# 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.

Last updated date: 26th Mar 2023

•

Total views: 307.8k

•

Views today: 5.85k

Answer

Verified

307.8k+ 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

If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?