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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 nonzero 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 nonzero 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 nonzero 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 nonzero 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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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.
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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 nonzero 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.
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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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