Question

By using the formula, find the amount and compound interest on Rs. 10,000 for 2 years at 11% per
annum compounded annually​

Answer 1

Given:-

${\rm \red{\bullet \: Principal \: (P) \: = \: Rs. \: 10,000}}$

${\rm \pink{\bullet \: Time \: (n) \: = \: 2 \: years}}$

${\rm \blue{\bullet \: Rate \: of \: interest \: (r) \: = \: 11 \: \%}}$

To Find:-

${\rm \green{\bullet \: Amount}}$

${\rm \pink{\bullet \: Compound \: Interest}}$

Solution:-

We know the formula of Compound Interest is –

${\bf \boxed{\rm \red{A = P( 1 + \frac{r}{100})^n}}}$

A = 10,000 $( 1 + \frac{11}{100} )^2$

A = 10,000 $(\frac{111}{100})^2$

A = 10,000 × $\frac{111}{100}$ × $\frac{111}{100}$

A = Rs. 12321

Since, we know

${\bf \boxed{\rm \green{C.I. \: = \: Amount \: - \: Principal}}}$

$\therefore$ C.I. = Rs. (12321 – 10000) = Rs. 2321

${\bf \rm \pink{Compound \: Interest (C.I.) \: is \: Rs. \: 2321 \: and \: Amount \: is \: Rs. \: 12321 \: [Ans]}}$

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Answer 2

Solution!!

Principal (P) = Rs 10000

Rate of interest (R) = 11%

Time (n) = 2 years

Amount = $\sf P\left(1+\dfrac{R}{100}\right)^{n}$

Amount = $\sf 10000\left(1+\dfrac{11}{100}\right)^{2}$

Amount = 10000(1 + 0.11)²

Amount = 10000(1.11)²

Amount = 10000 × 1.2321

Amount = Rs 12321

We have the amount now. We can easily find out the compound interest.

CI = Amount - Principal

CI = Rs 12321 - Rs 10000

CI = Rs 2321

We can find it out without using the formula also.

For 1st year

P = Rs 10000

T = 1 year

R = 11%

Interest = (P × R × T)/100

Interest = (10000 × 11 × 1)/100

Interest = Rs 1100

Amount = P + I

Amount = Rs 10000 + Rs 1100

Amount = Rs 11100

For 2nd year

P = Rs 11100

T = 1 year

R = 11%

Interest = (P × R × T)/100

Interest = (11100 × 11 × 1)/100

Interest = Rs 1221

Amount = P + I

Amount = Rs 11100 + Rs 1221

Amount = Rs 12321

CI = Rs 1100 + Rs 1221

CI = Rs 2321