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Three Grades of milk are 1 percent, 2 percent and 3 percent

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by [email protected] » Thu Nov 11, 2010 1:57 am
I understand your solution but this is how i was doing it and for some reason i am getting a wrong answer. Can someone point out to me why i am wrong.

For every 1 unit of Z i would need to add 3 units of X to keep the ratio 1.5 and for every unit of y i would have to add equal amount of x.
converting it to equations,
3x=z
x=y
4x=y+z and hence answer is b.
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by Rezinka » Thu Nov 11, 2010 2:43 am
The way I did it :
ATQ :
1*x + 2*y + 3*z = 1.5*(x+y+z)
or, x + 2y + 3z = 1.5x + 1.5y + 1.5z
or, x - 1.5x = 1.5y - 2y + 1.5z - 3z
or, -0.5x = - 0.5y - 1.5z
or, x = y+3z (divide the above equation by -0.5)
ANS : A
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by GMATGuruNY » Thu Nov 11, 2010 3:47 am
lunarpower wrote:courtesy of user 'karenmeow'

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 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z
I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.
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by goyalsau » Thu Nov 11, 2010 9:17 am
lunarpower wrote:courtesy of user 'karenmeow'

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 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

1. y+ 3z
2. y+z/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z


Ron Please explain........ What's wrong with this. { I am confused. }

Just assume there are
100 ml of y means 2 ml of fat. ( 2 % fat. )
100 ml of z means 3 ml of fat. ( 3% fat. )
400 ml of x means 4 ml of fat. ( 1 % fat )

in all we have 600 ml of solution and 9 ml of fat.

This means 9/600 is 1.5 % of fat,

by this means Both the options are correct option 1 and 4.

1. y+ 3z 100 ml of y + 300 ml of z

4. 3y + z 300 ml of y + 100 ml of z
Saurabh Goyal
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by MAAJ » Thu Nov 11, 2010 10:36 am
I find it easier to solve this by doing the equations:

0.01x+0.02y+0.03z = 0.015 (x+y+z) [multiplied by 1000]
10x+20y+30z = 15 (x+y+z) [divided by 5]
2x+4y+6z = 3 (x+y+z)
2x+4y+6z = 3x+3y+3z
4y+6z-3y-3z = 3x-2x
y + 3z = x and voila!

Correct Answer: [spoiler](A)[/spoiler]
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by lunarpower » Sat Nov 20, 2010 3:02 am
goyalsau wrote:
lunarpower wrote:courtesy of user 'karenmeow'

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 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

1. y+ 3z
2. y+z/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z


Ron Please explain........ What's wrong with this. { I am confused. }

Just assume there are
100 ml of y means 2 ml of fat. ( 2 % fat. )
100 ml of z means 3 ml of fat. ( 3% fat. )
400 ml of x means 4 ml of fat. ( 1 % fat )

in all we have 600 ml of solution and 9 ml of fat.

This means 9/600 is 1.5 % of fat,

by this means Both the options are correct option 1 and 4.

1. y+ 3z 100 ml of y + 300 ml of z

4. 3y + z 300 ml of y + 100 ml of z
well ... nothing is wrong with that.

when you pick your own numbers, you run the risk of coincidence.
the correct answer will ALWAYS be correct. some incorrect answers will SOMETIMES be correct, if certain numbers are picked.


so:
* pick the numbers
* solve the problem
* eliminate anything that's incorrect
* keep anything that's correct
* if there are 2 or more choices left, pick OTHER numbers and do the problem AGAIN

... unless you don't have time; in that case, just guess.

--

here's a really stupid example, just to get the point across:

What is x + x?
(a) 2x
(b) x^2
obviously, the answer is (a), but if you plug in x = 0 or x = 2 you'll find that both of them work. in that case, just plug in another x and try again.
Ron has been teaching various standardized tests for 20 years.

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On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

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by RadiumBall » Fri Mar 25, 2011 10:32 pm
Sure this is easy. But I did this algebraically just as MAAJ.

However I have one question:
How do you find the wieghts here? I assumed the percents to be wieghts and got the whole thing wrong.

(x * 1 + y * 2 + z * 3) / (1 + 2 + 3) = 1.5
and I got x = 9 - 2y - 3z (wrong)

But if I did
(x * 1 + y * 2 + z * 3) / (x + y + z) = 1.5
then I get x = y + 3 z (correct)

Could anyone kindly clarify?
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by force5 » Sat Mar 26, 2011 4:53 am
yes correct A is the answer...x=y+3z
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by force5 » Sat Mar 26, 2011 4:53 am
yes correct A is the answer...x=y+3z
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by GmatKiss » Fri Nov 04, 2011 3:46 am
GMATGuruNY wrote:
lunarpower wrote:courtesy of user 'karenmeow'

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 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z
I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.
Hi Mitch,

What is the reason to choose Z=0 ? and X=Y=2
Missing something here! Please help.

TIA,
GK
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by GMATGuruNY » Fri Nov 04, 2011 6:40 am
GmatKiss wrote:
GMATGuruNY wrote:
lunarpower wrote:courtesy of user 'karenmeow'

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 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

1. y+ 3z
2. (y+z)/4
3. 2y+3z
4. 3y + z
5. 3y + 4.5z
I too would plug in values. It helps to recognize that since x=1% and y=2%, if we use the same amounts of x and y and none of z, the result will be a mixture that is 1.5%.

Plug in x=2, y=2, z=0. Since the question is asking for the value of x, our target answer is x=2.

A) y+3z = 2 + 3*0 = 2. Hold onto A.
B) (y+z)/4 = (2+0)/4 = 1/2. Eliminate B.
C) 2y + 3z = 2*2 + 3*0 = 4. Eliminate C.
D) 3y+z = 3*2 + 0 = 6. Eliminate D.
E) 3y + 4.5z = 3*2 + (4.5)*0 = 6. Eliminate E.

The correct answer is A.

The trick to plugging in is to choose numbers that make the math easy.
Hi Mitch,

What is the reason to choose Z=0 ? and X=Y=2
Missing something here! Please help.

TIA,
GK
X=1%, Y=2%, Z=3%.
It would involve quite a bit of trial and error to come up with values for X,Y and Z so that X+Y+Z yields a mixture that is 1.5%.
It is FAR easier to come up with values for only TWO of the ingredients.
Thus:
Let Z=0, implying that no Z will be in the mixture.
Thus, the mixture will consist only of X and Y.
Since 1.5% is HALFWAY between X (1%) and Y (2%), combining equal amounts of X and Y will yield a mixture that is 1.5%.
To make the math easy, I chose X=2 and Y=2.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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