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 with Growing Principal

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 

From Compound Interest with Periodic Deductions to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Method of H.C.F. |Highest Common Factor|Factorization &Division Method

    Apr 13, 24 05:12 PM

    HCF by Short Division Method
    We will discuss here about the method of h.c.f. (highest common factor). The highest common factor or HCF of two or more numbers is the greatest number which divides exactly the given numbers. Let us…

    Read More

  2. Factors | Understand the Factors of the Product | Concept of Factors

    Apr 13, 24 03:29 PM

    Factors
    Factors of a number are discussed here so that students can understand the factors of the product. What are factors? (i) If a dividend, when divided by a divisor, is divided completely

    Read More

  3. Methods of Prime Factorization | Division Method | Factor Tree Method

    Apr 13, 24 01:27 PM

    Factor Tree Method
    In prime factorization, we factorise the numbers into prime numbers, called prime factors. There are two methods of prime factorization: 1. Division Method 2. Factor Tree Method

    Read More

  4. Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18

    Apr 13, 24 12:41 PM

    Divisibility Rules
    To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4…

    Read More

  5. Even and Odd Numbers Between 1 and 100 | Even and Odd Numbers|Examples

    Apr 12, 24 04:22 PM

    even and odd numbers
    All the even and odd numbers between 1 and 100 are discussed here. What are the even numbers from 1 to 100? The even numbers from 1 to 100 are:

    Read More