Question

The difference between the compound interest and simple interest on a certain sum at 15% per annum for 2 years is Rs. 45. find the sum?

Answer 1

Step-by-step explanation:

Let the sum of money be x.

Given:

• Principal (P) = x
• Rate (r) = 15%
• Time (t) = 2 years
• Compound Interest (C.I.) - Simple Interest (S.I.)= ₹45

Solution :

We know that,

$\boxed{\rm C.I. = P\Bigg [ \Bigg (1+\dfrac{r}{100} \Bigg )^t -1\Bigg ] }$

And,

$\boxed{\rm S.I. = \dfrac{PRT}{100}}$

From Question,

CI - SI = ₹45

Hence,

$\rm \dashrightarrow P\Bigg [ \Bigg (1+\dfrac{r}{100} \Bigg )^t -1\Bigg ] - \dfrac{PRT}{100} = 45$

Substituting the values, we get,

$\rm \dashrightarrow x\Bigg [ \Bigg (1+\dfrac{15}{100} \Bigg )^2 -1\Bigg ] - \dfrac{x \times 15 \times 2}{100} = 45$

$\rm \dashrightarrow x\Bigg [ \Bigg (1+\dfrac{3}{20} \Bigg )^2 -1\Bigg ] - \dfrac{30x}{100} = 45$

$\rm \dashrightarrow x\Bigg [ \Bigg (\dfrac{23}{20} \Bigg )^2 -1\Bigg ] - \dfrac{30x}{100} = 45$

$\rm \dashrightarrow x\Bigg ( \dfrac{529}{400} -1\Bigg ) - \dfrac{30x}{100} = 45$

$\rm \dashrightarrow x\Bigg ( \dfrac{529-400}{400}\Bigg ) - \dfrac{30x}{100} = 45$

$\rm \dashrightarrow x\Bigg ( \dfrac{129}{400}\Bigg ) - \dfrac{30x}{100} = 45$

$\rm \dashrightarrow\dfrac{129}{400}x- \dfrac{30}{100}x = 45$

$\rm \dashrightarrow\dfrac{129-120}{400}x = 45$

$\rm \dashrightarrow\dfrac{9}{400}x = 45$

$\rm \dashrightarrow x = \dfrac{400 \times 45}{9}$

$\rm \dashrightarrow x = 400 \times 5$

$\bf \dashrightarrow x = 2000$

Hence, the sum of money is ₹2000.