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If x, y and z are integers greater than 1, what is the value

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If x, y and z are integers greater than 1, what is the value [#permalink] New post 09 Jun 2008, 06:43
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A
B
C
D
E

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If x, y and z are integers greater than 1, what is the value of x + y + z?
1) xyz = 70
2) x/yz = 7/10
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Re: DS: GMATPrep xyz [#permalink] New post 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

=> answer is A)
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Re: DS: GMATPrep xyz [#permalink] New post 09 Jun 2008, 07:03
what about 1*14*5 or 10*1*7

the answer is E
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Re: DS: GMATPrep xyz [#permalink] New post 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

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Re: DS: GMATPrep xyz [#permalink] New post 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.

Answer D

j allen morris
rino wrote:
what about 1*14*5 or 10*1*7

the answer is E
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Re: DS: GMATPrep xyz [#permalink] New post 09 Jun 2008, 13:18
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lexis wrote:
If x, y and z are integers greater than 1, what is the value of x + y + z?
1) xyz = 70
2) x/yz = 7/10


A.

1: only can be 7,2,5

2: could be 7/(2*5) or 14/(2*10)
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Re: DS: GMATPrep xyz [#permalink] New post 09 Jun 2008, 13:29
i too Get A..

2) is insuff because x/yz=7/10 can easily be x=14, yz=20... or x=7 yz=10
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Re: DS: GMATPrep xyz [#permalink] New post 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;))
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Re: DS: GMATPrep xyz [#permalink] New post 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!
Re: DS: GMATPrep xyz   [#permalink] 12 Jun 2008, 04:03
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