Aptitude problems on Simple Interest (S.I) for Competitive exams


Click here for Latest Sample Papers & Formulas



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counted while the day on which money is withdrawn is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

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

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



For Aptitude problems on Profit and Loss : Click here



If you have questions, please ask below

6 Comments

  1. Atul Solanki says:

    Rs.100000 at the rate of Rs.1500 per annum.Find how many intrest we get in 15 days?

  2. shivu says:

    divide rs 2500 into two parts such that the s.i on one at 4% for 5 years is double at 5% for 3 years.

  3. veena says:

    A company accepts deposits and pays 15% interest per year , if the deposit is made for a minimum of 5 years.

  4. Pranab Das says:

    A borrowed Rs 400 from B @ 6.25% pa on 1st February 2006. As soon as the interest became Rs 5, A repaid with interest. Find the date on which Ajay had repaid the money?

  5. Aninda De says:

    rs1500 is invested at a rate of 10 simple interest and interest is added to the principal after every 5 years.in how many days will it amount to rs2500?

Leave a Reply

If you have any questions headover to our forums

You can use these XHTML tags: <a href="" title=""> <abbr title=""> <acronym title=""> <blockquote cite=""> <code> <em> <strong>