Questions & Answers

Question

Answers

(a). 17652

(b). 18000

(c). 19000

(d). 18522

Answer

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Hint: Here, it is given that the interest is compounded half-yearly. Therefore the rate of interest will be halved from $r$ to $\dfrac{r}{2}$ and the time is doubled from $n$ to $2n$. Then we have to calculate the amount of interest for the principal amount P, by the formula:

$A=P{{\left( 1+\dfrac{\dfrac{r}{2}}{100} \right)}^{2n}}$

Complete step-by-step answer:

We are given that Mishael borrowed Rs.16000 from a finance company at 10% per annum compounded half-yearly. We have to calculate the amount of debt after $1\dfrac{1}{2}$ years.

Here we have:

Principal amount, $P=16000$

Rate of interest, \[r\text{ }=10%\] per annum

Time, $n=1\dfrac{1}{2}=\dfrac{3}{2}$ years

When the interest is compounded annually, we have the formula for amount, A. i.e.

$A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$

But, here the interest is compounded half-yearly. Therefore the number of years, $n$ is doubled and the rate of annual interest $r$ is halved. Hence we can say that:

$\begin{align}

& n=2n \\

& r=\dfrac{r}{2} \\

\end{align}$

Then the amount A, is calculated using the formula:

$A=P{{\left( 1+\dfrac{\dfrac{r}{2}}{100} \right)}^{2n}}$

Now, by substituting all the values we get:

$A=16000{{\left( 1+\dfrac{\dfrac{10}{2}}{100} \right)}^{2\times \dfrac{3}{2}}}$

Now, by cancellation we obtain:

$A=16000{{\left( 1+\dfrac{5}{100} \right)}^{3}}$

Next, by taking the LCM we get:

$\begin{align}

& A=16000{{\left( \dfrac{100+5}{100} \right)}^{3}} \\

& A=16000{{\left( \dfrac{105}{100} \right)}^{3}} \\

\end{align}$

Hence by cancellation we obtain:

$\begin{align}

& A=16000\times {{\left( \dfrac{21}{20} \right)}^{3}} \\

& A=16000\times \dfrac{21\times 21\times 21}{20\times 20\times 20} \\

\end{align}$

Now, again by cancellation we get:

$\begin{align}

& A=2\times 21\times 21\times 21 \\

& A=18522 \\

\end{align}$

Therefore, the amount of debt after $1\dfrac{1}{2}$ years is Rs.18,522.

Hence, the correct answer for this question is option (d).

Note: Here, you have to calculate the amount for 6 months not for one year. When the interest is compounded half yearly, the principal amount will be the same, only rate of interest and time changes.

$A=P{{\left( 1+\dfrac{\dfrac{r}{2}}{100} \right)}^{2n}}$

Complete step-by-step answer:

We are given that Mishael borrowed Rs.16000 from a finance company at 10% per annum compounded half-yearly. We have to calculate the amount of debt after $1\dfrac{1}{2}$ years.

Here we have:

Principal amount, $P=16000$

Rate of interest, \[r\text{ }=10%\] per annum

Time, $n=1\dfrac{1}{2}=\dfrac{3}{2}$ years

When the interest is compounded annually, we have the formula for amount, A. i.e.

$A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$

But, here the interest is compounded half-yearly. Therefore the number of years, $n$ is doubled and the rate of annual interest $r$ is halved. Hence we can say that:

$\begin{align}

& n=2n \\

& r=\dfrac{r}{2} \\

\end{align}$

Then the amount A, is calculated using the formula:

$A=P{{\left( 1+\dfrac{\dfrac{r}{2}}{100} \right)}^{2n}}$

Now, by substituting all the values we get:

$A=16000{{\left( 1+\dfrac{\dfrac{10}{2}}{100} \right)}^{2\times \dfrac{3}{2}}}$

Now, by cancellation we obtain:

$A=16000{{\left( 1+\dfrac{5}{100} \right)}^{3}}$

Next, by taking the LCM we get:

$\begin{align}

& A=16000{{\left( \dfrac{100+5}{100} \right)}^{3}} \\

& A=16000{{\left( \dfrac{105}{100} \right)}^{3}} \\

\end{align}$

Hence by cancellation we obtain:

$\begin{align}

& A=16000\times {{\left( \dfrac{21}{20} \right)}^{3}} \\

& A=16000\times \dfrac{21\times 21\times 21}{20\times 20\times 20} \\

\end{align}$

Now, again by cancellation we get:

$\begin{align}

& A=2\times 21\times 21\times 21 \\

& A=18522 \\

\end{align}$

Therefore, the amount of debt after $1\dfrac{1}{2}$ years is Rs.18,522.

Hence, the correct answer for this question is option (d).

Note: Here, you have to calculate the amount for 6 months not for one year. When the interest is compounded half yearly, the principal amount will be the same, only rate of interest and time changes.