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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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Question Stats: 35% (02:44) correct 65% (02:26) wrong based on 302 sessions

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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|?

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

Thanks,
Saquib
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e-GMAT Representative V
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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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Reserving this space to post the official solution. _________________
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GMAT 1: 770 Q49 V47 Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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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
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Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  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-zero integers. If |x| < 5 and  [#permalink]

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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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GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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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

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts  |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-zero integers. If |x| < 5 and  [#permalink]

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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-zero integers. If |x| < 5 and  [#permalink]

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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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|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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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|?

A. -12
B. -6
C. -4
D. -2
E. 0
Math Expert V
Joined: 02 Aug 2009
Posts: 7961
|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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kiran120680 wrote:
|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

To maximize -|xy|, we have to minimize |xy|

Possible cases |x + 3| = |y -4|
(I) $$x+3=y-4....x=y-7$$........|xy|=|y(y-7)|... |y(y-7)| will be at least 6 even if we take y as 1.
(II) $$-(x+3)=y-4....-x-3=y-4................y+x=1$$.....let us take minimum possible value for this. Both are non-zero, so one of them 2 and other -1 will fit and |xy|=|2(-1)|=|2|

so maximum value of -|xy|=-|2|. No need to check further as in the options the max value is -2.
0 cannot be a value as both x and y are non-zero.

D
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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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kiran120680 wrote:
|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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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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kiran120680 wrote:
|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

simplify the expression
|x + 3| = |y -4|
x+3= y-4
x-y=-7
or say
-x-3=y-4
-(x+y)=-1
x+y=1
so x=2 and y=-1
possible and it satisfies
|x + 3| = |y -4|
-lxyl ; -2
IMO D
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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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Archit3110 wrote:
kiran120680 wrote:
|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

simplify the expression
|x + 3| = |y -4|
x+3= y-4
x-y=-7
or say
-x-3=y-4
-(x+y)=-1
x+y=1
so x=2 and y=-1
possible and it satisfies
|x + 3| = |y -4|
-lxyl ; -2
IMO D

correct me if am wrong but the solution of eqution in red is x=-3 and y=4

I tried a simlar approach and was struck for quite a bit .

Press kudos if it helps!!
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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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chetan2u wrote:
kiran120680 wrote:
|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

To maximize -|xy|, we have to minimize |xy|

Possible cases |x + 3| = |y -4|
(I) $$x+3=y-4....x=y-7$$........|xy|=|y(y-7)|... |y(y-7)| will be at least 6 even if we take y as 1.
(II) $$-(x+3)=y-4....-x-3=y-4................y+x=1$$.....let us take minimum possible value for this. Both are non-zero, so one of them 2 and other -1 will fit and |xy|=|2(-1)|=|2|

so maximum value of -|xy|=-|2|. No need to check further as in the options the max value is -2.
0 cannot be a value as both x and y are non-zero.

D

hello Egmat/Chetan sir,

I tried solving the inequalities and landed up withe x=-3 and y=4.

Can you please suggest the right line of thinking when seeing such questions..I see your approach is clear...But seeing inequalities like this general tendency is to jump into solving them.
Here after seeing the options it didnt help rather just plugging in values from options(assuming |x|<4 and |y|<5 wold have been far more beneficial)
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|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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I think this type of question is not too difficult if we solve things step by step. Below is my approach:

First things first, we have |x| < 5 and |y| < 5, which mean -5 < x < 5 and -5 < y < 5

Then, solve the equation:|x + 3| = |y -4|

We apply this: |A| = |B| then A = B or A = -B

=> x+3 = y-4 or x+3 = 4-y
=> x - y = -7 or x + y = 1

The possible values of x and y may be:
1. x - y = -7 -> pairs of value (x,y) can be: (-4;3) or (-3;4)
2. x + y = 1 -> (-1;2) or (-2;3) or (-3;4)

Then -|xy| can get these following values: -2, -6, -12 => -2 is the highest => D |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and   [#permalink] 27 Apr 2019, 05:29
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