# Simple Interest

P- Principal

R- Rate of interest

T- Time

A – Amount

S.I- simple interest

Formulas:

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

II) A = P + S.I.

III) P =(100*A)/(100 + (RT))

IV) 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 ))

V) 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 = 1052 Rs

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:

Considrer 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 .

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