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Re: GRE Math Challenge #120-x < y < z [#permalink]
1
sandy 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]
2
josemu wrote:
sandy 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
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Re: GRE Math Challenge #120-x < y < z [#permalink]
GreenlightTestPrep wrote:
josemu wrote:
sandy 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]
josemu wrote:
sandy 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.
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Re: GRE Math Challenge #120-x < y < z [#permalink]
Hello from the GRE Prep Club BumpBot!

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Re: GRE Math Challenge #120-x < y < z [#permalink]
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