5 Mark Question

Question 15 Marks

Find the difference between the compound interest and simpal interest, On a sum of Rs. 50,000 at 10% per annum for 2 years.

Answer

Given:
$P=\text { Rs. } 50,000$
$R=10 \% \text { p.a. }$
$n=2 \text { years }$
We know that amount A at the end of n years at the rate $R \%$ per annum when the interest is compounded annually is given by $A = P \left(1+\frac{ R }{100}\right)^{ n }$.
$\therefore A=\text { Rs. } 50,000\left(1+\frac{10}{100}\right)^2$
$=\text { Rs. } 50,000(1.1)^2$
$=\text { Rs. } 60,500$
Also,
$C I=A-P$
$=\text { Rs. } 60,500-\text { Rs. } 50,000$
$=\text { Rs. } 10,500$
We know that:
$SI=\frac{PRT}{100}$
$=\frac{50,000 \times 10 \times 2}{100}$
$=\text { Rs. } 10,000$
$\therefore \text { Difference between CI and SI = Rs. } 10,500-\text { Rs. } 10,000$
$=\text { Rs. } 500$

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Question 25 Marks

The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs. 283.50. Find the sum.

Answer

Given:
CI − SI = Rs. 283.50
R = 15%
n = 3 years
Let the sum be Rs. x.
We know that:
$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{x}\Big(1+\frac{15}{100}\Big)^{3}$
$=\text{x}(1.15)^{3}\ ...(1)$
Also,
$\text{SI}=\frac{\text{PRT}}{100}=\frac{\text{x}(15)(3)}{100}=0.45\text{x}$
$\text{A}=\text{SI}+\text{P}=1.45\text{x}\ ...(2)$
Thus, we have:
$\text{x}(1.15)^{3}-1.45\text{x}=283.50$ [From (1) and (2)]
$1.523\text{x}-1.45\text{x}=283.50$
$0.070875\text{ x}=283.50$
$\text{x}=\frac{283.50}{0.070875}$
$=4,000$
Thus, the sum is Rs. 4,000.

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Question 35 Marks

Romesh borrowed a sum of Rs. 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years.

Answer

Given: P = Rs. 245,760 R = 12.5% p.a. n = 2 yearsWhen compounded annually, we have:
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$ $=\text{Rs. }245,760\Big(1+\frac{12.5}{100}\Big)^{2}$ $=\text{Rs. }311,040$ When the interest is compounded half−yearly, we have $\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$ $=\text{Rs. }10,000\Big(1+\frac{20}{200}\Big)^{4}$ $=\text{Rs. }10,000(1.1)^4$ $=\text{Rs. }14,641$ Difference = Rs. 14,641 − Rs. 14,400 = Rs. 241

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Question 45 Marks

Rakesh lent out Rs. 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?

Answer

Given:
P = Rs. 10,000
R = 20% p.a.
n = 2 years
$\text{A = P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }10,000\Big(1+\frac{20}{100}\Big)^{2}$
$=\text{Rs. }10,000(1.2)^2$
$=\text{Rs. }14,400$
When the interest is compounded half−yearly, we have
$\text{A = P}\Big(1+\frac{\text{R}}{200}\Big)^{\text{2n}}$
$=\text{Rs. }10,000\Big(1+\frac{20}{200}\Big)^{4}$
$=\text{Rs. }10,000(1.1)^4$
$=\text{Rs. }14,641$
Difference = Rs. 14,641 − Rs. 14,400
= Rs. 241

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Question 55 Marks

A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs. 482 more. Find the sum.

Answer

$\text{A}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
Also,
P = A - CI
Let the sum of money be Rs. x.
If the interest is compounded annually, then:
$\text{A}_{1}=\text{x}\Big(1+\frac{20}{100}\Big)^{2}$
$=1.44\text{x}$
$\therefore\ \text{CI}=1.44\text{x}-\text{x}$
$=0.44\text{x}\ ...(1)$
If the interest is compounded half−yearly, then:
$\text{A}_{2}=\text{x}\Big(1+\frac{10}{100}\Big)^{4}$
$=1.4641\text{x}$
$\therefore\ \text{CI}=1.4641\text{x}-\text{x}$
$=0.4641\text{x}\ ...(2)$
It is given that if interest is compounded half−yearly, then it will be Rs. 482 more
$\therefore\ 0.4641\text{x}=0.44\text{x}+482$ [From (1) and (2)]
$0.4641\text{x}-0.44\text{x}=482$
$0.0241\text{x}=482$
$\text{x}=\frac{482}{0.0241}$
$=20,000$
Thus, the required sum is Rs. 20,000.

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