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Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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Difficulty:   35% (medium)

Question Stats: 68% (01:23) correct 32% (01:48) wrong based on 135 sessions

### HideShow timer Statistics If $$x$$ and $$y$$ are integers and $$x + y = -12$$, which of the following must be true?

A. Both $$x$$ and $$y$$ are negative
B. $$xy \gt 0$$
C. If $$y \lt 0$$, then $$x \gt 0$$
D. If $$y \gt 0$$, then $$x \lt 0$$
E. $$x-y \gt 0$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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Official Solution:

If $$x$$ and $$y$$ are integers and $$x + y = -12$$, which of the following must be true?

A. Both $$x$$ and $$y$$ are negative
B. $$xy \gt 0$$
C. If $$y \lt 0$$, then $$x \gt 0$$
D. If $$y \gt 0$$, then $$x \lt 0$$
E. $$x-y \gt 0$$

Look at option D: if $$y$$ is positive, then $$x$$ must be negative in order the sum of $$x$$ and $$y$$ to be negative.

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Manager  Joined: 24 Nov 2013
Posts: 57

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given x+y=−12 and both x and y are integers.
Consider few examples that give a sum of -12

-8 + (-4) = -12
4 + (- 16) = -12
Now check the given options. Only d holds good.
Intern  Joined: 13 Feb 2014
Posts: 7
Location: India
Schools: HBS '18, ISB '16, IIMA
GMAT Date: 12-14-2014
GPA: 3.5
WE: Engineering (Consumer Electronics)

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d)-16(x) + 4(y)=-12---> x<0, y>0

c)4(x)+(-16)(y)=-12----->x>0,y<0

so there are two valid answer options in this question, correct me if i am wrong
Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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ramarao443 wrote:

d)-16(x) + 4(y)=-12---> x<0, y>0

c)4(x)+(-16)(y)=-12----->x>0,y<0

so there are two valid answer options in this question, correct me if i am wrong

The question asks which of the following must be true, not could be true. Only D must be true.
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Manager  Joined: 24 Nov 2013
Posts: 57

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1
ramarao443 wrote:

d)-16(x) + 4(y)=-12---> x<0, y>0

c)4(x)+(-16)(y)=-12----->x>0,y<0

so there are two valid answer options in this question, correct me if i am wrong

Option c (If y<0, then x>0) gets eliminated when we use these values

x = -8 and y = -4
-8 + (-4) = -12
Intern  Joined: 10 Oct 2012
Posts: 25

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this question is wrong I believe becuase what happens if

x=-12 and y=0

x+y=-12 and THE SAME HAPPENS viceversa

Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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mmelendez wrote:
this question is wrong I believe becuase what happens if

x=-12 and y=0

x+y=-12 and THE SAME HAPPENS viceversa

The question is fine, you just did not understand it.

The question asks which of the following must be true, not could be true.

If y is positive, then x must be negative in order the sum of x and y to be negative. So, D MUST be true.
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Retired Moderator G
Joined: 26 Nov 2012
Posts: 586

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Bunuel wrote:
Official Solution:

If $$x$$ and $$y$$ are integers and $$x + y = -12$$, which of the following must be true?

A. Both $$x$$ and $$y$$ are negative
B. $$xy \gt 0$$
C. If $$y \lt 0$$, then $$x \gt 0$$
D. If $$y \gt 0$$, then $$x \lt 0$$
E. $$x-y \gt 0$$

Look at option D: if $$y$$ is positive, then $$x$$ must be negative in order the sum of $$x$$ and $$y$$ to be negative.

Hi Bunuel,

If D is must be true option then even C will also satisfy .

I tried with C and D with some random numbers and both seemed correct to me.

I have gone through the other explanations which you gave but didn't get.

Can you please help me to understand much better that how we can get MUST case and COULD BE in such questions.
Current Student B
Status: Persevere
Joined: 09 Jan 2016
Posts: 118
Location: Hong Kong
GMAT 1: 750 Q50 V41 GPA: 3.52

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2
1
msk0657 wrote:
Bunuel wrote:
Official Solution:

If $$x$$ and $$y$$ are integers and $$x + y = -12$$, which of the following must be true?

