Question

Sneha borrowed a sum of ₹1,20,000 from a bank for 2 years at the rate of10% per annum, compounded annually. What is the compound interest she must pay? 
A) ₹25200
B) ₹24000
C) ₹42000
D) ₹145200​

Answer 1

Question :

Sneha borrowed a sum of ₹1,20,000 from a bank for 2 years at the rate of 10% per annum, compounded annually. What is the compound interest she must pay? 

Formula Used :

$\sf \small \star \: A = P(1 + \frac{r}{100} )^{n}$

Here,

• "A" refers Amount

• "P" refers Principal

• "r" refers rate of interest

• "n" refers time

$\sf \small \star \: C.I = A - P$

Here,

• "C.I" means Compound Interest

• "A" refers Amount

• "P" refers Principal

Given Data :

• Principal → ₹ 120000

• Rate → 10 % p.a. compounded annually

• Time → 2 years

Solution :

First we need to get the amount and then get C.I by subtracting principal from amount

$\sf \small \dashrightarrow \: A = P(1 + \frac{r}{100}) ^{n}$

Putting the values we get

$\sf \small \dashrightarrow \: A = 120000(1 + \frac{10}{100} )^{2}$

$\sf \small \dashrightarrow \: A = 120000 (1 + \frac{ \cancel{10}}{ \cancel{100}})^{2}$

$\sf \small \dashrightarrow \: A =120000 {(1 + \frac{1}{10} )}^{2}$

$\sf \small \dashrightarrow \: A = 120000 \times { (\frac{11}{10} )}^{2}$

$\sf \small \dashrightarrow \: A = 120000 \times \frac{121}{100}$

$\sf \small \dashrightarrow \: A = 1200 \cancel{00} \times \frac{121}{1 \cancel{00}}$

$\sf \small \dashrightarrow \: A = 1200 \times 121$

$\sf \small \dashrightarrow \: A = 145200$

So the amount is ₹ 145200

Now, To get the C.I from the amount and the principal

C.I = A - P

C.I. = ₹ 145200 - ₹ 120000

C.I. = ₹ 25200

Hence, the Compound Interest after 2 years is ₹ 25200

So the answer is option A i.e. ₹ 25200

Answer 2

Answer:

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}$

• ➳ Principle = Rs.120000
• ➳ Time = 2 years
• ➳ Rate of Interest = 10%

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To Find :}}}}}}\end{gathered}$

• ➳ Compound Interest

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Concept:}}}}}}\end{gathered}$

➳ Here the Concept of Compound Interest and Amount has been used . We see we are given the details of the exchange that is Principal, Time and Rate .Here the Principle is Rs.120000, the time is 2 years and the rate of Interest is 10% p.a. So firstly we will find the amount and after finding amount we will find the compound interest.!!

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Using Formulas :}}}}}}\end{gathered}$

$\quad{\dag{\underline{\boxed{\sf{A ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\quad\dag{\underline{\boxed{\sf{C.I = A- P }}}}$

Where

• ➽ A = Amount
• ➽ P = Principle
• ➽ R = Rate of Interest
• ➽ T = Time Period
• ➽ C.I = Compound Interest

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}$

${\bigstar \:{\underline{\underline{\pmb{\frak{{Firstly, calculating\: the \: Amount.. }}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}$

• Substituting the values

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg(1 + \dfrac{10}{100}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg(\dfrac{(1 \times 100) + 10}{100}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg(\dfrac{100 + 10}{100}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg(\dfrac{110}{100}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg( \cancel{\dfrac{110}{100}}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg(\dfrac{55}{50}{\bigg)}^{2}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg( \dfrac{55}{50} \times \dfrac{55}{50}{\bigg)}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg( \dfrac{3025}{2500}{\bigg)}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg( \cancel{\dfrac{3025}{2500}}{\bigg)}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000{\bigg( \dfrac{121}{100}{\bigg)}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {120000 \times \dfrac{121}{100}}}}}$

$\quad {: \longmapsto{\sf{Amount = {\cancel{120000}\times \dfrac{121}{\cancel{100}}}}}}$

$\quad {: \longmapsto{\sf{Amount = {1200 \times 121}}}}$

$\quad {: \longmapsto{\sf{Amount = {Rs.145200}}}}$

$\quad\dag{\underline{\boxed{\tt{\blue{Amount}={\purple{Rs.145200}}}}}}$

• Hence, The Amount is Rs.145200.

$\begin{gathered}\end{gathered}$

${\bigstar{\underline{\underline{\pmb{\frak{Now,Finding \: the \: Compound \: Interest}}}}}}$

$\quad{: \longmapsto{\sf{Compound \: Interest = {Amount - Principle }}}}$

• Substituting the values

$\quad{: \longmapsto{\sf{Compound \: Interest = {145200 - 120000}}}}$

$\quad{: \longmapsto{\sf{Compound \: Interest = {Rs.25200}}}}$

$\quad\dag{\underline{\boxed{\tt{\blue{Compound \: Interest} = \purple{Rs.25200}}}}}$

• Hence,The Compound Interest is Rs.25200..
• So,The option A) ₹25200 is the correct answer.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Learn More :}}}}}}\end{gathered}$

$\quad\odot{\underline{\boxed{\rm{\green{Amount = Principle + Interest}}}}}$

$\quad\odot{\underline{\boxed{\rm{\green{ P=Amount - Interest }}}}}$

$\quad\odot{\underline{\boxed{\rm{\green{ S.I = \dfrac{P \times R \times T}{100}}}}}}$

$\quad\odot{\underline{\boxed{\rm{\green{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\quad\odot{\underline{\boxed{\rm{\green{P = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

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