Compound Interest : Aptitude Test (65 Questions with Explanation)

1.

The difference between the simple interest on a certain sum at the rate of 10% p.a. for 2 years and compound interest which is compounded every 6 months is Rs.124.05 .What is the principal sum? 

   A.) Rs.12,000
   B.) Rs.8000
   C.) Rs.10,000
   D.) Rs.6000

Answer: Option 'B'

Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
=P(1+ (R / 2)100) 2T−P= P(1+(10 / 2)100) 2× 2− P
=P(1+120)4−P=P(21 / 20) 4− P
Simple Interest on P at 10% for 2 years
=P × R × T /100
=P×10×2100
= P / 5
Then P[(1+5 / 100)4-1] - P x 10 x 2/100 = 124.05
⇒ P[(21/20)4 - 1 - 1/4] = 124.05
⇒ P[(194481/160000) - (6/5)] = 12405 /100
⇒ P[194481-192000 / 160000] = 12405 /100
⇒ P = [(12405/100) x (160000/2481)]
= 124.05 x 64.490
= 7999.9845
= 8000.

2.

A person receives a sum of Rs. 210 as interest for investing some amount at 10% p.a compounding annually for 2 years. Find the amount invested at the beginning

   A.) 1050
   B.) 1000
   C.) 850
   D.) 950

Answer: Option 'B'

Given
Compound interest received by the person ( C.I ) = Rs. 210
Rate of interest ( r ) = 10 %
Number of years ( n ) = 2 years
To find, Amount invested at the beginning = principal ( p )
Compound interest ( C.I ) = Amount - Principal
Amount = p ( 1 + r / 100 )n

=> C.I = p [ ( 1 + r / 100 )n- 1 ]
=> 210 = p [ ( 1 + 10 / 100 )2- 1 ]
=> 210 = p [ ( 110 / 100 )2- 1 ]
=> 210 = p [ ( 11 / 10 )2- 1 ]
=> 210 = p [ ( 121 / 100 ) - 1 ]
=> 210 = p [ ( 121 - 100 ) / 100 ]
=> 210 = p [ 21 / 100 ]
=> 210 x ( 100 / 21 ) = p
=> 1000 = p
The amount invested at the beginning = p = Rs.1000

3.

What would be the compound interest accrued on an amount of 14000 Rs. at the end of 3 years at the rate of 5 % per annum?

   A.) 16206.75
   B.) 2206.75
   C.) 15216.75
   D.) 16216.75

Answer: Option 'B'

Given
principal = 14000
No. of years = 3
Rate of interest = 5
Amount = P x (1+r/100)n
we get Amount = 14000 x (1+5/100)3
= 14000 x (105 / 100)3
= 14000 x (21 / 20)3
= 14000 x (9261 / 8000)
= 64827 / 4
16206.75
compound interest = Amount - principal
=16206.75 -14000
=2206.75

4.

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs.12,000 after 3 years at the same rate?  

   A.) Rs.3972
   B.) Rs.2160
   C.) Rs.3120
   D.) Rs.6240

Answer: Option 'A'

Let the Principal be, P = Rs. 100
Given, S.I = 60% of 100 = Rs. 60, n = 6 years
Then, Rate of Interest, r = (S.I × 100 )/ (p × n)
=> r = (60 × 100) / (100 × 6)
=> r = 10 % p.a
Now, P = Rs. 12,000, n = 3 years, r = 10% p.a
C.I = P {[1 + (r/100)]n - 1}
= 12,000 × {[1 + (10/100)]3-1}
= 12,000 × [(11/10)3-1]
= 12,000 × [(1331/1000) - 1]
= 12,000 × (331/1000)
= 12 × 331
= 3972
Thus, Compound Interest = Rs. 3972

5.

John invested an amount of Rs. 20000 for 2 years at compound interest at the rate of 6 % per annum. Find the amount he receives at the end of 2 years.  

   A.) 22472
   B.) 22000
   C.) 22372
   D.) 22120

Answer: Option 'A'

Given
Principal : P = 20000 Rs.
Rate of Interest : r = 6 %
Number of years : n = 2
Amount = P x (1 + r/100)n
=> Amount = 20000 x (1+6/100)2
= 20000 x (1+3/50)2
= 20000 x (53/50) x (53/50)
22472
Therefore, Amount received by John at the end of two years = Rs. 22472

6.

Akarsh left a will of Rs. 16,400 for his two sons whose age are 17 and 18 years.They must get equal amounts when they are 20 years at 5% compound interest. Find the present share of the younger son.

