SOLUTION: Mitch invested $7500 for 1 year, part at 12% annual intrest and the rest at 10% annual intreast. His total intrest for one year was 890. How much money did he invest at 12%.

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Question 15069: Mitch invested $7500 for 1 year, part at 12% annual intrest and the rest at 10% annual intreast. His total intrest for one year was 890. How much money did he invest at 12%.
Answer by bam878s(77) About Me  (Show Source):
You can put this solution on YOUR website!
Let i = interest rate p = amount at end of year, s = amount at beginning of year, and t = time in years.
Now depending on whether you are using simple or compound interest the equations are
p = x%281%2Bi%29%5Et for compound interest and
p = x(1+it) for simple interest
Note these two formulas are the same for t = 1 year so it doesn't matter which one we use. I will use the simple interest formula since I don't have to deal with solving exponents.
Total interest = 890
x = part invested at 12%
y = part invested at 10%
Amount in account at end of year is 7500 + 890 = 8390.
So 8390 = 1.12x + 1.10y. (1)
and 7500 = x + y. (2)
multiply (2) by 1.12 TO GET
8400 = 1.12X + 1.10Y
Now subtract (1) from this to get
(8400-8390) = (1.12-1.12)x + (1.12-1.10)y
10 = .02y
now divide both sides by .02 to get
y = 500 = amount invested at 10%
Now plug this back into either (1) or (2) to solve for x
x + 500 = 7500
x = 7000 = amount invested at 12%
This would make since since you more than likely invest more money at the higher interest rate to yield a higher return.