Question

The difference between compound interest compounded annually and simple interest on a certain sum of money for 2 years at 5%p.a is Rs. 12.50. What is the compound interest on this sum for 2 years? (A) Rs. 262.50 (B) Rs. 525.00 (C) Rs. 250 (D) Rs. 512.50​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The difference between compound interest compounded annually and simple interest on a certain sum of money for 2 years at 5%p.a is Rs. 12.50.

Let assume that the sum of money is Rs P.

Rate of interest, r = 5 % per annum

Time, n = 2 years.

We know,

Compound Interest (CI) received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\sf{ \: \: CI \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - P \: \: }}$

And

Simple interest (SI) received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{ \: \: SI \: = \: \frac{P \times r \times n}{100} \: \: }}$

Now, According to statement, we have

$\rm \: CI \: - \: SI \: = \: 12.50 \:$

$\rm \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - P - \dfrac{Prn}{100} = 12.5$

On substituting the values of r and n, we get

$\rm \: P \: {\bigg[1 + \dfrac{5}{100} \bigg]}^{2} - P - \dfrac{P \times 2 \times 5}{100} = 12.5$

$\rm \: P \: {\bigg[1 + \dfrac{1}{20} \bigg]}^{2} - P - \dfrac{P }{10} = 12.5$

$\rm \: P \: {\bigg[\dfrac{20 + 1}{20} \bigg]}^{2} - P - \dfrac{P }{10} = 12.5$

$\rm \: P \: {\bigg[\dfrac{21}{20} \bigg]}^{2} - P - \dfrac{P }{10} = 12.5$

$\rm \: \: \dfrac{441}{400}P - P - \dfrac{P }{10} = 12.5$

$\rm \: \: \dfrac{441P - 400P - 40P}{400} = 12.5$

$\rm \: \: \dfrac{P}{400} = 12.5$

$\rm \: P = 400 \times 12.5$

$\rm\implies \:P \: = \: Rs \: 5000$

Now, we have

Principal, P = Rs 5000

Time, n = 2 years

Rate, r = 5 % per annum compounded annually

So,

$\rm \: CI = P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} - P$

$\rm \: CI = 5000 \: {\bigg[1 + \dfrac{5}{100} \bigg]}^{2} - 5000$

$\rm \: CI = 5000 \: {\bigg[1 + \dfrac{1}{20} \bigg]}^{2} - 5000$

$\rm \: CI = 5000 \: {\bigg[\dfrac{20 + 1}{20} \bigg]}^{2} - 5000$

$\rm \: CI = 5000 \: {\bigg[\dfrac{21}{20} \bigg]}^{2} - 5000$

$\rm \: CI = 5000 \times \dfrac{441}{400} - 5000$

$\rm \: CI = 5512.5 - 5000$

$\rm\implies \:CI \: = \: Rs \: 512.50$

So, option (D) is correct.

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Additional Information :-

1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\boxed{\sf{ \: \: \: Amount \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }}$

2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by

$\boxed{\sf{ \: \: \: Amount \: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\sf{ \: \: \: Amount \: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }}$

4. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{\sf{ \: \: \: Amount \: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }}$

Answer 2

Answer:

512.50 (D)

Step-by-step explanation:

☆Given:

The difference between compound intrest and

Simple intrest is 12.5

☆To find

Principle and compound intrest

☆Solution:

▪P = ?

▪I = 5%

▪N = 2

$P[(1+i)^n-1]$ - p × i × t = 12.5

$P[(1+0.05)^2-1]$ - p × (⁵/₁₀₀) × 2 = 12.5

$[(1.05)^2-1] = 0.1025$

0.1025p - $\frac{10p}{100}$ = 12.5

$\frac{10.25p-10p}{100}$ = 12.5

0.25p = 1250

P= 1250 ÷ .25 = 5,000

$CI = P[(1+i)^n-1]$

CI = 5000 [(1.05)^2-1]

CI = 5000 × 0.1025

CI = 512.50