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Re: DS: GMATPrep xyz [#permalink]
09 Jun 2008, 06:48
1) xyz = 70 decompose 70 into its prime factors: 70 = 2 x 5 x 7 this is the only product of 3 integers greater than 1 that gives 70 => x+y+z = 2+5+7 = 14 => 1) is sufficient
2) x/zy = 7/10 doesn't say much about x, y and z. Take any x, y and z that satisfy the equation above, multiply x by a number and z or y by the same number you would still have the equality above but a different sum. => 2 is not sufficient
Re: DS: GMATPrep xyz [#permalink]
09 Jun 2008, 07:23
The question stem says "integers greater than 1". 1 is not greater than 1. It would be possible if the question asked for positive integers, or integers >= 1, etc. But the way it is written, 1 is not an option.
Why wouldn't D be the answer, "Both statements independently are sufficient to answer the question."
If xyz = 70 is sufficient, and then #2 says \(\frac{x}{yz}=\frac{7}{10}\)
#2 is saying the same thing as #1. It tells you the value of x = 7. and yz = 10. The only possible integers > 1 with product of 10 is 2 & 5. #2 is sufficient as well.
Answer D
j allen morris
rino wrote:
what about 1*14*5 or 10*1*7
the answer is E
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: DS: GMATPrep xyz [#permalink]
09 Jun 2008, 13:09
Correct. 1 is not "greater than 1". x/yz = 7 / 10 doesn't say anything about the values of x, y and z.
Here's an example: x = 7, y = 2, z = 5 sum = 14 x/yz = 7 / (2x5) = 7 / 10 => fine
now pick x = 70, y = 2, z = 50 (notice i multiplied x and z by 10... which would be later on simplified) sum = 122 x/yz = 70 / (2x50) = 70/100 = 7 / 10
jallenmorris wrote:
The question stem says "integers greater than 1". 1 is not greater than 1. It would be possible if the question asked for positive integers, or integers >= 1, etc. But the way it is written, 1 is not an option.
Why wouldn't D be the answer, "Both statements independently are sufficient to answer the question."
If xyz = 70 is sufficient, and then #2 says \(\frac{x}{yz}=\frac{7}{10}\)
#2 is saying the same thing as #1. It tells you the value of x = 7. and yz = 10. The only possible integers > 1 with product of 10 is 2 & 5. #2 is sufficient as well.
Re: DS: GMATPrep xyz [#permalink]
09 Jun 2008, 13:44
GMATBLACKBELT,
Good call on that. I wasn't think of having to reduce x/yz to get 7/10. I was taking what they gave us and not thinking beyond that. That's a mistake on the GMAT (as long as we don't over-think it;)) _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: DS: GMATPrep xyz [#permalink]
12 Jun 2008, 04:03
businessm wrote:
Correct. 1 is not "greater than 1". x/yz = 7 / 10 doesn't say anything about the values of x, y and z.
Here's an example: x = 7, y = 2, z = 5 sum = 14 x/yz = 7 / (2x5) = 7 / 10 => fine
now pick x = 70, y = 2, z = 50 (notice i multiplied x and z by 10... which would be later on simplified) sum = 122 x/yz = 70 / (2x50) = 70/100 = 7 / 10
jallenmorris wrote:
The question stem says "integers greater than 1". 1 is not greater than 1. It would be possible if the question asked for positive integers, or integers >= 1, etc. But the way it is written, 1 is not an option.
Why wouldn't D be the answer, "Both statements independently are sufficient to answer the question."
If xyz = 70 is sufficient, and then #2 says \(\frac{x}{yz}=\frac{7}{10}\)
#2 is saying the same thing as #1. It tells you the value of x = 7. and yz = 10. The only possible integers > 1 with product of 10 is 2 & 5. #2 is sufficient as well.
Answer D
j allen morris
rino wrote:
what about 1*14*5 or 10*1*7
the answer is E
OA is A. Thanks for your explanation!
gmatclubot
Re: DS: GMATPrep xyz
[#permalink]
12 Jun 2008, 04:03