   A.) Rs. 5,000
   B.) Rs. 8,000
   C.) Rs. 7,000
   D.) Rs. 9,000

Answer: Option 'B'

Given, total amount (to be shared by two sons at the age of 20 on Compound interest) = Rs. 16,400
Let the Present share (Principal amount) for 17 year old son = "X"
Then the Present share (Principal amount) for 18 year old son = (16,400 - X)
To attain 20 years of age,
=> 17 year old son takes 3 years (N = 3 years on Compound interest)
=> 18 year old son takes 2 years (N = 2 years on Compound interest)
Given, Rate of interest (R) = 5%
Given that, at the age of 20, two sons get equal amount
=> Compound Amount of 17 year old son = Compound Amount of 18 year old son
W.K.T, Formula for Compound Amount = P [1 + (R/100)]N
=> X (1 + 5/100)3 = (16,400 - X) (1 + 5/100)2
=> X (1 + 5/100) = (16,400 - X)
=> (105/100) X = (16,400 - X)
=> [(105/100) X] + X = 16,400
=> 205 X = 16,400 × 100
=> X = 16,40,000 / 205
=> X = 8,000
Therefore, Present share for 17 year old son = Rs. 8,000

7.

What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 16 % per annum?

   A.) 1728
   B.) 6528
   C.) 6928
   D.) 6728

Answer: Option 'A'

Given principal = 5000
No. of years = 2
Rate of interest = 16
Amount = P x (1+r/100)n,
= 5000 x (1+16/100)2
= 5000 x (1 + 4/25)2
= 5000 x (29 / 25)2
= 5000 x (841 / 625)
=6728
=>Amount = Rs.6728
Compound Interest = Amount - Principal

= 6728 - 5000
=1728
Thus,Compound Interest = Rs.1728

8.

What would be the compound interest accrued on an amount of 6250 Rs. at the end of 2 years at the rate of 8 % per annum?  

   A.) 7280
   B.) 1040
   C.) 1065
   D.) 7390

Answer: Option 'B'

Given, principal = 6250
No. of years = 2
Rate of interest = 8
Amount = P x (1+r/100)n,
=> Amount = 6250 x (1+8/100)2
= 6250 x (108 / 100)2
= 6250 x (108 / 100) x (108 / 100)
= 7290
Compound Interest = Amount - Principal
= 7290 - 6250
= 1040
Therefore, Compound Interest = Rs. 1040

9.

What would be the compound interest accrued on an amount of 12500 Rs. at the end of 3 years at the rate of 10 % per annum?

   A.) 4537.5
   B.) 4137.5
   C.) 6637.5
   D.) 6647.5

Answer: Option 'B'

Given principal = 12500
No. of years = 3
Rate of interest = 10
Amount = P x (1+r/100)n
= 12500 x (1+10/100)3
= 12500 x (11/10)3
= 12500 x (11/10)x (11/10)x (11/10)
= 16637.5
Compound Interest, C. I = Amount - Principal = 16637.5 - 12500 = 4137.5

10.

What would be the compound interest accrued on an amount of 8000 Rs. at the end of 2 years at the rate of 5 % per annum?

   A.) 8920
   B.) 8820
   C.) 8780
   D.) 8810

Answer: Option 'A'

Given principal = 8000.
No. of years = 2
Rate of interest = 5.
Amount = P x (1+r/100)n,
we get Amount = 8000 x (1+5/100)2 .
Ans : 8820

11.

Hari lended a sum of Rs.8000 for 20% per annum at compound interest then the sum of the amount will be Rs.13824 is obtained. After how many years he will get that amount? 

   A.) 2 years
   B.) 1 year
   C.) 4 years
   D.) 3 years

Answer: Option 'D'

Let Principal = P, Rate = R% per annum, Time = n years
When interest is compounded annually, total amount can be calculated by using the formula
Compound Amount = P ( 1 + R / 100)n
Given that, P = Rs.8000, R = 20% per annum
Compound Amount = Rs. 13824
We have to find the time period during which the amount will be Rs.13824
=> Rs.13824 = 8000 x (1 + 20/100)n
=> (13824 /8000) = (120 / 100)n
=> (24 / 20)3 = (12 / 10)n
=> (12 /10)3 = (12 /10 )n
Therefore, n = 3.
Hence the required time period is 3 years.

12.

