Read time: 2 mins OR Video time: 1 min. Solve time: 5 mins
Lesson Objective
Solving questions that require finding the rate of interest being charged
#1.
A common question asked in exams is finding the rate of interest being charged for a given Principle (or Debt) and EMI details. Finding the rate for the case of CI, especially when the number of instalments is more than 2 is not feasible using just calculators and will require other resources. Hence such questions are limited to just the case of SI.
#2.
We do not need to learn anything new for this question type; all we need is to re-arrange the formula of EMI in case of SI, as done here:
$$ \text{EMI} = \frac{P \times \left(1 \;+\; r\% \times n \right)}{n\;+\;r\% \times \frac{\left(n-1\right) \times n}{2}} $$
$$ \text{EMI} \times n + \text{EMI} \times r\% \times \frac{\left(n-1\right) \times n}{2} = P + P \times r\% \times n $$
Re-arranging terms to isolate terms having r%,
$$ r\% \times \left(P \times n − \text{EMI} \times \frac{\left(n-1\right) \times n}{2}\right) = \text{EMI} \times n \;-\; P$$
$$ r\% = \frac{\text{EMI} \times n \;-\; P}{P \times n \;−\; \text{EMI} \times \frac{\left(n-1\right) \times n}{2}} $$
#3.
And if instead of P, we are given the debt, D, i.e. the amount due as on a future date, then it is all the more easier,
$$ \text{EMI} = \frac{D}{n\;+\;r\% \times \frac{\left(n-1\right) \times n}{2}} $$
$$ \text{EMI} \times n + \text{EMI} \times r\% \times \frac{\left(n-1\right) \times n}{2} = D$$
Re-arranging terms to isolate terms having r%,
$$ r\% \times \text{EMI} \times \frac{\left(n-1\right) \times n}{2} = D \;-\; \text{EMI} \times n$$
$$ r\% = \frac{D \;-\; \text{EMI} \times n}{\text{EMI} \times \frac{\left(n-1\right) \times n}{2}} $$
You dont need to mug these formulae, you can also re-arrange the earlier formula for EMI on the spot and solve for r%. Some examples of standard questions asked in exams are given below.
E.g. 1
A money lender offers a loan of Rs. 1,32,000 on the condition that Rs. 48,000 is paid to him at end of each successive year for next 4 years. Find the rate of interest (% p.a.) being charged, assuming it is a case of simple interest.
Answer
25
Hint
Plug the given data i.e. P = 132000, EMI = 48000 and n = 4 in \( r\% = \frac{\text{EMI} \times n \;-\; P}{P \times n \;−\; \text{EMI} \times \frac{\left(n-1\right) \times n}{2}} \) to find r.
Explanation
\( r\% = \frac{48000 \times 4 \;-\; 132000}{132000 \times 4 \;−\; 48000 \times \frac{3 \times 4}{2}} \)
Cancelling out 1000, since it appears in all the terms,
\( r\% = \frac{192 \;-\; 132}{528 \;−\; 288} \)
\( r\% = \frac{60}{240} = \frac{1}{4} \)
Thus, r = 1/4 × 100 = 25
E.g. 2
A credit card offers to convert a purchase of Rs. 40,000 into 9 monthly EMIs of Rs. 5,000 each. If it is a case of simple interest, find the rate (% p.a.) being charged.
Answer
33.33…
Hint
We find the rate in the same manner as in the previous question. But since it is a case of monthly EMIs, the found rate will be per month, whereas the question asks us to find the rate per annum.
Explanation
\( r\% = \frac{5000 \times 9 \;-\; 40000}{40000 \times 9 \;−\; 5000 \times \frac{8 \times 9}{2}} \)
Cancelling out 1000, since it appears in all the terms,
\( r\% = \frac{45 \;-\; 40}{360 \;−\; 180} \)
\( r\% = \frac{5}{180} = \frac{1}{36} \)
Thus, r = 1/36 × 100 will be the rate, but this will be the per month rate, since it is case of monthly EMIs. Whereas the question asks rate % per annum.
Thus, the answer will be 1/36 × 100 × 12 = 100/3 i.e. 33.33…