Compound Interest Questions | Compound Interest Questions with Solutions (2024)

Compound interest questions are provided here to help the students understand the applications of compound interest in our daily existence. As we know, compound interest is one of the important mathematical concepts that can be applied in many financial transactions.

Below are some situations where we can use the formula of CI to calculate the required results.

  • Increase or decrease in population
  • The growth of a bacteria (when the rate of growth is known)
  • The value of an item, if its price increases or decreases in the intermediate years

What is Compound Interest?

Interest is the additional money paid by organisations like banks or post offices on money deposited (kept) with them. Interest is also paid by people when they borrow money. When the interest is calculated on the previous year’s amount, the interest is called compounded or Compound Interest (C.I.).

The formula for finding the amount on compound interest is given by:

A = P[1 +(R/100)]n

This is the amount when interest is compounded annually.

Compound interest (CI) = A – P

Read more: Compound interest

Compound Interest Questions and Answers

1. Find the compound interest (CI) on Rs. 12,600 for 2 years at 10% per annum compounded annually.

Solution:

Given,

Principal (P) = Rs. 12,600

Rate (R) = 10

Number of years (n) = 2

A = P[1 +(R/100)]n

= 12600[1 + (10/100)]2

= 12600[1 + (1/10)]2

= 12600 [(10 + 1)/10]2

= 12600 × (11/10) × (11/10)

= 126 × 121

= 15246

Total amount, A = Rs. 15,246

Compound interest (CI) = A – P

= Rs. 15,246 – Rs. 12,600

= Rs. 2646

2. At what rate of compound interest per annum, a sum of Rs. 1200 becomes Rs. 1348.32 in 2 years?

Solution:

Let R% be the rate of interest per annum.

Given,

Principal (P) = Rs. 1200

Total amount after 2 years (A) = Rs. 1348.32

n = 2

We know that,

A = P[1 + (R/100)]n

Rs. 1348.32 = Rs. 1200[1 + (R/100)]2

1348.32/1200 = [1 + (R/100)]2

[1 + (R/100)]2 = 134832/120000[1 + (R/100)]2 = 2809/2500[1 + (R/100)]2 = (53/50)2

1 + (R/100) = 53/50

R/100 = (53/50) – 1

R/100 = (53 – 50)/50

R = 300/50

R = 6

Hence, the rate of interest is 6%.

Amount, when interest is compounded half-yearly, is

A = P[1 + (R/200)]2n

Here,

R/200 = half-yearly rate

2n = the number of half years

3. A TV was bought for Rs. 21,000. The value of the TV was depreciated by 5% per annum. Find the value of the TV after 3 years. (Depreciation means the reduction of value due to use and age of the item)

Solution:

Principal (P) = Rs. 21,000

Rate of depreciation (R) = 5%

n = 3

Using the formula of CI for depreciation,

A = P[1 – (R/100)]n

A = Rs. 21,000[1 (5/100)]3

= Rs. 21,000[1 – (1/20)]3

= Rs. 21,000[(20 – 1)/20]3

= Rs. 21,000 × (19/20) × (19/20) × (19/20)

= Rs. 18,004.875

Therefore, the value of the TV after 3 years = Rs. 18,004.875.

4. Find the compound interest on Rs 48,000 for one year at 8% per annum when compounded half-yearly.

Solution:

Given,

Principal (P) = Rs 48,000

Rate (R) = 8% p.a.

Time (n) = 1 year

Also, the interest is compounded half-yearly.

So, A = P[1 + (R/200)]2n

= Rs. 48000[1 + (8/200)]2(1)

= Rs. 48000[1 + (1/25)]2

= Rs. 48000[(25 + 1)/25]2

= Rs. 48,000 × (26/25) × (26/25)

= Rs. 76.8 × 26 × 26

= Rs 51,916.80

Therefore, the compound interest = A – P

= Rs (519,16.80 – 48,000)

= Rs 3,916.80

5. Find the compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

Solution:

Given,

Principal (P) = Rs. 8000

Rate of interest (R) = 15% p.a

Time (n) = 2 years 4 months

4 months = 4/12 years = 1/3 years

So,

A = P[1 + (R/100)]n

= Rs. 8000 [1 + (15/100)]2 [1 + (1/3) × (15/100)]

= Rs. 8000 [1 + (3/20)]2 [1 + (3/20 × 3)]

= Rs. 8000 [(20 + 3)/20]2 [(20 + 1)/20]

= Rs. 8000 × (23/20) × (23/20) × (21/20)

= Rs. 11,109

Therefore, the compound interest = A – P = Rs. 11,109 – Rs. 8000 = Rs. 3109

6. If principal = Rs 1,00,000. rate of interest = 10% compounded half-yearly. Find

(i) Interest for 6 months.

