Question

the compound interest on a sum of money for 2 years is 410 and simple interest on same sum for the same period is 400.Find the sum and rate of interest.​

Answer 1

AnswEr :

• Compound Interest = Rs. 410
• Simple Interest = Rs. 400
• Time = 2 Years
• Principal and Rate is Same.

$\boxed{ \bf{ Simple \: Interest= \dfrac{Prt}{100}}}$

$\longrightarrow \sf{SI = \dfrac{Prt}{100}}$

$\longrightarrow \sf{ \cancel{400} = \dfrac{Pr \times \cancel2}{100}}$

$\longrightarrow \sf{200 = \dfrac{Pr}{100}}$

$\longrightarrow \sf{Pr = 200 \times 100}$

$\longrightarrow \sf{Pr = 20000}$

$\longrightarrow \sf{P = \dfrac{20000}{r} }$

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$\boxed{ \bf{Compound \: Interest =P \bigg(1 + \dfrac{r}{100} \bigg)^{t} - 1 }}$

$\longrightarrow\sf{CI =P \bigg(1 + \dfrac{r}{100} \bigg)^{t} - 1 }$

$\longrightarrow\sf{410 =P \bigg(1 + \dfrac{r}{100} \bigg)^{2} - 1 }$

$\longrightarrow\sf{410 = \dfrac{20000}{r} \times \bigg(\dfrac{100 + r}{100} \bigg)^{2} - 1 }$

$\longrightarrow\sf{410 = \dfrac{20000}{r} \times \bigg(\dfrac{(100 + r)^{2} }{ {(100)}^{2} } \bigg) -1}$

$\longrightarrow\sf{410 = \dfrac{20000}{r} \times \bigg(\dfrac{10000+ r^{2} + 200r }{ {10000} } \bigg) -1}$

$\longrightarrow\sf{410 = \dfrac{2 \cancel{0000}}{r} \times \dfrac{ \cancel{10000}+ r^{2} + 200r - \cancel{10000} }{ \cancel{10000}}}$

$\longrightarrow\sf{410 = \dfrac{2}{r} \times ({r}^{2} + 200r) }$

$\longrightarrow\sf{ \cancel{410} = \dfrac{ \cancel2}{ \cancel{r}} \times \cancel{r}(r + 200) }$

$\longrightarrow\sf{205 = (r + 200)}$

$\longrightarrow\sf{205 - 200= r}$

$\longrightarrow \large\sf{r = 5\% \: p.a.}$

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• Now Calculation of Principal :

$\longrightarrow \sf{P = \dfrac{20000}{r}}$

$\longrightarrow \sf{P = \cancel\dfrac{20000}{5} }$

$\longrightarrow \large\sf{P = Rs.4000 }$

⠀

$\therefore$ Principal is Rs.4000 and Rate is 5% p.a.

Answer 2

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Explanation:

Compound Interest = Rs. 410

Simple Interest = Rs. 400

Time = 2 Years

Principal and Rate is

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• Now Calculation of Principal :

Principal is Rs.4000 and Rate is 5% p.a.