SOLUTION: A man invested part of $15,000 at 12% and the remainder at 8%. If his annual income from the two investments is $1456, how much does he have invested at each rate?

Question 1157106: A man invested part of $15,000 at 12% and the remainder at 8%. If his annual
income from the two investments is $1456, how much does he have invested at
each rate?
Found 3 solutions by mananth, MathTherapy, greenestamps:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
part at 12% ---------------- x
and the remainder at 8%.---------- y
Total investment =$15,000
1.00 x + 1.00 y = 15000.00

0.12 x + 0.08 y = 1456.00
Eliminate y
multiply (1)by -0.08
Multiply (2) by 1.00
-0.08 x -0.08 y = -1200.00
0.12 x 0.08 y = 1456.00
Add the two equations
0.04 x = 256.00
/ 0.08
x = 6400.00
plug value of x in (1)
1.00 x + 1.00 y = 15000.00
6400.00 + 1.00 y = 15000.00
1.00 y = 8600.00
y = 8600.00
Ans x = 6400.00
y = 8600.00
6400.00 @12%
8600.00 @8%

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
A man invested part of $15,000 at 12% and the remainder at 8%. If his annual
income from the two investments is $1456, how much does he have invested at
each rate?

Let amount invested at 12% be T
Then amount invested at 8% = 15,000 - T
We then get: .12T + .08(15,000 - T) = 1,456
.12T + 1,200 - .08T = 1,456
.12T - .08T = 1,456 - 1,200
.04T = 256
Amount invested at 12%, or
You should be able to find amount invested at 8%.
Is that DUDE for real? Has he actually convinced himself that his method is EASIER/QUICKER, and less complex? He MUST be KIDDING!!
Who'd really want to go through his method to get an answer? He MUST be BUGGING!!
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Here is a way to solve two-part mixture problems like this quickly without formal algebra. If you understand the method, it will get you to the answer much faster and with much less effort (and easier calculations) than the formal algebraic method.

(1) Determine what the interest would be if the whole amount were invested at each rate
$15,000 at 12% --> $1800 interest
$15,000 at 8% --> $1200 interest

(2) Determine where the actual interest lies between those two amounts (imagine the three percentages on a number line: 1200, 1456, 1800)
1200 to 1800 is a difference of 600
1200 to 1456 is a difference of 256

The actual interest is 256/600 = 64/150 of the way from 1200 to 1800

(3) The fraction of the total invested at the higher interest rate is that same fraction.
64/150 of $15,000 is $6400

ANSWER: $6400 invested at 12%; the other $8600 at 8%

CHECK:
.12(6400)+.08(8600) = 768+688 = 1456

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