Assuming she make no profit and no loss from the business.
Let the number of gallons of the $8 grade water used be x, that of the $3 grade water, y, and that of the $4.50 grade water be z, then:
x + y + z = 200 . . . (1)
8x + 3y + 4.5z = 200(5) = 1,000 . . . (2)
z = 2y . . . (3)
Putting equation (3) into equations (1) and (2), we have:
x + y + 2y = 200
or x + 3y = 200 . . . (4)
and
8x + 3y + 4.5(2y) = 1000
or 8x + 3y + 9y = 1000
or 8x + 12y = 1000 . . . (5)
Multiplying equation (4) by 4, we have:
4x + 12y = 800 . . . (6)
Subtracting equation (6) from equation (5), we have:
4x = 200
or x = 200 / 4 = 50
Substituting for x into equation (4), we have:
50 + 3y = 200
or 3y = 200 - 50 = 150
or y = 150 / 3 = 50
Substituting for z into equation (3) gives:
z = 2(50) = 100
Therefore, 50 gallons each of the $8 grade water and the $3 grade water should be used and 100 gallons of the $4.50 grade water.
Answer:
Step-by-step explanation:
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-7/10 -3/4 8/10 3/4
least/middle/middle/greatest
