Shankar takes a loan of Rs 10000 at a compound interest rate of 10% per annum (p.a.)?

byAmlesh Mahato •March 09, 2021

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Solution:

i. Find the compound interest after one year.

⇨Given

⇨p =10000

⇨ r =10%

⇨n =1

⇨Compound interest after 1 year will be

$Shankar takes a loan of Rs 10000 at a compound interest rate of 10% per annum (p.a.).i. Find the compound interest after one year.ii. Find the compoun$

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

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

⇨ $\large C.I= 10000\left \{ \left ( 1+\frac{10}{100} \right )^{1}-1 \right \}$

⇨ $\large C.I= 10000\left \{ \frac{100+10}{100}-1 \right \}$

⇨ $\large C.I= 10000\left \{ \frac{110}{100}-1 \right \}$

⇨ $\large C.I= 10000\left \{ \frac{110-100}{100} \right \}$

⇨ $\large C.I= \frac{100000}{100}$

⇨ $\large C.I=1000$ Rs

ii. Find the compound interest for 2 years.

⇨Given

⇨p =10000

⇨r =10%

⇨n=2
⇨ Compound interest after 2 year will be

$\large C.I=A-P$

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

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

$\large \Rightarrow C.I=10000\left \{ \left ( 1+\frac{10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( \frac{100+10}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \left ( \frac{110}{100} \right )^{2}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12100}{10000}-1 \right \}$

$\large \Rightarrow C.I=10000\left \{ \frac{12100-10000}{10000} \right \}$

$\large \Rightarrow C.I=10000\times \frac{2100}{10000}$

$\large \Rightarrow C.I=2100$ Rs

iii. Find the sum of money required to clean the debit at the end of 2 years.

⇨From Above, We have

⇨p =10000 Rs

⇨C.I. of two years=2100 Rs

⇨Therefore ,

⇨The sum of money required to clean the debit at the end of 2 years

Will be = p + C.I. of two years

= 10000 + 2100

=12100 Rs

iv. Find the differences between the compound interest and the simple interest at the same rate for 2 years.

⇨From Above C.I. of two years = 2100 Rs

⇨Given

⇨p=10000

⇨r=10

⇨n=2

⇨For Simple interest

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

$\large \Rightarrow S.I=\frac{10000\times 10\times 2}{100}$

$\large \Rightarrow S.I=\frac{200000}{100}$

$\large \Rightarrow S.I=2000$ Rs

⇨Therefore the differences between the compound interest and the simple interest at the same rate for 2 years will be

= C.I.- S.I
=2100 - 2000
=100 Rs

Tags: Math-Class-8

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