Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 10 Jul 2013
Posts: 334

A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
06 Aug 2013, 12:15
2
This post received KUDOS
3
This post was BOOKMARKED
Question Stats:
48% (02:27) correct
52% (01:29) wrong based on 127 sessions
HideShow timer Statistics
(x  a)(x  b)(x  c)(x  d) = 0 (y  a)(y  b)(y  c)(y  d) = 0 A is a set holding only all possible solutions of x, and B is a set holding only all possible solutions of y. If a, b, c and d are four unequal numbers, then is the number of elements in set A equal to the number of elements in set B ? (1) a and b are positive (2) c and d are negative My type of solution:
from question, we know, x=a or b or c or d. so A ={a,b,c,d} (possible 4 solutions)
Now from the latter part of the question, we know (│y│a) (│y│b) (│y│  c) (│y│ d ) = 0 , it means │y│ a = 0 ,so │y│= a , we can put y= a or a but both will provide the same solution which is positive a. so here we have two possible solutions. but we can't put y=0 because from statement(1) we know a and b are positive numbers. Thus we have four solutions +a,a,+b,b. from statement(2) we know, c and d are negative numbers, but │y│= c , and there is nothing that you can put instead of y and get a negative value, so here c can never be a negative one neither 0. that's why we have to eliminate +c,c,+d,d. After doing so we have only B = {+a,a,+b,b} (possible 4 solutions)
so they are in equal number thus both statement together have done the job this time and the Answer is (C) _________________
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Asif vai.....
Last edited by Bunuel on 30 Jan 2014, 23:19, edited 2 times in total.
Edited the question.



Director
Joined: 14 Dec 2012
Posts: 832
Location: India
Concentration: General Management, Operations
GPA: 3.6

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
06 Aug 2013, 16:05
1
This post received KUDOS
Asifpirlo wrote: (xa)(xb)(xc)(xd) = 0 (│y│a) (│y│b) (│y│  c) (│y│ d ) = 0 A is a set holding only all possible solutions of x, and B is a set holding only all possible solutions of y. if a , b ,c and d are four unequal numbers, then is the number of elements in set A equal to the number of elements in set B ? (1) a and b are positive (2) c and d are negative need discussions on this topic and then i will publish the solution done by me according to question set A :{a,b,c,d} set B : IF a,b,c,d they are negative then (│y│a) (│y│b) (│y│  c) (│y│ d )=0==>this will have no solution so set B depends whether a,b,c,d are +ve or ve now given a,b,c,d are different numbers (1) a and b are positivelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d= 4 ==> in this case SET A= (1,2,3,4) AND SET B=(1,2,3,4,1,2,3,4) HENCE B>A not sufficient (2) c and d are negativelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d=4 ==> in this case SET A=(1,2,3,4) AND SET B= NULL HENCE A>B not sufficient now combining both means a,b = +ve..and c,d = velet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=3 b=4 c=3 d=4 ==> in this case set A=(3,4,3,4) AND SET B =(3,4,3,4) SO COMBINIG BOTH number of elements of set A = number of elements of set B
hence C
_________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
GIVE VALUE TO OFFICIAL QUESTIONS...
GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabularylistforgmatreadingcomprehension155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmatanalyticalwritingassessment : http://www.youtube.com/watch?v=APt9ITygGss



Senior Manager
Joined: 10 Jul 2013
Posts: 334

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
06 Aug 2013, 23:55
blueseas wrote: Asifpirlo wrote: (xa)(xb)(xc)(xd) = 0 (│y│a) (│y│b) (│y│  c) (│y│ d ) = 0 A is a set holding only all possible solutions of x, and B is a set holding only all possible solutions of y. if a , b ,c and d are four unequal numbers, then is the number of elements in set A equal to the number of elements in set B ? (1) a and b are positive (2) c and d are negative need discussions on this topic and then i will publish the solution done by me according to question set A :{a,b,c,d} set B : IF a,b,c,d they are negative then (│y│a) (│y│b) (│y│  c) (│y│ d )=0==>this will have no solution so set B depends whether a,b,c,d are +ve or ve now given a,b,c,d are different numbers (1) a and b are positivelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d= 4 ==> in this case SET A= (1,2,3,4) AND SET B=(1,2,3,4,1,2,3,4) HENCE B>A not sufficient (2) c and d are negativelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d=4 ==> in this case SET A=(1,2,3,4) AND SET B= NULL HENCE A>B not sufficient now combining both means a,b = +ve..and c,d = velet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=3 b=4 c=3 d=4 ==> in this case set A=(3,4,3,4) AND SET B =(3,4,3,4) SO COMBINIG BOTH number of elements of set A = number of elements of set B
hence C Blueseas always has his own amazing solutions. really amazing. My type of solution: from question, we know, x=a or b or c or d. so A ={a,b,c,d} (possible 4 solutions) now from the latter part of the question, we know (│y│a) (│y│b) (│y│  c) (│y│ d ) = 0 , it means │y│ a = 0 ,so │y│= a , we can put y= a or a but both will provide the same solution which is positive a. so here we have two possible solutions. but we can't put y=0 because from statement(1) we know a and b are positive numbers. Thus we have four solutions +a,a,+b,b. from statement(2) we know, c and d are negative numbers, but │y│= c , and there is nothing that you can put instead of y and get a negative value, so here c can never be a negative one neither 0. that's why we have to eliminate +c,c,+d,d. After doing so we have only B = {+a,a,+b,b} (possible 4 solutions) so they are in equal number thus both statement together have done the job this time and the Answer is (C)
_________________
Asif vai.....



