SOLUTION: An investor invested a total of $800 in two mutual funds. One fund earned a 7% profit while the other earned a 2% profit. If the investor's total profit was $26, how
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-> SOLUTION: An investor invested a total of $800 in two mutual funds. One fund earned a 7% profit while the other earned a 2% profit. If the investor's total profit was $26, how
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Question 1199654: An investor invested a total of $800 in two mutual funds. One fund earned a 7% profit while the other earned a 2% profit. If the investor's total profit was $26, how much was invested in each mutual fund? Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
An investor invested a total of $800 in two mutual funds.
One fund earned a 7% profit while the other earned a 2% profit.
If the investor's total profit was $26, how much was invested in each mutual fund?
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Let x be the amount in dollars invested at 7%;
then the amount invested at 2% is the reat 800-x dollars.
Write the total annual interest equation
0.07x + 0.02*(800-x) = 26 dollars.
Simplify it and find x
0.07x + 16 - 0.02x = 26
0.07x - 0.02x = 26 - 16
0.05x = 16
x = 16/0.05 = 1600/5 = 320.
ANSWER. $320 invested at 7%, the rest, or $800 - $320 = $480 invested at 2%.
CHECK. 0.07*320 + 0.02*480 = 32 dollars, the total annual interest. ! correct !
Here is an quick and easy informal method for solving any 2-part "mixture" problem like this.
(1) All $800 invested at 2% would earn $16 interest; all at 7% would earn $56 interest.
(2) The actual interest, $26, was 10/40 = 1/4 of the way from $16 to $56. ($16 to $56 is a difference of 40; $16 to $26 is a difference of 10; 10/40 = 1/4.)
(3) That means 1/4 of the total was invested at the higher rate.
1/4 of $800 is $200; so
ANSWERS: $200 was invested at 7% and the other $600 at 2%.
