SOLUTION: Mike invested $1200, part at 8% and the rest at 6% per year. The interest earned after one year was $88. How much did mike invested at each rate?

Question 1141977: Mike invested $1200, part at 8% and the rest at 6% per year. The interest earned after one year was $88. How much did mike invested at each rate?
Found 2 solutions by VFBundy, greenestamps:
Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
8% rate investment:
Principal = p
Rate = 0.08
Interest = 0.08p

6% rate investment:
Principal = 1200 - p
Rate = 0.06
Interest = 0.06(1200 - p)

0.08p + 0.06(1200 - p) = 88

0.08p + 72 - 0.06p = 88

0.02p + 72 = 88

0.02p = 16

p = 800

8% rate investment:
Principal = p = $800

6% rate investment:
Principal = 1200 - p = 1200 - 800 = $400
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Here is an easy alternative to the standard algebraic solution method shown by the other tutor; this method can be used on a wide variety of "mixture" problems.

(1) $1200 all at 6% would give $72 interest; all at 8% would give $96 interest.

(2) The actual interest, $88, is 2/3 of the way from $72 to $96. (Picture the three interest amounts $72, $88, and $96 on a number line; 88 is 2/3 of the way from 72 to 96.)

(3) Therefore 2/3 of the total investment needs to be at the higher rate.

ANSWER: 2/3 of $1200 = $800 at 8%; 1/3 of $1200 = 400 at 6%.

CHECK: 400(.06)+800(.08) = 24+64 = 88
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