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

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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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26 May 2017, 04:21
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95% (hard)

Question Stats:

35% (02:46) correct 65% (02:29) wrong based on 381 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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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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26 May 2017, 04:43
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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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26 May 2017, 04:46
3
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-zero integers. If |x| < 5 and  [#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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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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26 May 2017, 05:25
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
Quant Expert
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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-zero integers. If |x| < 5 and  [#permalink]

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28 May 2017, 04:45
2
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
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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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19 Jun 2017, 15:08
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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
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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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29 Mar 2019, 11:40
1
|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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Posts: 8586
|x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and  [#permalink]

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29 Mar 2019, 15:54
1
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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29 Mar 2019, 23:29
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

Merging topics. Please search before posting.
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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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30 Mar 2019, 00:00
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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27 Apr 2019, 02:13
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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27 Apr 2019, 02:19
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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27 Apr 2019, 04:29
2
1
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
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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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24 Oct 2019, 21:34
Hi.. I think the question is tricky and twisted. However, the answer -2 is definitely incorrect.

Why?

Because -mod(xy) can have a maximum value of zero, right. Let's keep it as it will be useful.

Let's consider a pair of (x,y) as y=1, and x=0.

This pair definitely satisfies all the equations in the question.

So, a valid pair of (x,y) is (0,1), which will give a product of xy as zero.

Hence, -mod(x,y)=0, which is the maximum attainable value.

Any thoughts?
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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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27 Oct 2019, 05:19
Harshit Srivastava wrote:
Hi.. I think the question is tricky and twisted. However, the answer -2 is definitely incorrect.

Why?

Because -mod(xy) can have a maximum value of zero, right. Let's keep it as it will be useful.

Let's consider a pair of (x,y) as y=1, and x=0.

This pair definitely satisfies all the equations in the question.

So, a valid pair of (x,y) is (0,1), which will give a product of xy as zero.

Hence, -mod(x,y)=0, which is the maximum attainable value.

Any thoughts?

hi
it is given in the question that x and y are non-zero integers
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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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01 Nov 2019, 01:20
EgmatQuantExpert 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

Thanks,
Saquib
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Here is my approach to this question:

Information gathering:
We are given that the distance of x from -3 is equal to the distance of y from 4...(converting |x + 3| = |y -4| into a sentence)..(info 1)
We are also given that distance of x from 0 is less than 5..(|x| < 5)..means -5<x<5
and distance of y from 0 is less than 5...(|y| < 5) -5<y<5

To find: Maximum possible value of -|xy|

Method:
Plot points -5, -3, 0 , 4 and 5 on a number line...and start pairing numbers!
pair 1: x = -4..which is 1 distance away from -3..hence y would also be 1 distance away from 4 as per info 1. For (-4,3) -|xy| = -|12| = -12
Pair 2: x=-2...which is 1 distance away from -3..hence y would also be 1 distance away from 4 as per info 1. For (-2,3) -|xy| = -|6| = -6
Pair 3: x=-1.....which is 2 distance away from -3..hence y would also be 2 distance away from 4 as per info 1. For (-1,2) -|xy| = -|2| = -2
Since x and y are non-zero integers, there is no point testing x and y with 0 value. Among the 3 pairs, -2 is the greatest.
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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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01 Nov 2019, 02:13
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

EgmatQuantExpert 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

Thanks,
Saquib
Quant Expert
e-GMAT

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Re: |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and   [#permalink] 01 Nov 2019, 02:13