Question

Calculate the rate at which P = ₹6000, I = ₹1500 and T= 2years.

Answer 1

Hint:So, in this question we have to find the rate of interest for a simple interest of ₹1500 on ₹6000 over a period of 2 years. We will use the formula of Simple Interest which is given as:
$I = \dfrac{{P \times r \times t}}{{100}}$,
Where P is the Principal, r is the rate of interest and t is the time period.

Complete step-by-step solution:
Now, we are given a Principal sum of ₹6000 and an interest of ₹1500 calculated over a time of 2 years.
So we can say that, principal is ‘P’, rate of interest is ‘r’ and time period is ‘n’ years
So,
$
  P = 6000 \\
  r = ? \\
  t = 2years \\
  I = 1500 \\
 $
Now, the formula of Interest is given as: $I = \dfrac{{P \times r \times t}}{{100}}$
So now on substituting the values of P, r and t we will get the Interest as:
$1500 = \dfrac{{6000 \times r \times 2}}{{100}}$
So, we need to solve this equation to get the value of r,
So, first we will take the denominator 100 on the right hand side to the left hand side and multiply the same with 1500 to get:
$1500 \times 100 = 6000 \times r \times 2$
Now, we will take 6000 and 2 from the right hand side to the left hand side and divide the left hand side by 6000 and 2 and get
$\Rightarrow \dfrac{{1500 \times 100}}{{6000 \times 2}} = r$
Or,
$
\Rightarrow r = \dfrac{{1500 \times 100}}{{6000 \times 2}} \\
\Rightarrow r = \dfrac{{25}}{2}\% \\
\Rightarrow r = 12.5\% \\
 $
 Hence, the rate of interest comes out to be 12.5%.
Therefore, a rate of interest of 12.5% per annum compounded annually for a Principal of ₹6000 over a period of 2 years will yield an interest of ₹1500.

Note:We need to remember that simple interest is always calculated yearly and its value is the same for every consecutive year since the Principal is the same. While applying the formula for interest we have to be careful not to be confused with compound interest. Interest calculated is always simple interest unless mentioned or specified otherwise. The formula for CI is given as : $CI = P\left[ {{{\left( {1 + \dfrac{R}{{100}}} \right)}^N} - 1} \right]$ .