Simple Interest Tricks and formulae to crack competitive exam

Interest based questions can be seen in every aptitude and government question papers. These kind of sums are very easy to solve. Simple interest sums can yield you more marks in the exams as it is totally formula oriented. A student has to remember the only two formulae for Simple Interest in order to solve maximum questions.

Principal is always same for each year while calculating simple interest.

Formula :

$Simple Interest (SI) = \frac{P * T * R}{100}$

$P = A - I$

As per the above formula,
SI = Simple Interest
P = Principal or Loan amount
T = Time period
R = Rate
A = Amount
I = Interest Amount

Let’s dive into some questions.
Question 1 ) Joey took a loan from Chandler at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 as interest for the period What was the principal amount borrowed by Joey?
A) 18000 B) 15000 C) 12000 D) 16000

Solution: Given,
SI = 5400
Time = 3 years
Rate = 12%
Principal = x
Let us substitute the above things in our formula.

$SI = \frac{P*T*R}{100}$

$5400 = \frac{x*3*12}{100}$

By taking the 100 to the LHS.

$5400*100 = 36x$

$540000 = 36x$

$x = 15,000$

P = 15,000

Answer: B

Question 2 ) How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at the rate of 4.5% p.a. simple interest?
A) 5 years B) 3 years C) 4 years D) 6 years

Solution: Given,
SI = 81
P = 450
R = 4.5
T = x
Substitute these things in our formula

$SI = \frac{P*T*R}{100}$

$81 = \frac{450*x*4.5}{100}$

Taking 100 to the LHS.

$81*100 = 202.5x$

$x = \frac{8100}{202.5}$

$x = 4$

$T = 4$

Answer: C

Question 3 ) A sum of Rs. 800 amounts to Rs. 920 in 3 years at SI. If the interest rate is increased by 3% it would amount to how much?
A) 992 B) 800 C) 900 D) 920

Solution: First we will find out the R for the given and then we will add 3% more as described in the question.
Given,
SI = 920
P = 800
T = 3
R = x

$SI = \frac{P*T*R}{100}$

$920 = \frac{800*3*x}{100}$

Take 100 to the LHS

$920*100 = 2400x$

$x = \frac{92000}{2400}$

$x = 38.3$

Now, add 3 to it and find the value for the same given
The present R will be 38.3 + 3 = 41.3

$SI = \frac{800*3*41.3}{100}$

$SI = 991.2$

Answer: A

Question 4 ) A certain sum of money in simple interest amounts to Rs. 1008 in 2 years and to Rs. 1164 in 3 ½ years. Find the sum.
A) 208 B) 900 C) 804 D) 800

Solution: Two interest amounts are given. One is after 3 and a half years and the other is one year. From this we can find out half year interest amount. Let us do that first.

$I_3_._5 - I_2 = 1164 - 1008 = 156$

$I_1_._5 = 156$

We have found out the interest amount for one and a half year.
If we have to find out the interest amount for one year divide the interest amount with 1.5 as shown below.

$I_1 = \frac{156}{1.5}$

$I_1 = 104$

In the question we are given the amount after two years. So, in order to find the principal as given in the question we need to need to find the interest amount for 2 years and subtract it with two years amount.

$I_1 = 104$

Interest for 2 years = 104 * 2 = 208

$I_2 = 208$

The amount after two years is 1008.

P = A – I

$P = 1008 - 208 = 800$

$P = 800$

Answer: D

Question 5 ) In how many years will a sum double itself at 12.5% p.a. simple interest?
A) 4 B) 8 C) 10 D) 16

Solution: If the sum is ‘x’. How many years will it take to become ‘2x’.
The increase rate is 100%.
Every year it is increasing by 12.5% as given in the question
So,

$Time = \frac{100}{12.5} = 8$

Time Taken = 8 years.

Answer: B

Question 7 ) Sansa deposited a sum of Rs. 8000 in a bank paying simple interest. After one year, she withdraws Rs. 2000. At the end of 3 years, she received Rs. 7800. Find the rate of interest, assuming that she never collected her interests till then.
A) 9% B) 8% C) 10% D) 12%

Solution: She deposited 8,000 and she withdrew 2,000 after one year. Her interest is 7,800 at the end of three years. So,

$Total Interest = 7,800 - 6,000$

$I_1 + I_2_+_3 = 1,800$

Interest for all the three years is 1,800
She deposited 8,000 at the end of the first year and she had the remaining 6000 till the end of the third year. (6000 for the second year and the third year)

$\frac{8000*1*R}{100} + {6000*2*R}{100} = 1800$

After reducing,

$80R + 120R = 1800$

$200R = 1800$

$R = \frac{1800}{200}$

$R = 9$

R = 9%

Answer: A

Question 8 ) Mr White invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years is Rs. 3508, what was the amount invested in Scheme B?
A) Rs. 6400 B) Rs. 6500 C) Rs. 7200 D) Rs. 7500

Solution: Mr White must have invested as shown below.