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anujkm
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For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)


(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Answer D
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LM
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anujkm
Any help in solving these questions is appreciated ..


ANswer for the question 1 is straight away result frmo Wilson's Theroem. Please check Wilson's theorem in WIkipedia and remember it.
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anujkm
Any help in solving these questions is appreciated ..


Please remember this.... WILSON's THEOREM

p divides (p-1)!+1 if p is prime. The converse is also true. For instance, 7 is prime, and 7 divides (6!)+1=721





.
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LM
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joroivanov
I will skip the first Q as I have no explanation for the answer. It's crazy that you can get that kind of a Q so early:)

For #2:

Since students either like or dislike either food (i will abreviate them "LB" and "BS"), by knowing what they don't like you actually know what they like.

We need the # of students who dislike LB but like BS

You know that 2/3 of the students dislike LB and of those 2/5 like BS (1-3/5). So, in essence, all you need to know is the actual number of students who dislike LB.

(1) Sufficient. Knowing the total number of students in the cafeteria, you can find the # of the students who dislike LB. (120*2/3 = 80)

(2) Sufficient. Number of students who like LB = 1/3 of all students = 40.
You can use that to find the number of students who dislike LB (80).

Answer D


the key as you mentioned is by knowing what they don't like you actually know what they like.
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joroivanov
For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)


(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Answer D


Is there no possibility that Z might be odd number?? :roll:
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anujkm
Any help in solving these questions is appreciated ..


You've attached the same file twice. Please remove one of them.
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Question 1 has been extensively discussed in this forum. You can do a search and find the solutions.
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LM - I think not. :)

if zx is the product of an odd number z and an even number x, then what is stated in the question stem wouldn't hold true (z/y and y/x wouldn't be integers).

If we assume that z is odd, for zx to be even, x has to be even. But then for z/y to be an integer, y has to be odd. However, if y is odd, for y/x to be an integer x has to be odd, which contradicts our first assumption (that x is even).



LM
joroivanov
For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)


(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Answer D

Is there no possibility that Z might be odd number?? :roll:
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hsk
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Quote:
Question 1 has been extensively discussed in this forum. You can do a search and find the solutions.


does anyone have a link? I am having trouble finding the past discussion
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hsk
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can someone please explain number one or show me the link for the past discussions.



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