Read time: 2 mins OR Video time: 4 mins. Solving time: 6 mins
Lesson Objective
To learn the formula tying P, r%, n and EMI for the case of CI
#1.
If it is a case of CI, the instalment amount can be found as:
$$ \text{EMI} = \frac{P \times r\% \times \left( 1 + r\% \right)^n}{\left(1 + r\% \right)^n \;-\; 1} $$
If the question gives us the debt value i.e. the amount owed as on the last date of instalments, then the formula changes to:
$$ \text{EMI} = \frac{D \times r\% }{\left(1 + r\%\right)^n \;-\; 1} $$
#2.
It is easier to use r% as a fraction instead of a decimal to make the calculations easier. E.g. if rate of interest is 25%, use r% = 1/4 instead of 0.25. And (1 + r%) as 5/4 instead of 1.25.
Lets use this in the following examples:
E.g. 1
A washing machine costs Rs. 40,950. Ajaj Finance offers a scheme where I can take home the washing machine without paying anything now and instead I pay equal instalments, at the end of every 1 year from now, for three years. For this they are charging an interest rate of 20% p.a. compound interest. Find the instalment amount to be paid each year.
Answer
Rs. 19,440
Hint
Plug the given data i.e. P = 40,950, r% = 1/5, n = 3 in \( \text{EMI} = \frac{P \times r\% \times \left(1 + r\%\right)^n}{\left(1 + r\%\right)^n – 1} \) to find the EMI amount.
Explanation
\( \text{EMI} = \frac{40950 \times \frac{1}{5} \times \left( \frac{6}{5}\right)^3}{\left( \frac{6}{5}\right)^3 – 1} \)
40950/5 = 8190.
\( \text{EMI} = \frac{8190 \times \left( \frac{216}{125}\right)}{ \frac{216}{125}\ – 1} \)
\( \text{EMI} = \frac{8190 \times \left( \frac{216}{125}\right)}{ \frac{91}{125}} \)
The 125 will get cancelled from both the numerator and denominator, and we will get
\( \text{EMI} = \frac{8190 \times 216}{91} \)
8190/91 = 90.
EMI = 90 × 216 = 19,440
E.g. 2
I take a loan of Rs. 12,30,000 to be repaid at an interest rate of 5% p.a. compound interest. If the loan is repayable in two equal annual instalments, one to be paid at end of 1 year and second at end of 2 years from now, find the instalment amount, in Rs.
Answer
Rs. 6,61,500
Hint
Plug the given data i.e. P = 12,30,000, r% = 1/20, n = 2 in \( \text{EMI} = \frac{P \times r\% \times \left(1 + r\%\right)^n}{\left(1 + r\%\right)^n – 1} \) to find the EMI amount.
Explanation
\( \text{EMI} = \frac{1230000 \times \frac{1}{20} \times \left( \frac{21}{20}\right)^2}{\left( \frac{21}{20}\right)^2 – 1} \)
\( \text{EMI} = \frac{61500 \times \left( \frac{441}{400}\right)}{ \frac{441}{400}\ – 1} \)
\( \text{EMI} = \frac{61500 \times \left( \frac{441}{400}\right)}{ \frac{41}{400}} \)
The 400 will get cancelled from both the numerator and denominator, and we will get
\( \text{EMI} = \frac{61500 \times 441}{41} \)
61500/41 = 1500.
EMI = 1500 × 441 = 6,61,500
E.g. 3
I have an obligation of Rs. 12,30,000 at the end of 2 years from now. If I choose to save Rs. X at end of 1 year and Rs. X at end of 2 years from now, find the value of X such that I can meet my obligation. Assume an interest rate of 5% p.a. compound interest.
Answer
Rs. 6,00,000
Hint
Please notice that Rs. 12,30,000 is due 2 years from now i.e. this is the value of the term ‘debt’ as used on questions. So you need to use second formula given i.e. plug the given data i.e. D = 12,30,000, r% = 1/20, n = 2 in \( \text{EMI} = \frac{D \times r\% }{\left(1 + r\%\right)^n – 1} \) to find the EMI amount.
Explanation
\( \text{EMI} = \frac{1230000 \times \frac{1}{5} }{\left( \frac{21}{20}\right)^2 – 1} $
\( \text{EMI} = \frac{61500}{ \frac{441}{400}\ – 1} \)
\( \text{EMI} = \frac{61500 }{ \frac{41}{400}} \)
61500/41 = 1500.
EMI = 1500 × 400 = 6,00,000