# 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 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 = $$\frac{20}{3}$$% p.a. and T = 3 years.

sum = 100 × SI / R × T

= $(100 × 6750 × ³/₂₀ × 1/3 ) =$ 33750.

Now, P = $33750, R = $$\frac{20}{3}$$% 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) = $7210. ### 2. The difference between the compound interest, compounded annually and the simple interest on a certain sum for 2 years at 6% per annum is$ 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} = $($$\frac{2809}{2500}$$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. ### 4. 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.

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