Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A is a set holding only all possible solutions of x, and B [#permalink]
06 Aug 2013, 11:15

1

This post received KUDOS

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

45% (02:23) correct
55% (01:07) wrong based on 77 sessions

(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 ?

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) _________________

Re: A is a set holding only all possible solutions of x, and B [#permalink]
06 Aug 2013, 15:05

1

This post received 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 ?

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 _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: A is a set holding only all possible solutions of x, and B [#permalink]
06 Aug 2013, 22:55

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 ?

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) _________________

Re: A is a set holding only all possible solutions of x, and B [#permalink]
31 Aug 2013, 22:18

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

Re: A is a set holding only all possible solutions of x, and B [#permalink]
01 Sep 2013, 17:09

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 ?

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)

Re: A is a set holding only all possible solutions of x, and B [#permalink]
02 Jan 2014, 07:48

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 _________________

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".

gmatclubot

Re: A is a set holding only all possible solutions of x, and B
[#permalink]
02 Jan 2014, 07:48