Problems on Compound Interest are included with answers in this article. Compound interest questions and answers provided here are helpful for all the competitive exams. So, carefully solve all the practice questions, multiple-choice questions, and problems here check the answers to know your preparation. It is easy to score good marks if you know the tricks to solve the compound interest problems. To know the complete concept of compound interest, check out 10th Grade Math articles available on our website.
Also, See:
- Practice Test on Compound Interest
- Introduction to Compound Interest
- Compound Interest as Repeated Simple Interest
Compound Interest Problems with Solutions PDF | Maths Compound Interest Questions with Answers
Question 1. Calculate the compound interest on $ 75000 at 8 % per annum for 2 years which is compounded annually.
Solution:
Given that the compound interest on $ 75000 at 8 % per annum for 2 years.
Principal amount = $ 75000
Rate of Interest = 8 %
Time = 2 years
The formula to find the compound interest is Compound Interest = P(1 + \(\frac { r }{ 100 } \))n – P where P is the Principal, r is the rate of interest, and n is the time.
Substitute the given details in the above formula.
Compound Interest = $ 75000(1 + \(\frac { 8 }{ 100 } \))2 – $ 75000
CI = $ 75000{(\(\frac { 108 }{ 100 } \))2 – $1}
CI = $ 75000{(\(\frac { 27 }{ 25 } \))2 – 1}
Compound Interest = $ 75000 {\(\frac { 729 }{ 625 } \) – 1}
CI = $ 75000{\(\frac { 104 }{ 625 } \)} = $12480
Therefore, the Compound Interest is $ 12480.
Question 2. What sum invested for 3.5 years amounts to Rs. 668 in 3 ½ years compounded half-yearly at the rate of 2% p.a.?
Solution:
Given that Amount = Rs. 668
Rate of interest (r%) = 2% p.a. or 1% half yearly
Time n = 3 ½ years or 7 half years.
Amount = P{(1 + \(\frac { r }{ 100 } \))n}
Substitute the given details in the above formula.
Rs. 668 = P{(1 + \(\frac { 1 }{ 100 } \))7}
Rs. 668 = P{(\(\frac { 101 }{ 100 } \))7}
Rs. 668 = P{(1.01)7}
P = Rs. 1329.612
Hence, the principal amount is Rs. 623.0556
Therefore, the answer is Rs. 623.0556
Question 3. At what rate percent per annum will a sum of Rs 13689 amount to Rs 2500 in 2 years, compounded annually?
Solution:
Let the rate percent be r.
Given that at r rate percent per annum will a sum of Rs 13689 amount to Rs 2500 in 2 years.
Sum = Rs 13689
Amount = Rs 2500
Time = 2 years
The formula to find the rate percent is Amount A = P{(1 + \(\frac { r }{ 100 } \))n}where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
Rs 13689 = Rs 2500{(1 + \(\frac { r }{ 100 } \))2}
Rs 13689/Rs 2500 = {(1 + \(\frac { r }{ 100 } \))2}
(117/50)2 = (1 + \(\frac { r }{ 100 } \))2
117/50 = 1 + \(\frac { r }{ 100 } \)
117/50 – 1 = \(\frac { r }{ 100 } \)
67/50 = \(\frac { r }{ 100 } \)
1.34 = \(\frac { r }{ 100 } \)
r = 1.34 * 100 = 134%
At 134 rate percent per annum will a sum of Rs 13689 amount to Rs 2500 in 2 years, compounded annually.
Therefore, the answer is 134 %.
Question 4. Calculate the compound interest accrued on Rs.9,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 10%, and 15% respectively.
Solution:
Given that Principal = Rs. 9,000
r1 = 5%, r2 = 10%, and r3 = 15%.
Time n = 3 years
Amount = P{(1 + \(\frac { r1 }{ 100 } \))* (1 + \(\frac { r2 }{ 100 } \)) * (1 + \(\frac { r3 }{ 100 } \))}
Substitute the given details in the above formula.
Amount = 9000{(1 + \(\frac { 5 }{ 100 } \))* (1 + \(\frac { 10 }{ 100 } \)) * (1 + \(\frac { 15 }{ 100 } \))}
Amount = 9000{(\(\frac { 105 }{ 100 } \))* (\(\frac { 110 }{ 100 } \)) * (\(\frac { 115 }{ 100 } \))} = 11954.25
Compound Interest C.I. = Amount – Principal
Compound Interest C.I. = Rs. 11954.25 – Rs. 9000 = Rs. 2954.25
Therefore, the answer is Rs. 2954.25
Question 5. James deposits $ 10000 in a bank at 12% per annum compound interest for a certain time is $ 2544. Find the time period?
Solution:
Let the period be n years.
Given that Michael deposits $ 10000 in a bank at 12% per annum compound interest for a certain time is $ 2544.
Then, amount after n years = $(10000 + 2544) = $12544
Principal = $10000
Rate of Interest = 8%
Compound Interest = $ 2544.
The formula to find the time is the amount after n years = P{(1 + \(\frac { r }{ 100 } \))n} where P is the Principal, r is the rate of interest, and n is the time.
Substitute the given details in the above formula.
$ 12544 = $ 10000{(1 + \(\frac { 12 }{ 100 } \))n}
$ 12544/$ 10000 = {(1 + \(\frac { 12 }{ 100 } \))n}
(12544/10000) = (\(\frac { 112 }{ 100 } \))n
(112/100)2 = (\(\frac { 112 }{ 100 } \))n
n = 3 years.
Therefore, the time is 3 years.
