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I think question is asking w/x + y/z odd and not w/x + w/z odd or not? If this is the case then here is my explanation. Rephrased question is wz+xy/xz is odd or not?

Statement 1: wx + yz is odd, this implies one pair is odd and other pair is even. As even + odd = odd. But we are not sure which pair is even or odd. So this is not sufficient and answer cannot be A or D.

Statement 2; wz + xy is odd, this numerator of question is odd. As given in question w/x + y/z is integer so wz+xy/xz is not a fraction and this implies wz+xy is divisible by xz. Only way a odd number divisible by another number is that divisor has to be odd as well. So odd divided by odd will yield odd. So question is answered.

Re: If w, x, y, and z are integers such that w/x and y/z are [#permalink]

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13 Mar 2012, 12:06

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japped187 wrote:

If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + w/z odd?

(1) wx + yz is odd (2) wz + xy is odd

IMO A is also satisfactory .... wx + yz is odd

Case 1 wx is odd & yz is even wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd yz is even ==> There can be 3 cases y is even and z is odd = Not possible as we are given y/z is integer y is odd and z is even = Not possible as we are given y/z is integer y is even and z is even = Possible ==> y/z is even w/x (odd) + y/z (even) = Odd

Case 2 wx is even & yz is odd can be proved in a similar manner.

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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13 Mar 2012, 14:24

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If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

Given: \(\frac{w}{x}=integer\) and \(\frac{y}{z}=integer\). Hence, \(\frac{w}{x}+\frac{y}{z}=\frac{wz+yx}{xz}=integer\) and the question is whether this integer is odd.

(1) wx + yz is odd --> if \(w=x=1\) and \(y=z=2\) then \(\frac{w}{x}+\frac{y}{z}=2=even\) but if \(w=x=1\) and \(y=2\), \(z=1\) then \(\frac{w}{x}+\frac{y}{z}=3=odd\). Not sufficient.

(2) wz + yx is odd --> \(\frac{wz+yx}{xz}=\frac{odd}{xz}=integer\) --> \(odd=(xz)*integer\) --> all multiple must be odd in order the product to be odd, hence \(integer =odd\). Sufficient.

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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20 Sep 2012, 06:33

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Other way to look at the problem

As w/x is integer we can say w=xa+0 , where a= any integer------> xa/x = a Same way y/z is integer we can say y=zb+0 , where b= any integer------->zb/z = b

So the basically the question is, "Is a+b Odd"

1) wx + yz ----> x^2a + z^2b = odd------> a+b may be or may not be odd ---> Insufficient 2) wz + yx -----> xza + xzb ----> xz(a+b) = odd--->it means that both xz & (a+b) are odd ---->Sufficient

Answer B

Hope it helps. _________________

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Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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20 Sep 2012, 07:16

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rohitgoel15 wrote:

IMO A is also satisfactory .... wx + yz is odd

Case 1 wx is odd & yz is even wx is odd ==> w & x are odd (odd*odd = odd) ==> w/x is odd yz is even ==> There can be 3 cases y is even and z is odd = Not possible as we are given y/z is integer y is odd and z is even = Not possible as we are given y/z is integer y is even and z is even = Possible ==> y/z is even w/x (odd) + y/z (even) = Odd

Case 2 wx is even & yz is odd can be proved in a similar manner.

Please advice if i am wrong.

rohitgoel15 wrote:

Hi Bunuel,

I agree with ur exp .. but is there a problem with my algebraic method ?

Yes there is a problem in statement: y is even and z is odd = Not possible as we are given y/z is integer Even/Odd can be integer, Consider eg Y=6 , z=3.. Hence that solution is incorrect. _________________

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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02 Aug 2013, 08:33

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Hi All,

Problem helps us to understand the following: For \(\frac{w}{x}\) to be an integer, both w and x need to be either even or odd. Same is the case with y and z, both either need to be even or odd. We need, w/x + y/z to be of the following format:

ODD + EVEN, so of the two terms, one of them has to be even and the other to be odd.

Now, lets look the statements..

Statement (1): \(wx+yz\) is odd

From this statement we can deduce that either wx is odd and yz is even OR wx is even and yz is odd. Lets assume that wx is odd and yz is even. For the term wx to be odd and w/x to be an integer, both w and x needs to be odd. However, even though yz is even since both y and z are even, it doesn't guarantee that y/z will be even. (6/2 = 3)

So from statement 1, what we get is ODD + EVEN/ODD, hence not sufficient.

Statement (2): \(wz+yx\) is odd

We can simply multiply and divide by xz as below:

(w/x + y/z) * xz = ODD

Now since we know that only ODD * ODD = ODD, we can be certain that w/x + y/z is odd, hence sufficient.

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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29 Dec 2013, 04:43

Expert's post

Paris75 wrote:

HI,

Just for further discussion,

if AB is odd, is A/B also odd? What is the rule on this type of construction?

Thanks

You could try some examples to answer your own question.

For integers a and b, ab is odd only if both are odd. Now, odd/odd can be odd or not an integer at all. For example, 3/1=3=odd but 1/3 is not an integer at all. _________________

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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29 Dec 2013, 20:27

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japped187 wrote:

If w, x, y and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1) wx + yz is odd (2) wz + yx is odd

Plug in approach requiring minimal thinking:

Take the case that is to be proved i.e, w/x + y/z is odd and then take the contrary case i.e, w/x + y/z is even for both the statements

(i) Case to be proved : wx+ yz is odd and w/x + y/z is odd.

The former is satisfied by odd + even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wx and w/x are both odd and y/z and yz are both even. an example is w=15 , x=3 and y=4 and z=2

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wx and w/x are both odd and yz is even and y/z is odd. An example is w=9 x=3 and y=6 z=2

since the case to be proved and the contrary are both satisfied (i) is not sufficient

(ii) Case to be proved: wz+yx is odd and w/x and y/z is odd

The former is satisfied by odd+ even and the latter again by odd+ even

Thus this will be satisfied say, if the following is satisfied: wz and w/x are both odd and y/z and yx are both even. an example is w=15 , x=3 and y=12 and z=1

How about the contrary case for (i). i.e., wx+ yz is odd but w/x + y/z is even.

The former is satisfied by odd+ even and the latter by odd+odd or even +even

This will be satisfied say, if wz and w/x are both odd and yx is even and y/z is odd.

We find this cannot be satisfied

Since the contrary case cannot be proved for (ii) , this alone is sufficient and the answer is B. _________________

Re: If w, x, y and z are integers such that w/x and y/z are [#permalink]

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