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The operation # is defined in the following way for any two numbers

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The operation # is defined in the following way for any two numbers  [#permalink]

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New post 08 Mar 2018, 02:05
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The operation # is defined in the following way for any two numbers: p#q = (p - q)(q - p). If p#q = -1, then which of the following are true?

I. p could equal 5 and q could equal 4
II. p could equal 4 and q could equal 5
III. p could equal 1 and q could equal -1
IV. p could equal -1 and q could equal 1

(A) I and II only
(B) I and III only
(C) II and IV only
(D) III and IV only
(E) I, II, III, IV

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Re: The operation # is defined in the following way for any two numbers  [#permalink]

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New post 08 Mar 2018, 02:10
Bunuel wrote:
The operation # is defined in the following way for any two numbers: p#q = (p - q)(q - p). If p#q = -1, then which of the following are true?

I. p could equal 5 and q could equal 4
II. p could equal 4 and q could equal 5
III. p could equal 1 and q could equal -1
IV. p could equal -1 and q could equal 1

(A) I and II only
(B) I and III only
(C) II and IV only
(D) III and IV only
(E) I, II, III, IV



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Re: The operation # is defined in the following way for any two numbers  [#permalink]

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New post 08 Mar 2018, 03:48
Since the operation # is defined as p#q = (p - q)(q - p),
and we are given that p#q = -1, then testing the following values, we get

I. p could equal 5 and q could equal 4 -> (p - q)(q - p) = (5-4)(4-5) = -1
II. p could equal 4 and q could equal 5 -> (p - q)(q - p) = (4-5)(5-4) = -1
III. p could equal 1 and q could equal -1 -> (p - q)(q - p) = (1-(-1))(-1-1) = -4
IV. p could equal -1 and q could equal 1 > (p - q)(q - p) = (-1-1)(1-(-1)) = -4

Therefore, Option A(I and II only) is always true for the operation # such that p#q = -1
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Re: The operation # is defined in the following way for any two numbers  [#permalink]

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New post 08 Mar 2018, 23:46
Bunuel wrote:
The operation # is defined in the following way for any two numbers: p#q = (p - q)(q - p). If p#q = -1, then which of the following are true?

I. p could equal 5 and q could equal 4
II. p could equal 4 and q could equal 5
III. p could equal 1 and q could equal -1
IV. p could equal -1 and q could equal 1

(A) I and II only
(B) I and III only
(C) II and IV only
(D) III and IV only
(E) I, II, III, IV

IMO A
IF p#q = (p - q)(q - p)
AND P#Q=-1
THEN
I 1*-1 yes
II -1*1 yes
III 2*-2 no
and no need to check last one as we can see there is no option with I, II and IV
So A is the answer
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Re: The operation # is defined in the following way for any two numbers   [#permalink] 08 Mar 2018, 23:46
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