# Arif took a loan of Rs.80,000 from a bank. If the rate of interest is $10%$ per annum, find the difference in amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is:

A) Compounded annually.

B) Compounded half yearly.

Answer

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Hint: The given question is based on compound interest. Try to recall the formulae related to compound interest, where the interest is compounded annually as well as half yearly.

Complete step-by-step answer:

Before proceeding with the problems, we must know the formula used in compound interest.

The formula for compound interest is given as $A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$, where $A=$ Amount after $n$ compounding, $P=$ principal amount, $r=$ rate of interest, $n=$ number of times the interest is compounded in the given time.

If $T$ is the total time for which interest is to be calculated, then:

1) $n=T$, if the interest is compounded annually.

2) $n=2T$, if the interest is compounded half-yearly

3) $n=4T$, if the interest is compounded quarterly.

While solving the problem, we will also consider the fact that when the interest is compounded half-yearly, then the rate of interest is also halved (if the rate is given in per annum).

Now, in the question, the given principal amount is $P=Rs.80000$, Rate of interest is $r=10%$ per annum and time is $T=1\dfrac{1}{2}=\dfrac{3}{2}$ years.

First, let’s consider the case when the interest is compounded annually.

Here, time is $1\dfrac{1}{2}$ years. So, for the first year, the interest will be compounded annually. But, for the remaining $6$ months, it would be compounded half-yearly.

So, for the first year, $A=80000{{\left( 1+\dfrac{10}{100} \right)}^{1}}$

$=80000\left( 1.1 \right)$ $=88000$

Now, this amount will be the principal for the next $6$ months. So, $n=2\times \dfrac{1}{2}=1$ , and $r=\dfrac{10}{2}=5%$

So, amount after $1\dfrac{1}{2}$ years is given as $A=88000{{\left( 1+\dfrac{5}{100} \right)}^{1}}$

$\Rightarrow A=88000\times 1.05=Rs.92400$

Hence, the amount to be paid after $1\dfrac{1}{2}$ years, when the interest is compounded annually, is $Rs.92400$ .

Now, when the interest is compounded half yearly, $n=\dfrac{3}{2}\times 2=3$, and the rate of interest $r=\dfrac{10}{2}=5%$ .

So, amount after $1\dfrac{1}{2}$ years is given as $A=80000{{\left( 1+\dfrac{5}{100} \right)}^{3}}$

$=80000{{\left( 1.05 \right)}^{3}}$

$=80000\times 1.157625$

$=92610$

Hence, the amount to be paid after $1\dfrac{1}{2}$ years , when the interest is compounded half yearly, is $Rs.92610$ .

So, the difference between the amounts is given as $Rs.\left( 92610-92400 \right)$

$=Rs.210$

Therefore, the difference between the amount to be paid after $1\dfrac{1}{2}$ years, when the interest is compounded half yearly and annually, is $Rs.210$.

Note: When the interest is compounded annually, students generally make a mistake of taking $n$ as $\dfrac{3}{2}$, which is wrong.

Complete step-by-step answer:

Before proceeding with the problems, we must know the formula used in compound interest.

The formula for compound interest is given as $A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$, where $A=$ Amount after $n$ compounding, $P=$ principal amount, $r=$ rate of interest, $n=$ number of times the interest is compounded in the given time.

If $T$ is the total time for which interest is to be calculated, then:

1) $n=T$, if the interest is compounded annually.

2) $n=2T$, if the interest is compounded half-yearly

3) $n=4T$, if the interest is compounded quarterly.

While solving the problem, we will also consider the fact that when the interest is compounded half-yearly, then the rate of interest is also halved (if the rate is given in per annum).

Now, in the question, the given principal amount is $P=Rs.80000$, Rate of interest is $r=10%$ per annum and time is $T=1\dfrac{1}{2}=\dfrac{3}{2}$ years.

First, let’s consider the case when the interest is compounded annually.

Here, time is $1\dfrac{1}{2}$ years. So, for the first year, the interest will be compounded annually. But, for the remaining $6$ months, it would be compounded half-yearly.

So, for the first year, $A=80000{{\left( 1+\dfrac{10}{100} \right)}^{1}}$

$=80000\left( 1.1 \right)$ $=88000$

Now, this amount will be the principal for the next $6$ months. So, $n=2\times \dfrac{1}{2}=1$ , and $r=\dfrac{10}{2}=5%$

So, amount after $1\dfrac{1}{2}$ years is given as $A=88000{{\left( 1+\dfrac{5}{100} \right)}^{1}}$

$\Rightarrow A=88000\times 1.05=Rs.92400$

Hence, the amount to be paid after $1\dfrac{1}{2}$ years, when the interest is compounded annually, is $Rs.92400$ .

Now, when the interest is compounded half yearly, $n=\dfrac{3}{2}\times 2=3$, and the rate of interest $r=\dfrac{10}{2}=5%$ .

So, amount after $1\dfrac{1}{2}$ years is given as $A=80000{{\left( 1+\dfrac{5}{100} \right)}^{3}}$

$=80000{{\left( 1.05 \right)}^{3}}$

$=80000\times 1.157625$

$=92610$

Hence, the amount to be paid after $1\dfrac{1}{2}$ years , when the interest is compounded half yearly, is $Rs.92610$ .

So, the difference between the amounts is given as $Rs.\left( 92610-92400 \right)$

$=Rs.210$

Therefore, the difference between the amount to be paid after $1\dfrac{1}{2}$ years, when the interest is compounded half yearly and annually, is $Rs.210$.

Note: When the interest is compounded annually, students generally make a mistake of taking $n$ as $\dfrac{3}{2}$, which is wrong.

Last updated date: 20th Sep 2023

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