5 Mark Question

Question 15 Marks

The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs. 93. Find the sum.

Answer

$\therefore$ Simple interest (S.I) $=\frac{\text{prt}}{100}$
$=\frac{100\times10\times3}{100}=\text{Rs. }30$
Amount (A) $=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$=\text{Rs. }100\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }100\times\Big(\frac{11}{10}\Big)^3$
$=\text{Rs. }100\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}$
$=\text{Rs. }\frac{1331}{10}$
$\therefore$ C.I = A - P $=\text{Rs. }\frac{1331}{10}-\text{Rs. }100$
$=\text{Rs. }\frac{331}{10}$
Difference between C.I and S.I
$=\text{Rs. }\frac{1331}{10}-\text{Rs. }30=\text{Rs. }\frac{31}{10}$
If difference is $=\text{Rs. }\frac{31}{10},$ then sum
and if difference is Rs. 93 then sum
$=\text{Rs. }\frac{100\times93\times10}{31}$
$=\text{Rs. }3000$

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Question 25 Marks

Neeraj lent Rs. 65536 for 2 years at $12\frac{1}{2}\%$ per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?

Answer

Amount of loan = Rs. 65536
Rate of interest (R) $=12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a. or }\frac{25}{4}\%$ half yearly
Period (n) = 2 year or 4 half years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }65536\Big(1+\frac{25}{4\times100}\Big)^4$
$=\text{Rs. }65536\times\Big(\frac{17}{16}\Big)^4$
$=\text{Rs. }65536\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}\times\frac{17}{16}$
$=\text{Rs. }83521$
If the interest in compounded annually, then
Amount $=\text{Rs. }65536\Big(1+\frac{25}{2\times100}\Big)^2$
$=\text{Rs. }65536\times\Big(\frac{9}{8}\Big)^2$
$=\text{Rs. }65536\times\frac{9}{8}\times\frac{9}{8}=\text{Rs. }82944$
$\therefore$ More interest recieved
= Rs. 183521 - 82944
= Rs. 577

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Maths 5 Mark Question

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