Simple Interests Important Facts and Formulae: Principal or Sum:- The money borrowed or lent out for a certain period is called Principal or the Sum. Interest:- Extra money paid for using others money is called Interest. Simple Interest:- If the interest on a sum borrowed for a certain period is reckoned uniformly,then it is called Simple Interest. Formulae: Principal = P Rate = R% per annum Time = T years. Then, (i)Simple Interest(S.I)= (P*T*R)/100 (ii) Principal(P) = (100*S.I)/(R*T) Rate(R) = (100*S.I)/(P*T) Time(T) = (100*S.I)/(P*R) Simple Problems 1.Find S.I on Rs68000 at 16 2/3% per annum for 9months. Sol:- P=68000 R=50/3% p.a T=9/12 years=4/3 years S.I=(P*R*T)/100 =(68000*(50/3)*(3/4)*(1/100)) =Rs 8500 Note:If months are given we have to converted into years by dividing 12 ie., no.of months/12=years

2.Find S.I on Rs3000 at 18% per annum for the period from 4th Feb to 18th April 1995 Sol:- Time=(24+31+18)days =73 days =73/365=1/5 years P= Rs 3000 R= 18% p.a S.I = (P*R*T)/100 =(3000*18*1/5*1/100) =Rs 108 Remark:- The day on which money is deposited is not counted while the day on which money is withdrawn is counted. 3. In how many years will a sum of money becomes triple at 10% per annum. Sol:- Let principal =P S.I = 2P S.I = (P*T*R)/100 2P = (P*T*10)/100 T = 20 years Note: (1) Total amount = Principal + S.I (2) If sum of money becomes double means Total amount or Sum = Principal + S.I = P + P = 2P Top Medium Problems 1.A sum at Simple interest at 13 1/2% per annum amounts to Rs 2502.50 after 4 years.Find the sum. Sol:- Let Sum be x. then,

S.I = (P*T*R)/100 = ((x*4*27)/(100*2)) = 27x/100 Amount = (x+(27x)/100) = 77x/50 77x/50 = 2502.50 x = (2502.50*50)/77 = 1625 Sum = 1625 2. A some of money becomes double of itself in 4 years in 12 years it will become how many times at the same rate. Sol:- 4 yrs - - - - - - - - - P 12 yrs - - - - - - - - - ? (12/4)* P =3P Amount or Sum = P+3P = 4 times 3. A Sum was put at S.I at a certain rate for 3 years. Had it been put at 2% higher rate ,it would have fetched Rs 360 more .Find the Sum. Sol:- Let Sum =P original rate = R T = 3 years If 2% is more than the original rate ,it would have fetched 360 more ie., R+2 (P*(R+2)*3/100) - (P*R*3)/100 = 360 3PR+ 6P-3PR = 36000 6P = 36000 P = 6000 Sum = 6000. 4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest rate is increased by 3%, it would amount to how much?

Sol:- S.I = 920 - 800 = 120 Rate = (100*120)/(800*3) = 5% New Rate = 5 + 3 = 8% Principal = 800 Time = 3 yrs S.I = (800*8*3)/100 = 192 New Amount = 800 + 192 = 992 5. Prabhat took a certain amount as a loan from bank at the rate of 8% p.a S.I and gave the same amount to Ashish as a loan at the rate of 12% p.a . If at the end of 12 yrs, he made a profit of Rs. 320 in the deal,What was the original amount? Sol:- Let the original amount be Rs x. T = 12 R1 = 8% R2 = 12% Profit = 320 P = x (P*T*R2)/100 - (P*T*R1)/100 =320 (x*12*12)/100 - (x*8*12)/100 = 320 x = 2000/3 x = Rs.666.67 6. Simple Interest on a certail sum at a certain rate is 9/16 of the sum . if the number representing rate percent and time in years be equal ,then the rate is. Sol:- Let Sum = x .Then, S.I = 9x/16 Let time = n years & rate = n% n = 100 * 9x/16 * 1/x * 1/n n * n = 900/16 n = 30/4 = 7 1/2%

