*Thursday, November 5th, 2020*

Nowadays, loan has become crucial part of our life. We all have learnt living our life on credit. Whether be it a businessman taking loans to run his business or a household to buy a car, everyone has become dependent on sustaining their life and fulfilling their wishes with the help of these loans. But, when the amount has been borrowed then it has to be returned too and now not just the principal loan amount but some interest as well. Interest plays a very significant role in our life. It is a deciding factor whether or not loan has to be taken or not as higher the interest then higher the amount that has to repaid. Now, after the loan has been taken it could either be returned along with the interest in a lump-sum after some specified period of time or it can also be recovered in form of installments of some kind in which some amount of interest along with principal sum is repaid at some time intervals. Currently, all major finance lending institutions such as banks etc. recover their loans through EMIâ€™s i.e. Equated monthly installments. Today, in this blog we will discuss the how to calculate these installments considering various different factors and cases.

Interest charged on the loan can be of any type either Simple Interest or Compound Interest. Though we have discussed regarding it but for revisionâ€™s sake.

Simple interest is a the one where interest once credited does not earn interest on it.

**SI = (P * R * T)/ 100**

Compound Interest is where interest earns itself interest. It is the most common form of interest that is being charged nowadays.

**CI = P(1+r/100) ^{n}**

Suppose Ravi bought a T.V. worth â‚¹20000 on EMIâ€™s and every month a fix installment has to be for next n months where interest is charged @ r% per annum on simple interest.

Now, if the loan is for n months then Ravi will pay end the of 1^{st} month interest for (n-1) months, at the end of second month heâ€™ll pay interest for (n-2) months, at the end of 3^{rd} month heâ€™ll pay interest for (n-3) months and similarly, at the end of n^{th }month heâ€™ll pay no interest i.e.

Therefore, total amount paid by Ravi = [x+ (x* (n-1) * r)/ 12* 100] + [x+ (x* (n-2) * r)/ 12* 100] + [x+ (x* (n-3) * r)/ 12* 100] â€¦ [x+ (x* 1* r)/ 12* 100] + x

This will be equal to the total interest charged for n months i.e. [P+ (P* n* r)/ 12* 100].

Thus, [P+ (P* n* r)/ 12* 100] = [x+ (x* (n-1) * r)/ 12* 100] + [x+ (x* (n-2) * r)/ 12* 100] + [x+ (x* (n-3) * r)/ 12* 100] â€¦ [x+ (x* 1* r)/ 12* 100] + x

Simplifying and generalizing the above equation we get the following formula, **x = P (1 + nr/100)/ (n + n(n-1)/2 * r/100))**

And instead of principal sum total amount (Principal + Interest) to be repaid is given then, **x = 100A/ 100n + n(n-1) r/2 **

Let a person takes a loan from bank at r% and agrees to pay loan in equal installments for n years. Then, the value of each installment is given by

**P (1 + r/100) ^{n} = X (1 + r/100)^{n-1} + X (1 + r/100)^{n-2} + X (1 + r/100)^{n-3}+â€¦.+ X (1 + r/100)**

Using the Present Value Method,

**P = X/ (1 + r/100) ^{n}â€¦â€¦…X/ (1 + r/100)^{2} + X/ (1 + r/100)**

**Installments on Simple Interest and Compound Interest Case 1: **To calculate the installment when interest is charged on SI

A mobile phone is available for â‚¹2500 or â‚¹520 down payment followed by 4 monthly equal installments. If the rate of interest is 25%p.a. SI, calculate the installment.

**Installments on Simple Interest and Compound Interest Sol: **This is one basic question. You have to just use the above formula and calculate the amount of installment.

Therefore, x = **P (1 + nr/100)/ (n + n(n-1)/2 * r/100))**

Here P = 2500 â€“ 520 = 1980

R = 25% p.a.

T = 4 months

Hence, x = **1980(1 + 25*4/1200)/ (4 + (4*3/2)*25/1200)**

= **â‚¹520**

**Installments on Simple Interest and Compound Interest Case 2: **To calculate the installment when interest is charged on CI

What annual payment will discharge a debt of â‚¹7620 due in 3 years at 16 2/3% p.a. compounded interest?

**Installments on Simple Interest and Compound Interest Sol: **Again, we will use the following formula,

**P (1 + r/100) ^{n} = X (1 + r/100)^{n-1} + X (1 + r/100)^{n-2} + X (1 + r/100)^{n-3}+â€¦.+ X (1 + r/100)**

Here P = â‚¹7620

n = 3 years

r = 16 ^{2}/_{3}% p.a.

**7620(1+ 50/300) ^{3} = x (1 + 50/300)^{2 }+ x (1 + 50/300) + x**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 12100.2778 = x (1.36111 + 1.1667 + 1)**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â XÂ Â Â Â Â Â Â = Â â‚¹3430**

**Installments on Simple Interest and Compound Interest Case 3: **To calculate loan amount when interest charged is Compound Interest

Ram borrowed money and returned it in 3 equal quarterly installments of â‚¹17576 each. What sum he had borrowed if the rate of interest was 16 p.a. compounded quarterly?

