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Re: If x  y = x+y and xy does not equal to 0, which of [#permalink]
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03 May 2014, 00:05
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divakarbio7 wrote: If x  y = x+y and xy does not equal to 0, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y > 0 D. xy > 0 E. xy < 0 Let us plug in and eliminate: x = 2 and y = 2 OR x = 2 and y = 2 then 2  2 = 22, let us check the options A. ELIMINATED (2  2>0) B. ELIMINATED (2  (2) <0) C. ELIMINATED (2  2 > 0) D. ELIMINATED (2(2) < 0 ) E. The only one which stays Hence answer is E.
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Re: PS  Number system [#permalink]
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03 May 2014, 04:22
PathFinder007 wrote: Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\). Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. HI Bunnel, 3. x0y: x<0<y > xy=xy and x+y=xy > xy={xy} could you please clarify above statement. here y is positive so x+y should be x+y Please clarify Thanks. y is positive but x is negative and x is further from zero than y, so x+y is negative hence x+y=(x+y). For example, x=5 and y=1 > x+y=4=negative.
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Re: If x  y = x+y and xy does not equal to 0, which of [#permalink]
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06 May 2014, 21:00
My Analysis : xy # 0 mean x and y might have either positive or negative Numbers ( Integers) Concepts : Absolute value , Modules My Steps : As xy # 0 , So x and y have either positive or negative numbers By Plugging x and y as positive numbers x  y = x+y the Equation never becomes Equal So One Number has to be Negative Ex : x = 6 , y = 3 6  3 = 6 + 3 6  3 = 3  3 = 3 So xy = 6*3 = 18 Mean , xy < 0 E If Any thing is Wrong Pls Reply
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Re: PS  Number system [#permalink]
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18 Jun 2014, 06:57
Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\) Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. Hi Bunuel , Can you please explain how \(xy=xy\) > \(xy\leq{0}\),



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Re: PS  Number system [#permalink]
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18 Jun 2014, 08:13
gauravsoni wrote: Bunuel wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\) Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. Hi Bunuel , Can you please explain how \(xy=xy\) > \(xy\leq{0}\), Absolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). Hope this helps.
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Re: If x  y = x+y and xy does not equal to 0, which of [#permalink]
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18 Jun 2014, 12:41
Bunuel wrote: gauravsoni wrote: divakarbio7 wrote: if lxl  lyl = lx+yl anf xy does , not equal to o, which of the following must be true?
A. xy > 0 B. xy < 0 C. x+y >0 D. xy>0 E. xy<0 \(xy=x+y\) > square both sides > \((xy)^2=(x+y)^2\) > note that \((x+y)^2=(x+y)^2\) > \((xy)^2=(x+y)^2\) > \(x^22xy+y^2=x^2+2xy+y^2\) > \(xy=xy\) > \(xy\leq{0}\) Answer: E. Another way:Right hand side, \(x+y\), is an absolute value, which is always nonnegative, but as \(xy\neq{0}\), then in this case it's positive > \(RHS=x+y>0\). So LHS must also be more than zero \(xy>0\), or \(x>y\). So we can have following 4 scenarios: 1. 0yx: \(0<y<x\) > \(xy=xy\) and \(x+y=x+y\) > \(xy\neq{x+y}\). Not correct. 2. y0x: \(y<0<x\) > \(xy=x+y\) and \(x+y=x+y\) > \(x+y={x+y}\). Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. 4. xy0: \(x<y<0\) > \(xy=x+y\) and \(x+y=xy\) > \(x+y\neq{xy}\). Not correct. So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) > \(x\) and \(y\) have opposite signs > \(xy<0\). Answer: E. Hope it helps. Hi Bunuel , Can you please explain how \(xy=xy\) > \(xy\leq{0}\), Absolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). Hope this helps.[/quote] Thanks a lot Bunuel.



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Re: If x  y = x+y and xy not equal zero , which of the following [#permalink]
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23 Nov 2014, 00:19
Hi, How did we get this? Quote: 3. x0y: x<0<y > xy=xy and x+y=xy > xy={xy}. Correct.



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Re: If x  y = x+y and xy not equal zero , which of the following [#permalink]
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23 Nov 2014, 05:21
aj0809 wrote: Hi, How did we get this? Quote: 3. x0y: x<0<y > xy=xy and x+y=xy > xy={xy}. Correct. 3. x0y: \(x<0<y\) > \(xy=xy\) and \(x+y=xy\) > \(xy={xy}\). Correct. For this case x < 0 < y and x is further from 0 than y is. Now, since x < 0, then x = x and since y > 0, then y = y. Also, since x is negative and further from 0 than y, then x + y= (x + y)=  x  y (for example, consider x = 2 and y =1). Does this make sense?
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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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If x  y = x+y and xy not equal zero , which of the following [#permalink]
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25 Jul 2016, 09:37
Pedros wrote: If x  y = x+y and xy not equal zero , which of the following must be true ?
A. xy> 0 B. xy< 0 C. x+y> 0 D. xy>0 E. xy<0 Whatever comes out of x will be positive, whatever comes out of y will be positive since LHS =RHS and RHS is positive we can say for LHS : positive  positive = smaller positive Smaller positive = x+y {we are comparing LHS with RHS now as per the equation} Smaller Positive = ??? WHAT KIND OF POSITIVE ..obviously smaller SO, The positive on RHS also has to be a smaller positive. Not a big positive This can be only possible if either x is negative or y is negative because if both x and y have same polarity then subtracting xy will be smaller positive and x+y will be bigger ; but in that case LHS will not be equal to RHS so we can infer that both x and y do not have the same polarity either x or y is negative therefore their product xy will be negative hence xy<0 ANSWER is E
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Re: If x  y = x+y and xy not equal zero , which of the following [#permalink]
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25 Dec 2017, 09:51
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Re: If x  y = x+y and xy not equal zero , which of the following
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