Solution:
Given,
We have,
Principal or the sum = Rs P
Rate of Compound Interest = R% per annum
Now,
In Case I,
Duration of time (T) = 2 years
Compound Interest (CI) = Rs 5560
So, CI = P [ \left ( 1+\frac{R}{100} \right )^T -1]
or, 5560 = P [ \left ( 1+\frac{R}{100} \right )^2 -1]
or, 5560 = P [ \left ( \frac{100+R}{100} \right )^2 -1]
or, 5560 = P \left ( \frac{(100+R)^2 -100^2}{100^2} \right)
or, P = \frac{5560 * 100^2}{(100+R)^2-100^2} ----- (i)
And,
In Case II,
Duration of time (T) = 4 years
Compound Interest (CI) = Rs 12066.60
So, CI = P [ \left ( 1+\frac{R}{100} \right )^T -1]
or, 12066.60 = P [ \left ( 1+\frac{R}{100} \right )^4 - 1]
or, 12066.60 = P [ \left ( \frac{100+R}{100} \right )^4 -1]
or, 12066.60 = P \left ( \frac{(100+R)^4 -100^4}{100^4} \right)
or, P = \frac{12066.60 * 100^4}{(100+R)^4-100^4} ------ (ii)
Since, values of P are equal, the equation (i) and (ii) can be written as:
\frac{5560 * 100^2}{(100+R)^2-100^2} = \frac{12066.60 * 100^4}{(100+R)^4-100^4}
or, \frac{5560 * 100^2}{(100+R)^2-100^2} = \frac{12066.60 * 100^4}{\{(100+R)^2-100^2\}\{(100+R)^2+100^2\}}
or, \frac{\{(100+R)^2-100^2\}\{(100+R)^2+100^2\}}{\{(100+R)^2-100^2\}} = \frac{12066.60 * 100^4}{5560 * 100^2}
or, (100+R)^2 +100^2 = 21702.51
or, 100^2 +200R +R^2 +100^2 = 21702.51
or, R^2 +200R +20000 = 21702.51
or, R^2 +200R - 1702.51 = 0 ----- (iii)
Comparing equation (iii) with ax² +bx +c = 0
We get,
a = 1, b = 200, c = -1702.51, x = R
Using formula of quadratic equation;
x = \frac{-b +_{-} \sqrt{b^2-4ac}}{2a}
Since, our rate will we positive, we remove the '-' sign,
or, R = \frac{-200 + \sqrt{200^2-4(1)(-1702.51)}}{2(1)}
= \frac{ -200 + \sqrt{46810.04}}{2}
= \frac{ -200 + 216.35}{2}
So, R = 8.17
\therefore R = 8.17%
Again,
Put value of R in equation (i)
P = \frac{5560 * 100^2}{(100+8.17)^2-100^2}
= Rs 32691.5
Therefore, the required rate of interest for the given cases is 8.17% per annum and the sum of money or principal is equal to Rs 32691.5.
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