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# A is a set holding only all possible solutions of x, and B

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A is a set holding only all possible solutions of x, and B [#permalink]

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06 Aug 2013, 12:15
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Question Stats:

48% (02:28) correct 52% (01:29) wrong based on 126 sessions

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(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

[Reveal] Spoiler:
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)
_________________
[Reveal] Spoiler: OA

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Asif vai.....

Last edited by Bunuel on 30 Jan 2014, 23:19, edited 2 times in total.
Edited the question.
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Re: A is a set holding only all possible solutions of x, and B [#permalink]

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06 Aug 2013, 16:05
1
KUDOS
Asifpirlo wrote:
(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

[Reveal] Spoiler:
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 positive
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
not sufficient
(2) c and d are negative
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= NULL HENCE A>B
not sufficient

now combining both means a,b = +ve..and c,d = -ve
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=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
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Senior Manager
Joined: 10 Jul 2013
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Kudos [?]: 275 [0], given: 102

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

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06 Aug 2013, 23:55
blueseas wrote:
Asifpirlo wrote:
(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

[Reveal] Spoiler:
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 positive
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
not sufficient
(2) c and d are negative
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= NULL HENCE A>B
not sufficient

now combining both means a,b = +ve..and c,d = -ve
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=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.....

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Kudos [?]: 21 [0], given: 29

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

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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
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Joined: 18 Aug 2013
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Kudos [?]: 4 [0], given: 6

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

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01 Sep 2013, 18:09
blueseas wrote:
Asifpirlo wrote:
(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

[Reveal] Spoiler:
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 positive
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
not sufficient
(2) c and d are negative
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= NULL HENCE A>B
not sufficient

now combining both means a,b = +ve..and c,d = -ve
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=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
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Re: A is a set holding only all possible solutions of x, and B [#permalink]

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02 Jan 2014, 08:48
Eq 1 . (x-a) (x-b) (x-c) (x-d) = 0 thus possible solution set for x A={a,b,c,d}
Eq 2 . (|y|-a) (|y|-b) (|y|-c) (|y|-d) =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
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Re: A is a set holding only all possible solutions of x, and B [#permalink]

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