Question

The simple interest and the compound interest on a certain sum for 2 years is 1250 and 1475 respectively. Find the rate of interest. ​

Answer 1

Solution:

Let Principal amount be P.

Rate of interest be R%.

Given:

• Simple Interest (SI) = Rs. 1250
• Compound interest (CI) = Rs. 1475

• Time = 2 years

To Find:

• Rate of interest.

Formula used:

• $\tt{Simple\;Interest=\dfrac{PRT}{100}}$

• $\tt{Compound\;Interest=Amount-Principal\;amount}$

• $\tt{Amount=P\Bigg(1+\dfrac{R}{100}\Bigg)^{t}}$

$\rule{200}{2}$

Now,

$\tt{\implies Simple\;Interest=\dfrac{PRT}{100}}$

$\tt{\implies 1250=\dfrac{P\times R\times 2}{100}}$

$\tt{\implies 1250=\dfrac{PR}{50}}$

$\tt{\implies PR=1250\times 50}$

$\tt{\implies PR = 62500}$

$\tt{\implies R=\dfrac{62500}{P}\;\;\;\;\;.............(1)}$

$\rule{200}{2}$

$\tt{\implies Compound\;Interest=Amount-Principal\;amount}$

$\tt{\implies CI = P\Bigg(1+\dfrac{R}{100}\Bigg)^{t}-P}$

Now, put the values in the formula,

$\tt{\implies 1475=P\Bigg(1+\dfrac{62500}{100P}\Bigg)^{2}-P\;\;\;\;\;\Bigg[\therefore R=\dfrac{62500}{P}\Bigg]}$

$\tt{\implies 1475=P\Bigg(1+\dfrac{625}{P}\Bigg)^{2}-P}$

Now, by using (a + b)² = a² + b² + 2ab

$\tt{\implies 1475=P\Bigg(1+\dfrac{390625}{P^{2}}+\dfrac{1250}{P}\Bigg)-P}$

$\tt{\implies 1475=P+\dfrac{390625P}{P^{2}}+\dfrac{1250P}{P}-P}$

$\tt{\implies 1475=P+\dfrac{390625}{P}+1250-P}$

$\tt{\implies 1475=\dfrac{390625}{P}+1250}$

$\tt{\implies 1475=\dfrac{390625+1250P}{P}}$

$\tt{\implies 1475P=390625+1250P}$

$\tt{\implies 1475P-1250P=390625}$

$\tt{\implies 225P=390625}$

$\tt{\implies P=\dfrac{390625}{225}}$

$\tt{\implies P=Rs.\;1736.11}$

Hence, Principal amount = Rs. 1736.11

$\rule{200}{2}$

Now, put the value of "P" in equation (1),

$\tt{\implies R=\dfrac{62500}{P}\;\;\;\;\;.............(1)}$

$\tt{\implies R=\dfrac{62500}{1736.11}}$

$\tt{\implies R=36\%\;(Approx)}$

Hence, Rate of interest is 36%.

Answer 2

Question : The simple interest and the compound interest on a certain sum for 2 years is 1250 and 1475 respectively. Find the rate of interest.

Solution :

Let Pricipal be x and rate of interest be y%.

Given :

• Simple Interest (SI) = 1250
• Compound Interest (CI) =1475
• Time (T) = 2 years

To Find :

• Rate of interest (R)

Formula used :

$SI = \frac{P \times R \times T}{100}$

$CI = P( { 1 + \frac{R}{100}) }^{T} - P$

$= > SI = \frac{P \times R \times T}{100} \\ = > 1250 = \frac{x \times y \times 2}{100} \\ = > 1250 = \frac{xy}{50} \\ = > 1250 \times 50 = xy \\ = > 62500 = xy \\ = > \frac{62500}{x} = y \\ = > y = \frac{62500}{x} \: \: \: \: \: \: \: \: ........(1)$

Now put the value of y in this formula :

$= >CI = P( { 1 + \frac{R}{100}) }^{T} -P \\ = > 1475 = x( {1 + \frac{62500}{100 x}) }^{2} - x\\ = > 1475 = x( { 1 + \frac{625}{x}) }^{2} - x \\ = >1475 = x( { 1 + \frac{390625}{ x^2 } + \frac{1250}{x} ) } - x \\ = > 1475 = x + \frac{390625}{x} + 1250 - x \\=>1475=\frac{390625}{x} + 1250\\=>1475=\frac{390625+1250x}{x}\\=>1475x=390625+1250x\\=>1475x-1250x=390625\\=>225x=390625\\=>x=\frac{390625}{225}\\=>x=1736.1$

Now we put the value of x in equation (1).

$=>y=\frac{62500}{x}\\=>y=\frac{62500}{1736.1}\\=>36\% (approx)$

Hence, Rate of interest = 36% (approx)

Note : We can also take value of x in approx.