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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink]
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 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]
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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]
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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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If x and y are integers, and x + y < 0, is x — y > 0 [#permalink]
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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 ?
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink]
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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]
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08 Apr 2018, 12:34
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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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink]
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08 Apr 2018, 12:38
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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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests?
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Re: If x and y are integers, and x + y < 0, is x — y > 0 [#permalink]
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10 Apr 2018, 05:54
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
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