Question

Find the compound interest on Rupees 75000 at 12% per annum for 9 months if the interest is compounded quarterly.​

Answer 1

$\bf{\Huge{\boxed{\tt{\green{ANSWER\::}}}}}$

$\bf{Given}\begin{cases}\sf{Principal,[P]\:=\:Rs.75000}\\ \sf{Rate,[R]\:=\:12\%}\\ \sf{Time,[n]\:=\:9\:month}\\ \sf{Time\:[n]\:=\cancel{\dfrac{9}{12} }=\dfrac{3}{4}\:years }\end{cases}}$

$\bf{\Large{\underline{\bf{To\:Find\::}}}}$

The compound Interest.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

We know that formula of the compound Interest (C.I.):

$\leadsto\sf{Amount(A)\:-\:Principal(P)}$

Or [compounded quarterly]

$\bf{\Large{\boxed{\sf{A\:=\:P\bigg(1+\dfrac{R}{100*4} \bigg)^{4n} }}}}$

Therefore,

$\bigstar\implies\sf{A\:=\:75000\bigg(1+\dfrac{12}{100*4} \bigg)^{\cancel{4}*\dfrac{3}{\cancel{4}}}}$

$\bigstar\implies\sf{A\:=\:75000\bigg(1+\cancel{\dfrac{12}{400}} \bigg)^{3} }$

$\bigstar\implies\sf{A\:=\:75000\bigg(1+\dfrac{3}{100} \bigg)^{3} }$

$\bigstar\implies\sf{A\:=\:75000\bigg(\dfrac{100+3}{100} \bigg)^{3} }$

$\bigstar\implies\sf{A\:=\:75000\bigg(\dfrac{103}{100} \bigg)^{3} }$

$\bigstar\implies\sf{A\:=\:75\cancel{000}*\dfrac{103}{\cancel{100}} *\dfrac{103}{10\cancel{0}} *\dfrac{103}{100} }$

$\bigstar\implies\sf{A\:=\:\dfrac{75*103*103*103}{1000} }$

$\bigstar\implies\sf{A\:=\:Rs.\cancel{\bigg(\dfrac{81954525}{1000}\bigg)} }$

$\bigstar\implies\sf{\red{A\:=\:Rs.81954.52}}$

Thus,

The Compound Interest (C.I.) = A - P

The Compound Interest (C.I.) = Rs.(81954.52 - 75000)

$\bigstar\:\huge{\boxed{\sf{The\:compound\:Interest\:is\:\:Rs.6954.52}}}}$

Answer 2

Step-by-step explanation:

[tex]\bf{Given,}\begin{cases}\sf{Principal = ₹75000}\ \sf{Rate Of Interest = 12%}\ \sf{Time = 9 months}[\tex]