Complex Problems 1. A certain sum of money amounts t 1680 in 3yrs & it becomes 1920 in 7 yrs .What is the sum. Sol:- 3 yrs - - - - - - - - - - - - - 1680 7 yrs - - - - - - - - - - - - - 1920 then, 4 yrs - - - - - - - - - - - - - 240 1 yr - - - - - - - - - - - - - ? (1/4) * 240 = 60 S.I in 3 yrs = 3*60 = 18012 Sum = Amount - S.I = 1680 - 180 = 1500 we get the same amount if we take S.I in 7 yrs I.e., 7*60 =420 Sum = Amount - S.I = 1920 - 420 = 1500 2. A Person takes a loan of Rs 200 at 5% simple Interest. He returns Rs.100 at the end of 1 yr. In order to clear his dues at the end of 2yrs ,he would pay: Sol:- Amount to be paid = Rs(100 + (200*5*1)/100 + (100*5*1)/100) = Rs 115 3. A Man borrowed Rs 24000 from two money lenders.For one loan, he paid 15% per annum and for other 18% per annum. At the end of one year,he paid Rs.4050.How much did he borrowed at each rate? Sol:- Let the Sum at 15% be Rs.x & then at 18% be Rs (24000-x) P1 = x R1 = 15 P2 = (24000-x) R2 = 18

At the end of ine year T = 1 (P1*T*R1)/100 + (P2*T*R2)/100 = 4050 (x*1*15)/100 + ((24000-x)*1*18)/100 = 4050 15x + 432000 - 18x = 405000 x = 9000 Money borrowed at 15% = 9000 Money borrowed at 18% = (24000 - 9000) = 15000 Top 4.What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% Simple Interest ? Sol:- Let each instalment be Rs x (x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092 28x/25 + 31x/25 + x =1092 (28x +31x + 25x) = (1092 * 25) 84x = 1092 * 25 x = (1092*25)/84 = 325 Each instalement = 325 5.If x,y,z are three sums of money such that y is the simple interest on x,z is the simple interest on y for the same time and at the same rate of interest ,then we have: Sol:- y is simple interest on x, means y = (x*R*T)/100 RT = 100y/x z is simple interest on y, z = (y*R*T)/100 RT = 100z/y 100y/x = 100z/y y * y = xz 6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% and partly at 8% p.a Simple interest .The total interest received after 3 years was Rs.300.The ratio of the money

lent at 5% to that lent at 8% is: Sol:- Let the Sum at 5% be Rs x at 8% be Rs(1550-x) (x*5*3)/100 + ((1500-x)*8*3)/100 = 300 15x + 1500 * 24 - 24x = 30000 x = 800 Money at 5%/ Money at 8% = 800/(1550 - 800) = 800/750 = 16/15 7. A Man invests a certain sum of money at 6% p.a Simple interest and another sum at 7% p.a Simple interest. His income from interest after 2 years was Rs 354 .one fourth of the first sum is equal to one fifth of the second sum.The total sum invested was: Sol:- Let the sums be x & y R1 = 6 R2 = 7 T = 2 (P1*R1*T)/100 + (P2*R2*T)/100 = 354 (x * 6 * 2)/100 + (y * 7 * 2)/100 = 354 6x + 7y = 17700 ———(1) also one fourth of the first sum is equal to one fifth of the second sum x/4 = y/5 => 5x - 4y = 0 —— (2) By solving 1 & 2 we get, x = 1200 y = 1500 Total sum = 1200 +1500 = 2700 8. Rs 2189 are divided into three parts such that their amounts after 1,2& 3 years respectively may be equal, the rate of S.I being 4% p.a in all cases. The Smallest part is: Sol:- Let these parts be x,y and[2189-(x+y)] then, (x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100

4x/100 = 8y/100 x = 2y By substituting values (2y*1*4)/100 = (2189-3y)*3*4/100 44y = 2189 *12 y = 597 Smallest Part = 597 9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and the remainder at 10%.If his annual income is Rs.561. The capital is: Sol:- Let the capital be Rs.x Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1) + (5x/12 * 10/100 * 1) = 561 7x/300 + x/50 + x/24 = 561 51x = 561 * 600 x = 6600