Question 1190296: Phyllis invested
56000
dollars, a portion earning a simple interest rate of
5
percent per year and the rest earning a rate of
6
percent per year. After one year the total interest earned on these investments was
3200
dollars. How much money did she invest 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 5% makes 0.05x per year (dollars)
56000-x at 6% makes 3360-0.06x per year
they add to 3200
so 3200=-0.01x+3360
0.01x=-160
x=$16000 @5%=$800
56000-x=40000@6%=$2400
they add to $3200.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
A quick, informal solution method, if formal algebra is not required....
$56000 all invested at 5% would yield $2800 interest; all invested at 6% would yield $3360 interest. The ratio in which the total was split between the two investments is determined by where the actual interest of $3200 lies between $2800 and $3360.
(1) Look at the three interest amounts 2800, 3200, and 3360 on a number line and observe/calculate that 3200 is 400/560 = 5/7 of the way from 2800 to 3360.
(2) That means 5/7 of the total was invested at the higher rate.
ANSWER: 5/7 of the total $56,000, or $40,000, was invested at 6%; the other $16,000 at 5%.
CHECK: .06(40,000)+.05(16,000) = 2400+800 = 3200
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