Question

If in two years time a principal of Rs 100 amounts to Rs 121 when the interest at the rate of r% is compounded annually, then the value of r will be
A. \[10.5\]
B. 10
C. 15
D. 14

Answer 1

Hint: In the given question, we have given the principle, amount, and time.
We have to find the rate of interest.
As money is compounded annually at the rate of $r\% $. So, we use compound interest formula and the formula for amount is given by:
$ \Rightarrow A = P{(1 + \dfrac{r}{{100}})^t}$………….(1)
Here P is principal, r is the rate of interest, t is time and A is the amount.

Complete step by step answer:

Let us see what is given to us. We have given principle, amount, and time. i.e.
$ \Rightarrow P = Rs.100$
$ \Rightarrow A = Rs.121$
$ \Rightarrow T = 2years$
After that see what we have to find? We have to find the rate of interest i.e. $r\% $
So, by putting the values $ \Rightarrow P = Rs.100$, $A = Rs.121$ and $T = 2years$ in the formula (1) we get,
$ \Rightarrow A = P{(1 + \dfrac{r}{{100}})^t}$
$ \Rightarrow 121 = 100{(1 + \dfrac{r}{{100}})^2}$
By dividing 100 to both sides we get,
$ \Rightarrow \dfrac{{121}}{{100}} = {(1 + \dfrac{r}{{100}})^2}$
As 121 is the square of 11 and 100 is the square of 10, so we can write it as follows:
$ \Rightarrow \dfrac{{{{(11)}^2}}}{{{{(10)}^2}}} = {(1 + \dfrac{r}{{100}})^2}$
Taking square root on both sides we get,
$ \Rightarrow \dfrac{{11}}{{10}} = 1 + \dfrac{r}{{100}}$
By subtracting 1 from each side and cancel 1 from R.H.S we get,
$ \Rightarrow \dfrac{{11}}{{10}} - 1 = \dfrac{r}{{100}}$
By taking LCM of L.H.S we get,
$ \Rightarrow \dfrac{{11 - 10}}{{10}} = \dfrac{r}{{100}}$
By subtracting 10 from 11 we get,
$\begin{gathered}
   \Rightarrow \dfrac{{100}}{{10}} = r \\
   \Rightarrow r = 10\% \\
\end{gathered} $
So, the principle is compounded annually at 10% for 2 years.
So, the correct option is B.

Note: The students generally get confused about whether to apply the formula of simple interest and compound interest to find the amount. They have given that money is compounded annually i.e. every year it increases at the rate of $r\% $ then we apply the given formula for finding the amount.
$ \Rightarrow A = P{(1 + \dfrac{r}{{100}})^t}$
If they have given that it increases once at the rate of $r\% $ then we apply the simple interest formula i.e. as follows
$ \Rightarrow S.I = \dfrac{{P \times r \times t}}{{100}}$
And we can find the amount by adding P and S.I.
We can solve the given question by shortcut way i.e. without using the formula and the steps are as follows:
First, we will find the difference of principle or amount, we find the difference as $121 - 100 = 21$.
Which is further split into 10 and 11.
Now, apply the concept of the rate of interest is $10%$ then increase in the first year will be $10% $ of 100 is 10. Then the amount will become 100 plus 10 i.e. 110.
Now if the rate of interest is $10% $ then the increase in 2nd year will be 11. Then the amount will become 110 plus 11 we will get 121. Hence, $10% $ is the answer.