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

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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?

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

Originally posted by japped187 on 01 Jun 2008, 05:41.
Last edited by Bunuel on 21 Oct 2019, 10:24, edited 2 times in total.
Edited the question and added the OA
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post 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.

Answer: B.

Hope it's clear.
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post 01 Jun 2008, 09:00
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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.

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

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New post 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.

Please advice if i am wrong.
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post 14 Mar 2012, 01:02
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Hi Bunuel,

I agree with ur exp .. but is there a problem with my algebraic method ?
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post 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 integers, is  [#permalink]

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

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New post 31 Jul 2013, 07:33
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St 2 can also be proved in the following way.

w/x + y/z = something

=> (wz+xy)/xz = something
=> wz+xy = A -----------------------> Is xz*something odd ??

As per st 2 , wz+xy = odd .

Hence B

Hope there are no loopholes in my understanding.


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

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New post 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.

So my response, B
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post 27 Dec 2013, 07:48
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HI,

Just for further discussion,

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

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

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

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

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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


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

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New post Updated on: 05 Sep 2018, 14:08
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Hi,

Some important observations that could be helpful if we internalize this.

\(\frac{ODD}{EVEN}\) = never an integer and remainder is always odd.
\(\frac{3}{2}\) = never an integer value and reminder is 1,
\(\frac{191}{60}\) = never an integer remainder is 11.

\(\frac{ODD}{ODD}\) = either Odd Integer or never integer, then remainder can be either Odd or even
\(\frac{3}{1}\)= Odd integer remainder 0,
\(\frac{3}{5}\)= not an integer , remainder is 3,
\(\frac{5}{3}\)= not an integer and reminder is 2.


\(\frac{EVEN}{EVEN}\)= either Integer ( could be even or could be odd ) or could not be an integer, then remainder is always even.
\(\frac{6}{2}\)= 3 which is odd integer and remainder is 0 which is even .
\(\frac{12}{6}\)= 6 which is even integer and remainder is 0 which is even.
\(\frac{14}{4}\)= not an integer value and remainder is 2 which is even

\(\frac{EVEN}{ODD}\)= even integer or not an integer,then remainder is even or odd.
\(\frac{12}{3}\)= 4 even integer and remainder is 0
\(\frac{12}{5}\)= not an integer & reminder is 2 which is even
\(\frac{12}{7}\)= not an integer & remainder is 5 which is Odd


Now back to question,
we are given that w,x,y,z are integers and \(\frac {w}{x}\)and \(\frac{y}{z}\) are integers is \(\frac {w}{x}\)+ \(\frac{y}{z}\) = odd?

So lets say \(\frac{w}{x}\)= a , & \(\frac{y}{z}\) = b
then is a+b = odd integer . { note a, b are integers, and integer+integer= integer}
if we rearrange the terms \(\frac{wz+xy}{zx}\) = odd integer

we know that \(\frac{wz+xy}{zx}\)= integer,then we can write wz+xy= zx( integer). let this integer be denoted by k.

wz+xy= zx(k) -------(i)

Stm1 :wx+yz is odd.

adding wx+yz on both sides of equation(i)
then, wx+yz+wz+xy= k(zx)+wx+yz
so odd+ wz+xy= k(zx)+odd
or odd-odd+wz+xy= k(zx)
or even +wz+xy= k(zx)

so if wz+xy = odd then yes, if wz+xy = even then no.
So stmt 1 Insufficient .


Stmt 2: wz+xy = odd.
substituting in equation (i) we have
odd= k(zx) ,this implies k(zx) is odd { Note neither K, nor z or x is even}
So stmt 2 Sufficient .

Though , i did not use the observations above, but felt useful sharing as many Friends out here were using this approach.

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Originally posted by Probus on 29 Aug 2018, 23:20.
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Re: If w, x, y and z are integers such that w/x and y/z are integers, is  [#permalink]

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New post Updated on: 03 Jul 2019, 19:50
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


\(\frac{w}{x} + \frac{y}{z} = \frac{wz + xy}{xz}\)
Since \(\frac{w}{x}\) and \(\frac{y}{z}\) are integers, their SUM on the left side of the equation above must also be an integer.
Implication:
\(\frac{wz + xy}{xz}\) = INTEGER

Statement 1: wx + yz = odd
Case 1: w=1, x=1, y=2 and z=2, with the result that wx + yz = 1*1 + 2*2 = 5
In this case, \(\frac{w}{x} +\frac{y}{z} = \frac{1}{1} + \frac{2}{2} = 2\), so the answer to the question stem is NO.

