Question 1195905: A total of $9000 is invested: part at 5% and the remainder at 15%. How much is invested at each rate if the annual interest is $520? Found 2 solutions by math_tutor2020, greenestamps:Answer by math_tutor2020(3817) (Show Source):
Assuming simple interest is involved (instead of compound interest), we then use this formula
i = P*r*t
where,
P = principal aka deposit
r = interest rate in decimal form
t = number of years
The portion invested at 5% will get us this amount of interest
i = P*r*t
i = x*0.05*1
i = 0.05x
which is what we'll have at the end of the year from this one investment.
The other account yields...
i = P*r*t
i = (9000-x)*0.15*1
i = 1350-0.15x
These two interest subtotals add to the stated annual interest of $520
(0.05x) + (1350-0.15x) = 520
-0.10x + 1350 = 520
-0.10x = 520-1350
-0.10x = -830
x = -830/(-0.10)
x = 8300 dollars was invested at a 5% interest rate
9000-x = 9000 - 8300 = 700 dollars was invested at a rate of 15%
Answers:
Amount invested at 5% = $8300
Amount invested at 15% = $700
Here is an informal method for solving this kind of problem that is fast and easy if the numbers are "nice".
(1) All $9000 invested at 5% would earn $450 interest; all at 15% would earn $1350 interest.
(2) Perhaps using a number line, look where the actual interest of $520 lies between $450 and $1350: 450 to 520 is a difference of 70; 450 to 1350 is a difference of 900; 520 is 70/900 = 7/90 of the way from 450 to 1350.
(3) That means 7/90 of the total was invested at the higher rate.
ANSWER: 7/90 of $9000, or $700, was invested at 15%; the other $8300 at 5%.
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