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If w, x, y and z are integers such that w/x and y/z are
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Updated on: 13 Mar 2012, 13:00
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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
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Originally posted by japped187 on 01 Jun 2008, 04:41.
Last edited by Bunuel on 13 Mar 2012, 13:00, edited 1 time 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
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13 Mar 2012, 13:24




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Re: if w,x,y,and z are integers...DS
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01 Jun 2008, 08:00
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
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13 Mar 2012, 11:06
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
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14 Mar 2012, 00:02
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
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20 Sep 2012, 05:33
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
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20 Sep 2012, 06:16
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 integerEven/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
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19 Jul 2013, 00:22



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Re: If w, x, y and z are integers such that w/x and y/z are
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31 Jul 2013, 06:33
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. Cheers



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Re: If w, x, y and z are integers such that w/x and y/z are
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02 Aug 2013, 07:33
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
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10 Oct 2013, 06:16
Statement 1 is indeed tricky... Statement 2 is easy.



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Re: If w, x, y and z are integers such that w/x and y/z are
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27 Dec 2013, 06:48
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
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Re: If w, x, y and z are integers such that w/x and y/z are
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29 Dec 2013, 19:27
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
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25 Aug 2016, 21:32
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 Please find the solution as attached
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If w, x, y and z are integers such that w/x and y/z are
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Updated on: 05 Sep 2018, 13:08
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 oddodd+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.
Probus
Originally posted by Probus on 29 Aug 2018, 22:20.
Last edited by Probus on 05 Sep 2018, 13:08, edited 1 time in total.



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Re: If w, x, y and z are integers such that w/x and y/z are
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30 Aug 2018, 07:49
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 very hard. from 1. case 1; wx is odd and yz is even w, x is odd and y or z is even from here we can not know case 2 similar to case 1, so we can not know.



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Re: If w, x, y and z are integers such that w/x and y/z are
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28 Oct 2018, 10:51
Bunuel wrote: 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. I don't understand the following part: "all multiple must be odd in order the product to be odd" Posted from my mobile device
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Re: If w, x, y and z are integers such that w/x and y/z are
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29 Oct 2018, 03:07
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 another way to think about this problem is: Given w/x is an integer so w=x * a y/z is an integer so y=z * b is w/x + y/z odd? bt putting values w/x + y/z = (x*a)/x+(z*b)/z= (a+b)= odd? now the given statements a. wx + yz = odd putting these values (x *a) x+ (z*b)z is odd so this is a x^2 +bz^2 dont have any info about individual terms so not sufficient b. wz + xy = odd (x *a) z+ (z*b)x is odd so this is xza+xzb = odd xz(a+b)= odd now we know that Odd x Odd = Odd so (a+b) = odd hence sufficient Posted from my mobile device
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If w, x, y and z are integers such that w/x and y/z are
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29 Oct 2018, 03:12
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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