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Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink]

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01 Jan 2007, 07:58

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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?

A. y + 3z B. (y +z) / 4 C. 2y + 3z D. 3y + z E. 3y + 4.5z

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallon of 1 percent grade, y gallon of 2 percent grade, and z gallon of 3 percent grade are mixed to give x + y + z gallons of 1.5 percent 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

Is'nt it amazing that sometimes just because of the wording of the question easy questions become difficult. I can't believe I spent over 5 mins trying to solve it in my GMAT Prep and yet got it wrong. I hope I learn from it & don't commit such mistakes in the actual exam. _________________

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallon of 1 percent grade, y gallon of 2 percent grade, and z gallon of 3 percent grade are mixed to give x + y + z gallons of 1.5 percent 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

The question can be rewritten as:

(x/100+2/100*y+3/100*z)/(x+y+z) = 1.5/100

=> x+2y+3z = 1.5(x+y+z)

solving, we get x = y+3z

hsampath, shouldn't it be x+2y+3z = 1.5/(x+y+z)?

(x+2y+3z/100) * (x+y+z) = 1.5/100

We can get rid of /100 =>
(x+2y+3z)(x+y+z) = 1.5 =>

(x+2y+3z) = 1.5/(x+y+z)

Well, actually I am not sure. Please correct me if I am wrong. Thanks

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallon of 1 percent grade, y gallon of 2 percent grade, and z gallon of 3 percent grade are mixed to give x + y + z gallons of 1.5 percent 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

The question can be rewritten as:

(x/100+2/100*y+3/100*z)/(x+y+z) = 1.5/100

=> x+2y+3z = 1.5(x+y+z)

solving, we get x = y+3z

hsampath, shouldn't it be x+2y+3z = 1.5/(x+y+z)?

(x+2y+3z/100) * (x+y+z) = 1.5/100

We can get rid of /100 => (x+2y+3z)(x+y+z) = 1.5 =>

(x+2y+3z) = 1.5/(x+y+z)

Well, actually I am not sure. Please correct me if I am wrong. Thanks

Yes, Basically both side 100 doesn't make any sense. We can directly say x+2y+3z/(x+y+z)=1.5

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallon of 1 percent grade, y gallon of 2 percent grade, and z gallon of 3 percent grade are mixed to give x + y + z gallons of 1.5 percent 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

Soln: The resulting equation is

=> (.01x + .02y + .03z)/(x+y+z) = 1.5/100 => x + 2y + 3z = 1.5x + 1.5y + 1.5z taking x to one side and y and z to other side we get => x = y + 3z Ans is a

Re: I am stumped Three grades of milk are 1 percent, 2 [#permalink]

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25 Nov 2013, 08:56

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