MAC1105 Practice Question 113

Question

A​ sparkling-water distributor wants to make up 300 gal of sparkling water to sell for $6.00 per gallon. She wishes to mix three grades of water selling for ​$9​.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?

She should use _____gal of ​$9​.00, ___________ gal $3.00, and ________________

gal of​ $4.50.

Answer 1

Let:

-- x be the amount of $9 water

-- y be the amount of $3 water

-- z be the amount of $4.50 water

9.00x + 3.00y + 4.50z = 300(6.00) <-- Total cost based on x, y, and z gallons must be 300(6.00) = $1800

x + y + z = 300 <-- Total amount of gallons used must be 300

z = 2y <-- "She must use twice as much of the​ $4.50 water as the $3.00 water"

Substitute z = 2y into first and second equations

9x + 3y + 4.50(2y) = 1800

9x + 3y + 9y = 1800

9x + 12y = 1800

x + y + z = 300

x + y + 2y = 300

x + 3y = 300

Now we have a new system of equations to work with.

Use elimination method and solve for x

9x + 12y = 1800

x + 3y = 300

9x + 12y = 1800

4x + 12y = 1200

5x = 600

x = 120

Plug into the new equation

x + 3y = 300

120 + 3y = 300

3y = 180

y = 60

Plug y=60 into z=2y

z = 2y

z = 2(60)

z = 120

Therefore, she should use 120 gallons of $9 water, 60 gallons of $3 water, and 120 gallons of $4.50 water.

Hope this helped!

Answer 2

x @ $3

2x @ $4.50

300 - 3x @ $9

(3)(x) + (4.5)(2x) + (9)(900 - 3x) = (6)(300)

Can you solve for x and answer?