What would be the compound interest accrued on an amount of 2500 Rs. at the end of 3 years at the rate of 10 % per annum? 

   A.) 827.5
   B.) 3357.5
   C.) 895.6
   D.) 863.4

Answer: Option 'A'

Given principal = 2500
No. of years = 3
Rate of interest (r) = 10
Amount = P x (1+r/100)n
=>Amount = 2500 x (1+10/100)3
= 2500 x (1+1/10)3
= 2500 x (11/10)3
= 2500 x (1331 / 1000)
=3327.5
=>Amount = Rs.3327.5
Compound interest = Amount - Principal
= 3327.5 - 2500
=827.5
Thus, theCompound interest = Rs.827.5

13.

A person borrows a certain amount from his friend at the rate of 15% per annum compound interest, interest being compounded annually and agrees to return it in 2 equal yearly installments of Rs.529/- each. Find the amount borrow.

   A.) Rs.820/-
   B.) Rs.880/-
   C.) Rs.860/-
   D.) Rs.840/-

Answer: Option 'C'

Rs.860/-

14.

What would be the compound interest accrued on an amount of 5000 Rs. at the end of 2 years at the rate of 9 % per annum?

   A.) 940.5
   B.) 5940.5
   C.) 5990.5
   D.) 980.5

Answer: Option 'A'

Given principal = 5000
No. of years = 2
Rate of interest = 9 
Amount = P x (1+r/100)n,
We get Amount = 5000 x (1+9/100)2 = 5000 x (109/100) x (109/100) = 5940.5
Compound Interest, C. I = Amount - Principal = 5940.5 - 5000 = 940.5

15.

The difference between compound interest and simple interest compounded annually on a certain sum of money for 2 years at 4% p.a. is Re.1 The sum (in Rs) is:

   A.) 625
   B.) 525
   C.) 635
   D.) 685

Answer: Option 'A'

625

16.

Hari lended a sum of Rs.8000 for 20% per annum at compound interest then the sum of the amount will be Rs.13824 is obtained. After how many years he will get that amount? 

   A.) 2 years
   B.) 1 year
   C.) 3 years
   D.) 4 years

Answer: Option 'C'

Let Principal = P, Rate = R% per annum, Time = n years
When interest is compounded annually, total amount can be calculated by using the formula
Compound Amount = P ( 1 + R / 100)n
Given that, P = Rs.8000, R = 20% per annum
Compound Amount = Rs. 13824
We have to find the time period during which the amount will be Rs.13824
=> Rs.13824 = 8000 x (1 + 20/100)n
=> (13824 /8000) = (120 / 100)n
=> (24 / 20)3 = (12 / 10)n
=> (12 /10)3 = (12 /10 )n
Therefore, n = 3.
Hence the required time period is 3 years.

17.

What would be the compound interest accrued on an amount of 11500 Rs. at the end of 2 years at the rate of 10 % per annum ? 

   A.) 13905
   B.) 13915
   C.) 13925
   D.) 13965

Answer: Option 'B'

Given
Principal = 11500 No. of years = 2 Rate of interest = 10
Amount = P x (1+r/100)n
we get Amount = 11500 x (1+10/100)2
= 11500 x ( 110 / 100 )2
= 115 x 11 x 11
= 115 x 121
Ans : 13915

18.

Certain loan amount was repaid in two annual installments of Rs.1331/- each. If the rate of interest be 10% per annum Compounded annually the sum borrowed was?

   A.) Rs.121/-
   B.) Rs.2130/-
   C.) Rs.2310/-
   D.) Rs.1331/-

Answer: Option 'C'

Principal = (P.W of Rs. 1331/- due 1 year hence) + (P.W of Rs. 1331/- due 2 years hence) 
= [1331/(1 + 10/100) + 1331/(1 + 10/100)2
= [1331/(110/100) + 1331/(110/100 × 110/100)]
= 13310/11 + 133100/121 = 1210 + 1100 = Rs.2310/-

19.

Certain sum becomes 3 times it self at compound interest in 10 years. In how many years it becomes 9 times?

   A.) 25 years
   B.) 27 years
   C.) 30 years
   D.) 20 years

Answer: Option 'D'

P(1 + R/100)10 = 3P 
=> P(1 + R/100)10 = 3 
Let P(1 + R/100)n = 9P 
=> (1 + R/100)n = 9 
=> 32 = [(1 + R/100)10]2 
=> (1 + R/100)n => (1 + R/100)20 
=> n = 20 Years.

20.

