Compound Interest with Growing Principal

We will learn how to calculate the compound interest with growing principal.

If the interest which has become due at the end of a certain period (i.e., 1 year, half-year, ect. as given ) is not paid to the money lender, but is added to the some borrowed, the amount thus obtained becomes the principal for the next period of borrowing. This process goes on until the amount for the specified time is found.


Solved examples on compound interest with growing principal:

1. A man takes a loan of $ 10,000 at a compound interest rate of 10% per annum.

(i)  Find the amount after 1 year.

(ii) Find the compound interest for 2 years.

(iii) Find the sum of money required to clear the debt at the end of 2 years.

(iv) Find the difference between the compound interest and simple interest at the same rate for 2 years.

Solution:

(i) The interest for the first year = 10% of $10,000

                                            = $\(\frac{10}{100}\) × 10,000

                                            = $ 1,000

Therefore, the amount after 1 year = Principal + Interest

                                                = $10,000 + $ 1,000

                                                = $ 11,000

(ii) For the second year, the new principal is $ 11,000

Therefore, the interest for the 2nd year = 10% of $ 11,000

                                                      = $\(\frac{10}{100}\) × 11,000

                                                       = $ 1,100

Therefore, the compound interest for 2 years = the interest for the 1st year + the interest for the 2nd year

                                                             = $ 1,000 + $ 1,100

                                                             = $ 2,100

(iii) The required sum of money = Principal + compound Interest for 2 years

                                          = $ 10,000 + $ 2,100

                                          = $ 12,100

(iv) The simple interest for 2 years = \(\frac{P × R × T}{100}\)

                                               = $ \(\frac{10,000 × 10 × 2}{100}\)

                                               = $ 2,000

Therefore, the required difference = $ 2,100 - $ 2,000 = $ 100

 

2. At 4% per annum, the difference between simple and compound interest for 2 years on a certain sum of money is Rs. 80. Find the sum

Solution:

Let the sum of money be $ x,

The interest for the first year = 4 % of $x

                                        = $ \(\frac{4}{100}\) × x

                                        = $ \(\frac{4x}{100}\)

                                        = $ \(\frac{x}{25}\)

 

Therefore, the amount after 1 year = Principal + Interest

                                                = $ x + $ \(\frac{x}{25}\)

                                                = $ \(\frac{26x}{25}\)

For the second year, the new principal is $ \(\frac{26x}{25}\)

Therefore, the interest for the 2nd year = 4 % of $ \(\frac{26x}{25}\)

                                                      = $ \(\frac{4}{100}\) × \(\frac{26x}{25}\)

                                                      = $ \(\frac{26x}{625}\)

Compound interest for 2 years = $ \(\frac{x}{25}\) + $ \(\frac{26x}{625}\)

                                             = $ \(\frac{51x}{625}\)

At 4% rate simple interest for 2 years = $\(\frac{\frac{26x}{25} × 4 × T}{100}\)

                                                   = $\(\frac{x × 4 × 2}{100}\)

                                                   = $\(\frac{8x}{100}\)

                                                   = $\(\frac{2x}{25}\)

 

Now, according to the problem, we get

\(\frac{51x}{625}\) - \(\frac{2x}{25}\) = 80

x(\(\frac{51}{625}\) - \(\frac{2}{25}\)) = 80

\(\frac{x}{625}\) = 80

x = 80 × 625

x = 50000

The required sum of money is $ 50000


3. Find the amount and the compound interest on $10,000 at 8% per annum and in 1 year, interest will being compounded half-yearly.

Solution:

For first half-year principal = $ 10,000

Rate = 8%

Time = ½ year

The interest for the first half-year = \(\frac{P × R × T}{100}\)

                                               = \(\frac{10000 × 8 × 1}{100 × 2}\)

                                               = $ 400

Therefore, the amount after half- year = Principal + Interest

                                                     = $ 10,000 + $ 400

                                                     = $ 10,400

Therefore, at 8% rate the interest for the 2nd half-year = $\(\frac{10400 × 8 × 1}{100 × 2}\)

                                                                            = $ 416

The required sum of money = Principal + compound Interest

                                     = $10,400 + $ 416

                                     = $ 10,816

Therefore, the required amount = $ 10,816 and

the compound interest = Amount - Principal

                               = $ 10,816 - $ 10,000

                               = $ 816


From the above examples we conclude that:

(i) When the interest is compounded yearly, then the principal does not remain same every year.

(ii) When the interest is compounded half-yearly, then the principal does not remain same every 6 months.

Thus the principal changes at the end of every phases.

 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




8th Grade Math Practice 

From Compound Interest with Growing Principal to HOME PAGE




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.



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.




Share this page: What’s this?

Recent Articles

  1. 2nd grade math Worksheets | Free Math Worksheets | By Grade and Topic

    Dec 04, 24 01:30 AM

    2nd Grade Math Worksheet
    2nd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students.

    Read More

  2. Time Duration |How to Calculate the Time Duration (in Hours & Minutes)

    Dec 04, 24 01:07 AM

    Time Duration Example
    Time duration tells us how long it takes for an activity to complete. We will learn how to calculate the time duration in minutes and in hours. Time Duration (in minutes) Ron and Clara play badminton…

    Read More

  3. Worksheet on Subtraction of Money | Real-life Word Problems | Answers

    Dec 04, 24 12:45 AM

    Worksheet on Subtraction of Money
    Practice the questions given in the worksheet on subtraction of money by using without conversion and by conversion method (without regrouping and with regrouping). Note: Arrange the amount of rupees…

    Read More

  4. Worksheet on Addition of Money | Questions on Adding Amount of Money

    Dec 04, 24 12:06 AM

    Worksheet on Addition of Money
    Practice the questions given in the worksheet on addition of money by using without conversion and by conversion method (without regrouping and with regrouping). Note: Arrange the amount of money in t…

    Read More

  5. Worksheet on Money | Conversion of Money from Rupees to Paisa

    Dec 03, 24 11:37 PM

    Worksheet on Money
    Practice the questions given in the worksheet on money. This sheet provides different types of questions where students need to express the amount of money in short form and long form

    Read More