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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
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A. -12
B. -6
C. -4
D. -2
E. 0
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Gladiator59
Joined: 16 Sep 2016
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Status:It always seems impossible until it's done.
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Posts: 829
First of all, we need to realize this - Mod is always nonnegative, so maximum possible of Negative of a mod expression will be at the minimum possible value of the mod expression.
So we are looking at Min value of |xy|
|x| < 5 or
-5 < x < 5
|x| is also the distance of x from zero, and in general |x-a| is the distance of x from a.
Hence |x+3| would be the distance of x from -3: which would translate to x ranging from 5 units to left of-3 to 5 units to right of -3. Just imagine that we have shifted the origin to -3 and then applying mod function
therefore, -8 < x < 2
Similarly, -1 < y < 9
we also have |x+3| = |y-4| so this could happen at x = -1 and y = 2 or at any of the following (x,y) pairs = ( -2.3) (-3,4) (-4,5) (-5,6) ...( -7,8) you can check by putting the values in.
Min absolute value of xy would be when x = -1 and y = 2 as y is nonzero integer. Hence , max value of -|xy| will be -2.
Option (D) is our choice
Regards,
Gladi
|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
So we are looking at Min value of |xy|
|x| < 5 or
-5 < x < 5
|x| is also the distance of x from zero, and in general |x-a| is the distance of x from a.
Hence |x+3| would be the distance of x from -3: which would translate to x ranging from 5 units to left of-3 to 5 units to right of -3. Just imagine that we have shifted the origin to -3 and then applying mod function
therefore, -8 < x < 2
Similarly, -1 < y < 9
we also have |x+3| = |y-4| so this could happen at x = -1 and y = 2 or at any of the following (x,y) pairs = ( -2.3) (-3,4) (-4,5) (-5,6) ...( -7,8) you can check by putting the values in.
Min absolute value of xy would be when x = -1 and y = 2 as y is nonzero integer. Hence , max value of -|xy| will be -2.
Option (D) is our choice
Regards,
Gladi
|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
Posts: 48
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
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
General Discussion
GMAT 1: 770 Q49 V47
Posts: 18
EgmatQuantExpert
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.
Posts: 3,651
eliaslatour
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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Luckisnoexcuse
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EgmatQuantExpert
|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
ydmuley
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Schools: Haas '20 HBS '20 CBS '20 Wharton '20 Stanford '20 Kellogg '20
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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
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
shashankism
Joined: 13 Mar 2017
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Location: India
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GPA: 3.8
WE:Engineering (Energy)
Posts: 612
|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|?
|x| < 5
-5<x<5
-2<x+3<8
0<|x+3|<8
|y| < 5
-5<y<5
-9<y-4<1
0<=|y-4|<1
Now, |x+3| = |y-4|
Hence 0<=|x+3|<1
-1<x+3<1
-4<x<-2
x=-3
-5<y<5
y = -4,-3,-2...,3,4
-|xy| = -12
Answer A.-12
_________________
|x| < 5
-5<x<5
-2<x+3<8
0<|x+3|<8
|y| < 5
-5<y<5
-9<y-4<1
0<=|y-4|<1
Now, |x+3| = |y-4|
Hence 0<=|x+3|<1
-1<x+3<1
-4<x<-2
x=-3
-5<y<5
y = -4,-3,-2...,3,4
-|xy| = -12
Answer A.-12
_________________
Nothing is Impossible, Even Impossible says I'm possible.
Posts: 118
gmatbusters
The question is best solved by going through the options. Option E can be eliminated as x and y are non zero integers, so xy cannot be zero.
Choosing the next maximum value among the options, i.e -2
xy can be -2 or 2.
By plugging in values for xy to get -2, we can see x=-1 and y=2 solves the given equation. So -2 is the maximum value. No need to even check other options as they are lesser than -2 and question asks for the greatest value.
Hence D.
Cheers!
Rupesh1Nonly
Joined: 26 Nov 2018
Last visit: 21 Jun 2019
Posts: 18
Status:when you say,"I can or I can't", Both times you are right!
Location: India
Posts: 18
|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
would anyone please bring some light on this approach?
-5<x<5
hence, -2<x+3<8
-5<y<5
hence, -9<y-4<1
the question says, lx+3l=ly-4l; this relation is equal at 0 and -1
when we equate x+3 =0 or -1,we get -3 or -4 respectively
when we equate y-4 =0 or -1, we get 4 or 3 respectively
multiplying XY(in both cases) produces -12
Hence, It's A.
A.-12
B-6
C.-4
D.-2
E.0
would anyone please bring some light on this approach?
-5<x<5
hence, -2<x+3<8
-5<y<5
hence, -9<y-4<1
the question says, lx+3l=ly-4l; this relation is equal at 0 and -1
when we equate x+3 =0 or -1,we get -3 or -4 respectively
when we equate y-4 =0 or -1, we get 4 or 3 respectively
multiplying XY(in both cases) produces -12
Hence, It's A.