Mr. Joshua invested Rs 15,000 divided into two different schemes A and B at S.I of 5% and 10%. If the total amount of the simple interest earned in 2 years is 2500, What was the amount invested in scheme B. 

   A.) 25,000
   B.) 10,000
   C.) 15,500
   D.) 30,000

Answer: Option 'B'

Given Total Principal = Rs. 15,000
Number of years = 2 years
Total S.I at the end of 2 years = Rs. 2500

For scheme A, Amount invested = x Rs.
Rate of interest, r = 5%

For scheme B, Amount invested= (15,000 - x) Rs.
Rate of interest, r = 10%

W.K.T: S.I = p * n * r / 100
=> S.I for scheme A + S.I for scheme B = Rs. 2500
=> {(x * 2 * 5)/ 100} + {(15,000 - x) * 2 * 10/ 100} = 2500
=> (x / 10) + 2(15,000 - x)/ 10 = 2500
=> (x/ 10) + (30,000 - 2x) /10 = 2500
=> x + 30,000 - 2x = 2500 * 10
=> 30,000 - x = 25000
=> x = 30,000 - 25,000
=> x= 5,000

For scheme B, Amount invested = (15,000 - x) Rs.
= 15,000 - 5,000
= 10,000 Rs.

21.

What would be the compound interest accrued on an amount of 6500 Rs. at the end of 2 years at the rate of 15 % per annum?

   A.) 2046.25
   B.) 2056.25
   C.) 2096.25
   D.) 2076.25

Answer: Option 'C'

Given principal = 6500
No. of years = 2
Rate of interest = 15
Amount = P x (1+r/100)n,
we get Amount = 6500 x (1+15/100)2 = 8596.25
C.I = Amount - Principal = 8596.25 - 6500 = 2096.25

22.

A person receives a sum of Rs. 210 as interest for investing some amount at 10% p.a compounding annually for 2 years. Find the amount invested at the beginning

   A.) 1050
   B.) 850
   C.) 1000
   D.) 950

Answer: Option 'C'

Given
Compound interest received by the person ( C.I ) = Rs. 210
Rate of interest ( r ) = 10 %
Number of years ( n ) = 2 years
To find, Amount invested at the beginning = principal ( p )
Compound interest ( C.I ) = Amount - Principal
Amount = p ( 1 + r / 100 )n

=> C.I = p [ ( 1 + r / 100 )n- 1 ]
=> 210 = p [ ( 1 + 10 / 100 )2- 1 ]
=> 210 = p [ ( 110 / 100 )2- 1 ]
=> 210 = p [ ( 11 / 10 )2- 1 ]
=> 210 = p [ ( 121 / 100 ) - 1 ]
=> 210 = p [ ( 121 - 100 ) / 100 ]
=> 210 = p [ 21 / 100 ]
=> 210 x ( 100 / 21 ) = p
=> 1000 = p
The amount invested at the beginning = p = Rs. 1000 

23.

Find the simple interest on Rs. 2000 at 7 % per annum for 4 years

   A.)  Rs. 485
   B.)  Rs. 450
   C.)  Rs. 560
   D.)  Rs. 760

Answer: Option 'C'

Given
Principal : 2000
Rate of interest : 7
Number of years : 4
Simple Interest = pnr / 100 
= ( 2000 x 4 x 7 ) / 100
Rs. 560 

24.

An amount at compound interest sums to Rs.17640/- in 2 years and to Rs.18522/- in 3 years at the same rate of interest. Find the rate percentage?

   A.) 5%
   B.) 6%
   C.) 4%
   D.) 10%

Answer: Option 'A'

The difference of two successive amounts must be the simple interest in 1 year on
the lower amount of money. 
S.I = 18522/- - 17640/- = Rs. 882/- 
Rate of interest = (882/17640) × (100/1) => 8820/1764 = 5% 
Principal = Amount/(1 + R/100)n 
= 17640/(1 + 5/100)2 
= 17640/(21/20 × 21/20) 
= 17640/(1.05 × 1.05) 
= 17640/1.1025 
= 16000

25.

What would be the compound interest accrued on an amount of 2500 Rs. at the end of 2 years at the rate of 10 % per annum?

   A.) 3025
   B.) 3035
   C.) 3125
   D.) 3045

Answer: Option 'A'

Given principal = 2500
No. of years = 2
Rate of interest = 10
Amount = P x (1+r/100)n,
we get Amount = 2500 x (1+10/100)2 Ans : 3025


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