Case 2: w=1, x=1, y=2, and z=1, with the result that wx + yz = 1*1 + 2*1 = 3
In this case, \(\frac{w}{x} + \frac{y}{z} = \frac{1}{1} + \frac{2}{1} = 3\), so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2: wz + xy = ODD
Consider the following cases:
15 is an integer and \(\frac{15}{3}\) is an integer, so \(\frac{15}{3}\) is a FACTOR OF 15.
21 is an integer and \(\frac{21}{7}\)is an integer, so \(\frac{21}{7}\) is a FACTOR OF 21.
Using the same reasoning:
wz + xy is an integer and \(\frac{wz + xy}{xz}\) is an integer, so \(\frac{wz + xy}{xz}\) is a FACTOR of wz + xy.
Since wz + xy is odd, all of its factors must be odd.
Thus,\(\frac{wz + xy}{xz}\) must be odd.
Since \(\frac{wz + xy}{xz} = \frac{w}{x} + \frac{y}{z}\) = ODD, the answer to the question stem is YES.


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

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New post 03 Jul 2019, 19:44
Hi All,

We're told that W, X, Y and Z are INTEGERS such that W/X and Y/Z are INTEGERS. We're asked if W/X + Y/Z is ODD. This is a YES/NO question and can be solved with a mix of TESTing VALUES and Number Properties. While this question is certainly more complex than a typical DS prompt, the basic Number Property rules involved are just about multiplication, division and addition. Since we'll be doing a lot of 'dividing' in this question, it would likely be easiest to test numbers that divide easily into other numbers (such as 1 or 2) for the values of X and Z.

To start, since W/X is an INTEGER, we know that W is a MULTIPLE of X (by extension, X is a FACTOR of W). In that same way, since Y/Z is an INTEGER, we know that Y is a MULTIPLE of Z (by extension, Z is a FACTOR of Y).

For W/X + Y/Z to be ODD, we know that one of those fractions would have to be Even and the other would have to be Odd.

(1) (W)(X) + (Y)(Z) is ODD

With the information in Fact 1, we know that one of the products MUST be Even and the other MUST be ODD. The only way for that ODD product to occur is if the two variables in the product are BOTH ODD.

IF we use ...

(1)(1) + (2)(2), then the value of W/X + Y/Z = 1/1 + 2/2 = 2 and the answer to the question is NO.
(2)(1) + (3)(1), then the value of W/X + Y/Z = 2/1 + 3/1 = 5 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

(2) (W)(Z) + (Y)(X) is ODD

With the information in Fact 2, we know that one of the products MUST be Even and the other MUST be ODD. The only way for that ODD product to occur is if the two variables in the product are BOTH ODD. Thus, either 2 or 3 of the four variables MUST be ODD. We also have to keep in mind that W is a MULTIPLE of X and Y is a MULTIPLE of Z.

The second example we used in Fact 1 also fits Fact 2 (be careful to assign each value to the proper variable):

(2)(1) + (3)(1), then the value of W/X + Y/Z = 2/1 + 3/1 = 5 and the answer to the question is YES. This example has 3 ODD variables.

Even numbers do NOT divide into Odd numbers - and remember that W/X and Y/Z have to be INTEGERS, so if we had just 2 ODD variables, then they would have to be the values of X and Z. Consider how that logic impacts Fact 2:

(W)(Odd) + (Y)(Odd) = ODD. For this outcome to occur, either W or Y would have to be ODD... but W and Y MUST be EVEN in this situation. This ultimately means that there are NO examples in which there are exactly 2 Odd integers.

Thus, there MUST be 3 Odd integers and one Even integer (and the Even number would have to be in the numerator of one of those fractions. Thus, we'd have:

(Even)/(Odd) + (Odd)/(Odd) = Even + Odd = Odd. This means that the resulting calculation will ALWAYS be Odd and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: B

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