A. Both $$x$$ and $$y$$ are negative
B. $$xy \gt 0$$
C. If $$y \lt 0$$, then $$x \gt 0$$
D. If $$y \gt 0$$, then $$x \lt 0$$
E. $$x-y \gt 0$$

Look at option D: if $$y$$ is positive, then $$x$$ must be negative in order the sum of $$x$$ and $$y$$ to be negative.

Hi Bunuel,

If D is must be true option then even C will also satisfy .

I tried with C and D with some random numbers and both seemed correct to me.

I have gone through the other explanations which you gave but didn't get.

Can you please help me to understand much better that how we can get MUST case and COULD BE in such questions.

As per the question, we know:
i. $$x$$ and $$y$$ are integers, and
ii. $$x + y = -12$$

From this, we get following five situations/ scenarios:
1) $$x < 0$$, $$y < 0$$ (eg. $$x = -5$$, $$y = -7$$ or $$x = -7$$, $$y = -5$$)
2) $$x < 0$$, $$y > 0$$ (eg. $$x = -13$$, $$y = 1$$)
3) $$x > 0$$, $$y < 0$$ (eg. $$x = 1$$, $$y = -13$$)
4) $$x = 0$$, $$y < 0$$ ($$x = 0$$, $$y = -12$$)
5) $$x < 0$$, $$y = 0$$ ($$x = -12$$, $$y = 0$$)

Let us now consider each of the options to determine whether they must be true for each of the above scenarios:
Option A:
Both $$x$$ and $$y$$ are negative
in other words: $$x <0$$, $$y < 0$$. This is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option B:
$$xy>0$$, this is only possible when $$x>0$$ and $$y>0$$ or when $$x<0 and y<0$$. Again, this is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option C:
If $$y<0$$, then $$x>0$$.
If we look at all the 5 scenarios, $$y<0$$ in scenario 3 and 4. However, this statement is true for scenario 3 but not true for scenario 4. This is because, if $$y = -12$$ (i.e. $$y<0$$), then $$x = 0$$ (neither negative nor positive). Hence, this could be true, but not definitely true.

Option D:
If $$y>0$$, then $$x<0$$.
If we look at all the 5 scenarios, $$y<0$$ only in scenario 2 and as per that scenario $$x>0$$. So this must be true because whenever $$y$$ is a negative integer, $$x$$ has to be a positive integer. This is the correct answer.

Option E:
$$x-y>0$$. This is true for scenario 4 but not true for scenario 5. Therefore, this could be true, but not definitely true. Remember, it is sufficient to consider 2 out of 5 scenarios, if we get different answers from them. We have to consider all the 5 scenarios only in cases where the statement is definitely true or must be true.

Hope this helps!!

Regards,
Nalin
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Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)

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1
nalinnair wrote:
msk0657 wrote:
Bunuel wrote:
Official Solution:

If $$x$$ and $$y$$ are integers and $$x + y = -12$$, which of the following must be true?

A. Both $$x$$ and $$y$$ are negative
B. $$xy \gt 0$$
C. If $$y \lt 0$$, then $$x \gt 0$$
D. If $$y \gt 0$$, then $$x \lt 0$$
E. $$x-y \gt 0$$

Look at option D: if $$y$$ is positive, then $$x$$ must be negative in order the sum of $$x$$ and $$y$$ to be negative.

Hi Bunuel,

If D is must be true option then even C will also satisfy .

I tried with C and D with some random numbers and both seemed correct to me.

I have gone through the other explanations which you gave but didn't get.

Can you please help me to understand much better that how we can get MUST case and COULD BE in such questions.

As per the question, we know:
i. $$x$$ and $$y$$ are integers, and
ii. $$x + y = -12$$

From this, we get following five situations/ scenarios:
1) $$x < 0$$, $$y < 0$$ (eg. $$x = -5$$, $$y = -7$$ or $$x = -7$$, $$y = -5$$)
2) $$x < 0$$, $$y > 0$$ (eg. $$x = -13$$, $$y = 1$$)
3) $$x > 0$$, $$y < 0$$ (eg. $$x = 1$$, $$y = -13$$)
4) $$x = 0$$, $$y < 0$$ ($$x = 0$$, $$y = -12$$)
5) $$x < 0$$, $$y = 0$$ ($$x = -12$$, $$y = 0$$)

Let us now consider each of the options to determine whether they must be true for each of the above scenarios:
Option A:
Both $$x$$ and $$y$$ are negative
in other words: $$x <0$$, $$y < 0$$. This is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option B:
$$xy>0$$, this is only possible when $$x>0$$ and $$y>0$$ or when $$x<0 and y<0$$. Again, this is true only for scenario 1) above. But this is not true for other 4 scenarios. Hence, this could be true, but not definitely true.

