SOLUTION: Ed invested a certain amount at 10% simple interest per year after 2 years the interest he recieved amounted to 3,000. how much did he invest?

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Question 1120101: Ed invested a certain amount at 10% simple interest per year after 2 years the interest he recieved amounted to 3,000.
how much did he invest?

Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
f = p * (1 + r) ^ n

f = future value
p = present value
r = interest rate per time period
n = number of time periods.

in your problem, this becomes:

p + 3000 = p * (1 + .10) ^ 2

subtract p from both sides of this equation to get:

3000 = p * (1 + .10) ^ 2 - p

factor out the p to get:

3000 = p * (1 + .10) ^ 2 - 1)

simplify to get:

3000 = p * .21

solve for p to get p = 3000 / .21 = 14285.71429.

that's your present value.

to confirm this is accurate, replace p in your equation with that to get:

p + 3000 = p * (1 + .10) ^ 2 becomes:

17285.71429 = 14285.71429 * (1 + .10) ^ 2, which becomes:

17285.71429 = 17285.71429, which is true.

this confirms the solution is correct.

Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website!
.

Ed invested a certain amount at 10% simple interest per year after 2 years the interest he received amounted to 3,000.
how much did he invest?
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It is about a SIMPLE INTEREST !

The other tutor solved another problem for a COMPOUND interest.

At simple interest, the interest after 2 years is I = 2*0.1*X, Where X is an unknown principal. 2*0.1*X = 3000 ====> X = = 15000. Answer. 15000 were invested under given conditions.