
Find the amount which Ram will get on\[Rs{\text{ }}4096\] , if he gave it for \[18\] month at $12\dfrac{1}{2}\% $ per annum, interest being compounded half-yearly.
A.$Rs.4913$
B.$Rs.4425$
C.$Rs.4814$
D.$Rs.4124$
Answer
600k+ views
Hint:Before proceeding further we need the formula to calculate the amount when the interest is compounded half-yearly.
\[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\].
Where,
$
A = {\text{Amount}} \\
P = {\text{Principal}} \\
R = {\text{Rate}} \\
t = {\text{time}} \\
$
Complete step by step solution
When we are calculating compound interest yearly there is no impact but in case of the half-yearly, there is impact on time as well as rate.
In this question we have,
$
A = {\text{?}} \\
P = {\text{4096}} \\
R = 12\dfrac{1}{2} = \dfrac{{12 \times 2 + 1}}{2} = \dfrac{{25}}{2} \\
t = {\text{18 month}} \\
$
Now we need to calculate time in a year.
$
12\,{\text{month}} = 1\,{\text{year}} \\
1\,{\text{month}} = \dfrac{1}{{12}} \times {\text{year}} \\
18\,{\text{month}} = \dfrac{{18}}{{12}}{\text{year}} \\
{\text{ = }}\dfrac{3}{2}{\text{year}} \\
$
In half-yearly, time is double and rate is half then we can rewrite the rate and time as,
$
t = \dfrac{3}{2} \times 2 = 3\,{\text{half yearly}} \\
R = \dfrac{{25}}{2} \times \dfrac{1}{2} = \dfrac{{25}}{4}\% \\
$
Now we have all the values except A which we need to calculate substitute all the value in the equation \[\left( 1 \right),\]
\[
A = 4096{\left( {1 + \dfrac{{25}}{4} \times \dfrac{1}{{100}}} \right)^3} \\
= 4096{\left( {1 + \dfrac{1}{4} \times \dfrac{1}{4}} \right)^3} \\
= 4096{\left( {1 + \dfrac{1}{{16}}} \right)^3} \\
= 4096{\left( {\dfrac{{17}}{{16}}} \right)^3} \\
= 4096 \times \left( {\dfrac{{4913}}{{4096}}} \right) \\
= 4913 \\
\]
So by the discussion, we see that the amount is\[Rs{\text{ }}4913\] .
Thus, the amount is \[Rs{\text{ }}4913\].
Note: Here the main point of concern is while calculating time and rate in case of half year.
We must note that if time is \[1\] year and rate is \[1\% \] then in case of half-year time will be \[2\] year and the rate will be \[0.5\% .\]
While calculating the time year from month we must follow a unitary method of calculation.
\[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\].
Where,
$
A = {\text{Amount}} \\
P = {\text{Principal}} \\
R = {\text{Rate}} \\
t = {\text{time}} \\
$
Complete step by step solution
When we are calculating compound interest yearly there is no impact but in case of the half-yearly, there is impact on time as well as rate.
In this question we have,
$
A = {\text{?}} \\
P = {\text{4096}} \\
R = 12\dfrac{1}{2} = \dfrac{{12 \times 2 + 1}}{2} = \dfrac{{25}}{2} \\
t = {\text{18 month}} \\
$
Now we need to calculate time in a year.
$
12\,{\text{month}} = 1\,{\text{year}} \\
1\,{\text{month}} = \dfrac{1}{{12}} \times {\text{year}} \\
18\,{\text{month}} = \dfrac{{18}}{{12}}{\text{year}} \\
{\text{ = }}\dfrac{3}{2}{\text{year}} \\
$
In half-yearly, time is double and rate is half then we can rewrite the rate and time as,
$
t = \dfrac{3}{2} \times 2 = 3\,{\text{half yearly}} \\
R = \dfrac{{25}}{2} \times \dfrac{1}{2} = \dfrac{{25}}{4}\% \\
$
Now we have all the values except A which we need to calculate substitute all the value in the equation \[\left( 1 \right),\]
\[
A = 4096{\left( {1 + \dfrac{{25}}{4} \times \dfrac{1}{{100}}} \right)^3} \\
= 4096{\left( {1 + \dfrac{1}{4} \times \dfrac{1}{4}} \right)^3} \\
= 4096{\left( {1 + \dfrac{1}{{16}}} \right)^3} \\
= 4096{\left( {\dfrac{{17}}{{16}}} \right)^3} \\
= 4096 \times \left( {\dfrac{{4913}}{{4096}}} \right) \\
= 4913 \\
\]
So by the discussion, we see that the amount is\[Rs{\text{ }}4913\] .
Thus, the amount is \[Rs{\text{ }}4913\].
Note: Here the main point of concern is while calculating time and rate in case of half year.
We must note that if time is \[1\] year and rate is \[1\% \] then in case of half-year time will be \[2\] year and the rate will be \[0.5\% .\]
While calculating the time year from month we must follow a unitary method of calculation.
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