Question

Find the compound interest if the amount of a certain principal after two years is 4036.80 at the rate of 16p.c.p.a
​

Answer 1

AnsWer :

First of all we need to write the known values which are provided in the question. So, let's write it first :

$\frak {\pink{Given}}\begin{cases} \sf{\green{Principal=?}}\\ \sf{\blue{Time=2\:Years}}\\ \sf{\orange{Rate=16\%}}\\ \sf{\red{Amount= 4036.80}}\\ \sf{\gray{Compound \: Interest=?}}\end{cases}$

We are asked to find the Compound Interest. But before that we need to calculate the Principal which is not given in the question. So, let's solve :

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\dashrightarrow\:\:\sf Amount = Principal \Bigg\lgroup 1 + \dfrac{Rate}{100}\Bigg\rgroup^{\tiny Time}$

$\dashrightarrow\:\:\sf 4036.80 = Principal \Bigg\lgroup 1 + \dfrac{16}{100}\Bigg\rgroup^{\tiny 2}$

$\dashrightarrow\:\:\sf 4036.80 = Principal \Bigg\lgroup \dfrac{100 \times 1 + 16}{100}\Bigg\rgroup^{\tiny 2}$

$\dashrightarrow\:\:\sf 4036.80 = Principal \Bigg\lgroup \dfrac{116}{100}\Bigg\rgroup^{\tiny 2}$

$\dashrightarrow\:\:\sf 4036.80 = Principal \times \bigg\lgroup 1.16\bigg\rgroup^{\tiny 2}$

$\dashrightarrow\:\:\sf 4036.80 = Principal \times 1.3456$

$\dashrightarrow\:\:\sf \dfrac{ 4036.80}{1.3456} = Principal$

$\dashrightarrow\:\:\sf Principal = \dfrac{ 4036.80}{1.3456}$

$\dashrightarrow\:\: \underline{ \boxed{\sf Principal =Rs. \: 3000 }}$

• Hence, the Principal is Rs. 3,000.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Now, we can calculate the Compound Interest :

$:\implies\sf Compound \: Interest = Amount - Principal$

$:\implies\sf Compound \: Interest = 4036.80 - 3000$

$:\implies \underline{ \boxed{\sf Compound \: Interest =Rs. \: 1036.8 }}$

• Hence,the Compound Interest is Rs. 1036.8

Answer 2

AnswEr-:

• $\dag{\underline {\red{\mathrm {Compound\:Interest= Rs.1036.80 }}}}$

Explanation -:

• $\mathrm{\bf{Given \:\: -:}} \begin{cases} \sf{\purple{Rate= \frak{16\% \:per\:annum}}} & \\\\ \sf{\blue {Amount \:=\:\frak{Rs.4036.80}}}& \\\\ \sf{\green{Time \:=\:\frak{2\:years}}}\end{cases}$

• $\mathrm{\bf{To\:Find \:\: -:}} \begin{cases} \sf{ The\: \bf{ Compound\:Interest\:or\:C.I.}} \end{cases}$

$\dag {\mathrm {\bf { Solution \:of\:Question-:}}}$

• $\underbrace{\mathrm { Understanding \: the \: Concept-:}}$

• We have to find the Compound Interest when Amount, Rate and Time is Given.

• First we have to find the Principal by Putting known Values in the Formula for Amount.

Then ,

• By Putting Principal and Amount in the Formula for Compound Interest.

• We can get the Compound Interest.

As , We know that ,

• $\underline {\boxed {\mathrm {\pink{ Amount = Principal \left( 1 + \dfrac{Rate}{100} \right)^{Time} }}}}$

• $\mathrm{\bf{Here \:\: -:}} \begin{cases} \sf{Rate= \frak{16\% \:per\:annum}} & \\\\ \sf{Amount \:=\:\frak{Rs.4036.80}}& \\\\ \sf{Time \:=\:\frak{2\:years}}& \\\\ \sf{Principal \:=\:\frak{??}}\end{cases}$

Now By Putting known Values in Formula for Amount-:

• $\longmapsto { \mathrm { Rs.4036.80 = Principal \left( 1 + \dfrac{16}{100} \right)^{2}}}$

• $\longmapsto { \mathrm { Rs.4036.80 = Principal \left( 1 + 0.16 \right)^{2}}}$

• $\longmapsto { \mathrm { Rs.4036.80 = Principal \left( 1.3456 \right)}}$

• $\longmapsto { \mathrm { \dfrac{4036.80}{1.3456} = Principal }}$

• $\longmapsto { \mathrm { \dfrac {\cancel {4036.80}}{\cancel{1.3456}} = Principal }}$

• $\underline{\boxed { \mathrm { Rs.3000 = Principal }}}$

Therefore,

• $\underline{\dag { \mathrm { Rs.3000 = Principal }}}$

___________________________________

As , We know that,

• $\underline {\boxed {\mathrm {\pink{ Compound \:Interest\:= Amount - Principal }}}}$

• $\mathrm{\bf{Here \:\: -:}} \begin{cases} \sf{Principal= \frak{Rs.3,000}} & \\\\ \sf{Amount \:=\:\frak{Rs.4036.80}}\end{cases}$

Now By Putting known Values in Formula for Compound Interest-:

• $\longmapsto { \mathrm {Compound\:Interest= 4036.80 -3000 }}$

• $\underline {\boxed{ \mathrm {Compound\:Interest= Rs.1036.80 }}}$

Hence ,

• $\dag{\underline {\red{\mathrm {Compound\:Interest= Rs.1036.80 }}}}$

________________________________________________