Compound Interest : Aptitude Test (65 Questions with Explanation)

46.

Rs. 10000 is borrowed at compound interest at the rate of 4 % per annum. What will be the amount to be paid after 2 years ?

   A.) 10816
   B.) 10808
   C.) 10800
   D.) 10826

Answer: Option 'A'

Principal : P = 10000 Rs.
Rate of Interest : r = 4 %
Number of years : n = 2
Amount = P x (1 + r/100)n
Amount = 10000 x (1+4/100)2
=10000 x (1+1/25)2 =10000 x (26/25) x (26/25) =10816

47.

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

48.

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/-

49.

Find the simple interest on Rs. 1920 at 45 % per annum for 3 months

   A.) Rs. 196
   B.) Rs. 216
   C.) Rs. 206
   D.) Rs. 306

Answer: Option 'B'

Given
Principal : 1920
Rate of interest : 45
Number of months : 3
Simple interest for 1 year = pnr / 100
= ( 1920 x 1 x 45 ) / 100
= 864
Simple interest for 3 months = ( 3 / 12 ) x SI for 1 year
= ( 3 / 12 ) x 864
216

50.

A person receives a sum of Rs. 2100 as interest for investing some amount at 10% p.a compounding annually for 2 years. Find the amount invested at the beginning

   A.) 10000
   B.) 9000
   C.) 10500
   D.) 9500

Answer: Option 'A'

Given Compound Interest = Rs.2100
Rate of Interest ( r ) = 10 % p.a
No.of years ( n ) = 2
To find , amount received at the beginning => principal
Compound Interest = P [ 1 + ( r / 100 )n- 1 ]
=> 2100 = P[ 1 + ( 10 / 100 )2- 1 ]
=> 2100 = P[ 1 + ( 1 / 10 )2- 1 ]
=> 2100 = P[ ( 11 / 10 )2- 1 ]
=> 2100 = P[ ( 121 / 100 ) - 1 ]
=> 2100 = P[ 21 / 100 ]
=> 2100 x ( 100 / 21 ) = P
Principal = Rs. 10000
Amount invested at the beginning = Rs. 10000

51.

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.) 4137.5
   B.) 4537.5
   C.) 4237.5
   D.) 4337.5

Answer: Option 'A'

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

52.

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. 7,000
   B.) Rs. 8,000
   C.) Rs. 5,000
   D.) Rs. 11,000

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

53.

What would be the compound interest accrued on an amount of 10000 Rs. at the end of 2 years at the rate of 4 % per annum?

   A.) 816
   B.) 10846
   C.) 10816
   D.) 10916

Answer: Option 'A'

Given principal = 10000
No. of years = 2
Rate of interest = 4
Amount = P [ 1 + ( r / 100 )n]
= 10000 x [ 1 +( 4 / 100 )2]
= 10000 x ( 104 / 100 )2
= 10000 x ( 104 / 100 ) x ( 104 / 100 )
= 104 x 104
= 10816
Compound Interest = Amount - Principal

= 10816 - 10000
= 816

54.

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.) 525
   B.) 575
   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,
=> Amount = 2500 x (1+10/100)2
= 2500 (11/ 10)2
= 2500 (121/ 100)
= 25 × 121
= 3025
So, Compound Amount = 3025
Compound Interest = Compound Amount - Principal

=> C.I = 3025 - 2500
=> C.I = 525 Rs.

55.

A sum of money is borrowed and paid back in two annual instalments of Rs.882 each allowing 5% compound interest .The sum borrowed was:

   A.) Rs.1640
   B.) Rs.1620
   C.) Rs.1680
   D.) Rs.1700

Answer: Option 'B'

Given
The sum borrowed
Present Worth of Rs.882 due 1 year + Present Worth of Rs.882 due 2 year
=> ( 882 ) / 1 + ( 5 / 100)1 + ( 882) / 1 + ( 5 / 100)1
=> (882 / 105 × 100 )1 + (882 / 105 × 100 )1
=> ( 882 /( 21 / 20 ) + ( 882 / (21 / 20)1
=> ( 882 × 20) / (21) + ( 882 × 20 × 20 / 21 × 21 )
=> 42 × 20 + 42 × 20 × 20 / 21
=> 840 + 2 × 20 × 20
=> 840 + 800
=> 1640
The sum borrowed = Rs.1640

56.

The population of a town is 196000. It increases by 7% in the 1st year and decreases by 5% in the 2nd year. What is the population of the town at the end of 2 years?

   A.) 199234
   B.) 201234
   C.) 200234
   D.) 189234

Answer: Option 'A'

Initial population is 196000
In First Year, population increasesby 7%
New population 
= (107/100) x 196000
= 107 × 196000 / 100
= 107 × 1960
209720
Population after 1 year = 209720
In second year, Population decreases by 5%,
New population
= (100 - 5)/100 x 209720
= (95/100) x 209720
= 19 * 10486
199234
Population after 2 years will be 199234.

57.

The Compound interest in a particular amount for the first year at 8% is Rs.50/-.The compound interest for 2 years at the same rate on the amount will be?

   A.) Rs.52/-
   B.) Rs.104/-
   C.) Rs.102/-
   D.) Rs.54/-

Answer: Option 'B'

Rs.104/-

58.

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

59.

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

60.

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.

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