Question
A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a different rate becomes three times in 12 years. The lower rate of interest per annum is = ?
Answer: Option D
Rate of interest
$$ = \frac{{100(x - 1)}}{t}\% $$
Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times.
Therefore, x = 3
Where t is the time taken by sum to become x times of itself. Here, t = 12 years
By the short trick approach, we get
$$\eqalign{
& = \frac{{100(3 - 1)}}{{12}} \cr
& = \frac{{200}}{{12}} \cr
& = \frac{{50}}{3} = 16\frac{2}{3}\% \cr} $$
Alternative Method :
Let the principal be P and in the 2nd scenario, SI = 2P
$$\eqalign{
& {\text{Rate}} = \frac{{{\text{SI}} \times 100}}{{{\text{Principal}} \times {\text{Time}}}} \cr
& = \frac{{2{\text{P}} \times 100}}{{{\text{P}} \times 12}} \cr
& = \frac{{50}}{3} \cr
& = 16\frac{2}{3}\% \cr} $$
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Rate of interest
$$ = \frac{{100(x - 1)}}{t}\% $$
Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times.
Therefore, x = 3
Where t is the time taken by sum to become x times of itself. Here, t = 12 years
By the short trick approach, we get
$$\eqalign{
& = \frac{{100(3 - 1)}}{{12}} \cr
& = \frac{{200}}{{12}} \cr
& = \frac{{50}}{3} = 16\frac{2}{3}\% \cr} $$
Alternative Method :
Let the principal be P and in the 2nd scenario, SI = 2P
$$\eqalign{
& {\text{Rate}} = \frac{{{\text{SI}} \times 100}}{{{\text{Principal}} \times {\text{Time}}}} \cr
& = \frac{{2{\text{P}} \times 100}}{{{\text{P}} \times 12}} \cr
& = \frac{{50}}{3} \cr
& = 16\frac{2}{3}\% \cr} $$
Was this answer helpful ?
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