|x + 3| = |y -4|, where x and y are non-zero integers. If |x|

Question

|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.

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Gladiator59 · Accepted Answer

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

gvvsnraju@1 · Answer

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

eliaslatour · Answer

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.

EgmatQuantExpert · Answer

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
Quant Expert
e-GMAT

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Luckisnoexcuse · Answer

|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 · Answer

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

shashankism · Answer

|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

Diwakar003 · Answer

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 · Answer

|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.

Bunuel · Answer

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|.

chetan2u · Answer

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

leminkhoa · Answer

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

nick1816 · Answer

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

Kinshook · Answer

|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

vasu1104 · Answer

|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

Kinshook · Answer

|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

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

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