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What is the value of |x| + |y| + xy, where x and y are integers?

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What is the value of |x| + |y| + xy, where x and y are integers?  [#permalink]

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New post 02 Jun 2016, 20:36
1
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A
B
C
D
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  35% (medium)

Question Stats:

64% (01:07) correct 36% (01:07) wrong based on 167 sessions

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What is the value of |x| + |y| + xy, where x and y are integers?

(1) \(xy=17\)
(2) \(x=\sqrt{17^2}\)


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Re: What is the value of |x| + |y| + xy, where x and y are integers?  [#permalink]

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New post 02 Jun 2016, 20:54
1
St1: xy = 17 is prime --> x * y = 1 * 17 or -1 * -17.

|x| + |y| + xy = 1 + 17 + 17 = 35.

Sufficient

St2: Clearly insufficient as we do not know the value of y.

Answer: A
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Re: What is the value of |x| + |y| + xy, where x and y are integers?  [#permalink]

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New post 03 Jun 2016, 10:54
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chetan2u wrote:
What is the value of |x| + |y| + xy, where x and y are integers?
(1) \(xy=17\)
(2) \(x=\sqrt{17^2}\)


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|x| + |y| + xy

(1) \(xy=17\)

Since x and y are integers x and y can have values -1, -17 or vice versa or 1,17 or vice versa.

Irrespective of the sign |x| and |y| will be +ve.

1+17+17= 35 Sufficient.

2) \(x=\sqrt{17^2}\)

We don't know the value of y. Not sufficient.

A is the answer
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Re: What is the value of |x| + |y| + xy, where x and y are integers?  [#permalink]

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New post 05 Jun 2016, 04:58
There are 2 variables ( x and y) in the original condition. Hence, there is high chance that C is the correct answer. Using both the condition 1) and the condition 2), we get x=17, y=1. The answers are unique and the conditions are sufficient. However, since this is one of key questions, an integer question, we need to apply the common mistake type 4(A).
In case of the condition 1), from (x,y)=(1,17),(17,1),(-1,-17),(-17,-1), the answer is always |x|+|y|+xy=35. The answer is unique and the condition is sufficient. Hence, the correct answer is A.
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Re: What is the value of |x| + |y| + xy, where x and y are integers?  [#permalink]

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New post 01 Feb 2018, 11:06
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Re: What is the value of |x| + |y| + xy, where x and y are integers? &nbs [#permalink] 01 Feb 2018, 11:06
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