P- Principal
R- Rate of interest
T- Time
A – Amount
S.I- simple interest
Formulas:
I) Simple interest = (P*R*T)/100Â
II) A = P + S.I.
III) P =(100*A)/(100 + (RT))Â
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 :Â Â 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 .
