5. How much was invested in an account that currently contains $13258.75 if it was compounded quarterly at 2%?
11. A 1/4-ton truck that cost $38,990 depreciates at a rate of 13% per year. What is the value of the truck after 5 years?
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5. How long has it been invested? Is the 2% the annual rate?
I'll assume it's been invested for one year, and that 2% is the annual rate.
In general, Balance = Principal * (1 + i/m)^(nm) where m is the compounding frequency per year & n is the number of years. Quarterly compounding implies that m=4. I already assumed that n = 1. We were told that the Balance = 13258.75
Plug it all in:
13258.75 = P * (1 + .02/4)^(1*4) = P (1.005)^4
P = 13258.75 / (1.005)^4 ≈ 12996.9
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11.
Start with 38990 and find 13% of it. "Of" means multiply in math.
38990 * .13 = 5068.7
Subtract that from 38990: 38990 - 5068.7 = 33921.3 is the value at the end of year 1.
Do it again for years 2-5.
[or] realize that all you're doing is starting with P and multiplying by (1-d) where d is the rate of depreciation.
End of Year 1 value = P(1-d)
End of Year 2 value = EOY 1 value * (1-d) = P(1-d)(1-d) = P(1-d)²
End of Year 3 value = EOY 2 value * (1-d) = P(1-d)²(1-d) = P(1-d)³
yadda yadda yadda
End of Year n value = P(1-d)^n
So, P was 38990, d was .13, n was 5
Value at end of year 5 = 38990(1-.13)^5 = 19433.4