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# |x + 3| = |y -4|, where x and y are non-zer

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Joined: 04 Jan 2015
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|x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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26 May 2017, 04:21
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11
00:00

Difficulty:

95% (hard)

Question Stats:

33% (02:47) correct 67% (02:02) wrong based on 263 sessions

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Q.

|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and |y| < 5, what is the maximum possible value of -|xy|?

A.-12
B. -6
C. -4
D. -2
E. 0

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Saquib
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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26 May 2017, 04:43
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Joined: 20 Apr 2017
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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26 May 2017, 04:46
1
EgmatQuantExpert wrote:
Q.

|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and |y| < 5, what is the maximum possible value of -|xy|?

A.-12
B. -6
C. -4
D. -2
E. 0

Thanks,
Saquib
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Well, I don't 100% understand the question, but I'll take a stab at it.

Since y<5, 1 is a possible value for y. Since |1-4|=3, x would have to be 0 or 6 and that's not possible.

So I'm going to try -1, which yields |-1-4|=5 so x could be 2. That will yield answer choice (D), and I don't see how you can reach zero since neither x nor y can be zero. I'm going with D.
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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26 May 2017, 04:58
eliaslatour wrote:
Well, I don't 100% understand the question, but I'll take a stab at it.

Since y<5, 1 is a possible value for y. Since |1-4|=3, x would have to be 0 or 6 and that's not possible.

So I'm going to try -1, which yields |-1-4|=5 so x could be 2. That will yield answer choice (D), and I don't see how you can reach zero since neither x nor y can be zero. I'm going with D.

Nice try eliaslatour.

Though I won't comment whether is correct or not, as I want the others to also give it a shot. I just wanted to point out one error in your analysis.

When we write |y| < 5 and we remove the modulus from y, the range of y is not y < 5, but -5 < y < 5.

Apart from that, kudos to your observation that since x and y are non-zero integers, the value of -|xy| can never be 0.

Thanks,
Saquib
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|x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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26 May 2017, 05:25
EgmatQuantExpert wrote:
Q.

|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and |y| < 5, what is the maximum possible value of -|xy|?

A.-12
B. -6
C. -4
D. -2
E. 0

Thanks,
Saquib
Quant Expert
e-GMAT

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|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and |y| < 5, what is the maximum possible value of -|xy|

-5<x<5, -5<y<5

We need to solve for values of x & y to keep XY to minimum using |x + 3| = |y -4|

by value substitution x =2 and y=-1 fits the criteria keeping XY to minimum

hence option D
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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28 May 2017, 04:45
1
I tried as mentioned below

Given :

-5<x<5, -5<y<5 & x and y are non zero integers.

Therefore; x = {-4,-3,-2,-1,1,2,3,4} & y = {-4,-3,-2,-1,1,2,3,4}

|x + 3| = |y -4|

Try plugging above values
&
above equation holds when
1. x=-4 & y = 3
2. x=-3 & y= 4
3. x=2 & y =-1
4. x = 3 & y=-2
5. x=4 & y = -3.

- |xy| will be max when x=2 & y = -1

Option D

This is how i arrived.

Thanx
Narayana raju
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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19 Jun 2017, 15:08
1
This is a good and tricky question.

Given: lx+3l = ly-4l - x & y are non-zero integers.

lxl < 5
lyl < 5

Max of -lxyl = ?

As we have to find the max value, and the final max number will be negative as -lxyl will be negative, we need to ensure that we get smallest value of negative possible which satisfies given equation.

As per given information values of x and y can range as per below:

-5 < x < 5
-5 < y < 5

As we have to take least numbers, we will plug numbers from this range.

Note we cannot take x & y as zero as per the given information, also it does not satisfy the given equation lx+3l = ly-4l.

Only values we can check for us -1 & 2 or 2 & -1

so maximum value we get is -2.

Hence, Answer is D = -2
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Re: |x + 3| = |y -4|, where x and y are non-zer  [#permalink]

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18 Oct 2018, 15:32
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Re: |x + 3| = |y -4|, where x and y are non-zer &nbs [#permalink] 18 Oct 2018, 15:32
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