RRB NTPC - Compound Interest : Aptitude Test (65 Questions with Explanation)

1.

What would be the compound interest on Rs.7700/- at 15 ¼% per annum for 2 years compounded annually

   A.) Rs.2725.75/-
   B.) Rs.2527.57/-
   C.) Rs.2227.57/-
   D.) Rs.2520.57/-

Answer: Option 'B'

Principal = Rs. 7700/- 
time = 2 years 
rate = 15 ¼% 
Amount = P(1+R/100)n 
= 7700 × (1 + 61/(4 × 100)2) 
= 7700 × [(1 + 61/400)2] 
= 7700 × [(461/400)2)] 
= 7700 × 461/400 × 461/400 
= 7700 × 1.1525 × 1.1525 
= 7700 × 1.32825625 
= 10227.573125 
C.I = 10227.573125 - 7700 = 2527.57/-

2.

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.

3.

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

4.

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/-

5.

The difference between the compound interest and simple interest on a certain sum of money at 5% per annum for 2 years is 45. Then the original sum is? 

   A.) Rs.16000/-
   B.) Rs.15000/-
   C.) Rs.18000/-
   D.) Rs.20000/-

Answer: Option 'C'

For 2 years = (1002D)/R2
= (1002 × 45)/(5 × 5) = (10000 × 45)/25 = Rs.18000/-

6.

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/-

7.

The present worth of Rs.9826/- due 2 years at 6 ¼% per annum compound interest is:

   A.) Rs.7804/-
   B.) Rs.8704/-
   C.) Rs.8760/-
   D.) Rs.8504/-

Answer: Option 'B'

Rs.8704/-

8.

The Compound interest of Rs.10240/- at 6 ¼% per annum for 2 years 292days is:

   A.) Rs.1898/-
   B.) Rs.1798/-
   C.) Rs.1688/-
   D.) Rs.1698/-

Answer: Option 'A'

Rs.1898/-

9.

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/-

10.

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.

11.

What will be the amount if sum of Rs.10,00,000 is invested at compound interest for 3 years with rate of interest 11%, 12% and 13% respectively?

   A.) Rs.14,04,816
   B.) Rs.16,00,816
   C.) Rs 12,14,816
   D.) Rs. 11, 13,816

Answer: Option 'A'

Given
Here, P = Rs.10,00,000, R1 = 11 , R2 = 12, R3 = 13.
Each rate of interest is calculated for one year.
Hence, N = 1 year.
Amount after 3 years,
= P(1 + R1/100) (1 + R2/100) (1 + R3/100)
= 10,00,000 × (1 + 11/100) × (1 + 12/100) × (1 + 13/100)
= 10,00,000 × (111/100) × (112/100) × (113/100)
= 111 x 112 x 113
= 14,04,816
Hence the total amount after 3 years is Rs.14,04,816

12.

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

13.

A sum of money is invested at 10% per annum compounding annually for 2 years. If the interest received is Rs. 210, find the principal.

   A.) 2500
   B.) 1000
   C.) 1900
   D.) 2100

Answer: Option 'B'

Given, r = 10%
n = 2 years
Compound interest, C.I = Rs. 210
Compound Interest = Amount – Principal
=> C.I = P {[1 + (r/100)]n - 1}
=> 210 = P {[1 + (10/100)]2 - 1}
=> 210 = P {[1 + (1/10)]2 - 1}
=> 210 = P {[(10 + 1)/10]2 - 1}
=> 210 = P {[11/10]^2 - 1}
=> 210 = P {[121 / 100] – 1}
=> 210 = P {(121 – 100) / 100}
=> 210 = P {21 / 100}
=> P = (210 × 100) / 21
=> P = 1000 Rs.
Thus, Principal = Rs. 1000

14.

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

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

15.

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

16.

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.

17.

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

18.

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

19.

The compound interest on a certain sum for 2 years at 10% p.a. is Rs.525.The simple interest on the same sum for double the time at the half the rate percent per annum is:  

   A.) Rs.600
   B.) Rs.500
   C.) Rs.400
   D.) Rs.800

Answer: Option 'B'

Given, n = 2 years
r = 10 %
Compound interest (C.I) = Rs. 525
Compound interest (C.I) = P {[1 + (r/100)]n – 1}
=> P {[1 + (10/100)]2– 1} = 525
=> P {[1 + (1/10)]2 – 1} = 525
=> P {[11/10]2 – 1} = 525
=> P {[121/100] – 1} = 525
=> 21 × P / 100 = 525
=> P = 2500
Now for Simple Interest, S.I = p × n × r / 100
P = Rs. 2500
n = 4 years
r = 5%
S.I = (2500 × 4 × 5) / 100
=> S.I = 500 Rs.

20.

What will be compounded interest on a sum of Rs.25,000 after 3 years at the rate of 12 p.c.p.a.?

   A.) Rs.9000.30
   B.) Rs.10123.20
   C.) Rs.9720
   D.) Rs.9820

Answer: Option 'B'

Rs.10123.20


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