# Compound Interest with Periodic Deductions

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

Compound Interest by Using Formula

Problems on Compound Interest

Practice Test on Compound Interest

Compound Interest - Worksheet

Worksheet on Compound Interest

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