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GRE Math Challenge #120x < y < z [#permalink]
15 May 2015, 11:23
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Question Stats:
68% (00:39) correct
32% (00:36) wrong based on 25 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
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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]
17 May 2015, 09:26
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sandy 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 zAnswer: Cheers, Brent
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Re: GRE Math Challenge #120x < y < z [#permalink]
11 Aug 2018, 11:51
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sandy 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 #120x < y < z [#permalink]
13 Aug 2018, 16:52
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 #120x < y < z [#permalink]
13 Aug 2018, 19:12
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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 zCheers 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 #120x < y < z [#permalink]
13 Aug 2018, 19:36
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 zCheers 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 #120x < y < z [#permalink]
15 Aug 2018, 17:19
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.




Re: GRE Math Challenge #120x < y < z
[#permalink]
15 Aug 2018, 17:19





