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

Question 1: The height of a plant is $\displaystyle 80 cm$ and it is expected to grow at the rate of $\displaystyle 20\%$ every month. What will be its height at the end of $\displaystyle 3 \text{ months }$ ?

Answer:

$\displaystyle P=80 \text{ cm; } r=20\%; n=3 \text{ months }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 80 \Big(1+ \frac{20}{100} \Big)^3 = 138.24 \text{ cm }$

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Question 2: The cost of a machine is supposed to depreciate each year by $\displaystyle 12\%$ of its value at the beginning of the year. If the machine is valued at $\displaystyle Rs.44,000$ at the beginning of $\displaystyle 2008$ , find its value:

At the end of $\displaystyle 2009$
At the beginning of $\displaystyle 2007$

Answer:

$\displaystyle P=44000 \text{ Rs.; } r=12\%; n=2 \text{ years }$

At the end of $\displaystyle 2009$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n = 44000 \Big(1- \frac{12}{100} \Big)^2 = 34073.6 \text{ Rs. per cm }$

At the beginning of $\displaystyle 2007$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 44000 \Big(1+ \frac{12}{100} \Big)^1 = 49280 \text{ Rs. per cm }$

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Question 3: The value of a machine is estimated to be $\displaystyle Rs.27000$ at the end of $\displaystyle 2004$ and $\displaystyle Rs.21,870$ at the beginning of $\displaystyle 2007$ . Supposing it depreciates at a constant rate per year of its value at the beginning of the year, calculate:

The rate of depreciation;
The value of the machine at the beginning of $\displaystyle 2004$ .

Answer:

$\displaystyle P=27000 \text{ Rs.; } A= 21870 \text{ Rs.; } r=r\%; n=2 \text{ years }$

At the end of $\displaystyle 2007$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 21870= 27000 \Big(1- \frac{r}{100} \Big)^2 \Rightarrow r=10\%$

At the beginning of $\displaystyle 2004$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 27000= A \Big(1- \frac{10}{100} \Big)^1 \Rightarrow A=30000 \text{ Rs. }$

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Question 4: The value of in article decreased for two years at the rate of $\displaystyle 10\%$ per year and then in the third year it increased by $\displaystyle 10\%$ . Find the original value of the article, if its value at the end of $\displaystyle 3 \text{ years }$ is $\displaystyle Rs.40095$ .

Answer:

$\displaystyle P=P \text{ Rs.; } A= A \text{ Rs.; } r=10\%; n=2 \text{ years }$

At the end of $\displaystyle 2^{nd} year$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow P= A \Big(1- \frac{10}{100} \Big)^2 \Rightarrow P=0.81A$

At the end of $\displaystyle 3^{rd} Year$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 40095= 0.81A \Big(1+ \frac{10}{100} \Big)^1 \Rightarrow A=45000 \text{ Rs. }$

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Question 5: According to a census taken towards the end of the year $\displaystyle 2009$ , the population of a rural town was found to be $\displaystyle 64000$ . The census authority also found that the population of this particular town had a growth of $\displaystyle 5\%$ per annum. In how many years after $\displaystyle 2009$ did the population of this town reach $\displaystyle 74088$ ?

Answer:

$\displaystyle P=64000. A= 74088; r=5\%; n=n \text{ years }$

At the end of $\displaystyle n^{nd} year$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 74088= 64000 \Big(1+ \frac{5}{100} \Big)^n \Rightarrow n=3$

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Question 6: The population of a town decreased by $\displaystyle 12\%$ during $\displaystyle 1998$ and then increased by $\displaystyle 8\%$ during $\displaystyle 1999$ . Find the population of the town, at the beginning of $\displaystyle 1998$ , if at the end of $\displaystyle 1999$ its population was $\displaystyle 2,85,120$ .

Answer:

At the end of $\displaystyle 1999 \text{ year }$

$\displaystyle P=P; A= 285120; r=8\%; n=1 \text{ years }$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 285120=P \Big(1+ \frac{8}{100} \Big)^1 \Rightarrow P=264000$

At the beginning of $\displaystyle 1998 Year$

$\displaystyle P=P; A= 26400; r=8\%; n=1 \text{ years }$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 264000= P \Big(1- \frac{12}{100} \Big)^1 \Rightarrow P=300000$

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Question 7: A sum of money, invested at compound interest, amounts to $\displaystyle Rs.16,500$ in $\displaystyle 1 \text{ year }$ and to $\displaystyle Rs.19965$ in $\displaystyle 3 \text{ years }$ . Find the rate per cent and the original sum of money.

