Question 57088: Helen Weller invested $15,000 in an account that pays 11% simple interest. How much additional money must be invested in an account that pays 14% simple interest so that the average return on the two investments amounts to 12%?
Found 2 solutions by Earlsdon, stanbon:
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let x = the additional amount to be invested at 14%
The interest on the initial $15,000 at 11% would be 0.11($15,000)
The interest on the additional amount, $x at 14% would be 0.14($x)
The sum of these two amounts of interest should come to 12% of ($15,000 + $x) or 0.12($15,000+$x). So...putting it all together. you can write:
0.11($15000) + 0.14($x) = 0.12($15000+$x) Simplify this and solve for x.
$1650 + 0.14x = 1800 + 0.12x Subtracting 0.12x from both sides, you get:
1650 + 0.02($x) = $1800 Subtract $1650 from both sides.
0.02($x) = $150 Finally, divide both sides by 0.02
x = $7500
So Helen Weller should invest an additional $7,500 at 14% interest to obtain an average rate of return of 12% on the entire amount.
Check:
11% of $15,000 = 0.11($15,000) = $1,650
14% of $7,500 = 0.14($7,500) = $1,050
The sum: $1,650 + $1,050 = $2,700 total interest.
12% of ($15,000 + $7,500) = 0.12($22,500) = $2,700
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Helen Weller invested $15,000 in an account that pays 11% simple interest. How much additional money must be invested in an account that pays 14% simple interest so that the average return on the two investments amounts to 12%?
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Interest #1: 15,000(0.11)= 1650
Interest #2: x(0.14)= 0.14x
EQUATIOH:
I think that word "average" can be misinterpreted on this problem.
Using it in the normal sense gives a negative value of "x".
Ignoring it you can get something meaningful, as follows:
(int#1+int#2) = 0.12(15000+x)
1650+0.14x = 1800+0.12x
0.02x=150
x=15000/2=$7500
That's the amount invested at 14% required so the total investment
will EQUAL 12% of all the money invested.
Cheers,
Stan h.
Cheers,
Stan H.
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