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If x and y are integers, and x + y < 0, is x — y > 0

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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 09:41
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If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?
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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 12:13
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jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   10 Apr 2018, 05:54
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jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   29 Jan 2020, 02:13
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jabhatta@umail.iu.edu wrote:
VeritasKarishma wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)


Hi karishma --

on the first highlight in red -- any reason why the following cases were not considered

Case 1) not considered) X and Y are both negative
Case 1, Scenario 1 when both variables are negative) |Y| > |x|
Case 1, Scenario 2 when both variables are negative) |x| > |Y|



Case 2) not considered) X and Y can be zero
Case 2, Scenario 1, x = 0 and Y is negative
Case 2, Scenario 2 y = 0 and X is negative


As per the highlighted part: x + y < 0 implies at least one of x and y is negative

When we say AT LEAST one of x and y is negative, it includes the case in which both are negative. It also includes the case in which one of them is 0 and the other negative. Also, it includes the case in which one of them is positive and the other negative such that the negative one has higher absolute value.
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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 11:50
hmm if interested in testing cases -- here is the link to the OA.....you should be able to find it using test cases ...but it takes too long imo

Has to be a simpler way

https://www.manhattanprep.com/gmat/blog ... flowchart/
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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   Updated on: 08 Apr 2018, 12:33
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.


Wow -- this is really well done ..

Bunuel - do you think its easier to

a) test cases in this case or
b) do you prefer arithmetic

Do you think you could have done your method in 3 mins time pressure ?

Originally posted by jabhatta2 on 08 Apr 2018, 12:30.
Last edited by jabhatta2 on 08 Apr 2018, 12:33, edited 1 time in total.
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 12:32
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.


Hi Bunuel - a question on the highlighted portion (bullet b and c)

How is it bullet B tells us x = -1 whereas bullet c tells us x = 1

Dont bullet b and bullet C almost contradict each other in a way ?
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 12:34
Expert Reply
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.


Wow -- this is really well done ..

Bunuel - do you think its easier to

a) test cases in this case or
b) do you prefer your method in terms -- Do you think you could have done your method in 3 mins


I'd done this way but it's really subjective.
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   08 Apr 2018, 12:38
Expert Reply
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.


Hi Bunuel - a question on the highlighted portion (bullet b and c)

How is it bullet B tells us x = -1 whereas bullet c tells us x = 1

Dont bullet b and bullet C almost contradict each other in a way ?


Those are different cases for which x^y = 1:
b. x = -1 and y = even. For example, x= -1 and y = 2 --> x^y = 1.
c. x = 1 and y = any integer. For example, x= 1 and y = 3 --> x^y = 1.
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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink] New post   28 Jan 2020, 17:21
VeritasKarishma wrote:
jabhatta@umail.iu.edu wrote:
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?


You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)


Hi karishma --

on the first highlight in red -- any reason why the following cases were not considered

Case 1) not considered) X and Y are both negative
Case 1, Scenario 1 when both variables are negative) |Y| > |x|
Case 1, Scenario 2 when both variables are negative) |x| > |Y|



Case 2) not considered) X and Y can be zero
Case 2, Scenario 1, x = 0 and Y is negative
Case 2, Scenario 2 y = 0 and X is negative
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Re: If x and y are integers, and x + y < 0, is x y > 0 [#permalink] New post   25 Nov 2022, 02:34
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Re: If x and y are integers, and x + y < 0, is x y > 0 [#permalink]
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