Question

The simple interest on a sum of money for2 years at 6 percent per annum is ₹ 900. what will be the compound interest at thatvsum at the same rateqnd for the same period?​

Answer 1

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

$\mathfrak{Suppose,} \\ \\ \mathtt{</p><p>The \: \: sum \: \: of \: \: money \: \: be \: \: Rs. \: P} \\ \\ \underline{ \mathfrak{ \: Given, \: }} \\ \\ \mathtt{Rate \: \: of \: \: Interest, r = 6\%} \\ \\ \mathtt{No. \: \: of \: \: years, n = 2 \: \: years.} \\ \\ \mathtt{S.I. = Rs. 900 }$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: }} \\ \\ \: \: \: \: \: \: \: \: \mathtt{S.I. = \frac{P \times r \times n}{100} } \\ \\ \mathtt{\Longrightarrow \: 900 = \frac{ P\times 6 \times 2}{100} } \\ \\ \mathtt{\Longrightarrow \:900 = \frac{12P}{100} } \\ \\ \mathtt{\Longrightarrow \:12P = 90000 } \\ \\ \mathtt{\Longrightarrow \: P= \frac{90000}{12} } \\ \\ \mathtt{\therefore \: \: \: P = 7500} \:$

$\bold{Hence,} \\ \: \: \: \: \: \: \: \bold{The \: \: sum \: \: of \: \: money = Rs. \: 7500}</p><p>$

$\mathfrak{Now,} \\ \\ \underline{ \mathfrak{ \: \: We \: \: know \: \: that, \: \: }} \\ \\ \mathtt{C.I. = A - P } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = P(1 + \frac{r}{100}) {}^{n} - P } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ =P(1 + \frac{r}{100} ) {}^{n} - 1 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 7500 \times (1 + \frac{6}{100}) {}^{2} - 1 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 7500 \times ( \frac{106}{100} ) {}^{2} - 1} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 7500 \times ( \frac{11236}{10000} - 1)} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 7500 \times ( \frac{11236 - 10000}{10000} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 7500 \times \frac{1236}{10000} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = \frac{9,270,000}{10000} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = 927}$

$\bold{Hence,} \\ \: \: \: \bold{Compound \: \: interest \: \: of \: \: the \: \: sum \: \: = Rs. \: 927}</p><p>$

Answer 2

Correct Question :-

The simple interest on a sum of money for 2 years at 6 percent per annum is ₹ 900. what will be the compound interest at The sum at the same Rate and for the same period?

$\rule{300}{2}$

Answer

$\small \underline{\underline{\texttt{\purple{compound\: interest = 927}}}}$

$\bf{Given}\begin{cases}\bf{ Rate \:of \:interest \:= 6percentage } \\ \sf{Year=2} \\\tt{simple \:interest =900}\\\tt{period =?}\end{cases}$

$\rule{300}{2}$

$\small\red{\boxed{\bold{ To\: Find }}}$

$\small\tt\underline{ We\: have\: to\: find \:period \:(p) }$

$\large\underline\textsf{Explanation:-}$

$\bf s.i \: \rightarrow \: \frac{ p \times r \times n}{100} \\ \: \: \\ \bf \rightarrow900 = \frac{p \times 6 \times 2}{100 } \\ \\ \bf \rightarrow 900 = \frac{12p}{100} \\ \\ \bf\rightarrow 12p = 90000 \\ \\ \bf\rightarrow p = \frac{90000}{12} \\ \\ \bf\rightarrow p = 7500$

$\because \huge\sf\underline{ period = 7500}$

$\rule{300}{2}$

$\large\because\sf\underline{ Compound \:interest :- }$

$\sf c.i = a - p \\ \sf\rightarrow p(1 + \frac{r}{100}) {}^{n} - 1 \\ \\ \sf\rightarrow p(1 + \frac{r}{100} ) {}^{n} - 1$

$\sf\rightarrow 7500 \times (1 + \frac{6}{100}) {}^{2} - 1 \\ \\ \sf\rightarrow 7500 \times (106) ^{2} - 1$

$\tt\rightarrow 7500 \times ( \frac{\cancel11236 - 1\cancel0\cancel0\cancel0\cancel0}{\cancel1\cancel0\cancel0\cancel0\cancel0} ) \\ \\ \tt\rightarrow 7500 \times \frac{1236}{10000}$

$\bf\rightarrow \frac{927\cancel0\cancel0\cancel0\cancel0}{\cancel1\cancel0\cancel0\cancel0\cancel0} \\ \\ \bf\rightarrow 927$

$\large{\boxed{\bold{ Compound \:interest = 927}}}$