**Installments on Simple Interest and Compound Interest Sol: **In this case, we will use present value method as we need to find the original sum borrowed by Ram.

Since, **P = X/ (1 + r/100) ^{n}â€¦â€¦…X/ (1 + r/100)^{2} + X/ (1 + r/100)**

Therefore, **P = 17576/ (1 + 4/100) ^{3} +**

**= 17576 (0.8889 + 0.92455 + 0.96153)**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 17576 * 2774988**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 48773.1972**

**Installments on Simple Interest and Compound Interest Case 4: **Gopal borrows â‚¹1,00,000 from a bank at 10% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are â‚¹10,000, â‚¹20,000, â‚¹30,000 and â‚¹40,000 respectively, what amount should be paid at the end of the fifth year to clear the debt?

**Installments on Simple Interest and Compound Interest Sol: **Total principal amount left after 5^{th} year = 100000 â€“ (10000 + 20000 + 30000 + 40000) = 100000 â€“ 100000 = 0

Therefore, only interest component has to be paid in the last installment.

Hence, Interest for the first year = **(100000 * 10 * 1) /100 =â‚¹10000**

Interest for the second year = **(100000 â€“ 10000) * 10/ 100 = â‚¹9000**

Interest for the third year = **(100000 â€“ 10000 â€“ 20000) * 10/ 100 = â‚¹7000**

Interest for the fourth year = **(100000 â€“ 10000 â€“ 20000 â€“ 30000) * 10/ 100 = â‚¹4000**

Thus, Amount that need to paid in the fifth installment = (**10000 + 9000 + 7000 + 4000) = â‚¹30000**

**Installments on Simple Interest and Compound Interest Case 5: **An amount of â‚¹12820 due in 3 years, hence is fully repaid in three annual installments starting after 1 year. The first installment is Â½ the second installment and the second installment is ^{2}/_{3} of the third installment. If rate of interest is 10% p.a. Find the first installment.

**Installments on Simple Interest and Compound Interest Sol: **Let the third installment be x.

Since, second installment is ^{2}/_{3} of the third, it will be ^{2}/_{3}x. And finally, 1^{st} installment will be Â½ * ^{2}/_{3} *x

Now proceeding in the similar fashion as we did earlier and using the compound interest formula to calculate the installment amount.

**P (1 + r/100) ^{n} = X (1 + r/100)^{n-1} + X (1 + r/100)^{n-2} + X (1 + r/100)^{n-3}+â€¦.+ X (1 + r/100)**

**12820 (1 + 10/100) ^{3} = **

**Â Â Â Â Â Â Â Â Â Â Â 12820(1.1) ^{3} = **

**Â Â Â Â Â Â Â Â Â Â Â ÂÂÂ17063.42 = x(0.40333 + 0.55 + 1)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Â Â Â Â Â Â Â Â Â Â Â 17063.42 = x* 1.953333**

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ÂÂÂÂXÂ Â Â = â‚¹8735.53**

Therefore, the amount of first installment will be â…” *Â½ *x, i.e. **â…“ *****8735.53 **= **â‚¹2911.84**

**Installments on Simple Interest and Compound Interest Case 6: **Ravi lent out â‚¹9 to Sam on the condition that the amount is payable in 10 months by 10 equal installments of â‚¹1 each payable at the start of every month. What is the rate of interest per annum if the first installment has to paid one month from the date the loan is availed.

**Installments on Simple Interest and Compound Interest Sol:** The value of money coming in should equal the value of the money going out for the loan to be completely paid off. Therefore,

â‚¹9 + Interest on â‚¹9 for 10 months = (â‚¹1 + interest on â‚¹1 for 9 months) + (â‚¹1 + interest on â‚¹1 for 8 months) + (â‚¹1 + interest on â‚¹1 for 7 months) + (â‚¹1 + interest on â‚¹1 for 6 months) + â€¦. + (â‚¹1 + interest on â‚¹1 for 2 months) + (â‚¹1 + interest on â‚¹1 for 1 month) + â‚¹1

â‚¹9 + interest on â‚¹1 for 90 months = â‚¹10 +Â interest on â‚¹1 for 45 months

**Â Â Â Â Â Â Â Â Â Â Â **Â interest on â‚¹1 for 90 months –Â interest on â‚¹1 for 45 months = â‚¹10 – â‚¹9

**Â Â Â Â Â Â Â Â Â Â Â **Â interest on â‚¹1 for 45 months = â‚¹1 (i.e. money would double in 45 months)

**Â Â Â Â Â Â Â Â Â Â Â Hence, the rate of interest = 100%/ 45 = 2.2222%**

I hope you are now clear with the installment concept and can now numerous other these and varied type of questions!

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Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,â€¦â€¦..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + â€¦.+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + â€¦. + an ) > 1830?

Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

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Quantitative Aptitude – Modern Maths – P&C – Q5

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didnt undderstood last case …can u pls explain

I want to solve question

Great explanation

I am very depresed

I waste my 10 days on this topic thank you very much

Appreciate it

Answer for case 6 is wrong I believe.

Thank you for such a brief and beautiful explanation