Question

if the difference between simple interest and compound interest on a sum of money for 2years at 25/2 per annum is $150. the sum is ​

Answer 1

Given,

Difference between CI and SI = $150

Rate = 25/2

Time = 2 years

Acccording to the question,

P[(1+R/100) - 1]^T - P×R×T/100 = $150

P[(1+25/2/100) - 1]^2 - P×25×2/2×100 = $150

After simplifying,

we get

17P - 16P/64 = $150

P = $960

Answer 2

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

$\huge{ \underline{ \mathfrak{ \: Given : \mapsto}}} \\ \\ \: \: \: \: \: \: \: \text{The \: difference \: between \: simple \: interest \: } \\ \text{and \: compound \: interest \: is \: Rs. \: 150.} \\ \\ \\ \huge{ \underline{ \mathfrak{Suppose, }}} \\ \\ \text{Principal = P} \\ \\ \text{Time, n = 2 years} \\ \\ \mathsf{</p><p>Rate \: \: of \: \: interest = \frac{25}{2} \%}$

$\huge{ \underline { \mathfrak{ \: We \: \: know \: \: that, \: }}}</p><p> \: \\ \\ \\ \bold{S.I. = \frac{P \times r \times n}{100} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = \frac{P \times \frac{25}{2} \times 2}{100} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: = \frac{P \times 25}{100} }$

$\huge{ \underline{\mathfrak{ \: Again, \: }}} \\ \\ \: \: \: \: \: \: \: \: \bold{C.I. = A - P } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P(1 + \frac{r}{100} ) {}^{n} -P } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P[(1 + \frac{ \frac{25}{2} }{100}) {}^{2} - 1]} \\ \\ \bold{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P[(1 + \frac{25}{200} ) {}^{2} - 1]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =P \: [( \frac{225}{200} ) {}^{2} - 1]} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P \times (\frac{50625}{40000} - 1) } \\ \\ \bold{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P \times ( \frac{50625 - 40000}{40000}) } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = P \times \frac{10625}{40000} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{P \times 10625}{40000} }$

$\huge{ \mathbb{ \underline{ \: A.T.Q. \: }}} \\ \\ \\ \\ </p><p></p><p> \bold{ \: \: \: \: C.I. - S.I. = 150} \\ \\ \bold{\Longrightarrow \frac{P \times 10625}{40000} - \frac{P \times 25}{100} =150 } \\ \\ \bold{\Longrightarrow \frac{(10625 \times P) -( 10000 \times P )}{40000} =150 } \\ \\ \bold{ \Longrightarrow 625 \times \: P = 6000000} \\ \\ \bold{\Longrightarrow P = \frac{6000000}{625} } \\ \\ \bold{ \therefore \: \: \underline{ \: P = 9600 \: \: }}$

$\text{Hence,} \\ \: \: \: \: \: \: \: \text{The \: sum \: of \: money \: is \: Rs. 9600}$