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Grades of milk

Expert replies
by srn » Sun Apr 05, 2009 6:05 am
Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of 2% grade and z gallons of 3% grade are mixed to give x+y+z gallons of a 1.5 % grade, what is x in terms of y and z.
a. y+3z
b.(y+z)/4
c.2y+3z
d.3y+z
e.3y+4.5z

Ans - y+3z.
Please explain how you came to the conclusion.
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Source: — Problem Solving |

by kapsii » Sun Apr 05, 2009 6:54 am
content of fat in x gallons of 1% grade milk => 0.01*x
content of fat in y gallons of 2% grade milk => 0.02*y
content of fat in z gallons of 3% grade milk => 0.03*z

total content of fat in the resulting mixture of 1.5% grade milk = grade*volume = 0.015 * (x+y+z)

so, our equation is:
0.015 (x+y+z) = 0.01*x + 0.02*y + 0.03*z
multiplying everything by 100, we get
1.5(x+y+z) = x + 2y + 3z
or 1.5x + 1.5y + 1.5z = x + 2y + 3z
or 0.5x = 0.5y + 1.5z
or x = y + 3z
Cheers,
Dubes
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by sanju09 » Wed Apr 08, 2009 3:09 am
Fat in milk is x*1%, y*2% and z*3%, respectively.

So we have the equation: x*1% + y*2% + z*3% =(x + y +z)*1.5%

Simplify the equation, we can obtain that x = y + 3 z
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by Brent@GMATPrepNow » Thu Oct 17, 2019 8:32 am
srn wrote:Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of 2% grade and z gallons of 3% grade are mixed to give x+y+z gallons of a 1.5 % grade, what is x in terms of y and z.
a. y+3z
b. (y+z)/4
c. 2y+3z
d. 3y+z
e. 3y+4.5z
Let's start with a word equation and slowly turn it into an algebraic expression:

Total fat in mixture = 1.5% of (x+y+z)
(1% of x) + (2% of y) + (3% of z) = 0.015(x+y+z)
Rewrite as: 0.01x + 0.02y + 0.03z = 0.015x + 0.015y + 0.015z
Multiply both sides by 100: 1x + 2y + 3z = 1.5x + 1.5y + 1.5z
Rearrange and simplify: 0.5y + 1.5z = 0.5x
Multiply both sides by 2 to get: y + 3z = x

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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