
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
522k+ views
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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Explain the system of Dual Government class 8 social science CBSE

What is Kayal in Geography class 8 social science CBSE

Who is the author of Kadambari AKalidas B Panini C class 8 social science CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Advantages and disadvantages of science

Write the smallest number divisible by both 306 and class 8 maths CBSE
