9514 1404 393
Answer:
3. $6000 at 8%; $5000 at 5%
4. 66 2/3 gallons of 9%; 133 1/3 gallons of 6%
Explanation:
For many mixture problems, it works well to let a variable represent the amount of the largest contributor used in the mix.
3. Let x represent the amount invested at 8%. Then the total interest is ...
8%x +5%(11000-x) = 730
3%x = 730 -550
x = 180/0.03 = 6000
$6000 was invested at 8%; $5000 was invested at 5%.
__
4. Let x represent the number of gallons of 9% content. Then the alcohol content of the mix is ...
9%x +6%(200 -x) = 7%(200)
3%x = 1%(200) . . . . . subtract 6%(200)
x = 200/3 = 66 2/3 . . . . . divide by 3%
200 -x = 133 1/3
66 2/3 gallons of 9% wine and 133 1/3 gallons of 6% wine should be used.