Question

37. On what principal will the compound interest be 4 more than the simple
interest for 2 years at 10 p.c.p.a.?
the answer is 400 but how pls solve my question I have asked two times but nobody answers​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \red{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \bold{ \: \: Given : \mapsto}} \\ \\ \text{No. \: of \: years, n = 2 \: years } \\ \\ \text{Rate \: of \: interest, r = 10\%} \\ \\ \text{Compound \: interest is \: Rs. 4 \: more \: } \\ \text{than \: simple \: interest.} \\ \\ \text{So,} \\ \text{ \: \: \: Difference \: between \: compound \: } \\ \text{interest \: and \: simple \: interest = Rs. 4}$

$\text{Let,} \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: principal \: \: amount \: \: be \: \: P}$

$\\ \text{ \underline{ \: In \: case \: of \: Simple \: Interest \: \: }}\\ \\ \bold{S.I. = \frac{ P \times r \times n}{100} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \:= \frac{ P \times 10\times 2}{100} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = \frac{2 P}{10} }$

$\\ \text{ \underline{ \: \: In \: case \: of \: Compound \: Interest \: \: }} \: \\ \\ \bold{C.I. = A - P} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P(1 + \frac{r}{100} ) {}^{n} - P} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P[ \:(1 + \frac{r}{100} ) {}^{n} - 1 ]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P[ \:(1 + \frac{10}{100} ) {}^{2} - 1 ]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P[ \:( \frac{110}{100} ) {}^{2} - 1 ]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P[ \:( \frac{ \: \: 12100 \: \: }{10000} ) - 1 ]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P \times ( \frac{12100 - 10000}{10000} }) \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = P \times \frac{2100}{10000} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = \frac{ \: 21 \: P \: }{100} }$

$\underline{ \bold{ \: \: A.T.Q. \: \: }} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \text{C.I. - S.I. = 4} \: \: \: \\ \\ \bold{ \Rightarrow \: \frac{ \: 21 \: P \: }{100} - \frac{2 \: P}{10} = 4 } \\ \\ \bold{ \Rightarrow \: \frac{21 \: P - 20 \: P}{100} = 4} \\ \\ \bold{ \Rightarrow \: \frac{P}{100} = 4 } \\ \\ \bold{ \therefore \: \:P = 400 }$

$\text{The \: principal \: amount \: = \underline{\: Rs. 400 \:}}$