Maths Tricks

# Maths Tips Simple Interest, Formulas, Problems

Here we are listed All the Important terms, Formulas, Problems of Simple Interest on General Aptitude. Those Maths Tips will help you get maximum marks from this section. This file will be updated regularly, So please bookmark this page on your browser to get latest updates. You can also download the PDF file of the following below.

Contents in Short

#### Important Terms

 P- Principal R- Rate of interest T- Time A – Amount S.I- simple interest

#### Formulas:

 Simple interest = (P*R*T)/100 A = P + S.I. P =(100*A)/(100 + (RT)) The annual payment that will discharge a debt of Rs. X due in T years at the rate of interest R% per annum is (100X)/ (100T + ((RT(T-1))/2 )) If sum of money become x times in T years at SI , the rate of interest is given by (100X(X-1)/T)%

#### Problems:

1) Rs.800 becomes Rs.965 in 3 years at a certain simple interest. If the rate of interest is increased by 4%, what amount will Rs.800 become in 3 years?

Solution:

increase in interest in 3 years due to increase in rate by 4%

Simple interest = (P*R*T)/100

=(800*3*4)/100  = 96

Total amount, at the end of 3 years = 965+96 = Rs 1052

2) Raj borrowed some money at the rate of 4p.c.p.a for the first 3 years, at the rate of 8p.c.p.a for the next 2 years and the rate of 9p.c.p.a for the period beyond 5 years . if he pays a total simple interest of Rs.19,550 at the end of 7 years, how much money did he borrow?

Solution:

((P*4*3)/100) + ((P*8*2)/100) + ((P*9*2)/100) =19550

P=(19550*100)/46 = 42,500

3) Arjun borrowed a sum of Rs.30,000. He took a part of it at 12% per annum rate of simple interest and the remaining at 10% per annum. At the end of two years, he returned Rs.36,480 and discharge his loan. What was the sum borrowed at 12% per annum rate of interest?

Solution:

Consider one part of money as x

((x*12*2)/100) + (((30000-X)*10*2)/100) = 6480

(4X/100) = 6480 -6000

X=12000

4) A sum of Rs. 2900 amount to Rs.3422 in 3 years at simple interest. If the interest rate were increased by 3%. What would be it amount in same period?

Solution:

A = P + S.I.

SI = 3422-2900 = 522

R=(522*100)/(3*2900) = 6%

Increase in interest rate = 3+6 = 9%

SI =(2900*3*9)/100  = 783

Amount = 2900+783 = 3683

Sample problems are here, more problems will be added soon.