Class 10: Compound Interest – Exercise 1(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the money, invested at $\displaystyle 10\%$ compounded annually, on which the sum of interest for the first year and the third year is $\displaystyle Rs. 1768$ .

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = \text{ Rs. } 0.1x$

$\displaystyle \text{and, Amount } = x+0.1x = \text{ Rs. } 1.1x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.1x \times 10 \times 1}{100} = \text{ Rs. } 0.11x$

$\displaystyle \text{and, Amount } = 1.10x + 0.11x = \text{ Rs. } 1.21x$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 1.21x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{1.21x \times 10 \times 1}{100} = \text{ Rs. } 0.121x$

$\displaystyle \text{and, Amount } = 1.21x + 0.121x = \text{ Rs. } 1.331x$

$\displaystyle \text{Given } 0.1x+0.121x= 1768 \Rightarrow x= \text{ Rs. } 8000$

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Question 2: A sum of money is invested at compound interest payable annually. The interest in first two successive years is $\displaystyle Rs. 1350$ and $\displaystyle Rs. 1440$ respectively. Find;

The rate of interest
The original Sum
The interest earned in the third year.

Answer:

Difference between the Compound interest of two successive years

$\displaystyle = 1440-1350 = \text{ Rs. } 90$

$\displaystyle \Rightarrow \text{ Rs. } 90 \text{ is the interest on Rs. } 1350$

$\displaystyle \therefore \text{Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 90}{1350 \times 1} \% = 6 \frac{2}{3} \%$

$\displaystyle \text{For 1st year: } P = \text{ Rs. } P; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest } = \text{ Rs. } 1350 = P \times \frac{6\frac{2}{3}}{100} \times 1 \Rightarrow P= 20250$

$\displaystyle \text{and, Amount } = 20250 + 1350x = \text{ Rs. } 21600$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 21600; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{21600 \times 6\frac{2}{3} \times 1}{100} = \text{ Rs. } 1440$

$\displaystyle \text{and, Amount } =21600 +1440 = \text{ Rs. } 23040$

$\displaystyle \text{For 3rd year: } P=Rs. 23040; R=6 \frac{2}{3} \% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Therefore Interest } = \frac{23040 \times 6\frac{2}{3} \times 1}{100} = \text{ Rs. } 1536$

$\displaystyle \text{and, Amount } =23040 +1536 = \text{ Rs. } 24576$

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Question 3: A sum of money amounts to $\displaystyle Rs. 46305$ in $\displaystyle 1$ year and to $\displaystyle Rs. 48620.25$ in $\displaystyle 1\frac{1}{2}$ years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.

Answer:

$\displaystyle \text{Interest earned in 3rd half year } =48620.25-46305 = \text{ Rs. } 2315.25$

$\displaystyle P=Rs. 46305; I = \text{ Rs. } 2315.25 T = \frac{1}{2} ; R=x\%$

$\displaystyle \text{Interest }= \text{ Rs. } 2315.25=46305 \times \frac{x}{100} \times \frac{1}{2} \Rightarrow x= 10\%$

For 1st half year: $\displaystyle P = \text{ Rs. } P; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Interest } = P \times \frac{10}{100} \times \frac{1}{2} = 0.05P$

$\displaystyle \text{and, Amount } = P+0.05P = 1.05P$

For 2nd half year: $\displaystyle P = \text{ Rs. } 1.05P; R=10\% \text{ and } T= \frac{1}{2} \text{ year }$

$\displaystyle \text{Interest } = 1.05 \times \frac{10}{100} \times \frac{1}{2} = 0.0525P$

$\displaystyle \therefore \text{Amount }= 1.05P+0.0525P = 1.1025P = 46305 \Rightarrow P = \text{ Rs. } 42000$

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Question 4: The cost of machine in $\displaystyle Rs. 32,000$ . Its value depreciates at the rate of $\displaystyle 5\%$ every year. Find the total depreciation in its value by the end of $\displaystyle 2$ years.

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 32000; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Depreciation } = 32000 \times \frac{5}{100} \times 1 = \text{ Rs. } 1600$

and, $\displaystyle \text{Value } = 32000-1600 = \text{ Rs. } 30400$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 30400; R=5\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Depreciation } = 30400 \times \frac{5}{100} \times 1 = \text{ Rs. } 1520$

and, $\displaystyle \text{Value } = 30400-1520 = \text{ Rs. } 28880$

$\displaystyle \therefore \text{Total Depreciation } = 1600+1520 = \text{ Rs. } 3120$

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Question 5: Find the sum, invested at $\displaystyle 10\%$ compounded annually, on which the interest for the third year exceeds the interest of the first year by $\displaystyle Rs. 252$ .

