Compound Interest Problems  with Solution, Question and Answers pdf
Compound Interest Problems
Compound Interest Shortcuts and
Tricks are used in aptitude exams to solve problems based on Compound Interest.
So, it becomes very necessary to know basics of CI and know who to solve Compound Interest Problems with
shortcuts. In this page Compound interest problems are presented along with
detailed solutions to help the individual so, that they can easily understand
it. To begin with Compound Interest Question and Answers you must know Compound
Interest Formula’s which are provided below.
There are several types of
interest problems but in this article we will deals with solving problems on
compound Interest. When, the interest is calculated more than once in a year,
then it is called as “Compound Interest”. Go through the below section
of page, that is prepared by the expert team members of www.privatejobshub.in
through which you can learn how to solve Compound Interest Problems. Let’s go
ahead and get to know how you can solve problems on compound interest in an
easy way.
Compound Interest Formula
Simple Interest Formula:

Simple
Interest = Principal * Time * Rate of interest / 100
Abbreviated as SI = PTR/100
Sometimes, the interest is also
calculated halfyearly or quarterly.

When compounded semiannually or halfyearly:

Amount
= P [1 + (R/2)/100]2t

When compounded quarterly:

Amount
= P[1 + (R/4)/100]4t
Present worth of Principal P due t
years hence is given by:
P/ [1+ R/100] t

Compound Interest
Question and Answers
Question 1: The simple interest on a sum of money for 3 years at
6²/₃
% per annum is $ 6750. What will be the compound interest on the same sum at
the same rate for the same period, compounded annually?
Solution:
Given, SI = $ 6750, R = 203203% p.a. and T = 3 years.
Given, SI = $ 6750, R = 203203% p.a. and T = 3 years.
Sum = 100 × SI / R × T
= $ (100 × 6750 × ³/₂₀ × 1/3) = $ 33750.
Now, P = $ 33750, R
= 203203% p.a. and T = 3 years.
Therefore, amount after 3
years
= $ {33750 × (1 + (20/3 × 100)}³ [using
A = P (1 + R/100)ᵀ]
= $ (33750 × 16/15 × 16/15 × 16/15)
= $ 40960.
Thus, amount = $ 40960.
Hence, compound interest = $
(40960  33750) = $ 7210.
Question 2: The difference between the compound interest,
compounded annually and the simple interest on a certain sum for 2 years at 6%
per annum is $ 18. Find the sum.
Solution:
Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)²  100}
= $ {(100 × 53/50 × 53/50)  100} = $ (2809/25  100) = $ 309/25
Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)²  100}
= $ {(100 × 53/50 × 53/50)  100} = $ (2809/25  100) = $ 309/25
Therefore, (CI)  (SI) = $ (309/25 – 100) = $ 9/25
If the difference between the CI and SI is $ 9/25, then the sum = $ 100.
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18)
= $ 5000.
Hence, the required sum is $ 5000.
If the difference between the CI and SI is $ 9/25, then the sum = $ 100.
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18)
= $ 5000.
Hence, the required sum is $ 5000.
Alternative method
Let the sum be $ P.
Then, SI = $ (P × 6/100 × 2) = $ 3P/25
And, CI = $ {P × (1 + 6/100)²  P}
= $ {(P × 53/50 × 53/50)  P} = $ (2809250028092500P  P) = $ (309P/2500)
Let the sum be $ P.
Then, SI = $ (P × 6/100 × 2) = $ 3P/25
And, CI = $ {P × (1 + 6/100)²  P}
= $ {(P × 53/50 × 53/50)  P} = $ (2809250028092500P  P) = $ (309P/2500)
(CI)  (SI) = $ (309P/2500 – 3P/25) = $ (9P/2500)
Therefore, 9P/2500 = 18
⇔ P = 2500 × 18/9
⇔ P = 5000.
Hence, the required sum is $ 5000.
⇔ P = 5000.
Hence, the required sum is $ 5000.
Question 3: A man deposited $1000 in a bank. In return he got
$1331. Bank gave interest 10% per annum. How long did he keep the money in the
bank?
Solution:
Let the required time be n years.
Then,
Amount = $ {1000 × (1 + 10/100)ⁿ}
= $ {1000 × (11/10)ⁿ}
Therefore, 1000 × (11/10)ⁿ =
1331 [since, amount = $ 1331 (given)]
⇒ (11/10)ⁿ = 1331/1000 = 11 × 11 ×
11/ 10 × 10 × 10 = (11/10)³
⇒ (11/10)ⁿ = (11/10)³
⇒ n = 3.
Thus, n = 3.
Hence, the required time is 3
years.
Question 4: Find the compound interest on $10,000 if Ron took loan
from a bank for 1 year at 8 % per annum, compounded quarterly
Solution:
Here, P = principal amount (the
initial amount) = $ 10,000
Rate of interest (r) = 8 % per
annum
Number of years the amount is
deposited or borrowed for (n) = 1 year
Using the compound interest when
interest is compounded quarterly formula, we have that
A = P(1 + r4100r4100)4n4n
= $ 10,000 (1
+ 8410084100)4∙14∙1
= $ 10,000 (1
+ 21002100)44
= $ 10,000 (1
+ 150150)44
= $ 10,000 ×
(51505150)44
= $ 10,000
× 51505150 × 51505150 × 51505150 × 51505150
= $ 10824.3216
= $ 10824.32
(Approx.)
Therefore, compound interest $
(10824.32  $ 10,000) = $ 824.32
Question 5: A certain sum amounts to $ 72900 in 2 years at 8% per
annum compound interest, compounded annually. Find the sum.
Solution:
Let the sum be $ 100. Then,
amount = $ {100 × (1 + 8/100)²}
= $ (100 × 27/25 × 27/25) = $ (2916/25)
If the amount is $ 2916/25 then the sum = $ 100.
If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500.
Hence, the required sum is $ 62500.
Let the sum be $ 100. Then,
amount = $ {100 × (1 + 8/100)²}
= $ (100 × 27/25 × 27/25) = $ (2916/25)
If the amount is $ 2916/25 then the sum = $ 100.
If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500.
Hence, the required sum is $ 62500.
Alternative method
Let the sum be $ P. Then,
amount = $ {P × (1 + 8/100)²}
= $ {P × 27/25 × 27/25} = $ (729P/625)
Let the sum be $ P. Then,
amount = $ {P × (1 + 8/100)²}
= $ {P × 27/25 × 27/25} = $ (729P/625)
Therefore, 729P/625 = 72900
⇔ P = (72900 × 625)/729
⇔ P = 62500.
⇔ P = (72900 × 625)/729
⇔ P = 62500.
Hence, the required sum is $ 62500.
Take
a Test Now

