Question

If 10,000 is invested at an interest rate of 6% per annum , what is the amount after 4 years if the compounding of interest is done .

Answer 1

$\sf{\bold{\purple{\underline{\underline{Given : }}}}}$

• Principle = Rs. 10,000
• Rate = 6%
• Time = 4 years

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$\sf{\bold{\green{\underline{\underline{To\:Find :}}}}}$

CI = ??

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$\sf{\bold{\red{\underline{\underline{Solution :}}}}}$

$\sf{\blue{\boxed{\bold{Amount = P \bigg\lgroup 1 + \dfrac{rate}{100}\bigg\rgroup^{time}}}}}$⠀⠀⠀⠀

$\sf :\implies\: {\bold{ Amount = 10000 \bigg\lgroup 1 + \dfrac{6}{100} \bigg\rgroup^{4} }}$

$\sf :\implies\: {\bold{ Amount = 10000\times \dfrac{106\times \times 106\times 106\times 106}{100\times 100\times 100\times 100}}}$

$\sf :\implies\: {\bold{ Amount =\dfrac {126247696}{10000}}}$

$\sf :\implies\: {\bold{ Amount = Rs. 12624.7696}}$⠀⠀⠀

$\sf{\orange{\boxed{\bold{CI = Amount - Principle }}}}$

$\sf :\implies\: {\bold{ CI = Rs. 12624.7696 - Rs. 10000}}$

$\sf :\implies\: {\bold{CI = Rs.2624.77 }}$

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$\sf{\bold{\pink{\underline{\underline{Answer}}}}}$

Rs. 10000 will earn Rs. 2624.77 as Compound Interest in 4

years at 6% per annum compounded annually..

Answer 2

Given:-

• Principal = 10000
• Rate = 6%
• Time = 4 years

To find:-

Compound Interest after 4 years.

Solution:-

We know,

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

= $\sf{A = 10000\bigg(1+\dfrac{6}{100}\bigg)^4}$

= $\sf{A = 10000\bigg(\dfrac{106}{100}\bigg)^4}$

= $\sf{A = 10000\bigg(\dfrac{106}{100}\bigg)\bigg(\dfrac{106}{100}\bigg)\bigg(\dfrac{106}{100}\bigg)\bigg(\dfrac{106}{100}\bigg)}$

= $\sf{A = \dfrac{126247696}{10000}}$

= $\sf{A = 12624.7696}$

$\sf{CI = A-P}$

= $\sf{CI = 12624.7696 - 10000}$

= $\sf{CI = 2624.7696}$

=> $\sf{CI = Rs.2624.77}$

Therefore CI after 4 years will be Rs.2624.77

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Formula Used:-

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

$\sf{CI = A-P}$. [Where A = Amount]

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Extras:-

$\sf{A = P\bigg(1+\dfrac{r}{200}\bigg)^{2t}}$

When compounded annually.

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Note:-

A = Amount

P = Principal

t = Time

r = Rate

CI = Compound Interest.

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