(ii) Amount after 6 months.

(iii) Interest for the next 6 months.

(iv) Amount after one year.

Solution:

Given,

P = Rs 1,00,000

R = 10%

(i) A = P[1 + (R/200)]2n

Here, 2n is the number of half years.

Let us find the interest compounded half-yearly for 6 months, i.e., one half year.

So, A = Rs. 1,00,000 [1 + (10/200)]1

= Rs. 1,00,000 [(20 + 1)/20]

= Rs. 1,00,000 × 21/20

= Rs. 1,05,000

Compounded interest for 6 months = Rs. 1,05,000 – Rs. 1,00,000 = Rs. 5000

(ii) Amount after 6 months = Rs. 1,05,000

(iii) To find the interest for the next 6 months, we should consider the principal amount as Rs. 1,05,000.

Thus, A = Rs. 1,05,000 [1 + (10/200)]1

= Rs. 1,05,000 × (21/20)

= Rs. 1,10,250

Compound interest for next 6 months = Rs. 1,10,250 – Rs. 1,05,000 = Rs. 5250

(iv) Amount after one year = Rs. 1,10,250

7. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(i) Find the population in 2001.

(ii) What would be its population in 2005?

Solution:

(i) Let P be the population in the year 2001.

Thus, population in the year 2003 = A = 54000 (given)

R = 5%

Also, n = 2

A = P[1 + (R/100)]n

54000 = P[1 + (5/100)]2

54000 = P[1 + (1/20)]2

54000 = P × [(20 + 1)/20]2

54000 = P × (21/20) × (21/20)

P = 54000 × (20/21) × (20/21)

P = 48979.6

The population in 2001 = 48980 (approx.)

(ii) Given that the population in the year 2003 = P = 54000

R = 5%

n = 2

A = P[1 + (R/100)]n

= 54000[1 + (5/100)]2

= 54000[1 + (1/20)]2

= 54000 × [(20 + 1)/20]2

= 54000 × (21/20) × (21/20)

= 59535

Therefore, the population in 2005 = 59535

Amount, when interest is compounded quarterly, is

A = P[1 + (R/400)]4n

Here,

R/400 = Quarterly rate

4n = The number of quarters

8. What is the difference between the compound interests on Rs. 5000 for 1 ½ year at 4% per annum compounded yearly and half-yearly?

Solution:

Given,

P = Rs. 5000

R = 4%

Time (n) = 1 ½ years

When the interest is compounded yearly,

A = P[1 + (R/100)]n

= Rs. 5000 [1 + (4/100)] [1 + (1/2 × 4/100)]

= Rs. 5000 [1 + (1/25)] [1 + (1/50)]

= Rs. 5000 [(25 + 1)/25] [(50 + 1)/50]

= Rs. 5000 × (26/25) × (51/50)

= Rs. 5304

CI = A – P = Rs. 5304 – Rs. 5000 = Rs. 304

When the interest is compounded half-yearly,

n = 1 ½ years = 3 half-years

A = P[1 + (R/200)]2n

Here, 2n = 3

A = Rs. 5000 [1 + (4/200)]3

= Rs. 5000 [1 + (1/50)]3

= Rs. 5000 [(50 + 1)/50]3

= Rs. 5000 × (51/50) × (51/50) × (51/50)

= Rs. 5306.04

CI = A – P = Rs. 5306.04 – Rs. 5000 = Rs. 306.04

Difference between compound interest = Rs. 306.04 – Rs. 304 = Rs. 2.04

9. The population of a town decreased every year due to migration, poverty and unemployment. The present population of the town is 6,31,680. Last year the migration was 4%, and the year before last, it was 6%. What was the population two years ago?

Solution:

Given,

The present population of the town (A) = 631680

Last year migration rate was 4%, and the year before, the previous migration rate was 6%.

Let P be the population of a town, two years ago.

