Reena and Ruchira borrowed Rs 60000 and Rs 50000 respectively for a period of 3 years. Reena paid simple interest at the rate of 10 % p.a., while Ruchira paid compound interest at the rate of 10 % p.a. Compounded annually.Who paid more interest and by how much ?

byAmlesh Mahato •March 20, 2021

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Solution:

⇨For Reena

⇨Given

⇨p = 60000

⇨r = 10 %

⇨n = 3 p.a.

⇨Therefore the Simple interest for Reena will be

$\large S.I=\frac{p\times r\times n}{100}$

$\large \Rightarrow S.I=\frac{60000\times 10\times 3}{100}$

$\large \Rightarrow S.I=\frac{600000\times 3}{100}$

$\large \Rightarrow S.I=6000\times 3$

$\large \Rightarrow S.I=18000$ Rs

⇨For Ruchira

⇨Given

⇨p = 50000

⇨r = 10 %

⇨n = 3 p.a.

⇨Therefore the Compound interest for Ruchira will be

$\large C.I=A-P$

$\large \Rightarrow C.I=p\left ( 1+\frac{r}{100} \right )^{n}-p$

$\large \Rightarrow C.I=p\left \{ \left ( 1+\frac{r}{100} \right )^{n}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left ( 1+\frac{10}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left (\frac{100+10}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \left (\frac{110}{100} \right )^{3}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \frac{1331000}{1000000}-1 \right \}$

$\large \Rightarrow C.I=50000\left \{ \frac{1331000-1000000}{1000000} \right \}$

$\large \Rightarrow C.I=50000\times \frac{331000}{1000000}$

$\large \Rightarrow C.I=5\times 3310$ Rs

$\large \Rightarrow C.I=16550$

⇨From above Reena paid more interest than Ruchira by

= Interest paid by Reena - Interest paid by Ruchira

= 18000 - 16550
= 1450 Rs

Tags: Math-Class-8