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Exercise 7 - Example: Compound Interest 		Home
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Example 1: Find the amount and compound interest on $5000 in 3 years at 5% per annum. Let P1, P2 and P3 represent the principal for the first, second and third year. Whereas I1, I2 and I3 represent the interest for the first, second and third year.
P1 = $5000
I1 at 5% = $250 		I1 = 5000 x (5/100) = $250
P2 = $5000 + $250 = $5250.0
I2 at 5% = $262.5 		I2 = 5250 x (5/100) = $262.5
P3 = $5250 + $262.5 = $5512.50
I3 at 5% = $275.63 		I3 = 5512.50 x (5/100) = $275.63
Amount after 3 years = $5512.50 + $275.63 = $5788.13
Principal = $5000
Compound Interest = $5788.13 - $5000 = $788.13

Formula for Compound Interest:
A = P[1 + (r/100)]n
CI = A - P

Where P is the principal, r is the rate of interest, n is the number of years and CI is the compound interest

Example 1 can be solved using the formula: A = P[1 + (r/100)]n
P = $5000, r = 5% and n = 3 years
A = 5000[1 + (5/100)]3
A = 5000[105/100]3
A = $5788.13
P = $5000
Compound Interest = $788.13

Example 2: Find the Principal that amounts to $6742 in 2 years at 6% per year.
A = P[1 + (r/100)]n
A = $6742, r = 6% and n = 2 years
6742 = P[1 + (6/100)]2
6742 = P[106/100]2
6742 = P[1.1236]
P = 6742/1.1236 = 6000
Principal = $6000

Example 3: Find the Principal that yields CI of $742 in 2 years at 6% per year.
CI = A - P
A = P[1 + (r/100)]n
CI = P[1 + (r/100)]n - P
CI = P{[1 + (r/100)]n - 1}
742 = P{[1 + (6/100)]2 - 1}
742 = P{[106/100]2 - 1}
742 = P{1.1236 - 1}
742 = P{0.1236}
P = 742/0.1236 = 6000
Principal = $6000


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