Answer:

$\displaystyle P=16500; A= 19965; r=r\%; n=2 \text{ years }$

$\displaystyle A=P \Big(1- \frac{r}{100} \Big)^n \Rightarrow 19965=16500 \Big(1+ \frac{r}{100} \Big)^2 \Rightarrow r=10\%$

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Question 8: The difference between C.I. and S.I. on $\displaystyle Rs.7500$ for two years is $\displaystyle Rs.12$ at the same rate of interest per annum. Find the rate of interest.

Answer:

Simple Interest

$\displaystyle P=7500 \text{ Rs.; } T=2 \text{ years; } r=x\%$

$\displaystyle S.I=7500 \times \frac{x}{100} \times 2 = 150x$

Compound Interest

$\displaystyle P=7500; A= A; r=x\%; n=2 \text{ years }$

$\displaystyle A= 7500 \times \Big(1+ \frac{x}{100} \Big)^2$

$\displaystyle C.I. = 7500 \times \Big(1+ \frac{x}{100} \Big)^2 - 7500$

$\displaystyle \text{Given } C.I. - S.I. = 12 \Rightarrow 7500 \times \Big(1+ \frac{x}{100} \Big)^2 - 7500 - 150x = 12$

$\displaystyle \Rightarrow 7500+7500 \times \frac{x^2}{10000} + 7500 \times 2 \times \frac{x}{100} - 7500 -150x = 12$

$\displaystyle \Rightarrow 0.75x^2=12$

$\displaystyle \Rightarrow x= 4%$

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Question 9: A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in $\displaystyle 10 \text{ years }$ . Find in how many years will the money become twenty-seven times of itself of the same rate of interest p.a.

Answer:

After 10 Years

$\displaystyle P=3A; A= A; r=x\%; n=10 \text{ years }$

$\displaystyle 3A=A \times \Big(1+ \frac{x}{100} \Big)^{10}$

$\displaystyle \Big(1+ \frac{x}{10} \Big)=3^{\frac{1}{10}} ... ... ... ... ... ... ... ... i)$

After n Years

$\displaystyle P=27A; A= A; r=x\%; n=n \text{ years }$

$\displaystyle 27A=A \times \Big(1+ \frac{x}{100} \Big)^n$

$\displaystyle 27=1 \times \Big(1+ \frac{x}{100} \Big)^n$

Substituting from i)

$\displaystyle 27=1 \times \Big(3^{\frac{1}{10}} \Big)^n \Rightarrow n=30 \text{ years }$

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Question 10: Sharma borrowed a certain sum of money at $\displaystyle 10\%$ per annum compounded annually. If by paying $\displaystyle Rs.19360$ at the end of the second year and $\displaystyle Rs.31,944$ at the end of the third he clears the debt; find the sum borrowed by him.

Answer:

$\displaystyle P=x; A= A; r=10\%; n=2 \text{ years }$

$\displaystyle A= x \times \Big(1+ \frac{10}{100} \Big)^2 \Rightarrow A= 1.21x$

Third year

$\displaystyle P=(1.21x-19360); A=31944 ; r=10\%; n=1 \text{ years }$

$\displaystyle 31944= (1.21x-19360) \times \Big(1+ \frac{10}{100} \Big)^1 \Rightarrow x= 40000 \text{ Rs. }$

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Question 11: The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at $\displaystyle 10\%$ for a year is $\displaystyle Rs.15$ . Find the sum of money lent out. [1998]

Answer:

Simple Interest

$\displaystyle P=x \text{ Rs.; } T=1 \text{ years; } r=10\%$

$\displaystyle S.I=x \times \frac{10}{100} \times 1 = 0.1x$

Compound Interest

$\displaystyle P=x; A= A; r=10\%; n=2 \text{ half years }$

$\displaystyle A= x \times \Big(1+ \frac{10}{200} \Big)^2 = x \times \Big( \frac{21}{20} \Big)^2$

$\displaystyle C.I. = x \times \Big( \frac{21}{20} \Big)^2 - x$

$\displaystyle \text{Given } C.I. - S.I. = x \times \Big( \frac{21}{20} \Big)^2 - x - 0.1x = 15$

$\displaystyle \Rightarrow x= 6000 \text{ Rs. }$

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Question 12: The ages of Person 1 and Person 2 are $\displaystyle 16 \text{ years }$ and $\displaystyle 18 \text{ years }$ respectively. In what ratio must they invest money at $\displaystyle 5\%$ p.a. compounded yearly so that both get the same sum on attaining the age of $\displaystyle 25 \text{ years }$ ?