We will learn how to calculate the compound interest with periodic deductions or additions to the amount.

Solved examples on compound interest with periodic deductions:

**1.** Ron borrows $ 10,000 at a compound interest rate of 8% per annum. If he repays $ 2000 at the end of each year, find the sum outstanding at the end of the third year.

**Solution:**

For the first year:

Principal = $ 10,000

Rate = 8 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{10000 × 8 × 1}{100}\)

= $\(\frac{80000}{100}\)

= $ 800

Therefore, the amount of loan after 1 year = Principal + Interest

= $ 10,000 + $ 800

= $ 10,800

Ron pays back $ 2,000 at the end of the first year.

So, the new principal at the beginning of the second year = $ 10,800 - $ 2,000 = $ 8,800

Therefore, for the second year:

Principal = $ 8,800

Rate = 8 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{8,800 × 8 × 1}{100}\)

= $\(\frac{70400}{100}\)

= $ 704

Therefore, the amount of loan after 2 year = Principal + Interest

= $ 8,800 + $ 704

= $ 9504

Ron pays back $ 2,000 at the end of the second year.

So, the new principal at the beginning of the third year = $ 9504 - $ 2,000

= $ 7504

Therefore, for the third year:

Principal = $ 7504

Rate = 8 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{7504 × 8 × 1}{100}\)

= $\(\frac{60032}{100}\)

= $ 600.32

Therefore, the amount of loan (outstanding sum) after 3 year = Principal + Interest

= $ 7504 + $ 600.32

= $ 8104.32

**2.** Davis invests $ 20,000 at the beginning of every year in a bank and earns 10 % annual interest, compounded at the end of the year. What will be his balance in the bank at the end of three years.

**Solution:**

For the first year:

Principal = $ 20,000

Rate = 10 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{20000 × 10 × 1}{100}\)

= $\(\frac{200000}{100}\)

= $ 2000

Therefore, the amount at the end of the 1 year = Principal + Interest

= $ 20,000 + $ 2000

= $ 22,000

Davis deposits $ 20,000 at the beginning of the second year.

So, the new principal for the second year = $ 22,000 + $ 20,000

= $ 42,000

Therefore, for the second year:

Principal = $ 42,000

Rate = 10 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{42000 × 10 × 1}{100}\)

= $\(\frac{420000}{100}\)

= $ 4,200

Therefore, the amount at the end of the 2 year = Principal + Interest

= $ 42,000 + $ 4,200

= $ 46,200

Davis deposits $ 20,000 at the beginning of the third year.

So, the new principal for the third year = $ 46,200 + $ 20,000

= $ 66,200

Therefore, for the third year:

Principal = $ 66,200

Rate = 10 %

Time = 1 year

Therefore, interest = $\(\frac{P × R × T}{100}\)

= $\(\frac{66200 × 10 × 1}{100}\)

= $\(\frac{662000}{100}\)

= $ 6620

Therefore, the amount at the end of the 3 year = Principal + Interest

= $ 66,200 + $ 6,620

= $ 72,820

Therefore, the balance in the bank at the end of thee years will be $ 72,820.

**From the above examples**, we observe that how the principal does not remain same always; at the end of every phase, the principal changes. There is a direct relation between the principal and the compound interest or amount.

● **Compound Interest**

Compound Interest with Growing Principal

Compound Interest by Using Formula

Practice Test on Compound Interest

● **Compound Interest - Worksheet**

Worksheet on Compound Interest

**8th Grade Math Practice****From Compound Interest with Periodic Deductions to HOME PAGE**

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