## Problems on Compound Interest

More solved problems on compound interest using formula are shown below.

### 1. The simple interest on a sum of money for 3 years at 6^{2}/_{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 =

^{20}/

_{3} % p.a. and T = 3 years.

sum = 100 × SI / R × T

= $ (100 × 6750 ×

^{3}/

_{20} × 1/3 ) = $ 33750.

Now, P = $ 33750, R =

^{20}/

_{3}% p.a. and T = 3years.

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.

### 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

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.

**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} = $ (

^{2809}/

_{2500} P - 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.

### 3. 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.

**Alternative method**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.

Hence, the required sum is $ 62500.

### In this question the formula is when the interest is compounded annually to solve this problem on compound interest.

### 4. At what rate per cent per annum will Ron lends a sum of $2000 to Ben. Ben returned after 2 years $2205, compounded annually?

**Solution:** Let the required rate be R% per annum.

Here, A = $ 2205, P = $ 2000 and n = 2 years.

**Using the formula A = P(1 + R/100)**^{n}, 2205 = 2000 × ( 1 + R/100)

^{2}⇒ (1 + R/100)

^{2} = 2205/2000 = 441/400 = (21/20)

^{2} ⇒ ( 1 + R/100) = 21/20

⇒ R/100 = (21/20 – 1) = 1/20

⇒ R = (100 × 1/20) = 5

Hence, the required rate of interest is 5% per annum.

### 5. A man deposited $1000 in a bank. In return he got $1331. Bank gave interest 10% per annum. How long did he kept 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.

**Compound Interest****Compound Interest**

**Compound Interest by Using Formula**

**Problems on Compound Interest**

**Practice Test on Compound Interest**

**Compound Interest - Worksheet****Worksheet on Compound Interest**

8th Grade Math Practice

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