SOLUTION: A man invested $16,000.00. He invested a portion at 6% and the balance at 8%. If he receives $260.00 every 3 months, how much did he invest at each rate?

Question 4130: A man invested $16,000.00. He invested a portion at 6% and the balance at 8%. If he receives $260.00 every 3 months, how much did he invest at each rate?
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
If the interest for 3 months is $260, then the total interest for the year is 4 times $260 or $1040.

Let x = amount invested at one of the rates, say 6%. This leaves the rest of the $16,000 or (16000-x) to be invested at the other rate, at 8%.

The equation is basically that the interest earned for 1 year at 6% plus the interest earned for the year at 8% is equal to the total interest for the year, or $1040.

.06x + .08(16000-x) = 1040
.06x + .08(16000) - .08x = 1040
-.02x + 1280 = 1040

Subtract 1280 from each side:
-.02x = 1040 - 1280
-.02x = -240

To solve for x, divide both sides by -.02, remembering that a negative divided by a negative is a negative.

x = 240/.02

Either divide with a calculator, or multiply numerator and denominator by 100 to eliminate the decimal.

x = 24000/2 = $12000 at 6%
16000-x = 16000-12000 = $4000 at 8%
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