Question

3. If the difference between the simple interest and
compound interest on some principal amount at 20%
per annum for 3 years is * 48, then the principle
amount must be
(a) 550
(b) 500
(c) 375
(d) 3400​

Answer 1

Given :

• Rate (R)= 20%
• Time (T)=3 years
• Difference between the compound interest (CI) and simple interest (SI) = Rs 48

To find :

• Principal (P).

Solution :-

Let the Principal (P) be Rs x.

In First Case :

$\bigstar {\boxed {\tt{SI=\dfrac{P\times R\times T}{100} }}}\bigstar$

$:\implies {\sf{SI=\dfrac{P\times R\times T}{100}} }\\ \\ :\implies {\sf{SI=Rs\:(\dfrac{x\times 20\times 3}{100})}}\\ \\ :\implies{\boxed {\tt{SI=Rs\:\dfrac{3x}{5}}}}$

In Second Case :

$\bigstar{ \boxed{ \tt{CI = P( 1+\dfrac{R}{100})^{T} -P }}} \bigstar$

$: \implies{\sf{CI = P(1 +\dfrac{R}{100})^{T} -P}} \\ \\ : \implies{\sf{CI = Rs\:[ x( 1+ \dfrac{20}{100})^{3}-x] }} \\ \\ : \implies{\sf{CI = Rs\:[ x( \dfrac{100+20}{100})^{3} -x] }} \\ \\ : \implies{\sf{CI = Rs\: (x \times \dfrac{120}{100} \times \dfrac{120}{100} \times\dfrac{120}{100}-x)}} \\ \\ : \implies{\sf{CI = Rs\: ( \dfrac{216x}{125} -x)}} \\ \\ : \implies{\sf{CI = Rs\: ( \dfrac{216x - 125x}{125} )}} \\ \\ : \implies{ \boxed{\tt{CI = Rs\: \dfrac{91x}{125} }}}$

Difference between the compound interest (CI) and simple interest (SI) :

$: \implies{ \sf{CI - SI = 48}} \\ \\ : \implies{ \sf{ \dfrac{91x}{125} - \dfrac{3x}{5}= 48}} \\ \\ : \implies{ \sf{ \dfrac{91x - 75x}{125}= 48}} \\ \\ : \implies{ \sf{ \dfrac{16x}{125}= 48}} \\ \\ : \implies{ \sf{ 16x= 48 \times 125}} \\ \\ : \implies{ \sf{x= \dfrac{48 \times 125}{16}}} \\ \\ : \implies{ \sf{x= 3 \times 125}} \\ \\ : \implies{ \boxed{ \tt{x=375}}}$

$\underline{\bf {\therefore{Principal =Rs\:375.}}}$

Answer 2

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

$\green{\tt{\therefore{Principal=375\:rupees}}}$

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

$\begin{gathered}\green{\underline \bold{Given :}} \\ \tt: \implies Rate\%(r) = 20\% \\ \\ \tt: \implies Time(t) = 3 \: years \\ \\ \tt: \implies C.I- S.I = 48 \: rupees \\ \\ \red{\underline \bold{to \: find:}}\\ \tt: \implies Principal(p) = ?\end{gathered}$

• According to given question :

$\begin{gathered}\bold{As \: we \: know \: that} \\ \tt: \implies S.I= \frac{p \times r \times t}{100} \\ \\ \tt: \implies S.I= \frac{p \times 20 \times 3}{100} \\ \\ \tt: \implies S.I= \frac{3p}{5} - - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies A =p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies A = p(1 + \frac{20}{100} )^{3} \\ \\ \tt: \implies A=p(1 + 0.2)^{3} \\ \\ \tt: \implies A =p \times 1.728\\ \\ \bold{For \: C.I} \\ \tt: \implies C.I =A - p \\ \\ \tt: \implies C.I=1.728p - p \\ \\ \tt: \implies C.I=0.728p - - - - - (2) \\ \\ \bold{As \: according \: to \: question} \\ \tt: \implies C.I - S.I= 48 \\ \\ \tt: \implies 0.728p - 0.6p = 48 \\ \\ \tt: \implies 0.128p =48 \\ \\ \tt: \implies p = \frac{48}{0.128} \\ \\ \green{\tt: \implies p = 375 \: rupees}\end{gathered} </p><p>$