Simple and Compound Interest is an important segment of the Arithmetic section under Quantitative Aptitude. To help you all understand these topics better, **we are sharing study notes with an example so that you can tackle questions based on Simple Interest easily.**

In this article, we will discuss the concepts of Simple Interest and talk about how to solve and approach the questions based on this topic.

**Example 2:** If the principal is 100 Rs. Difference of Simple Interest for 4yrs and 6yrs is Rs 8. Calculate the rate of simple interest.

Solution: In simple interest questions, interest always remains same for a year if the principal, rate of interest is constant for the same.

Let Interest for 4 yrs is I then interest for 6 yrs is (I+8)

interest for 2 yrs is Rs. 8

interest for 1 yr = 4

rate of interest = (4/100) × 100 = 4%**Example3:** If the amount is (10/9) times of Principal and rate of interest and time both are numerically equal. Then, what is the rate of interest per annum?

Solution: Let Principal is P. Given, numerically R = T

Interest = Amount – principal

I = (10/9)P – P

I = P/9 (Interest is in the multiples of Principal)

Now, I =[(P×R×T)/100]

P/9 = (P× R× T)/100

R^{2} = 100/9 (using, R=T)

R = (10/3)%

We can also say the time period is (10/3)years.

**Short approach:** Whenever Interest is in multiple of Principal and Rate of Interest and Time period is equal.

**Annual Instalments for Simple Interest:**

**Let's discuss a real example to understand instalment concepts:**

A person deposit Rs.140 to bank every year up to 5 yrs . The bank gives him 5% rate of interest simple annually. And at the end of 5 yrs he get total amount of Rs.770

So, 140 is the instalment, time is 5 years rate of interest is 5% and the amount or debt is Rs.770

This Instalment is also known as annual payment. Debt is total amount, so don’t confuse between these two terms.

**Installment =**

**where A = debt, r = rate of interest and t = time period**

**Example4**: What annual payment will discharge a debt of Rs.848 in 4yrs at 4% per annum simple interest?

In case if you forget formula then how to approach this question.

Let installment is X. There are 4 installments and rate of interest is also 4%

Debt (A) = four installments + (r%) × installments × (0+1+2+… (t-1))

So, 848 = 4X + (4%)(X)(0+1+2+3)

848 = 4X+

848 = 4X+

848 = 424X/100

X = 200

**Some Important examples based on Simple Interest.**

**Example5:** A sum amounts to Rs. 702 in 2 years and Rs. 783 in 3 years. Calculate the sum, rate of interest and the amount after 5 years?

Solution:

Amount for 2 years(A_{2}) = 702

Amount for 3 years (A_{3})= 783

Interest for 1 year (I) = 783-702 = 81

So Sum = A_{2} – 2I = 702 – 2×81

= 702-162 = 540

rate of interest = (81/540)×100

= 15%

Amount after 5 years = Sum+5I

= 540+ 5×81

= 945

**Example6:** A sum of money doubles itself in 3 yrs at a simple interest. In how many yrs will it amount to 8 times itself?

Solution: Doubles in 3 yrs

3 times in 3× 2 = 6yrs

4 times in 3× 3 = 9yrs

8 times in 3× 7 = 21yrs

**Example7:** Atul and Vijay are friends. Atul borrowed a sum of Rs.400 at 5% per annum simple interest from Vijay. He returns the amount with interest after 2 yrs. Vijay returns to Atul 2% of the total amount returned. How much did Atul receive?

Solution: After 2 yrs, amount returned to Vijay = 400+ (400*5*2)/100 = Rs 440

Amount returned to Atul = 2% of 440 = 8.8

**Example8:** Rs.4000 is divided into two parts such that if one part be invested at 3% and the other at 5%, the annual interest from both the investments is Rs. 144. Find each part.

Solution: Let the amount lent at 3% rate be Rs.X, then amount lent at 5% rate is 4000-X

So, 3% of X + 5% of (4000-X) = 144

5% of 4000 – 2% of X = 144

200 – 2% of X = 144

2% of X = 56

X = (56/2)×100

X = 2800

And 4000 -X = 1200.

**How to solve this Question by Alligation Method:**

First we will calculate net rate of interest for Rs. 144 on 4000

So, net rate = (144/4000)× 100 = 3.6%

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