Question

A sparkling-water distributor wants to make up 200 gallons of sparkling water to sell for $5.00 per gallon. She wishes to mix three grades of water selling for $8.00, $3.00, and $4.50 per gallon, respectively. She must use twice as much of the $4.50 water as the $3.00, water. How many gallons of each should she use?

Answer 1

The distributor should mix approximately 107.69 gallons of $8.00 water, 30.77 gallons of $3.00 water, and 61.54 gallons of $4.50 water to create 200 gallons of sparkling water that can be sold for $5.00 per gallon.

Step-by-step explanation:

A sparkling-water distributor wants to mix three grades of water to make 200 gallons of sparkling water to sell for $5.00 per gallon. Let's denote the gallons of $8.00 water as x, the gallons of $3.00 water as y, and the gallons of $4.50 water as 2y (as she must use twice as much of the $4.50 water as the $3.00). The total volume of the mix should equal 200 gallons, while the total cost of the mix should equal 200 times $5.00, which is $1000.

Setting up the equations based on the total volume and the total cost, we have:

• x + y + 2y = 200 (Total volume equation)
• 8x + 3y + 4.5(2y) = 1000 (Total cost equation)

The total volume equation simplifies to x + 3y = 200. Multiplying this equation by 8 gives 8x + 24y = 1600. To find the value of y, we can subtract the total cost equation from this, leaving us with 19.5y = 600. Thus, y = 600 / 19.5, which equals approximately 30.77 gallons. Consequently, 2y equals approximately 61.54 gallons. Now, we substitute y back into the total volume equation to find x: x + 3(30.77) = 200, solving this gives x as approximately 107.69 gallons.

Therefore, the distributor should use approximately 107.69 gallons of the $8.00 water, 30.77 gallons of the $3.00 water, and 61.54 gallons of the $4.50 water.

Answer 2

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.