# Quant Questions – Simple Interest & Compound Interest Set-3

Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions
1).Jackie invested Rs. 12000 in a bank at 15% p.a., C.I. He pays 40% of his income from interest to a charitable trust every year. How much money does Jackie have after 4 years ?
a)  Rs. 16583.16
b)  Rs. 15683.6
c)  Rs. 16843
d)  None of these
e)  Cannot be determined

2).A mobile phone can be purchased on cash payment of Rs. 7500. The same phone can also be purchased by making a down payment of Rs. 1750 and rest can be paid in 3 equal installments of Rs. 2000 each for next 3 months. The rate of simple interest charged by the shopkeeper is
a)  23.67%
b)  19.67%
c)  17.67%
d)  None of these
e)  16.27%

3).C.I and S.I on a certain amount of money for 2 years is Rs. 525 and Rs. 500 respectively. If the amount is invested such that S.I is Rs. 4050 while no. of years is double the rate per cent per annum, then find the time (in years) –
a)  16 years
b)  18 years
c)  20 years
d)  None of these
e)  24 years

4).Abhimanyu deposited RS. 55935 in each of the two banks for 5 years. On one he got 20% p.a. S.I while on the other he got 20% C.I p.a By how much % is the CI greater than the SI for the given period?
a)  38%
b)  48.8%
c)  52%
d)  40%
e)  None of these

5).A car depreciates at a certain rate. The current value of the car is estimated to be Rs. 32.768. Also, the ratio of decrease in the value for 4th year and 5th year is 5 : 4. Find the value of the car 5 years ago.
a)  Rs. 1,00,000
b)  Rs. 1,50,000
c)  Rs. 2,00,000
d)  None of these
e)  Cannot be determined

6).Basanti took a loan of Rs. 12,500 at 10% p.a. compounded annually, which is to be repaid in 3 equal installments, paid at the end of each year for 3 years. Find the value of each installment.
a)  Rs. 2056.4
b)  Rs. 5026.4
c)  Rs.4620.4
d)  Rs.6002.4
e)  None of these

7).A certain sum of money lent out at simple interest amounts to Rs. 1380 in 3 years and Rs. 1500 in 5 years. Find the rate per cent, per annum.
a)  3
b)  3.5
c)  4
d)  5
e)  None of these

8).The simple interest on a sum of money is 1/16 of the sum. If the number of years is numerically equal to the rate percent per annum, then the rate percent per annum is
a)  3(1 / 3)
b)  6(2 / 3)
c)  2(1 / 2)
d)  7(1 / 2)
e)  None of these

9).If x, y, z are three sums of money such that y is the simple interest on x and z is the simple interest on y for the same time and at the same rate of interest, then we have
a)  z2 = xy
b)  xyz = 1
c)  x2 = yz
d)  y2 = zx
e)  None of these
10).The difference between the interest received from two different banks on Rs. 500 for 2 years is Rs. 2.50. The difference between their rate is:
a)  0.5%
b)  2.5%
c)  0.25%
d)  1%
e)  None of these

1). d) 2). d) 3). b) 4). b) 5). a) 6). b) 7). d) 8). c) 9). d) 10). c)
Solution:

1).Finally, Jackie is earning an interest at the rate of 9% p.a only
15% – 40% of 15% = 15% – 6% = 9%
Amount = 12000(1 + (9 / 100) )4 = 12000 × (1.09)4
= 12000 × 1.411 = Rs. 16938.6

2).Loan amount = 7500 – 1750 = 5750
5750 + 5750 × (3 / 12) × (r / 100)
= [2000 + 2000 × (2 / 12) × (r / 100)] + [2000 + 2000 × (1 / 12) × (r / 100)] + 2000
=  5750 + (575 / 40) r = 6000 + (2000 / 1200) r [1 + 2]
(575 / 40)r = 250 + 5r
(375 / 40)r = 250àr = (250 × 40) / 375 = 26.67%

3).According to the information given, Rs. 25 is the interest on interest of first year. S.I for 2 years = Rs. 500, So, S.I for each year = Rs. 250, Hence Rs. 25 is the interest on Rs. 250
r = (25 / 250) × 100 = 10%
So. S.I = P × (10 / 100) × 2
P = (500 × 100) / (10 × 2) = 2500
Now, we are given that t = 2r
(2500 × r × 2r) / 100 = 4050
50r2= 4050 àr2 = 81
r = 9% and t = 18 years

4).Let us assume that Abhimanyu has deposited Rs. 100
Then, SI = 100 × (20 / 100) × 5 = 100
C.I. = 100(1 + (20 / 100) )5 – 100 = 248.83 – 100 = 148.8
C.I is greater than S.I by [ (148.83 – 100) / 100] × 100 = 48.83%

5).Ratio of decrease in the value for the 4th year an 5thyear = 5 : 4.
This means rate of depreciation = 20%
5 – (20 / 100) × 5 = 4, let us assume decrease in value for 4th year = 25& that of 5th year = 20 in ratio is 5 : 4
So, 25 – (20 / 100) × 25 = 20
Here, r = 20%
Let P be value 5 years ago, so
P[1 – (20 / 100) ]5 = 32768
P = 32768 × (105 / 85) = 105 = Rs. 100000

6).Let the value of each installment = Rs. x, then
12500 = {x / [ 1 + (10 / 100) ]1 } + {x / [ 1 + (10 / 100) ]2 } + {x / [ 1 + (10 / 100) ]3 }
12500 = (100 / 110)x + (100 / 110)2x + (100 / 110)3x
12500 = (10 / 11)x [ 1 + (10 / 11) + (100 / 121)
(12500   × 11) / 10 = x[ ( 121 + 110 + 100) / 121 ]
x = 12500 × (11 / 10) × (121 / 331) = Rs. 5026.4

7).If principal = Rs. x and rate = r% per annum then
1380 = x + (x × 3 × r) / 100                          ..(i)
1500 = x + (x × 5 × r) / 100                                    ..(ii)
S.I for two years = 1500 – 1380 = Rs. 120
(x × 2 × r) / 100 = 120
xr / 100 = 60                                                          ..(iii)
From equation (i)
1380 = x + 60 × 3
x = 1380 – 180 = Rs. 1200
From equation (iii)
(1200 × r) / 100 = 60
r = 6000 / 1200 = 5% per annum

8).S.I = (1 / 16) P.
Hence (1 / 16) P = (P × r × r) / 100
1 / 16 = r2 / 100
r = 10 / 4 = 2(2 / 4)% =2(1 / 2)%

9).SI = (P × R × T) / 100
y = (x × T × R) / 100
z = (y × T × R) / 100
(y / z) = (x / y)
y2= zx