Name: Calculation of Compounded Amount
Uploaded: 2019-07-19
Duration: 9 min 18 s
Description: The two types of interest are - simple and compound. Let's calculate the total amount for a principal when the interest is compounded annually and half yearly.

Question

A sum of money amounts to $₹ 1, 210$ in $2$ years and to $₹ 1, 610.51$ in $5$ years. Find the principal and the rate of interest when interest is compounded annually.

Accepted Answer

Let principle be $P$

and Rate of interest $= R %$ per annum

Amount $= R s 1210$ when time is $2$ years

$\Rightarrow 1210 = P {(1 + \frac{R}{100})}^{2} . . . . . (1)$

and amount $= R s 1610.51$ after $5$ years

$1610.51 = P {(1 + \frac{R}{100})}^{5} . . . . (2)$

divide equation $2 & 1$

$\frac{1610.51}{1210} = \frac{P {(1 + \frac{R}{10})}^{5}}{P {(1 + \frac{R}{100})}^{2}}$

$\Rightarrow \frac{161051}{121000} = {(1 + \frac{R}{100})}^{3}$

$\Rightarrow \frac{1331}{1000} = {(1 + \frac{R}{100})}^{3}$

$\Rightarrow {(\frac{11}{10})}^{3} = {(1 + \frac{R}{100})}^{3}$

$\Rightarrow 1 + \frac{R}{100} = \frac{11}{10}$

$\Rightarrow \frac{R}{100} = \frac{11}{10} - 1 = \frac{1}{10}$

$\Rightarrow R = \frac{100}{10} = 10 %$

Now from equation $1$

$\Rightarrow 1210 = P {(1 + \frac{10}{100})}^{2}$

$\Rightarrow 1210 = P {(1 + \frac{1}{10})}^{2}$

$\Rightarrow 1210 = P {(\frac{11}{10})}^{2}$

$\Rightarrow \frac{1210 \times 100}{121} = P$

$\Rightarrow P = 1000$

Hence principal is $R s$ $1000$ and rate of interest is $R s 10 %$ per annum

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