Question

Simple interest on a sum of money 2 years at 4% growth rate is ₹450. Find compound interest on the same sum and at the rate for 1 year, if the interest is reckoned half- yearly.​

Answer 1

Correct Question:-

Simple Interest on a sum of money for 2 years at 4% growth rate is Rs.450. Find the Compound Interest on the same sum and at the same rate for 1 year, if the interest is reckoned half-yearly.

Given:-

• Rate = 4%
• Time = 2 years
• SI = Rs.450

To find:-

Compound Interest for 1 year compounded (or reckoned) half-yearly.

Solution:-

Firstly We need yo find the Principal.

We know,

$\sf{P = \dfrac{SI\times 100}{R \times T}}$

Substituting the values,

$\sf{P = \dfrac{450\times 100}{4\times 2}}$

= $\sf{P = 275\times 25}$

= $\sf{P = Rs.6875}$

Now,

For finding CI

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

= $\sf{A = 6875\bigg(1+\dfrac{4}{200}\bigg)^{2\times 1}}$

= $\sf{A = 6875\bigg(\dfrac{200+4}{200}\bigg)^{2}}$

= $\sf{A = 6875\bigg(\dfrac{204}{200}\bigg)^2}$

= $\sf{A = 6875\bigg(\dfrac{204}{200}\bigg)\bigg(\dfrac{204}{200}\bigg)}$

= $\sf{A = Rs.7152.75}$

Now,

CI = A - P

$\sf{CI = Rs.(7152.75 - 6875)}$

= $\sf{CI = Rs.277.75}$

Therefore the Compound Interest would be Rs.277.75

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Additional Information:-

Use:-

$\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^n}$ When the Principal is compounded annually.

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Note:-

n stands for time which is also denoted by 't'.

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Answer 2

Answer:

Solution:-

$P= \frac{SI \times 100}{R \times T}$

Substituting the value

$P = \frac{450 \times 100}{4 \times 2}$

$P = 275 \times 25$

$P = rs.6875$

Now,

For CI

$A=P(1 + \frac{r}{200} ) {}^{2n}$

$A=6875(1 + \frac{4}{200} ) {}^{2 \times 1}$

$A=6875( \frac{200 + 4}{200} ) {}^{2}$

$A=6875( \frac{204}{200} ) {}^{2}$

$A=6875( \frac{204}{200} )( \frac{204}{200} )$

$A=rs.7152.75$

Now,

$CI=A-P$

$CI=rs.(7152.75 - 6875)$

$CI=rs.277.75$

Therefore the compound Interest would be Rs.277.75