Question

the difference between the compound interest (compounded annually) and and the simple interest on a sum of Rs. 1000 at a certain rate of interest for 2 years is Rs. 10. Find the rate of interest per annum.
Explain with complete explanation.​

Answer 1

✯ Rate of interest = 10% ✯

Step-by-step explanation:

Given:

• Difference between compound interest (C.I) and simple interest (S.I) is Rs.10
• Principal (P) = Rs. 1000
• Time (n) = 2 years

To find:

• Rate of interest per annum (r).

Solution:

We know that,

${ \boxed{ \sf{C.I = p \bigg(1 + \dfrac{r}{100} \bigg) ^{n} - p}}}$

${ \boxed{ \sf{S.I = \dfrac{pnr}{100} }}}$

${\underline{\sf{According\:to\:the \:question:-}}}$

• The difference between the compound interest (compounded annually) and and the simple interest on a sum of Rs. 1000 at a certain rate of interest for 2 years is Rs. 10.

$:\implies\sf{C.I-S.I=10}$

$:\implies \sf \: \bigg \{p \bigg(1 + \dfrac{r}{100} \bigg)^{n} - p \bigg \} - \bigg( \dfrac{pnr}{100} \bigg) = 10 \\ \\ :\implies \sf \: p \bigg \{ \bigg(1 + \dfrac{r}{100} \bigg)^{n} - 1 \bigg \} - \bigg( \dfrac{pnr}{100} \bigg) = 10 \\ \\ :\implies \sf \: 1000 \bigg \{ \bigg(1 + \dfrac{r}{100} \bigg) ^{2} - 1 \bigg \} - \bigg( \dfrac{1000 \times 2 \times r}{100} \bigg) = 10 \\ \\ :\implies \sf \: 1000 \bigg(1 + \dfrac{r}{100} + 1 \bigg) \bigg(1 + \dfrac{r}{100} - 1 \bigg) - 20r = 10 \\ \\ :\implies \sf \: 1000 \bigg( \dfrac{200 + r}{100 } \bigg) \bigg( \dfrac{r}{100 } \bigg) - 20r = 10 \\ \\ :\implies \sf \: 1000 \bigg( \dfrac{200r + {r}^{2} }{10000} \bigg) - 20r = 10 \\ \\ :\implies \sf \: \dfrac{200r + {r}^{2} }{10} - 20r = 10 \\ \\ : \implies \sf \: \dfrac{200r + {r}^{2} - 200r }{10} = 10 \\ \\ : \implies \sf \: \dfrac{ {r}^{2} }{10} = 10 \\ \\ : \implies \sf \: {r}^{2} = 100 \\ \\ : \implies \sf \: r = 10$

Therefore, the rate of interest per annum is 10%.

________________