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

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

Answer:

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

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

and, $Amount = x+0.1x = Rs. \ 1.1x$

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

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

and, $Amount = 1.10x + 0.11x = Rs. \ 1.21x$

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

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

and, $Amount = 1.21x + 0.121x = Rs. \ 1.331x$

Given $0.1x+0.121x= 1768 \Rightarrow x= Rs. \ 8000$

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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 $Rs. \ 1350$ and $Rs. \ 1440$ respectively. Find;

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

Answer:

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 1440-1350 = Rs. \ 90$

$\Rightarrow Rs. \ 90$ $\ is \ the \ interest \ on \ Rs. \ 1350$

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

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

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

and, $Amount = 20250 + 1350x = Rs. \ 21600$

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

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

and, $Amount =21600 +1440 = Rs. \ 23040$

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

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

and, $Amount =23040 +1536 = Rs. \ 24576$

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

Answer:

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

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

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

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

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

and, $Amount = P+0.05P = 1.05P$

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

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

$\therefore Amount = 1.05P+0.0525P = 1.1025P = 46305 \Rightarrow P = Rs. \ 42000$

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

Answer:

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

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

and, $Value = 32000-1600 = Rs. \ 30400$

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

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

and, $Value = 30400-1520 = Rs. \ 28880$

$\therefore Total \ Depreciation = 1600+1520 = Rs. \ 3120$

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

Answer:

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

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

and, $Amount = x+0.1x = Rs. \ 1.1x$

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

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

and, $Amount = 1.1x+0.11x = Rs. \ 1.21x$

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

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

and, $Amount = 1.21x+0.121x = Rs. \ 1.331x$

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

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

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

Answer:

$Difference \ between \ the \ Compound \ interest \ of \ two \ successive \ years$

$= 10648-9680 = Rs. \ 968$

$\Rightarrow Rs. \ 968$ is the interest on $Rs. \ 9680$

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

Let the sum of money $= Rs. \ 100$

Therefore Interest on it for 1st Year $= 10\% \ of \ Rs. \ 100 = Rs. \ 10$

$\Rightarrow Amount \ in \ one \ year = 100 + 10 = Rs. \ 110$

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

$\Rightarrow Amount \ in \ 2nd \ year = 110 + 11 = Rs. \ 121$

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

$\Rightarrow \ When \ interest \ of \ 2nd \ year= Rs. \ 9680$ , then

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

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

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

Answer:

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

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

and, $Amount = 10000+1000 = Rs. \ 11000$

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

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

and, $Amount = 8000+800 = Rs. \ 8800$

$Amount left = 8800-3000 = Rs. \ 5800$

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

Answer:

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

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

and, $Amount = 10000+1000 = Rs. \ 11000$

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

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

and, $Amount = 9000+900 = Rs. \ 9900$

$Amount \ left = 9900-2000 = Rs. \ 7900$

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

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

Answer:

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

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

and, $Population = x+0.1x = Rs. \ 1.1x$

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

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

and, $Population = 1.1x+0.11x = Rs. \ 1.21x$

Given $1.21x= 72600 \Rightarrow x= 60000$

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

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

and, $Population = 72600+7260 = Rs. \ 79860$

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

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

and, $Population = 79860+7986 = Rs. \ 87846$

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