# NIELIT 2017 OCT Scientific Assistant A - Section A: 1

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The difference between the compound interest and the simple interest earned at the end of $3^{rd}$ year on a sum of money at a rate of $10\%$ per annum is Rs. $77.5.$ What is the sum?

1. Rs. $3,500$
2. Rs. $2,500$
3. Rs. $3,000$
4. Rs. $2,000$

$SI = \dfrac{PRT}{100}$

$\implies SI= \dfrac{P \times 10 \times 3}{100}$

$\implies SI= \dfrac{30P}{100}$

$\implies SI = 0.3P \quad \rightarrow(1)$

$A = P\left(1+\dfrac{R}{100}\right)^{T}$

$A = P\left(1+\dfrac{10}{100}\right)^{3} = P(1.1)^{3} = 1.331\;P$

Now, $CI = A-P = 1.331\;P – P = 0.331\;P$

And, $CI – SI = 77.5$

$\implies 0.331P – 0.3P = 77.5$

$\implies 0.031 P = 77.5$

$\implies P = Rs.\;2,500.$

$\textbf{(OR)}$

${\color{Magenta}{CI – SI = \dfrac{Pr^{2}}{100^{2}}\;\;\;(\text{for 2 years})}}$

${\color{Teal}{CI – SI = \dfrac{Pr^{2}(300+r)}{100^{3}}\;\;\;(\text{for 3 years})}}$

$\implies 77.5 = \dfrac{P \times 10^{2}(300 + 10)}{100^{3}}$

$\implies 77.5 = \dfrac{P \times 100(310)}{100^{3}}$

$\implies 77.5 = \dfrac{P \times 310}{100^{2}}$

$\implies 77.5 = \dfrac{P \times 310}{10000}$

$\implies 77.5 = \dfrac{P \times 31}{1000}$

$\implies 31P = 77.5 \times 1000$

$\implies 31P = 77500$

$\implies P = Rs.\;2500.$

So, the correct answer is $(B).$
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