Simple Interest problems
Posted by
Ravi Kumar at Monday, November 24, 2008
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Simple Interest problems
Simple Interest problems
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counted while the day on which money is withdrawn is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
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+ 6P3PR = 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 (24000x)
P1 = x R1 = 15
P2 = (24000x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000x)*1*18)/100 = 4050
15x + 432000  18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000  9000)
= 15000
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(1550x)
(x*5*3)/100 + ((1500x)*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 â€”â€”â€”(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x  4y = 0 â€”â€” (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 = (21893y)*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