Bill's Two Investments
33) With $35,000, Bill made two investments. Since one had paid 8% interest,
and the other 12%, his total intererest-earning from both was $3,920, for the
year. How much had he initially invested at each percent?
x = 8% investment
y = 12% investment
x + y = 35,000 (The two investments and their total amount.)
.08(x) + .12(y) = $3,920 (The interest-earning of each investment,
and their combined annual yield.)
8x + 12y = 392,000 (We eliminated the decimal places by multiplying
by 100.
x + y = 35,000
8x + 12y = 392,000
-8x -8y = -280,000 (We multiplied the above equation by -8.)
8x +12y = 392,000
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4y = 112,000
y = 28,000 (We divided both sides by 4 to find the value of y.)
Plugging the value of y (28,000) into the above formula, we can
now solve for the value of x:
x + 28,000 = 35,000
x = 7,000
So, now we know that Bill had invested as follows:
$28,000 (at 12%) = $3,360 (interest earned)
$7,000 (at 8%) = $560 (interest earned)
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$35,000 $3,920 (combined interest earned)
We can also divide $3,920 by 35,000 to find his interest average
for both investments: 11.2%