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest } = x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.1x$

$\displaystyle \text{and, Amount } = x+0.1x = \text{ Rs. } 1.1x$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest } = 1.1x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.11x$

$\displaystyle \text{and, Amount } = 1.1x+0.11x = \text{ Rs. } 1.21x$

$\displaystyle \text{For 3rd year: } P = \text{ Rs. } 1.21x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest } = 1.21x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.121x$

$\displaystyle \text{and, Amount } = 1.21x+0.121x = \text{ Rs. } 1.331x$

$\displaystyle \text{Given } 0.121x-0.1x=252 \Rightarrow x= \frac{252}{0.021} = \text{ Rs. } 12000$

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Question 6: The compound interest, compounded annually, on a certain sum is $\displaystyle Rs. 9680$ in second year and is $\displaystyle Rs. 10648$ in third year. Calculate;

The rate of interest, sum borrowed
The interest of $\displaystyle 1^{st}$ year.

Answer:

Difference between the Compound interest of two successive years

$\displaystyle = 10648-9680 = \text{ Rs. } 968$

$\displaystyle \Rightarrow \text{ Rs. } 968$ is the interest on $\displaystyle Rs. 9680$

$\displaystyle \therefore \text{Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 968}{9680 \times 1} \% = 10\%$

Let the sum of money $\displaystyle = \text{ Rs. } 100$

Therefore Interest on it for 1st year $\displaystyle = 10\% \text{ of Rs. } 100 = \text{ Rs. } 10$

$\displaystyle \Rightarrow \text{Amount in one year } = 100 + 10 = \text{ Rs. } 110$

$\displaystyle \therefore \text{Interest on it for } 2^{nd} \text{Year } = 10\% \text{ of } 110 = \text{ Rs. } 11$

$\displaystyle \Rightarrow \text{Amount in 2nd year } = 110 + 11 = \text{ Rs. } 121$

When interest of 2nd year $\displaystyle = \text{ Rs. } 11, \text{sum is Rs. } 100$

$\displaystyle \Rightarrow \text{When interest of 2nd year }= \text{ Rs. } 9680$ , then

$\displaystyle \text{sum }= \frac{100}{11} \times 9680 = \text{ Rs. } 88000$

$\displaystyle \text{Interest of 1st year } = 88000 \times \frac{10}{100} \times 1 = \text{ Rs. } 8800$

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Question 7: A man borrows $\displaystyle Rs. 10,000$ at $\displaystyle 10\%$ compound interest compounded yearly. At the end of each year, he pays back $\displaystyle 30\%$ of the sum borrowed. How much money is left unpaid just after the second year?

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest }= 10000 \times \frac{10}{100} \times 1 = \text{ Rs. } 1000$

$\displaystyle \text{and, Amount } = 10000+1000 = \text{ Rs. } 11000$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } (11000-3000)=8000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest }= 8000 \times \frac{10}{100} \times 1 = \text{ Rs. } 800$

$\displaystyle \text{and, Amount } = 8000+800 = \text{ Rs. } 8800$

$\displaystyle \text{Amount left }= 8800-3000 = \text{ Rs. } 5800$

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Question 8: A man borrows $\displaystyle Rs. 10000$ at $\displaystyle 10\%$ compound interest compounded yearly. At the end of each year, he pays back $\displaystyle 20\%$ of the amount for that year. How much money is left unpaid just after the second year?

Answer:

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 10000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest }= 10000 \times \frac{10}{100} \times 1 = \text{ Rs. } 1000$

$\displaystyle \text{and, Amount } = 10000+1000 = \text{ Rs. } 11000$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } (11000-2000)=9000; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Interest } = 9000 \times \frac{10}{100} \times 1 = \text{ Rs. } 900$

$\displaystyle \text{and, Amount } = 9000+900 = \text{ Rs. } 9900$

$\displaystyle \text{Amount left } = 9900-2000 = \text{ Rs. } 7900$

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Question 9: The population of a town increases $\displaystyle 10\%$ every $\displaystyle 3$ years. If the present population of the town is $\displaystyle 72600$ , calculate:

Its population after $\displaystyle 2$ years;
Its population $\displaystyle 2$ years ago;

Answer:

$\displaystyle \text{For 2 years ago: } P = \text{ Rs. } x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Increase } = x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.1x$

$\displaystyle \text{and, Population } = x+0.1x = \text{ Rs. } 1.1x$

$\displaystyle \text{For 1 year ago: } P = \text{ Rs. } 1.1x; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Increase } = 1.1x \times \frac{10}{100} \times 1 = \text{ Rs. } 0.11x$

$\displaystyle \text{and, Population } = 1.1x+0.11x = \text{ Rs. } 1.21x$

$\displaystyle \text{Given } 1.21x= 72600 \Rightarrow x= 60000$

$\displaystyle \text{For 1st year: } P = \text{ Rs. } 72600; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Increase } = 72600 \times \frac{10}{100} \times 1 = \text{ Rs. } 7260$

$\displaystyle \text{and, Population } = 72600+7260 = \text{ Rs. } 79860$

$\displaystyle \text{For 2nd year: } P = \text{ Rs. } 79860; R=10\% \text{ and } T=1 \text{ year }$

$\displaystyle \text{Increase } = 79860 \times \frac{10}{100} \times 1 = \text{ Rs. } 7986$

$\displaystyle \text{and, Population } = 79860+7986 = \text{ Rs. } 87846$

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Question 10: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 1320$ and for the third year is $\displaystyle Rs. 1452$ . Calculate the rate of interest and the original sum of money. [2014]