Question

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x y z gallons of a 1.5 percent grade, what is x in terms of y and z

Answer 1

To determine the gallons of 1% milk (x) needed in the mixture to achieve a 1.5% fat concentration when mixing with 2% (y) and 3% (z) milk, we set up an equation based on the total fat content, which leads to the formula x = -(y + 3z).

Step-by-step explanation:

We have three grades of milk with different fat percentages. To find how many gallons of the 1 percent milk (x) are needed when mixed with y gallons of 2 percent milk and z gallons of 3 percent milk for a mix with a 1.5 percent fat content, we can set up an equation based on the total fat in the mixture:

Total fat from 1% milk + Total fat from 2% milk + Total fat from 3% milk = Total fat in final mixture

0.01x + 0.02y + 0.03z = 0.015(x + y + z)

This simplifies to:

0.01x + 0.02y + 0.03z = 0.015x + 0.015y + 0.015z

Rearranging terms to solve for x gives:

0.005x = 0.015y + 0.015z - 0.02y - 0.03z

Which further simplifies to:

0.005x = -0.005y - 0.015z

So:

x = -(y + 3z)

Hence, the number of gallons of 1 percent milk needed (x) in terms of gallons of the other two milks (y and z) is x = -(y + 3z).

Answer 2

x=y+3z

Step-by-step explanation:

If we mixed the three grades of milk, the total percentage of fat by volume is given by:

(1%*x)+(2%*y)+(3%*z)

Where x is the number of gallons of 1% grade, y is the number of gallons of 2% grade and z is the number of gallons of 3%

We want that these percentage be equal to 1.5% of the total volume, so we create the following equation:

1.5%( x + y + z ) = (1%*x) + (2%*y )+ (3%*z)

Solving for x, we get:

1.5%*x + 1.5%*y + 1.5%*z = 1%*x + 2%*y + 3%*z

1.5%*x - 1%*x = 2%*y + 3%*z - 1.5%*y - 1.5%*z

0.5%*x = 0.5%*y + 1.5%*z

x = y + 3z

Then x in terms of y and z is x=y+3z