Question 1120418:
$5400
is invested, part of it at 10
%
and part of it at 8
%.
For a certain year, the total yield is $nbsp 492.00
.
How much was invested at each rate?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x at 10%=.10x
5400-x at 8%=.08(5400-x)=432-.08x
That sum, .10x+432-.08x=492
.02x=60
x=$3000 at 10% or $300 ANSWER
2400 at 8% or $192 ANSWER
That adds to $492.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
See if you understand this alternative to the traditional algebraic method for solving mixture problems like this. If you understand it, you will solve this kind of problem much faster and with much less effort than with algebra.
The total of $5400 all invested at 10% would yield $540 interest; all invested at 8% would yield $432 interest.
Where the actual interest of $492 lies between those two numbers determines the ratio in which the money must be split between the two investments.
I will first show just the required calculations, to show how fast and easy the method is. Then I will explain the method.
540-432 = 108
492-432 = 60
60/108 = 5/9
Answer: 5/9 of the money should be invested at the higher rate.
5/9 of $5400 = 5*$600 = $3000 at 10%; the other 4/9 = $2400 at 8%.
Quick explanation....
Think of the three numbers on a number line -- the amount of interest if all at 8%, the actual amount of interest, and the amount of interest if all at 10%: 432, 492, and 540.
The three quick calculations shown above show that the actual number, 492, is 5/9 of the way from the lower number, 432, to the higher number, 540.
That means 5/9 of the money must be invested at the higher rate.
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