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
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