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Chapter 3- Compound Interest [Using Formula | Exercise 3(A)

Question 6

A person invests Rs. 5000 for three years at a certain tare of interest compounded annually. At the end of two years this sum amounts to Rs. 6272. Calculate:

(i) The rate of interest per annum

(ii) The amount at the end of the third year.

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Given: P=Rs5,000; A=Rs6,272 and n= 2years

(i) A=P\left(1+\frac{r}{100}\right)^{n}

\begin{array}{l} \Rightarrow 6,272=5,000\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow \frac{6,272}{5,000}=\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow \frac{784}{625}=\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow\left(\frac{28}{25}\right)^{2}=\left(1+\frac{r}{100}\right)^{2} \end{array}

on comparing,

\frac{28}{25}=1+\frac{r}{100}

on solving we get,

r=12%

(ii) Amount at the third year

\begin{array}{l} =5,000\left(1+\frac{12}{100}\right)^{3} \\ =5,000\left(\frac{28}{25}\right)^{3} \\ =\operatorname{Rs} 7,024.64 \end{array}

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Question 6

A person invests Rs. 5000 for three years at a certain tare of interest compounded annually. At the end of two years this sum amounts to Rs. 6272. Calculate:

(i) The rate of interest per annum

(ii) The amount at the end of the third year.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given: P=Rs5,000; A=Rs6,272 and n= 2years

(i) A=P\left(1+\frac{r}{100}\right)^{n}

\begin{array}{l} \Rightarrow 6,272=5,000\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow \frac{6,272}{5,000}=\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow \frac{784}{625}=\left(1+\frac{r}{100}\right)^{2} \\ \Rightarrow\left(\frac{28}{25}\right)^{2}=\left(1+\frac{r}{100}\right)^{2} \end{array}

on comparing,

\frac{28}{25}=1+\frac{r}{100}

on solving we get,

r=12%

(ii) Amount at the third year

\begin{array}{l} =5,000\left(1+\frac{12}{100}\right)^{3} \\ =5,000\left(\frac{28}{25}\right)^{3} \\ =\operatorname{Rs} 7,024.64 \end{array}

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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