Question 101094: Colin invested $37,000, part at 16% and part at 3%. If the total interest at the end of the year is $3,320, how much did he invest at 16%?
Found 2 solutions by checkley71, tutorcecilia:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! .16X+.03(37,000-X)=3,320
.16X+1,110-.03X)=3,320
.16X-.03X=3,320-1,110
.13X=2,210
X=2,210/.13
X=17,000 INVESTED @16%.
37,000-17,000=20,000 INVESTED @ 3%.
PROOF
.16*17,000+.03*20,000=3,320
2720+600=3,320
3,320=3,320
Answer by tutorcecilia(2152)  (Show Source):
You can put this solution on YOUR website! Let x+y=37,000
Than y=37,000-x
.
.16x + .03y = 3,320
So, .16x + .03(37,000-x)=3,320 [substitute y=37,000-x in the equation]
.16x + 1110- .03x=3,320 [sovle for the x-term]
.16x-.03x=3,320-1110
.13x=2210
.13x/.13=2210/.13
x=$17,000 invested at 16%
.
y=37,000-x
y=37,000-17,000=20,000
.
Check by plugging all of the values back into the original equation and solve:
.16x + .03y = 3,320
.16(17,000) + .03(20,000) = 3,320
2720+600=3320
3320=3320
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