Question

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1. Find the compound interest on 5000 at 12% p.a. for 1 year , compounded half yearly .

2. Find the amount and the compound interest on 16000 for 3/2 years at 10% p.a. the interest being compounded half yearly .
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Answer 1

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Answer 2

Solution 1 :-

Given that :-

★ Principal = ₹ 5000

★ Rate = 12% = $\frac{12}{2}$ = 6%

★ Time = 1 year = 2 half years

According to the question,

Using formula,

$\sf \star \fbox \orange{Amount = Principal {(1 + \frac{Rate}{100})}^{Time}}$

$Amount = 5000 \times {(1 + \frac{6}{100})}^{2}$

$Amount = 5000 \times ({1 + \frac{3}{50})}^{2}$

$Amount = 5000 \times {( \frac{50}{50} + \frac{3}{50})}^{2}$

$Amount = 5000 \times {( \frac{53}{50})}^{2}$

$Amount = 5000 \times \frac{53}{50} \times \frac{53}{50}$

$Amount = 50 \times \frac{53}{5} \times \frac{53}{5}$

$Amount = 2 \times 53 \times 53$

$\fbox{Amount = 5618}$

Therefore,

Compound interest = Amount - principal

Compound interest = 5618 - 5000

$\sf \fbox \orange{Compound \: interest =Rs \: 618}$

Solution 2 :-

Given that,

★ Principal = ₹16000

★ Rate = 10% = $\frac{10}{2}$ = 5%

★ Time = $\frac{3}{2}$years = 3 half years

According to the question,

Using formula,

$\sf \star \fbox \orange{Amount = Principal {(1 + \frac{Rate}{100})}^{Time}}$

$Amount = 16000 \times {(1 + \frac{5}{100})}^{3}$

$Amount = 16000 \times {(1 + \frac{1}{20})}^{3}$

$Amount = 16000 \times {( \frac{20}{20} + \frac{1}{20})}^{3}$

$Amount = 16000 \times {( \frac{21}{20})}^{3}$

$Amount = 16000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$

$Amount = 16 \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$

$Amount = 2 \times {21} \times {21} \times {21}$

$\fbox{Amount = 18522}$

Therefore,

Compound interest = Amount - principal

Compound interest = 18522 - 16000

$\sf \fbox \orange{Compound \: interest =Rs \: 2522}$