Simple Interest
Simple Interest
Simple Interest
SIMPLE INTEREST
IMPORTANT FACTS AND FORMULAE
1.. Principal : The money borrowed or lent out for a certain period is called the principal or the sum. 2. Inte rest: Extra money paid for using other's money is called interest. 3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest. Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then, (i) (ii) S.I. = (P*R*T )/100 P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)
SOLVED EXAMPLES
Ex. 1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.
Sol. P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years. S.I. = (P*R*T)/100 = Rs. 68,000*(50/3)*(3/4)*(1/100) = Rs.8500
Ex. 2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the pe riod from 4th Feb., 2005 to 18th April, 2005. Sol. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years. P = Rs.3000 and R = 6 %p.a = 25/4%p.a S.I. = Rs. 3,000*(25/4)*(1/5)*(1/100) = Rs.37.50. Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted . Ex. 3. A sum at simple interests at 13 % per annum amounts to Rs.2502.50 after 4 years find the sum. Sol. Let sum be Rs. x then , S.I.=Rs. x*(27/2) *4*(1/100)
) = Rs.27x/50
) = Rs.77x/50
= 1625
Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere interest rate is increased by 8%, it would amount to bow mucb ? Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs. _ . R = (100 x 120)/(800*3)
) % = 5%.
New rate = (5 + 3)% = 8%. New S.l. = Rs. (800*8*3)/100 = Rs. 192. : New amount = Rs.(800+192) = Rs. 992.
Ex. 5. Adam borrowe d some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. 1 he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ? Sol. Let the sum borrowed be x. Then, (x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400
95x/100 = 11400
x = (11400*100)/95
Ex. 6. A ce rtain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 years. Find the sum and rate of interests.
Sol.. S.I. for 1 years = Rs.(1164-1008) = Rs.156. S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208 Principal = Rs. (1008 - 208) = Rs. 800. Now, P = 800, T = 2 and S.l. = 208. Rate =(100* 208)/(800*2)% = 13%
Ex. 7. At what rate percent pe r annum will a sum of money double in 16 years. Sol.. Let principal = P. Then, S.l. = P and T = 16 yrs. Rate = (100 x P)/(P*16)% = 6 % p.a. Ex. 8. The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal. Sol. Let sum = Rs. x. Then, S.l. = Rs. 4x/9 Let rate = R% and time = R years. Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3. Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months. Ex. 9. The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40 less tban the simple inte rest on the same sum for 3 yea rs at 10% per annum. Find the sum. Sol. Let the sum be Rs. x Then, (7x/20)-(3x/10)=40 x = (40 * 20) = 800. Hence, the sum is Rs. 800. Ex. 10. A sum was put at simple interest 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 and original rate = R. Then,
((x*10*7)/(100*2)) ( (x*12*5)/(100*2)) = 40
Ex. 11. What annual instalme nt will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest? . Sol .Let each Instalment be Rs. x Then,
(28x+31x+25x)=(1092*25)
Ex. 12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate. Sol. Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x).