|
Question 54864: 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?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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?
:
Let x = amt of $9 water;
Let y = amt of $3 water;
Let z = amt of $4.50 water;
:
Wants a total of 300 gallons
x + y + z = 300
:
Write an equation for:
"must use twice as much of the $4.50 water as the $3.00 water"
z = 2y
:
In the 1st equation substitute 2y for z
x + y + 2y = 300
x + 3y = 300
:
Write an equation for:
"mix three grades of water selling for $9.00, $3.00 and $4.50 per gallon,"
for a total of 300 gal at $6 a gallon.
9(x) + 3(y) + 4.5(z) = 6(300)
:
Substitute 2y for z in the above equation:
9(x) + 3(y) + 4.5(2y) = 6(300)
;
Get rid of the brackets and you have:
9x + 3y + 9y = 1800
9x + 12y = 1800
:
Use the elimination method, two equations with two unknowns:
x + 3y = 300
9x + 12y = 1800
:
Mult the 1st equation by 4 and subtract, eliminating y
4x + 12y = 1200
9x + 12y = 1800
----------------- subtract
-5x + 0y = -600
x = -600/-5
x = +120 gal of $9 water
:
Use the equation x + 3y = 300 to find y
120 + 3y = 300
3y = 300 -120
y = 180/3
y = 60 gal of $3 water
:
Remember "must use twice as much of the $4.50 water as the $3.00 water"
Therefore we know that here must be 2 * 60 = 120 gal
:
z = 120 gal of the $4.50
:
Check:
9(120) + 3(60) + 4.5(120) = 6(300)
1080 + 180 + 540 = 1800 + 1800
|
|
|
| |