Manager
Joined: 29 Jun 2011
Posts: 161
WE 1: Information Technology(Retail)

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
31 Aug 2013, 23:18
Can you please elaborate on how you calculated the solutions?
let say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d= 4 ==> in this case SET A= (1,2,3,4) AND SET B=(1,2,3,4,1,2,3,4) HENCE B>A



Intern
Joined: 18 Aug 2013
Posts: 18

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
01 Sep 2013, 18:09
blueseas wrote: Asifpirlo wrote: (xa)(xb)(xc)(xd) = 0 (│y│a) (│y│b) (│y│  c) (│y│ d ) = 0 A is a set holding only all possible solutions of x, and B is a set holding only all possible solutions of y. if a , b ,c and d are four unequal numbers, then is the number of elements in set A equal to the number of elements in set B ? (1) a and b are positive (2) c and d are negative need discussions on this topic and then i will publish the solution done by me according to question set A :{a,b,c,d} set B : IF a,b,c,d they are negative then (│y│a) (│y│b) (│y│  c) (│y│ d )=0==>this will have no solution so set B depends whether a,b,c,d are +ve or ve now given a,b,c,d are different numbers (1) a and b are positivelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d= 4 ==> in this case SET A= (1,2,3,4) AND SET B=(1,2,3,4,1,2,3,4) HENCE B>A not sufficient (2) c and d are negativelet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=1 b=2 c=3 d=4 ==> in this case SET A=(1,2,3,4) AND SET B= NULL HENCE A>B not sufficient now combining both means a,b = +ve..and c,d = velet say a = 1 b= 2 c= 1 d=2 ==>in this case set A=(1, 2,1,2) AND SET B=(1,2,1,2) HENCE A=B now let say a=3 b=4 c=3 d=4 ==> in this case set A=(3,4,3,4) AND SET B =(3,4,3,4) SO COMBINIG BOTH number of elements of set A = number of elements of set B
hence C For statement 1&2 combined, what about the case where a=1, b=2, c=3, d=4 > A=(1,2,3,4) and set B=(1,2,3,4,1,2,3,4) Am I missing something? Thanks



Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 961
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
02 Jan 2014, 08:48
Eq 1 . (xa) (xb) (xc) (xd) = 0 thus possible solution set for x A={a,b,c,d} Eq 2 . (ya) (yb) (yc) (yd) =0 thus possible solution set for x B={a,a,b,b,c,c,d,d} 1. a and b are positive. for Eq.1 x=a,b are sure true factors and for Eq.2 x=a,a,b,b are sure true but we do not know about signs of c and d. 2. c and d are negative. For E.q 1 new sure solution set is A={c,d} not sure about {a,b} for E.q 2 x=b,c,c,d,d wont be able to subtract negative c and d therefore these are not possible factor of x and remaining factors are B={a,a,b,b} Combine 1 + 2 A={a,b,c,d) B={a,a,b,b) These are the possible factors 4 in each set, therefore Ans: C
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15978

Re: A is a set holding only all possible solutions of x, and B [#permalink]
Show Tags
26 Feb 2015, 21:12
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: A is a set holding only all possible solutions of x, and B
[#permalink]
26 Feb 2015, 21:12








Similar topics 
Author 
Replies 
Last post 
Similar Topics:


1


What is the variance of set X, which contains only consecutive even in

Bunuel 
4 
29 Nov 2016, 22:00 

2


If a and b are constants, what is one solution of the equation (x + a)

Bunuel 
2 
29 Mar 2016, 04:14 

1


Set X only contains all possible solutions of equation (I) above

nishith17 
3 
28 Feb 2016, 05:59 

4


Set A consists of all positive integers less than 100; Set B

mohnish104 
3 
14 Oct 2016, 02:08 

20


If x is a member of the set {a, b, c, d}

tulsa 
9 
22 Jun 2016, 22:45 



