Question 213021: I'ts been a while since I've solved these kinds of questions (ok 20+ years!) so I need a refresher. I looked a similar problems on-line, but am struggling with how to set it up. Here's the problem:
Sam invested $5000 for one year, part at 9% annual interest and the rest at 12% annual interest. The interest from the investment at 9% was $198 more than the interest from the investment at 12%. How much money did she invest at 9%?
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Sam invested $5000 for one year, part at 9% annual interest and the rest at 12% annual interest.
The interest from the investment at 9% was $198 more than the interest from the investment at 12%.
How much money did she invest at 9%?
:
Let x = amt invested at %9
Total invested given as 5000, therefore
(5000-x) = amt invested at 12%
:
.09x - 298 = .12(5000-x)
.09x - 298 = 600 - .12x
.09x + .12x = 600 + 298
.21x = 898
x = 
x = $4,276.19 invested at 9%
then
5000 - 4276.19 = $723.08 invested at %12
:
:
Is this true? find the interest of each
.09*4276.19 = 384.86
.12 * 723.08 = 86.86
-------------------
difference = 298; confirms these solutions
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
Let the amount invested at 9% be N
Then the amount invested at 12% is 5,000 - N
Now, the interest earned at 9%, less $198 equals the einterest earned at 12%
Therefore, we'll have: .09N - 198 = .12(5,000 - N)
.09N - 198 = 600 - .12N
.09N + .12N = 600 + 198
.21N = 798
 = $3,800
We can conclude that N, or the amount invested at 9% is: $
--------
Check
--------
Amount invested at 9% = $3,800, and interest earned at 9% = (.09 * 3,800) = $342
Amount invested at 12% = $1,200 (5,000 - 3,800), and interest earned at 12% = (.12 * 1,200) = $144
Difference: $198 ($342 - $144), with 9% interest earned ($342) being $198 more than 12% interest earned ($144).
|
|
|
