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GRE Math Challenge #120-x < y < z

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GRE Math Challenge #120-x < y < z [#permalink] New post 15 May 2015, 11:23
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Question Stats:

61% (00:32) correct 38% (00:43) wrong based on 39 sessions
\(x < y < z\)

Quantity A
Quantity B
\(\frac{(x+y+z)}{3}\)
\(z\)



• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Last edited by Carcass on 11 Aug 2018, 22:49, edited 1 time in total.
Edited by Carcass
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Re: GRE Math Challenge #120 [#permalink] New post 17 May 2015, 09:26
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sandy Array[WROTE]:
\(x < y < z\)

Quantity A: (x+y+z)/3
Quantity B: z


Nice question!

IMPORTANT: Notice that (x+y+z)/3 is equal to the AVERAGE of x, y and z.
Since x < y < z, the average of these 3 values must be greater than x and less than z.
Since the average of the 3 values must be less than z, we can conclude that (x+y+z)/3 is definitely less than z

Answer:
[Reveal] Spoiler:
B


Cheers,
Brent
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Re: GRE Math Challenge #120-x < y < z [#permalink] New post 11 Aug 2018, 11:51
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sandy Array[WROTE]:
\(x < y < z\)

Quantity A: (x+y+z)/3
Quantity B: z

• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined



x<y<z. we don't know whether they are consecutive or not. Even if they are consecutive average will always be lower than the highest numbers. So, z is greater.

The best answer is B.
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Re: GRE Math Challenge #120-x < y < z [#permalink] New post 13 Aug 2018, 16:52
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sandy Array[WROTE]:
\(x < y < z\)

Quantity A
Quantity B
\(\frac{(x+y+z)}{3}\)
\(z\)



• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined


Hi Guys,

I don´t know if I am missing something on the statement of the question. I understand the reasoning mentioned regarding the average, nonetheless, if we pick the numbers: -1<1<9
wouldn't the answer be D since (-1+1+9)/3 = 3 which is the same as z

Cheers and thanks for any explanation I am missing
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Re: GRE Math Challenge #120-x < y < z [#permalink] New post 13 Aug 2018, 19:12
2
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Expert's post
josemu Array[WROTE]:
sandy Array[WROTE]:
\(x < y < z\)

Quantity A
Quantity B
\(\frac{(x+y+z)}{3}\)
\(z\)



• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined


Hi Guys,

I don´t know if I am missing something on the statement of the question. I understand the reasoning mentioned regarding the average, nonetheless, if we pick the numbers: -1<1<9
wouldn't the answer be D since (-1+1+9)/3 = 3 which is the same as z

Cheers and thanks for any explanation I am missing


Be careful. In your example, z = 9, and quantity A = 3

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com
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Re: GRE Math Challenge #120-x < y < z [#permalink] New post 13 Aug 2018, 19:36
GreenlightTestPrep Array[WROTE]:
josemu Array[WROTE]:
sandy Array[WROTE]:
\(x < y < z\)

Quantity A
Quantity B
\(\frac{(x+y+z)}{3}\)
\(z\)



• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined


Hi Guys,

I don´t know if I am missing something on the statement of the question. I understand the reasoning mentioned regarding the average, nonetheless, if we pick the numbers: -1<1<9
wouldn't the answer be D since (-1+1+9)/3 = 3 which is the same as z

Cheers and thanks for any explanation I am missing


Be careful. In your example, z = 9, and quantity A = 3

Cheers,
Brent


Hi Brent,

You are absolutely right and now that I am coming back at it, everything makes sense (about the average rule). It is what happens when you try too much at something and forget sometimes the "obvious" and you can´t see past an idea. It is always good to take a step back and look at it. Thanks !
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Re: GRE Math Challenge #120-x < y < z [#permalink] New post 15 Aug 2018, 17:19
josemu Array[WROTE]:
sandy Array[WROTE]:
\(x < y < z\)

Quantity A
Quantity B
\(\frac{(x+y+z)}{3}\)
\(z\)



• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined


Hi Guys,

I don´t know if I am missing something on the statement of the question. I understand the reasoning mentioned regarding the average, nonetheless, if we pick the numbers: -1<1<9
wouldn't the answer be D since (-1+1+9)/3 = 3 which is the same as z

Cheers and thanks for any explanation I am missing


In your example, the value of z is 9, while the average is 3.
Re: GRE Math Challenge #120-x < y < z   [#permalink] 15 Aug 2018, 17:19
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