Question

Ashima took a loan of Rs. \[1,00,000\] at \[12\% \] p.a., compounded half yearly. She paid Rs.\[1,12,360.\] If \[{(1.06)^2}\] is equal to \[1.1236\], then the period for which she took the loan is
A.2 years
B.1 year
C.6 months
D.11.2 years

Answer 1

Hint: Here it is not a straight forward answer because in this question, the interest is compounded half yearly. So, care has to be given to this point as many students get this wrong because of this error. The given rate of interest is \[12\% \] p.a. but it is evaluated half yearly. So, we have to calculate it at half of \[12\% \], i.e., \[6\% \].

Formula Used:
We have halved the rate of interest so now we can directly calculate by using the basic formula for compound interest –
\[A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t}{\rm{ }}...(i)\]
And we have also been given the result of a power:
\[{(1.06)^2} = 1.1236{\rm{ }}...(ii)\]
We will use a base formula for calculation too:
If, \[{a^n} = {a^m}{\rm{ }}...{\rm{(}}iii{\rm{)}}\]
then, \[n = m\]

Complete step-by-step answer:
Given Principal \[P = 1,00,000\]
Rate of interest, for the one calculated half-yearly compound, \[R = 6\% \]
Amount, \[A = 1,12,360\]
Using the formula from \[(i)\] and putting in the above values of \[A,P,R\], we have
$\Rightarrow$ \[1,12,360 = 1,00,000{\left( {1 + \dfrac{6}{{100}}} \right)^t}\]
Now, taking \[100000\] to the other side and solving the bracket,
$\Rightarrow$ \[\dfrac{{112360}}{{100000}} = {\left( {\dfrac{{106}}{{100}}} \right)^t}\]
Bringing the Left Hand Side to the lowest term by dividing by \[40\] and simplifying the Right Hand Side,
$\Rightarrow$ \[1.1236 = {\left( {1.06} \right)^t}\]
Now, we use the value from \[(ii)\] and equate,
so we have
$\Rightarrow$ \[{(1.06)^2} = {(1.06)^t}\]
Now, finally using the \[\left( {iii} \right)\] equation and calculating the time,
$\Rightarrow$ \[t = 2\]
Now, this 2 is years in half-yearly, i.e., 1 year.
Hence, the answer of the question is B. 1 year.

Note: We must be careful while reading the question and check if it’s given half-yearly, quarterly, yearly. Then we have to accordingly calculate the result. Also, when comparing the powers, it must be done carefully, and the unknown quantity has to be expressed in the required form carefully too. Like in this question, though the time from the calculation came out to be 2, but it wasn’t the answer as it had to be calculated half-yearly, so it was 2 half-years or 1 full year.