Question 113986: Bob invested $20,000, part at 14% and part at 13%. If the total interest at the end of the year is $2,720, how much did he invest at 14%?
Found 2 solutions by checkley71, BrittanyM:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! .14X+.13(20,000-X)=2,720
.14X+2600-.13X=2,720
.01X=2,720-2,600
.01X=120
X=120/.01
X=$12,000 IS THE AMOUNT INVESTED @ 14%.
PROOF:
.14*12,000+.13(20,000-12,000)=2,720
1,680+.13*8,000=2,720
1,680+1,040=2,720
2,720=2,720
Answer by BrittanyM(80) (Show Source):
You can put this solution on YOUR website! We can solve this by setting up a sysrem:
Since we know the the total investment was 20,000, 100% of both rates is written as
x + y = 20,000
And since the given rates of the investment, 14% and 13%, yielded 2,720 we have
0.14x + 0.13y = 2,720
From the first equation, let
x = -y + 20,000
So let's plug this into the second equation:
0.14(-y + 20,000) + 0.13y = 2,720
-0.14y + 2,800 + 0.13y = 2,720
-0.01y + 2,800 = 2,720
-0.01y = -80
y = 8,000
Now, we can subtract our y value from 20,000 in order to find x:
x = -y + 20,000
x = -8,000 + 20,000
x = 12,000
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