Question

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

Answer 1

Solution:

Given details are,

Rate 15 % per annum

Compound Interest (CI) - Simple nterest (S)= Rs 45

Time (t) = 2 years

By using the formula,

CI - SI = 45

P[(1 +R/100)^n - 1] - (PTR)/100 = 45

P[(1+ 15/100)² - 1] - P(2)(15)/100 = 45

P[1.3225 - 1] - (30P)/100 = 45

0.3225P - 0.30P = 45

0.0225P = 45

P = 45/0.0225

p = 2000

The sum is Rs 2000

Answer 2

Answer:

Given:

• Principal (P) = x

• Rate (r) = 15%

• Time (t) = 2 Years

• Compound Interest (C.I.) - Simple Interest (S.I.)= ₹45

To Find : -

• find the sum

Solution :

• Let the sum of money be x.

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.