Question 1162082: Mr. Wilson invested money in two accounts. His total investment was $29,000. If one account pays 4% in interest and the other pays 7% in interest, how much did he invest in each account if he earned a total of $1,550 in interest in 1 year?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
A setup using the traditional algebraic approach would look something like this:
let x = amount invested at 4%
then 29000-x = amount invested at 7%
The interest from one investment is .04(x); from the other is .07(29000-x).
The total interest was $1160:
Solve using basic algebra. I leave that to you.
I find this alternative method faster and easier....
(1) Determine how the actual interest compares to the interest that would have been earned if the whole $29000 had been invested at each rate.
all at 4%: .04(29000) = 1160
actual interest: 1550
all at 7%: .07(29000) = 2030
(2) Where the actual interest lies between the two extremes exactly determines what fraction of the total was invested at each rate.
The difference between 1160 and 2030 is 870
The difference between 1160 and 1550 is 390
The actual interest 1550 is 390/870 = 39/87 = 13/29 of the way from 1160 to 2030.
That means 13/29 of the total was invested at the higher rate.
ANSWER: 13/29 of $29,000, or $13,000, at 7%; the other $16,000 at 4%.
CHECK: .07(13000)+.04(16000) = 910+640 = 1550
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
0.07x + 0.04(29000-x) = 1550 dollars.
x =  = 13000.
ANSWER. $13000 invested at 7% and the rest, 29000-13000 = 16000 dollars invested at 4%.
CHECK. 0.07*13000 + 0.04*16000 = 1550 dollars. ! Precisely correct !
Solved.
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