SOLUTION: Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%? I got 3000.00 is this correct?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%? I got 3000.00 is this correct?      Log On


   



Question 55083: Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%? I got 3000.00 is this correct?
Answer by Hook(36) About Me  (Show Source):
You can put this solution on YOUR website!
Big picture...why didn't Stan invest all his money at 17%? Don't take investment advice from Stan.

Ok..let's make up some variables to start:
We'll call x the amount that Stan invested at 8% and y the amount that he invested at 17%.
We know he invested 5 bills, so we get the following equation:
x+%2B+y+=+5000
Also, his combined interest income was $490. So 8% of x plus 17% of y must be $490. In math-speak:
.08x%2B.17y+=+490.
Here, we've got the following system of equations:
x+%2B+y+=+5000

.08x%2B.17y+=+490.
I'll use the substitution method and solve for x in the first equation
x+%2B+y+=+5000
x+=+5000-y
Now, I'll plug that into the second equation
.08%285000-y%29%2B.17y+=+490
400-.08y%2B.17y+=+490
400%2B.09y+=+490
.09y+=+90
y+=+1000
Alright! y = 1000!
if we plug that into x+=+5000-y, we get
x+=+5000-1000
x+=+4000
Ok. My answers are different than yours. I'll check my work.
CHECK:
Using the second original equation:
.08x%2B.17y+=+490
.08%284000%29+%2B.17%281000%29+=+490
320+%2B+170+=+490
490+=+490 Check!
Stan invested $4000 at 8%. Dumbass. Don't let Stan run your investments.