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# p, q, r, and s are integers such that p > q > r. what is the value of

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p, q, r, and s are integers such that p > q > r. what is the value of  [#permalink]

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Updated on: 17 Apr 2018, 06:36
1
2
00:00

Difficulty:

85% (hard)

Question Stats:

30% (00:47) correct 70% (02:04) wrong based on 20 sessions

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p, q, r, and s are integers such that p > q > r. what is the value of (r + s)?

(1) [m]x^4-x^2=(x-p)(x-q)(x-r)(x-s)[/m]
(2) s < q

Source :-expertsglobal

EDIT:- A correct DS question will NOT give two different answers through the two statements..

Originally posted by rishabhmishra on 27 Mar 2018, 21:16.
Last edited by chetan2u on 17 Apr 2018, 06:36, edited 1 time in total.
Flawed question
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Joined: 21 Aug 2013
Posts: 1374
Location: India
Re: p, q, r, and s are integers such that p > q > r. what is the value of  [#permalink]

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27 Mar 2018, 21:59
2
rishabhmishra wrote:
Q p,q,r, and s are integers such that p>q>r.
what is the value of (r+s)?
1.$$x^4-x^2=(x-p)(x-q)(x-r)(x-s)$$
2. s<q
Source :-expertsglobal

Statement 1:
Lets break down the left hand side into its factors, and then compare it with right hand side.

x^4 - x^2 = x^2 (x^2 - 1) = x^2 (x-1) (x+1). This can further be written as:
x*x*(x-1)*(x+1) = (x-0)*(x-0)*(x-1)*(x-(-1)) (I have written x+1 as x-(-1) so that I can easily compare it to right hand side)

So we have: (x-0)*(x-0)*(x-1)*(x-(-1)) = (x-p)(x-q)(x-r)(x-s)
Thus comparing the two sides, we can say that p, q, r, s will take the values of 0, 0, 1 and -1 in some order. And two of them will take the same value of '0'.
We are already given that p > q > r. This means p=1, q=0 and r=-1; that's the only possibility here. This leaves s=0.
So we can calculate r+s = -1+0 = -1.
Sufficient.

Statement 2:
s < q. Clearly this is not sufficient.

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Re: p, q, r, and s are integers such that p > q > r. what is the value of  [#permalink]

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16 Apr 2018, 22:06
amanvermagmat, Bunuel, chetan2u

In this question, statement 1 says s=q.
While as per statement 2, s<q.

Is it possible to statement contradict each other in GMAT?

Please provide insights on the quality of question.

amanvermagmat wrote:
rishabhmishra wrote:
Q p,q,r, and s are integers such that p>q>r.
what is the value of (r+s)?
1.$$x^4-x^2=(x-p)(x-q)(x-r)(x-s)$$
2. s<q
Source :-expertsglobal

Statement 1:
Lets break down the left hand side into its factors, and then compare it with right hand side.

x^4 - x^2 = x^2 (x^2 - 1) = x^2 (x-1) (x+1). This can further be written as:
x*x*(x-1)*(x+1) = (x-0)*(x-0)*(x-1)*(x-(-1)) (I have written x+1 as x-(-1) so that I can easily compare it to right hand side)

So we have: (x-0)*(x-0)*(x-1)*(x-(-1)) = (x-p)(x-q)(x-r)(x-s)
Thus comparing the two sides, we can say that p, q, r, s will take the values of 0, 0, 1 and -1 in some order. And two of them will take the same value of '0'.
We are already given that p > q > r. This means p=1, q=0 and r=-1; that's the only possibility here. This leaves s=0.
So we can calculate r+s = -1+0 = -1.
Sufficient.

Statement 2:
s < q. Clearly this is not sufficient.

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Joined: 22 Feb 2018
Posts: 397
Re: p, q, r, and s are integers such that p > q > r. what is the value of  [#permalink]

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16 Apr 2018, 22:11
1
Bunuel chetan2u
Is this question acceptable as GMAT question?
Statement 1 and Statement 2 are contradicting each other.
From Statement 1, we are getting q=s=0
Statement 2 :s < q
As per GMATinsight 's earlier comment on below post
https://gmatclub.com/forum/in-the-xy-plane-l-and-m-are-two-lines-3-4-is-a-point-only-on-263149.html
It is a disqualified question.
Quote:
a disqualified question as per GMAT standard as statement 1 and 2 are contradictory

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Joined: 21 Aug 2013
Posts: 1374
Location: India
Re: p, q, r, and s are integers such that p > q > r. what is the value of  [#permalink]

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16 Apr 2018, 22:20
gmatbusters wrote:
amanvermagmat, Bunuel, chetan2u

In this question, statement 1 says s=q.
While as per statement 2, s<q.

Is it possible to statement contradict each other in GMAT?

Please provide insights on the quality of question.

amanvermagmat wrote:
rishabhmishra wrote:
Q p,q,r, and s are integers such that p>q>r.
what is the value of (r+s)?
1.$$x^4-x^2=(x-p)(x-q)(x-r)(x-s)$$
2. s<q
Source :-expertsglobal

Statement 1:
Lets break down the left hand side into its factors, and then compare it with right hand side.

x^4 - x^2 = x^2 (x^2 - 1) = x^2 (x-1) (x+1). This can further be written as:
x*x*(x-1)*(x+1) = (x-0)*(x-0)*(x-1)*(x-(-1)) (I have written x+1 as x-(-1) so that I can easily compare it to right hand side)

So we have: (x-0)*(x-0)*(x-1)*(x-(-1)) = (x-p)(x-q)(x-r)(x-s)
Thus comparing the two sides, we can say that p, q, r, s will take the values of 0, 0, 1 and -1 in some order. And two of them will take the same value of '0'.
We are already given that p > q > r. This means p=1, q=0 and r=-1; that's the only possibility here. This leaves s=0.
So we can calculate r+s = -1+0 = -1.
Sufficient.

Statement 2:
s < q. Clearly this is not sufficient.

Hello

As per my understanding (from reading various Bunuel's posts over past some time) this is a flawed question in that the two statements contradict each other, just as you mentioned. Thanks for raising this point. Maybe Bunuel/Chetan/Karishma would've something to add to this.

So if we can just change the second statement slightly to make it consistent with the first statement (lets say we make it s < p instead of s < q), the question can be very easily rectified.
Re: p, q, r, and s are integers such that p > q > r. what is the value of &nbs [#permalink] 16 Apr 2018, 22:20
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