Read time: 2 mins OR Video time: 2 mins. Solving time: 5 mins
Lesson Objective
Solving questions that ask us to find the Interest paid when we purchase on EMIs
#1.
The relevant four terms in questions on instalments are P (or D), r%, n and EMI. We have already seen questions asking us to find EMI or r%. Lastly, we see a couple of questions where P or D is the unknown.
We do not need any new relation to find P (or D). The same formula, as learnt for calculating EMIs at CI or that at SI, is used. This time the unknown will be P (or D). Just be careful to note whether it is a case of SI or CI and accordingly use the correct formula.
Just one new term needs to be introduced before we can directly solve questions.
#2.
Occasionally a question is asked in exams that require us to find the Interest paid because of buying on EMIs instead of outright purchase. And this Interest amount can be found as:
Interest = EMI × n − P
#3. If it was the case of a debt at a future date being discharged by paying early instalments, then in this case one would be saving on the Interest earned on EMIs paid before the debt was due. And the savings due to the interest will be:
Interest Saved = Debt − EMI × n
E.g. 1
It takes 5 equal annual instalments of Rs. 20,000 to repay a loan completely. If the rate of interest is 5% p.a. simple interest, find the loan taken.
Answer
Rs. 88,000
Hint
Plug the given data i.e. EMI = 20,000, r% = 1/20, n = 5 in the formula for EMI in case of SI, \( \text{EMI} \times \left(n + r\% \times \frac{\left(n-1\right) \times n}{2} \right) = P \times \left( 1 \;+\; r\% \times n\right) \) to find P.
Explanation
\( 2000 \times \left(5 + \frac{1}{20} \times \frac{4 \times 5}{2}\right) = P \times \left( 1 \;+\; \frac{1}{20} \times 5\right) \)
\( 1000 \times 11 = P \times \frac{5}{4} \)
P = 1000 × 11 × 4/5 = 88,000
E.g. 2
I bought a bike under an instalment scheme where I had to pay 4 annual instalments of Rs. 36,450 with the first instalment to be paid right away and 3 further instalments to be paid at the end of each successive years. Find the total interest paid, in Rs, if the rate of interest is 12.5% p.a. compound interest.
Answer
Rs. 22,550
Hint
Hopefully you did not get confused with the first instalment of Rs. 36,450 to be paid right away. This is same as a down-payment. And one could consider the situation as that of 3 EMIs of Rs. 36,450. With n = 3, EMI = 36450 and r% = 1/8, we could find P.
Dont forget that this is a case of CI and hence use the correct formula, just because the last few questions were of SI, dont use the SI formula.
Once P is found, Interest can be found as the difference between the sum of EMIs paid versus the amount one would have to pay if purchased outright i.e. (EMI × n − P)
Explanation
\( 36450 \times \left( \left(1 + \frac{1}{8} \right)^3 \;-\; 1\right) = P \times \frac{1}{8} \times \left(1 + \frac{1}{8} \right)^3 \)
\( 36450 \times \left(\frac{729}{512} \;-\;1\right) = P \times \frac{1}{8} \times \frac{729}{512} \)
\( 36450 \times \frac{217}{512} = P \times \frac{1}{8} \times \frac{729}{512} \)
The 512 gets cancelled from both sides. Also 36450 can be successively divided by 9 thrice i.e. by 729, to get 50.
P = 50 × 217 × 8 = 400 × 217 = 86800
To find Interest paid, it does’nt matter if you do …
Loan of just 86,800 and 3 EMIs of 36450 i.e. (36450 × 3 − 86800)
OR
Price of bike = Down-payment + loan = 36450 + 86800 = 123250 and then 4 EMIs of 36450 i.e. or (36450 × 4 − 123250).
Both are the same, in the latter you needlessly have to do one extra addition. And the answer is Rs. 22,550.
With this we finish Level 1 of Instalments. Just this much should help you solve 95% of questions asked in exams. If you want to try questions with non-standard settings (e.g. increasing instalments or terminating the arrangement by repaying the pending principle) or you need to understand the logic behind the formula we used, you can proceed to Level 2.
However we recommend finishing off Level 1 of ALL topics before moving to Level 2 of any topic. This way you will ensure adequate and uniform coverage of all topics instead of in-depth coverage of one topic at expense of zero coverage of other topics.