$ C.I. for the remaining one year on the amount due after $\displaystyle 2 \text{ years }$ , the interest being compounded semi-annually. Find the total amount to be paid at the end of the three years.

Answer:

Simple interest for the first two years

$\displaystyle S.I. = 16000 \times \frac{20}{100} \times 2 = 6400 \text{ Rs. }$

Amount $\displaystyle = 16000+6400 = 22400 \text{ Rs. }$

Compound interest for the remainder of the term

$\displaystyle P=10000 \text{ Rs.; } r=20\% \text{ ; Compounded half yearly } n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{1 \times 2} = 22400 \Big(1+ \frac{20}{2 \times 100} \Big)^{1 \times 2} = 27104 \text{ Rs. }$

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Question 4: What sum of money will amount to Rs. $\displaystyle 27783$ in one and half years at $\displaystyle 10\%$ per annum compounded half-yearly?

Answer:

$\displaystyle P=x \text{ Rs.; } r=10\% \text{ ; Compounded half yearly } n= \frac{3}{2} \text{ year; } A=27783 \text{ Rs. }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{n \times 2}$

$\displaystyle 27783=x \Big(1+ \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} \Rightarrow 27783 = 1.157625x \Rightarrow x= 2400 \text{ Rs. }$

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Question 5: A invests a certain sum of money at $\displaystyle 20\%$ per annum, interest compounded yearly. $\displaystyle B$ invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If $\displaystyle B$ gets Rs. $\displaystyle 33$ more than $\displaystyle A$ in $\displaystyle 18$ months, calculate the money invested by each.

Answer:

A’s investment: Compounded Yearly

$\displaystyle P=x \text{ Rs.; } r=20\% \text{ ; Compounded yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{1 \times 100} \Big)^{1}. \Big(1+ \frac{r}{2 \times 100} \Big)^{\frac{1}{2} \times 2}$

$\displaystyle A=x \Big(1+ \frac{20}{1 \times 100} \Big)^{1}. \Big(1+ \frac{20}{2 \times 100} \Big)^{\frac{1}{2} \times 2} = 1.32x \text{ Rs. }$

Compounded Half Yearly

$\displaystyle P=x \text{ Rs.; } r=20\% \text{ ; Compounded half yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = x \Big(1+ \frac{20}{2 \times 100} \Big)^{\frac{3}{2} \times 2} = 1.331x \text{ Rs. }$

Difference $\displaystyle 1.331x-1.32x=33 \Rightarrow x= 3000 \text{ Rs. }$

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Question 6: At what rate of interest per annum will a sum of Rs. $\displaystyle 62500$ earn a compound interest of Rs. $\displaystyle 5100$ in one year? The interest is to be compounded half-yearly.

Answer:

Compounded Half Yearly

$\displaystyle P=62500 \text{ Rs.; } A=(62500 + 5100) = 67600 \text{ Rs.; } \\ \\ r=x\% \text{ ; Compounded half yearly } n= \frac{2}{2} \text{ year }$

$\displaystyle 67600=62500 \Big(1+ \frac{x}{2 \times 100} \Big)^{\frac{2}{2} \times 2} \Rightarrow 1.0816 = \Big(1+ \frac{x}{200} \Big)^2 \Rightarrow x=8\%$

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Question 7: In what time will Rs. $\displaystyle 1500$ yield Rs. $\displaystyle 496.50$ as compound interest at $\displaystyle 20\%$ per year compounded semi-annually?

Answer:

Compounded Half Yearly

$\displaystyle P=1500 \text{ Rs.; } A=(1500 + 496.50) = 1996.50 \text{ Rs.; } \\ \\ r=20\% \text{ ; Compounded half yearly } n=n \text{ year }$

$\displaystyle 1996.50=1500 \Big(1+ \frac{20}{2 \times 100} \Big)^{n \times 2} \Rightarrow 1.331 = \Big(1+ \frac{20}{200} \Big)^{2n} \Rightarrow n = \frac{3}{2} \text{ years }$

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Question 8: Calculate the C.I. on Rs. $\displaystyle 3500$ at $\displaystyle 6\%$ per annum for $\displaystyle 3 \text{ years }$ , the interest being compounded half-yearly.

Answer:

Compounded Half Yearly

$\displaystyle P=3500 \text{ Rs.; } r=6\% \text{ ; Compounded half yearly } n=3 \text{ year }$

$\displaystyle A=3500 \Big(1+ \frac{6}{2 \times 100} \Big)^{3 \times 2} \Rightarrow A= 4179.18 \text{ Rs. }$

$\displaystyle C.I. = 4179.18-3500 = 679.18 \text{ Rs. }$

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Question 9: Find the difference between compound interest and simple interest on Rs. $\displaystyle 12,000$ and in $\displaystyle 1$ $\displaystyle \frac{1}{2}$ at $\displaystyle 10\%$ compounded yearly.

Answer:

Compounded Yearly

$\displaystyle P=12000 \text{ Rs.; } r=10\% \text{ ; Compounded yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{1 \times 100} \Big)^{1}. \Big(1+ \frac{r}{2 \times 100} \Big)^{\frac{1}{2} \times 2}$

$\displaystyle A=12000 \Big(1+ \frac{10}{1 \times 100} \Big)^{1}. \Big(1+ \frac{10}{2 \times 100} \Big)^{\frac{1}{2} \times 2} = 13860 \text{ Rs. }$

Simple interest for $\displaystyle 1.5 \text{ years }$

S.I. $\displaystyle = 12000 \times \frac{10}{100} \times \frac{3}{2} = 1800$ Rs.

Amount $\displaystyle = 16000+6400 = 22400 \text{ Rs. }$

Difference $\displaystyle = (13860-13000)-1800 = 60 \text{ Rs. }$

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Question 10: The simple interest on a sum of money for $\displaystyle 3 \text{ years }$ at $\displaystyle 5\%$ per annum is Rs. $\displaystyle 900$ . Find:

The sum of money and
The compound interest on this sum for $\displaystyle 1.5 \text{ years }$ payable half-yearly at double the rate per annum.

Answer:

Simple interest for $\displaystyle 3 \text{ years }$

$\displaystyle 900 = x \times \frac{5}{100} \times 3 \Rightarrow x= 6000 \text{ Rs. }$

Amount $\displaystyle = 16000+6400 = 22400 \text{ Rs. }$

Compounded Half Yearly

$\displaystyle P=6000 \text{ Rs.; } r=10\% \text{ ; Compounded half yearly } n= \frac{3}{2} \text{ year }$

$\displaystyle A=6000 \Big(1+ \frac{10}{2 \times 100} \Big)^{\frac{3}{2} \times 2} \Rightarrow A= 6945.75 \text{ Rs. }$

Compound interest $\displaystyle = 6945.75-6000 = 945.75 \text{ Rs. }$

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Question 11: The compound interest in one year on a certain sum of money at $\displaystyle 10\%$ per annum compounded half-yearly exceeds the simple interest on the same sum at the same rate and for the same period by Rs. $\displaystyle 30$ . Calculate the sum.