Question 6: Find the amount and the compound interest on $ 7,500 in
2 years and at 6% compounded yearly.
Solution:
Here,
Principal (P) = $ 7,500
Number of years (n) = 2
Rate of interest compounded
yearly (r) = 6%
A = P(1 + r100r100)nn
= $ 7,500(1
+ 61006100)22
= $ 7,500 ×
(106100106100)22
= $ 7,500
× 11236100001123610000
= $ 8,427
Therefore, the required amount =
$ 8,427 and
Compound interest = Amount 
Principal
= $ 8,427  $ 7,500
= $ 927
Question 7: Find the rate of compound interest for $ 10,000 amounts
to $ 11,000 in two years.
Solution:
Let the rate of compound interest
be r% per annum.
Principal (P) = $ 10,000
Amount (A) = $ 11,000
Number of years (n) = 2
Therefore,
A = P(1 + r100r100)nn
⟹ 10000(1
+ r100r100)22 = 11664
⟹ (1
+ r100r100)22 = 11664100001166410000
⟹ (1
+ r100r100)22 = 729625729625
⟹ (1 + r100r100)22 =
(27252725)
⟹ 1
+ r100r100 = 27252725
⟹ r100r100 = 27252725 
1
⟹ r100r100 = 225225
⟹ 25r = 200
⟹ r = 8
Therefore, the required rate of
compound interest is 8 % per annum.
Question 8: A sum of money becomes $ 2,704 in 2 years at a compound
interest rate 4% per annum. Find
(i) The sum of money at the
beginning
(ii) The interest generated.
Solution:
Let the sum of money at the
beginning = $ P
Here,
Amount (A) = $ 2,704
Rate of interest compounded
yearly (r) = 4
Number of years (n) = 2
(i) A = P(1 + r100r100)nn
⟹ 2,704 = P(1
+ 41004100)22
⟹ 2,704 = P(1 + 125125)22
⟹ 2,704 = P(26252625)22
⟹ 2,704 = P × 676625676625
⟹ P = 2,704
× 625676625676
⟹ P = 2,500
Therefore, the sum of money at
the beginning was $ 2,500
(ii) The interest generated =
Amount – Principal
= $2,704  $2,500
= $ 204
Question 9: Find the amount and the compound interest on $ 4,000
is 11212 years at 10 % per annum compounded halfyearly.
Solution:
Here, the interest is compounded
halfyearly. So,
Principal (P) = $ 4,000
Number of years (n) =
11212 × 2 = 3232 × 2 = 3
Rate of interest compounded
halfyearly (r) = 102102% = 5%
Now, A = P (1 + r100r100)ⁿ
⟹ A = $ 4,000(1
+ 51005100³
⟹ A = $ 4,000(1 + 120120)³
⟹ A = $
4,000 × (21202120)³
⟹ A = $
4,000 × 9261800092618000
⟹ A = $ 4,630.50 and
Compound interest = Amount 
Principal
= $ 4,630.50  $ 4,000
= $ 630.50
Therefore, the amount is $
4,630.50 and the compound interest is $ 630.50
Question 10: Find the amount and the compound interest on $ 8,000
at 10 % per annum for 11212 years if the interest is compounded
halfyearly.
Solution:
Here, the interest is compounded
halfyearly. So,
Principal (P) = $ 8,000
Number of years (n) =
11212 × 2 = 3232 × 2 = 3
Rate of interest compounded
halfyearly (r) = 102102% = 5%
Now, A = P (1 + r100r100)ⁿ
⟹ A = $ 8,000(1 + 51005100)³
⟹ A = $ 8,000(1 + 120120)³
⟹ A = $ 8,000 × (21202120)³
⟹ A = $
8,000 × 9261800092618000
⟹ A = $ 9,261 and
Compound interest = Amount 
Principal
= $ 9,261  $ 8,000
= $ 1,261
Therefore, the amount is $ 9,261
and the compound interest is $ 1,261
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