Question 6. James borrowed Rs. 5,500 from Mic at 4 % per annum compound interest. After 2 years James gave Rs. 246 and an electronic device to clear his account. Find the cost of an electronic device.
Solution:
Given that James borrowed Rs. 5,500 from Mic at 4 % per annum compound interest. After 2 years James gave Rs. 246 and an electronic device to clear his account.
Sum = Rs. 5,500
Rate = 4%
Time = 2 years
Sum after 2 years = P{(1 + \(\frac { r }{ 100 } \))n} = Rs. 5,500 {(1 + \(\frac { 4 }{ 100 } \))2} = Rs. 5,500 * \(\frac { 104 }{ 100 } \) * \(\frac { 104 }{ 100 } \) = 5948.8
After 2 years Alex gave Rs. 246 and an electronic device to clear his account.
Cost of the electronic device = Rs. 5948.8 – Rs. 246 = 5702.8
Therefore, the Cost of the electronic device is about Rs. 5702.8
Question 7. What is the principle that amounts to $ 19652 in 3 years at \(\frac { 20 }{ 3 } \) % per annum compound interest, compounded annually?
Solution:
Given that the principal that amounts to $ 19652 in 3 years at \(\frac { 20 }{ 3 } \) % per annum compound interest.
Then, amount A = $ 19652
Time n = 3 years
Rate of Interest = \(\frac { 20 }{ 3 } \) %
The formula to find the compound interest is A = P{(1 + \(\frac { r }{ 100 } \))n} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
19652 = P{(1 + \(\frac { 20 }{ 300 } \))3}
19652 = P{(1 + \(\frac { 1 }{ 15 } \))3}
19652 = P{ (\(\frac { 16 }{ 15 } \))3}
19652 = P(\(\frac { 4096 }{ 3375 } \))
P = 16192.749.
Therefore, the principal amount is 16192.749.
Question 8. On a sum of $ 60000 for 2 years, if the difference between compound interest and simple interest is $ 186. Find the rate of interest percent per annum?
Solution:
Given that on a sum of $ 60000 for 2 years, if the difference between compound interest and simple interest is $ 186.
Then, compound interest (C.I.) = P{(1 + \(\frac { r }{ 100 } \))n – 1}
Simple Interest = Prt/100 = (60000 * r * 2)/100 = 1200r
C.I. = 60000{(1 + \(\frac { r }{ 100 } \))2 – 1} = 60000{1 + \(\frac { r2 }{ 1000 } \) + \(\frac { 2r }{ 100 } \) – 1} = 60r2 + 1200r
C.I. – S.I. = 186
60r2 + 1200r – 1200r = 186
60r2 = 186
r = 1.7606
Therefore, the rate of interest percent per annum is about 1.7606%.
Question 9. If the simple interest on a sum of money at 4% per annum for 2 years is $ 1200, then find the compound interest on the same sum for the same period at the same rate.
Solution:
Given that the simple interest on a sum of money at 4% per annum for 2 years is $ 1200.
Then, Simple Interest (S.I.) = $ 1200
Time n = 2 years
Rate of Interest = 4 %
The formula to find the compound interest on the same sum for the same period at the same rate will be is A = P{(1 + \(\frac { r }{ 100 } \))n} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Principal = \(\frac { 100 * 1200}{ 2 * 4 } \) = 15000
Substitute the given details in the above formula.
Amount = 15000 {(1 + \(\frac { 4 }{ 100 } \))2}
Amount = 15000 {(\(\frac { 104 }{ 100 } \))2}
A = 15000 {(\(\frac { 26 }{ 25 } \))2}
A = 15000 { (\(\frac { 676 }{ 625 } \))}
Amount = 16224.
C.I. = 16224 – 15000 = $1224
Therefore, the answer is $1224.
Question 10. A certain sum of money gives $ 2040 as compound interest 3¹/₂ % per annum for 2 years. Find the simple interest on the same sum of money at the same rate for the same period of time.
Solution:
Given that a certain sum of money gives $ 2040 as compound interest 3¹/₂ % per annum for 2 years.
Then, compound interest (C.I.) = $ 2040
Time n = 2 years
Rate of Interest = 3¹/₂ %
The formula to find the principal is CI = P{(1 + \(\frac { r }{ 100 } \))n} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
Compound interest = P{(1 + \(\frac { r }{ 100 } \))n} – P
2040 = P{(1 + \(\frac { 7 }{ 200 } \))2} – P
2040 = P{(\(\frac { 207 }{ 200 } \))2} – P
2040 = P{\(\frac { 42849 }{ 40000 } \)} – P
2040 = \(\frac { 2849P }{ 40000 } \)
P = 28641.6286
Simple Interest = (Principal * rate * time)/100 = 2004.914
Therefore, the Simple Interest is about 2004.914.
Question 11. Find the compound interest on Rs 24,000 for one year at 4% per annum when compounded half-yearly.
Solution:
Given that Principal Amount P = $ 24000
Time n = 1 year
Rate of Interest = 4%
The formula to find the principal is CI = P{(1 + \(\frac { r }{ 100 } \))n} where P is the Principal, r is the rate of interest, A is the amount, and n is the time.
Substitute the given details in the above formula.
Compound interest = P{(1 + \(\frac { r }{ 100 } \))n} – P
Compound interest = 24000 {(1 + \(\frac { 4 }{ 200 } \))2 – 1}
Compound interest = 24000 {(\(\frac { 204 }{ 200 } \))2 – 1}
C.I. = 24000{1.0404 – 1}
C.I. = 969.6
Therefore, the Compound Interest is 969.6.