Thus, R1 = 4%

R2 = 6%

According to the given situation, the total population is:

A = P[1 – (R1/100)] [1 – (R2/100)]

631680 = P [1 – (4/100)] [1 – (6/100)]

631680 = P [1 – (1/25)] [1 – (3/50)]

631680 = P[(25 – 1)/25] [(50 – 3)/50]

631680 = P × (24/25) × (47/50)

P = 631680 × (25/24) × (50/47)

P = 700000

Therefore, the population of the town, two years ago = 700000

10. Find the amount and the compound interest on Rs. 1,00,000 compounded quarterly for 9 months at the rate of 4% per annum.

Solution:

Given,

P = Rs. 1,00,000

R = 4%

Time = 9 months

A = P[1 + (R/400)]4n

Here, R/400 is the quarterly interest rate.

4n = 9 months = 3 quarters

So, A = Rs. 1,00,000 [1 + (4/400)]3

= Rs. 1,00,000 [1 + (1/100)]3

= Rs. 1,00,000 [(100 + 1)/100]3

= Rs. 1,00,000 × (101/100) × (101/100) × (101/100)

= Rs. 103030.10

Practice Questions on Compound Interest

  1. The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000.
  2. Find the compound interest on Rs. 16,000 at 20% per annum for 9 months, compounded quarterly.
  3. Vasudevan invested Rs. 60,000 at an interest rate of 12% per annum, compounded half-yearly. What amount would he get-
    (i) after 6 months?
    (ii) after 1 year?
  4. Kamala borrowed Rs. 26,400 from a bank to buy a scooter at a rate of 15% p.a., compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
  5. Find CI paid when a sum of Rs. 10,000 is invested for 1 year and 3 months at 8 1/2 % per annum, compounded annually.
Compound Interest Questions | Compound Interest Questions with Solutions (2024)

FAQs

What will be the compound interest on $25,000 after 3 years at 12 per annum? ›

Rate of interest = 12% p.a. ∴ The compound interest is Rs. 10123.20.

How long will it take for $5000 to accumulate to $8000 if it is invested at an interest rate of 7.5 %/a compounded annually? ›

To calculate how long it will take for $5000 to grow to $8000 with an annual compound interest rate of 7.5%, we use the compound interest formula, and solve for time 't', which is approximately 6.5 years. Therefore, the correct answer is option c. 6.5 years.

How much is $1000 worth at the end of 2 years if the interest rate of 6% is compounded daily? ›

Hence, if a two-year savings account containing $1,000 pays a 6% interest rate compounded daily, it will grow to $1,127.49 at the end of two years.

What is the trick for compound interest? ›

If a sum A is compounded annually becomes A1 in t years and A2 in (t+1) years, then the principal can be calculated using: P = A1 (A1/A2) In two years, the difference between compound interest and simple interest can be calculated using: P x (R)2/ (100)

What is the compound interest on $25,000 at 10% per annum for 3 years? ›

∴ CompoundInterest=Rs. 4775.40.

What is the simple interest on $8000 for 4 years at 2% per annum? ›

Answer. So, the simple interest on 8000 naira for 4 years at a rate of 2% per annum is 160 naira.

What is $5000 invested for 10 years at 10 percent compounded annually? ›

The future value of the investment is $12,968.71. It is the accumulated value of investing $5,000 for 10 years at a rate of 10% compound interest.

How much will $10,000 be worth in 20 years? ›

The table below shows the present value (PV) of $10,000 in 20 years for interest rates from 2% to 30%. As you will see, the future value of $10,000 over 20 years can range from $14,859.47 to $1,900,496.38.

How much will $100,000 invested be in 20 years? ›

If you invest $100,000 at an annual interest rate of 6%, at the end of 20 years, your initial investment will amount to a total of $320,714, putting your interest earned over the two decades at $220,714.

How many years it will take you to double your money if you invest $500 at an interest rate of 8% per year? ›

For example, if an investment scheme promises an 8% annual compounded rate of return, it will take approximately nine years (72 / 8 = 9) to double the invested money.

What is $15000 at 15 compounded annually for 5 years? ›

The time period T = 5 years. A = $30,170.36 hence, the total amount after 5 year will be $30,170.36.

What is the compound interest when Rs 100000 is invested for 9 months at 6 per annum compounded quarterly? ›

Answer. So, the compound interest when Rs 100000 is invested for 9 months at 6% per annum, compounded quarterly, is approximately Rs 4552.29.

What is the compound interest on 60000 for 2 years and 4 months at 8 per annum? ›

∴ The compound interest is Rs. 6,150.

What will be the compound interest on 8000 at the 15% rate per annum for 2 years and 4 months? ›

Compound interest = ₹ 11109 - ₹ 8000 = ₹ 3109. Q. Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

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