Question

Find the compound interest on rs4000 for 1 year at 10% per annum, the interest being compounded quarterly.​

Answer 1

QUESTION :-

Find the compound interest on rs4000 for 1 year at 10% per annum, the interest being compounded quarterly.

ANSWER :-

• $\small{ \mathrm{ Principal (P)=Rs.4000}}$
• $\small{ \mathrm{Rate \: of \: Interest \: (r)=10\%}}$
• $\small{ \mathrm{Time =2 \: years \: and \: 3 \: months}}$

$\small{ \mathrm{Here \: first \: we \: take \: n=2 \: years. }}\\ { \small{ \mathrm{</p><p>⇒ Amount \: for \: first \: 2 \: years (A)=P{(1+ \frac{r}{100} ) }^{n} }}} \\ { \small{ \mathrm{</p><p> </p><p></p><p> =4000{(1+ \frac{10}{100} ) }^{2} }}} \\ { \small{ \mathrm{</p><p> </p><p></p><p> =4000 {( \frac{11}{10} )}^{2} }}} \\ { \small{ \mathrm{</p><p> </p><p></p><p> =4000× \frac{121}{100} }}}</p><p> </p><p> \\ { \small{ \mathrm{</p><p></p><p> =Rs.4840}}}$

$\small{ \mathrm{⇒ The \: amount \: after \: two \: years =Rs.4840}} \\ { \small{ \mathrm{</p><p>⇒ Now, \: Principal =Rs.4840}}}</p><p> \\ { \small{ \mathrm{Simple \: interest \: for \: last \: 3 \: months }}} \\ { \small{ \mathrm{i.e. </p><p>41 years = \frac{PRT}{100} }}}</p><p> </p><p> \\ { \small{ \mathrm{ </p><p></p><p> = \frac{4840×10×1}{100 \times 4}}}} \\ </p><p> </p><p>{ \small{ \mathrm{ \underline</p><p></p><p> { = \: Rs.121}}}}</p><p>$

$\small{ \mathrm{⇒ Amount \: after \: 2 \: years \: and \: 3 months }} \\{ \small{ \mathrm{ =Rs.4840+Rs.121=Rs.4961}}} \\ { \small{ \mathrm{</p><p>⇒ C.I.=A−P}}} \\ { \small{ \mathrm{</p><p> =Rs.4961−Rs.4000}}} \\ { \small{ \mathrm{ \bold{ \underline{</p><p> =Rs.961}}}}}$

• $\small{ \bold{ \mathrm{ \underline{∴ \: The \: compound \: interest \: is \: Rs.961 \: }}}}$