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

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Intern
Joined: 10 Apr 2006
Posts: 33
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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35% (medium)

Question Stats:

74% (02:17) correct 26% (01:24) wrong based on 313 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: three-grades-of-milk-are-1-percent-2-percent-and-3-percent-fat-126122.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Nov 2013, 09:06, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Director
Joined: 28 Dec 2005
Posts: 919

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01 Jan 2007, 08:12
5
KUDOS
vmalaiya wrote:
I am stumped

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
Intern
Joined: 10 Apr 2006
Posts: 33

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01 Jan 2007, 22:04
Thanks hsampath this helped... I messed up in translating the question into an equation
Manager
Joined: 20 Dec 2004
Posts: 177

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16 Jan 2007, 02:07
3
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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.
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Subhen

Manager
Joined: 09 Nov 2006
Posts: 128

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16 Jan 2007, 05:47
hsampath wrote:
vmalaiya wrote:
I am stumped

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
Intern
Joined: 10 Apr 2006
Posts: 33

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16 Jan 2007, 08:21

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

A good way to understand such problem is to create a table -

Vol of Sol % of milk Amount of milk
1. x 1 1*x/100
2. y 2 2*y/100
3. z 3 3*z/100

For new mixture -
x+y+z 1.5 1.5(x+y+z)/100

Since the amount of milk in the mixture is the same -

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

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

which gives
x = y + 3z
Intern
Joined: 10 Apr 2006
Posts: 33

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16 Jan 2007, 08:25
6
KUDOS
A better formatted table....

A good way to understand such problem is to create a table -

----Vol of Sol------------% of milk----------- Amount of milk

-------x---------------------1----------------------- 1*x/100
-------y---------------------2------------------------2*y/100
-------z---------------------3------------------------3*z/100

For new mixture -
---x+y+z------------------1.5------------------1.5(x+y+z)/100

Since the amount of milk in the mixture is the same -

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

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

which gives
x = y + 3z
Intern
Joined: 24 Jun 2007
Posts: 1

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24 Jun 2007, 13:17
Thanks everyone for the contributions, I was stumped on this one, too.
CIO
Joined: 09 Mar 2003
Posts: 463

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24 Jun 2007, 15:34
1
KUDOS
vmalaiya wrote:
A better formatted table....

A good way to understand such problem is to create a table -

----Vol of Sol------------% of milk----------- Amount of milk

-------x---------------------1----------------------- 1*x/100
-------y---------------------2------------------------2*y/100
-------z---------------------3------------------------3*z/100

For new mixture -
---x+y+z------------------1.5------------------1.5(x+y+z)/100

Since the amount of milk in the mixture is the same -

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

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

which gives
x = y + 3z

This is exactly how I do mixture problems, too. I find it works best and lines up all the elements perfectly.
Director
Joined: 25 Oct 2008
Posts: 596
Location: Kolkata,India

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14 Aug 2009, 18:15
1
KUDOS
Mixture problems usually stump me but this was one of the easier ones !!:)
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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Manager
Joined: 11 May 2009
Posts: 125
Schools: Columbia Business School, Goizueta, Sloan

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15 Aug 2009, 07:01
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vmalaiya wrote:
x+2y+3z = 1.5(x+y+z)

which gives
x = y + 3z

I guess I am solving this the wrong way? Can you please write down how did you go from the first line above, to the second?
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Set your aims high, carpe diem

Manager
Joined: 09 Aug 2009
Posts: 51

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18 Aug 2009, 00:26
hsampath wrote:
vmalaiya wrote:
I am stumped

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

/Prabu
Manager
Joined: 10 Jul 2009
Posts: 127
Location: Ukraine, Kyiv

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23 Sep 2009, 10:07
very-very good Q!

(0.01x+0.02y+0.03z)/(x+y+z)=0.015
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Never, never, never give up

Manager
Joined: 27 Oct 2008
Posts: 185

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23 Sep 2009, 11:22
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
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Joined: 09 Sep 2013
Posts: 15915
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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Posts: 39589
Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink]

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25 Nov 2013, 09:07
OPEN DISCUSSION OF THIS QUESTION IS HERE: three-grades-of-milk-are-1-percent-2-percent-and-3-percent-fat-126122.html
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent   [#permalink] 25 Nov 2013, 09:07
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