Question 54273: If I invest $18,750 in two different accounts and one account earns 12% interest and the other earns 10% interest, and at the end of the year have a total $20,867 (original amount + interest), how much did I invest in each account?
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! INVESTING $ 18,750 AND OBTAINING $ 20,867 TOTAL FOR INTEREST FOR THE 10% & 12% INVESTMENTS.
WE HAVE $20,867-$18,750=$2,117. THUS THERE ARE TWO PARTS OF THE $18,750 BEING INVESTED X & ($18,750-X).
THUS THE INVESTMENTS ARE .12X+.10(18,750-X)=2,117 OR .12X+1,875-.10X=2,117 OR
.12X-.10X=2,117-1,875 OR .02X=242 OR X=242/.02 OR X=12,100 INVESTED @ 12% &
18,750-12,100=6,650 INVESTED @ 10%
PROOF .12*12,100+.10*6,650=2,117 OR 1,452+665=2,117 OR 2,117=2,117
& 18,750+2,117=20,867 OR 20,867=20,867
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If I invest $18,750 in two different accounts and one account earns 12% interest and the other earns 10% interest, and at the end of the year have a total $20,867 (original amount + interest), how much did I invest in each account?
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Let amount invested at 12% be "x".
Interest on this money is 0.12x.
Amount invested at 10% is "18750-x"
Interest on this money is 0.10(18750-x)=1875-0.1x
EQUATION:
interest + interest = 20,867-18750
0.12x +1875-0.1x = 2117
o.02x= 242
x=$12,110 (Amount invested at 12%)
18750-x= $6640 (Amount invested at 10%)
Chers,
Stan H.
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