SOLUTION: You invested $ 4000 between two accounts paying 7 % and 9 % annual​ interest, respectively. If the total interest earned for the year was $ 300 comma how much was invested at ea

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Question 1149229: You invested $ 4000 between two accounts paying 7 % and 9 % annual​ interest, respectively. If the total interest earned for the year was $ 300 comma how much was invested at each​ rate?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Using the standard formal algebraic solution....

7% of x, plus 9% of (4000-x) equals 300:

.07%28x%29%2B.09%284000-x%29+=+300
7x%2B9%284000-x%29+=+30000
7x%2B36000-9x+=+30000
6000+=+2x
x+=+3000

ANSWER: $3000 at 7%; $1000 at 9%.

CHECK:
.07(3000)+.09(1000) = 210+90 = 300

Here is a much faster path to solving a mixture problem like this if an algebraic solution is not required.

(1) $4000 all at 7% would yield $280 interest; all at 9% would yield $360 interest.
(2) The actual interest, $300, is 1/4 of the way from $280 to $360. (Picture the three numbers 280, 300, and 360 on a number line. 280 to 360 is a difference of 80; 280 to 300 is a difference of 20. 20 is 1/4 of 80.)
(3) Therefore, 1/4 of the total was invested at the higher rate.

ANSWER: 1/4 of $4000, or $1000, at 9%; the other $3000 at 7%.

Obviously the same answer as before....