Question

The difference between the compound interest, compounded annually and the simple interest on a certain sum for 2 years at 15% per annum is Rs 180. Find the sum.

Answer 1

★Answer :-

• Sum is Rs. 8000 .

★Step-by-step-Explaination :-

❒How to solve !!

See , a difference between compound interest, compounded annually and the simple interest on a certain sum for how many years is 2 years . Also rate per interest is provided that is 15% . Also , Difference between S.I. (Simple Interest) and C.I. (Compound Interest) is given that is Rs.180 .

❒Let's Proceed !!

Let ::

• Let the Principal be x .

★Let's Find S.I. First !!

We know that ,

$\rm \: S.I. \: = \dfrac{ P \times R \times T}{100}$

• Principal is x .
• Rate is 15% .
• Time is 2 years .

$\rm \implies \: S.I. \: = \dfrac{ x \times1 5 \times 2}{100}$

Let's cancel the values and bring it to simplest form .

$\rm \implies \: S.I. \: = \dfrac { x \times3}{10}$

$\rm \implies \: S.I. \: = \dfrac{ 3x}{10}$

Now , we also know that

Here , the formula for finding amount at the end of n years for Principal p and rate of interest R% per annum is

$\rm \: Amount = P \bigg(1 + \dfrac{R}{100} \bigg) ^{n}$

Putting the values given to us

$\rm \: \implies \: Amount = x \bigg(1 + \dfrac{15}{100} \bigg) ^{2}$

Solving this we get

$\rm \: \implies \: Amount = \dfrac{529}{400} x$

Then ,

Compound Interest is

$\rm \: Compound \: Interest = \dfrac{529}{400} x - x$

$\rm \: Compound \: Interest = \dfrac{529x - 400x}{400}$

$\rm \: Compound \: Interest = \dfrac{129}{400} x$

Now ,

★Given in the Question :-

The difference between the compound interest, compounded annually and the simple interest on a certain sum is 2 years interest is 15% and the difference is Rs. 180 .

So ,

$\rm \: C.I. - S.I. = 180$

Putting Values,

$\rm \: \implies \: \dfrac{129x - 120x}{400} = 180$

So , we will Subtract and 400 will go to R.H.S.

$\rm \: \implies \: {9x } = 180 \times 400$

$\rm \: \implies \: {x } = \dfrac{ 180 \times 400}{9}$

After cancelling ,

$\rm \: \implies \: {x } = 8000$

$\sf \: \underline{ \bigstar \bf \: x = 8000}$

Hence , the Sum is Rs. 8000 .

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Additional Information !!

$\rm S.I.=\dfrac{P\times R\times T}{100}$

$\rm P=\dfrac{100\times S.I.}{R\times T}$

$\rm R=\dfrac{100\times S.I.}{P\times T}$

$\rm T=\dfrac{100\times S.I.}{R\times P}$

$\rm A = P + I=P+\dfrac{P\times R\times T}{100}=\bigg(1+\dfrac{R \times T}{100}\bigg)$

Deducing a Formula for Compound Interest !!

Suppose a sum of rs. P is compounded annually at a rate of R% per annum for n years .

• The principal for first year be P

$\rm Interest \;for\; 1 st\; year = \dfrac{P\times R\times 1}{100}=\dfrac{PR}{100}$

$\rm Amount\;at\;the\;end\;of\; 1 st\; year =P+ \dfrac{P\times R\times 1}{100}=P+\dfrac{PR}{100}= P\bigg(1+\dfrac {R}{100}\bigg)$

$\rm Principal\;for\; 2nd\; year = P\bigg(1+\dfrac {R}{100}\bigg)$

$\rm Interest \;for\; 2nd\; year = \dfrac{P\bigg(1+\dfrac{R}{100}\bigg)\times R\times 1}{100}=\dfrac{PR}{100}\bigg(1+\dfrac R{100}\bigg)=P\bigg(1+\dfrac R{100}\bigg)\bigg(1+\dfrac R{100}\bigg)=P\bigg(1+\dfrac{R}{100}\bigg)^2$