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 by Using Formula
Practice Test on Compound Interest
● Compound Interest - Worksheet
Worksheet on Compound Interest
8th Grade Math Practice
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