1. 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?
SHOW ANSWER
Correct Ans:3 years
Explanation:
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.
2. 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.
SHOW ANSWER
Correct Ans:Rs. 8,000
Explanation:
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
3. 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?
SHOW ANSWER
Correct Ans:Rs.14,04,816
Explanation:
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
4. 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.
SHOW ANSWER
Correct Ans:1000
Explanation:
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
5. 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?
SHOW ANSWER
Correct Ans:4137.5
Explanation:
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
6. 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.
SHOW ANSWER
Correct Ans:22472
Explanation:
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
7. 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 ?
SHOW ANSWER
Correct Ans:10816
Explanation:
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
8. Find the simple interest on Rs. 2000 at 7 % per annum for 4 years
SHOW ANSWER
Correct Ans:560 Rs.
Explanation:
Solution is :
Given
Principal : 2000
Rate of interest : 7
Number of years : 4
Simple Interest = pnr / 100
= ( 2000 x 4 x 7 ) / 100
=560 Rs
9. 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?
SHOW ANSWER
Correct Ans:816
Explanation:
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
10. 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
SHOW ANSWER
Correct Ans:10000
Explanation:
Solution is :
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
11. 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 ?
SHOW ANSWER
Correct Ans:2096.25
Explanation:
Solution is :
Given principal = 6500
No. of years = 2
Rate of interest = 15
Amount = P [ 1 + ( r / 100 )^{n}]
= 6500 x [ 1 + ( 15 /100 )^{2}]
= 6500 x [ 1 + ( 3 / 20 )^{2}]
= 6500 x [ 23 / 20 ]^{2}
= 6500 x [ 529 / 400 ]
Amount =8596.25
Compound Interest = Amount - Principal
= 8596.25 - 6500
= 2096.25
12. What would be the compound interest accrued on an amount of 4500 Rs. at the end of 2 years at the rate of 10 % per annum ?
SHOW ANSWER
Correct Ans:945
Explanation:
Solution is :
Given principal = 4500
No. of years = 2
Rate of interest = 10
Amount = P [ 1 + ( r / 100 ) ]^{n}
= 4500 x [ 1 + ( 10 / 100 ) ]^{2
}= 4500 x [ 1 + ( 1 / 10 ) ]^{2
}= 4500 x [ 11 / 10 ]^{2}
= 4500 x [ 121 / 100 ]
Amount = 5445
Compound interest = Amount - principal
= 5445 - 4500
= 945 Rs
13. 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.
SHOW ANSWER
Correct Ans:10,000
Explanation:
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.
14. What would be the compound interest accrued on an amount of 8000 Rs. at the end of 3 years at the rate of 10 % per annum ?
SHOW ANSWER
Correct Ans:2648
Explanation:
Solution is
Given
principal = 8000
No. of years = 3
Rate of interest = 10
Amount = P x ( 1 + ( r / 100 ) )^{n}
= 8000 x ( 1 + ( 1 / 10 ) )^{3}
= 8000 x ( 11 / 10 )^{3}
= 8000 x ( 1331 / 1000 )
= 8 x 1331
Amount = 10648
Compound Interest = Amount - Principal
= 10648 - 8000
= 2648 Rs
&nb
15. Find the simple interest on Rs. 1920 at 45 % per annum for 3 months
SHOW ANSWER
Correct Ans:Rs. 216
Explanation:
Solution is:
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
16. A person receives a sum of Rs. 420 as interest for investing some amount at 10% p.a compounding annually for 2 years. Find the amount invested at the beginning
SHOW ANSWER
Correct Ans:2000
Explanation:
Solution is
Given compound interest ( C.I ) = Rs.420
Rate of interest ( r ) = 10 %
Number of years ( n ) = 2
To find , amount invested at the beginning i.e principal (P)
Amount = P [ 1 + ( r / 100 ) ]^{n}
Amount = C.I + P
=> Amount = C.I + P
=> P [ 1 + ( r / 100 ) ]^{n}= C.I + P
=> P [ 1 + ( 10 / 100 ) ]^{2}= 420 + P
=> P [ 110 / 100 ]^{2}= 420 + P
=> P [ 11 / 10 ]^{2}= 420 + P
=> P x 121 / 100 = 420 + P
=> ( 121 P / 100 ) - P = 420
=> ( 121 P - 100 P ) / 100 = 420
=> 21 P / 100 = 420
=> P = ( 420 * 100 ) / 21
= 2000
Amount invested at the beginning, P = 2000 Rs
17. Find the simple interest on Rs. 1300 at 10 % per annum for 5 years
SHOW ANSWER
Correct Ans:Rs. 650
Explanation:
Solution is
Given
Principal : 1300
Rate of interest : 10
Number of years : 5
Simple Interest = pnr / 100
Simple Interest = (1300 x 5 x 10) / 100
= 13 x 5 x 10
= 650
So,Simple Interest = 650
18. 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
SHOW ANSWER
Correct Ans:1000
Explanation:
Solution is
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 = 1000 Rs
19. Find the simple interest on Rs. 2000 at 7 % per annum for 4 years
SHOW ANSWER
Correct Ans:560 Rs.
Explanation:
Given
Principal = 2000
Rate of interest = 7
Number of years = 4
Simple Interest = pnr / 100
Simple Interest
= (2000 x 4 x 7) / 100
= 5600 / 100
= 560.
Answer is Rs 560
20. 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 ?
SHOW ANSWER
Correct Ans:4137.5
Explanation:
Solution is
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 ( 110 / 100 )^{3
}= 12500 x ( 11 / 10 )^{3}
= 12500 x ( 1331 / 1000 )
= 125 x ( 1331 / 10 )
Amount = 16637.5
Compound interest = Amount - Principal
= 16637.5 - 12500
&nbs
Are you seeking for good platform for practicing Compound Interest questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.
Online Test on Compound Interest @ Fresherslive
This page provides important questions on Compound Interest along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Compound Interest questions are framed with Latest concepts, so that you may get updated through these Compound Interest Online tests. Compound Interest Online Test questions are granted from basic level to complex level.
Why To Practice Compound Interest Test questions Online @ Fresherslive?
Compound Interest questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Compound Interest may improve your performance in the real Exams and Interview.
Time Management for answering the Compound Interest questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Compound Interest questions and answers, you can acquire all the essential idea to solve any difficult questions on Compound Interest in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.
Why Fresherslive For Compound Interest Online Test Preparation?
Most of the job seekers finding it hard to clear Compound Interest test or get stuck on any particular question, our Compound Interest test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Compound Interest, exercise these advanced Compound Interest questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Compound Interest Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.