Question

difference between the Cl. and S.l. on Rs 9500 for 2 years is Rs 95 at the same rate of interest per annum. Find the rate of interest
per annum​

Answer 1

Step-by-step explanation:

Given:-

• Difference between C.I and S.I is Rs.95

• Principal (P) = Rs. 9500

• Time (T) = 2 years

To Find:-

• The rate of interest per annum.

Solution:-

$\sf Let \: the rate \: of \: interest \: be \: r\%$

$\sf Simple\: Interest \: (SI) = \dfrac{PTR}{100}$

Here:-

• P = Rs. 9500 , T = 2 years , R = r

$\sf \implies SI =\dfrac{ 9500 \times 2 \times r}{100}$

$\sf \implies SI = Rs. 190r$

$\sf CI = Amount - Principal$

$\sf \implies CI = P\times \bigg( 1 +\dfrac{r}{100}\bigg)^{n} - P$

• P = Rs. 9500 , n = 2 years , R = r

$\sf \implies CI = P\times \bigg( 1 +\dfrac{r}{100}\bigg)^{n} - P$

$\sf \implies CI = 9500 \times \bigg( 1 +\dfrac{r}{100}\bigg)^{2} - 9500$

$\sf Given,\: CI - SI = Rs.95$

$\sf \implies \bigg[ 9500 \bigg( 1 +\dfrac{r}{100}\bigg)^{2} - 9500\bigg]-190r = 95$

$\sf \implies \bigg[ 9500 \bigg( 1 +\dfrac{r}{100}\bigg)^{2} - 9500\bigg]-190r = 95$

$\sf \implies \bigg[ 9500 \bigg( 1 +\dfrac{ {r}^{2} }{10000} + \dfrac{2r}{100} \bigg) - 9500\bigg]-190r = 95$

$\sf \implies \bigg[ 9500+\dfrac{ {9500r}^{2} }{10000} + \dfrac{9500(2r)}{100} - 9500\bigg]-190r = 95$

$\sf \implies 9500+\dfrac{ {95r}^{2} }{100} + {190r} - 9500-190r = 95$

$\sf \implies \dfrac{ {95r}^{2} }{100} = 95$

$\sf \implies {r}^{2} = 95 \times \dfrac{100}{95}$

$\sf \implies {r}^{2} = 100$

$\sf \implies {r} = \sqrt{100} = 10$

$\underline{\boxed{\purple{\rm \therefore Rate\: of \: interest = 10\%}}}$