Option C:
If $$y<0$$, then $$x>0$$.
If we look at all the 5 scenarios, $$y<0$$ in scenario 3 and 4. However, this statement is true for scenario 3 but not true for scenario 4. This is because, if $$y = -12$$ (i.e. $$y<0$$), then $$x = 0$$ (neither negative nor positive). Hence, this could be true, but not definitely true.

Option D:
If $$y>0$$, then $$x<0$$.
If we look at all the 5 scenarios, $$y<0$$ only in scenario 2 and as per that scenario $$x>0$$. So this must be true because whenever $$y$$ is a negative integer, $$x$$ has to be a positive integer. This is the correct answer.

Option E:
$$x-y>0$$. This is true for scenario 4 but not true for scenario 5. Therefore, this could be true, but not definitely true. Remember, it is sufficient to consider 2 out of 5 scenarios, if we get different answers from them. We have to consider all the 5 scenarios only in cases where the statement is definitely true or must be true.

Hope this helps!!

Regards,
Nalin

A bit lengthy but a good detailed solution. Would add a couple of observations.

For any MUST BE TRUE question, you can only be sure of your answer once you find that examples that will negate the other options. Any correct option in a MUST BE TRUE question, will be true for ALL possible cases.

x+y=-12

A. Both x and y are negative . x=-12, y = 0 negate this condition. Eliminate.
B. xy>0. x=-12, y = 0 negate this condition. Eliminate.
C. If y<0, then x>0. x=0, y = -12 negate this condition. Eliminate.
D. If y>0, then x<0
E. x−y>0. x=-12, y = 0 negate this condition. Eliminate.

Thus, D is the correct answer. you only had to use (-12,0) or (0,-12) to arrive at the correct answer.

Hope this helps.
Manager  S
Joined: 12 Apr 2017
Posts: 118
Location: United States
Concentration: Finance, Operations
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If x and y are integers and x+y=−12, which of the following must be true?

Use logic and test numbers

X = -22, y = 10
-22 + 10 = -12

A. Both x and y are negative, testing at least the solution above proves for this to not be true.
B. x>0 y>0, testing at least the solution above proves for this to not be true.
C. If y<0, then x>, testing at least the solution above proves for this to not be true.
D. If y>0, then x<0, do not need to prove any further that of the given answer choices this is the only one that can be correct.
E. x−y>0[/color], testing at least the solution above proves for this to not be true.
Manager  B
Joined: 05 Oct 2017
Posts: 66
Location: India
Schools: GWU '21, Isenberg'21

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As per the question we are given x+y=-12
and as per the option D , we have If y >0 then x <0

i followed the following approach instead of putting some random values, i put the value of y in the inequality (y>0). What we get after putting the value y from the given equation
Quote:
-12-x>0
.......(1)
On simplifying the equation 1 we get
Quote:
x<-12
means x will always be negative (or x<0).

this is our required answer (D)
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Intern  B
Joined: 14 Jul 2017
Posts: 21

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I don't agree with the explanation. If x=-24 and y=12 then option D is correct
If x=12 and y=-24, then option C is correct...
How are you judging that y has to be > 0
Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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twister68 wrote:
I don't agree with the explanation. If x=-24 and y=12 then option D is correct
If x=12 and y=-24, then option C is correct...
How are you judging that y has to be > 0

The question asks which of the following must be true, not could be true. Only D must be true (only D is always true).

As for C: if y = 1 and x = -13, then C is not true, so C is NOT always true.

Hope it's clear.
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Intern  B
Joined: 24 Dec 2017
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I think this is a poor-quality question and I don't agree with the explanation. It’s not mentioned which is greater x or y..

If x = -15 and y= 3 the statement is sufficient.

But if y = -15 and x = 3 , then also it is sufficient.

Posted from my mobile device
Math Expert V
Joined: 02 Sep 2009
Posts: 56260

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Nikii wrote:
I think this is a poor-quality question and I don't agree with the explanation. It’s not mentioned which is greater x or y..

If x = -15 and y= 3 the statement is sufficient.

But if y = -15 and x = 3 , then also it is sufficient.

Posted from my mobile device

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