Question

Compound interest on a certain sum of money for 2 years is Rs. 52 and
the simple interst for the same period at the same rate is Rs. 50. The rate
per cent is
(1)
2%
(2)
4%
(3)
6%
(4)
8%​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Rate=8\%}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green { \underline \bold{Given:}} \\ \tt: \implies Compound \: Interest(C.I) = 52 \\ \\ \tt: \implies Simple \: Interest(S.I) = 50 \\ \\ \tt: \implies Time(t) = 2 \: years \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Rate = ?$

• According to given question :.

$\bold{As \: we \: know \: that} \\ \tt: \implies Simple \: Interest = \frac{p \times r \times t}{100} \\ \\ \tt: \implies 50 = \frac{p \times r \times 2}{100} \\ \\ \tt: \implies r = \frac{2500}{p} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies C.I= A - p \\ \\ \tt: \implies 52 = A - p \\ \\ \tt: \implies A = 52 + p\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies A= p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies 52 + p = p(1 + \frac{ \frac{2500}{p} }{100} )^{2} \\ \\ \tt: \implies 52 + p = p(1 + \frac{2500}{p \times 100} )^{2} \\ \\ \tt: \implies 52 + p = p (\frac{p + 25}{p})^{2} \\ \\ \tt: \implies (52 + p)p = ( {p +25)}^{2} \\ \\ \tt: \implies 52p + {p}^{2} = {p}^{2} + 625 + 50p \\ \\ \tt: \implies 52p - 50p = 625 \\ \\ \tt: \implies 2p = 625 \\ \\ \tt: \implies p = \frac{625}{2} \\ \\ \green{\tt: \implies p = 312.5} \\ \\ \text{Putting \: value \: of \: p \: in \: (1)} \\ \tt: \implies r = \frac{2500}{312.5} \\ \\ \green{\tt:\implies r = 8\%}$