Rupesh1Nonly
You are missing cases there:
y - 4 is -1 and x + 3 is 1.
y - 4 is -2 and x + 3 is 2.
y - 4 is -3 and x + 3 is 3.
...
y - 4 is -7 and x + 3 is 7.
In all those cases |x + 3| = |y - 4|.
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kiran120680
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
_________________
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
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
This is 20 seconds problem imo
We need to maximize -|xy|; hence, check the given options from the bottom.
We know that x and y are nonzero integers, eliminate E
Now check for whether we can get -|xy|=-2, where -5<x<5 and -5<y<5
At x=-1 and y=2, |x + 3| = |y -4|
D
We need to maximize -|xy|; hence, check the given options from the bottom.
We know that x and y are nonzero integers, eliminate E
Now check for whether we can get -|xy|=-2, where -5<x<5 and -5<y<5
At x=-1 and y=2, |x + 3| = |y -4|
D
EgmatQuantExpert
Kinshook
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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|?
x = -4; |y-4| = 1; y=3; -|xy| = - 12
x = -3; |y-4| = 0; y=4; -|xy| = - 12;
x = -2; |y-4| = 1; y=3; -|xy| = - 6;
x = - 1; |y-4| = 2; y=2; -|xy| = -2
x = 1; |y-4| = 4; Not feasible
x = 2; |y-4| = 5; y = - 1; -|xy| = -2
x = 3; |y-4| = 6; y = -2; -|xy| = -6
x = 4; |y-4| = 7; y = -3; -|xy| = -12
Maximum possible value of -|xy| = -2
IMO D
x = -4; |y-4| = 1; y=3; -|xy| = - 12
x = -3; |y-4| = 0; y=4; -|xy| = - 12;
x = -2; |y-4| = 1; y=3; -|xy| = - 6;
x = - 1; |y-4| = 2; y=2; -|xy| = -2
x = 1; |y-4| = 4; Not feasible
x = 2; |y-4| = 5; y = - 1; -|xy| = -2
x = 3; |y-4| = 6; y = -2; -|xy| = -6
x = 4; |y-4| = 7; y = -3; -|xy| = -12
Maximum possible value of -|xy| = -2
IMO D
|x+3| = |y-4|
case 1= x+3 = -(y-4)
x+y = 1
case 2= x+3 = y-4
x-y = -7
since we have condition that |X| < 5 and |Y| <5
for case 1, the possible pairs (4,-3) , (3,-2) , (2,-1)
for case 2, (-4,3) , (-3,4), (-2,5), (-5,2)
so if we use these pairs to find max value of -|xy|
we get the pair (2,-1) = -2
option D
case 1= x+3 = -(y-4)
x+y = 1
case 2= x+3 = y-4
x-y = -7
since we have condition that |X| < 5 and |Y| <5
for case 1, the possible pairs (4,-3) , (3,-2) , (2,-1)
for case 2, (-4,3) , (-3,4), (-2,5), (-5,2)
so if we use these pairs to find max value of -|xy|
we get the pair (2,-1) = -2
option D
EgmatQuantExpert
Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 04 Sep 2026
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Schools: HBS '26 CBS '26 Wharton '26
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Products:
|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|?
x = -4; |x+3| = 1 = |y-4|; 1. y = 3; - |xy| = -12; 2. y = 5; -|xy| = -20
x = -3; |x+3| = 0 = |y-4|; y = 4; -|xy| = -12
x = -2; |x+3|= 1 = |y-4|; 1. y = 3; - |xy| = -6; 2. y = 5; -|xy| = -10
x = -1; |x+3|= 2 = |y-4|; y = 2; - |xy| = -2
x = 1; |x+3|= 4 = |y-4|; No value of y is possible
x = 2; |x+3|= 5 = |y-4|; y = -1; -|xy| = -2
x = 3; |x+3|= 6 = |y-4|; y = -2; -|xy| = -6
x = 4; |x+3|= 7 = |y-4|; y = -3; -|xy| = -12
Maximum value of -|xy| = -2
IMO D
If |x| < 5 and |y| < 5, what is the maximum possible value of -|xy|?
x = -4; |x+3| = 1 = |y-4|; 1. y = 3; - |xy| = -12; 2. y = 5; -|xy| = -20
x = -3; |x+3| = 0 = |y-4|; y = 4; -|xy| = -12
x = -2; |x+3|= 1 = |y-4|; 1. y = 3; - |xy| = -6; 2. y = 5; -|xy| = -10
x = -1; |x+3|= 2 = |y-4|; y = 2; - |xy| = -2
x = 1; |x+3|= 4 = |y-4|; No value of y is possible
x = 2; |x+3|= 5 = |y-4|; y = -1; -|xy| = -2
x = 3; |x+3|= 6 = |y-4|; y = -2; -|xy| = -6
x = 4; |x+3|= 7 = |y-4|; y = -3; -|xy| = -12
Maximum value of -|xy| = -2
IMO D
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