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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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16 Dec 2017, 00:05
TaN1213 wrote: Bunuel wrote: Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.
\(xy>xy\)?
(1) \(y<x\), 3 possible cases for \(xy>xy\):
A. \(0\)\(y\)\(x\), \(0<y<x\) > in this case \(xy>xy\) becomes: \(xy>xy\) > \(0>0\). Which is wrong; B. \(y\)\(0\)\(x\), \(y<0<x\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(y<0\). Which is right, as we consider the range \(y<0<x\); C. \(y\)\(x\)\(0\), \(y<x<0\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\). Which is right, as we consider the range \(y<x<0\).
Two different answers. Not sufficient.
(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(xy>xy\):
A. \(y\)\(0\)\(x\), \(y<0<x\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(y<0\). Which is right, as we consider the range \(y<0<x\); B. \(x\)\(0\)\(y\), \(x<0<y\) > in this case \(xy>xy\) becomes: \(x+y>xy\) > \(y>0\). Which is right, as we consider the range \(x<0<y\).
In both cases inequality holds true. Sufficient.
Answer: B. Hi Bunuel, Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt? In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y. Understood the explanation of st. 1 But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part)  Why have you not followed the process that you have followed in st 1 ? What did this difference in procedure depend on? Thank You in advance. When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\)(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(xy>xy\): B. \(x\)\(0\)\(y\). This means that \(x<0<y\). So, \(x  y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x  y < 0\), then \(xy=(xy)\), if \(x<0\), then \(x=x\) and if \(y>0\), then \(y=y\). Hope it's clear.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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16 Dec 2017, 10:56
Bunuel wrote: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\)
(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(xy>xy\):
B. \(x\)\(0\)\(y\).
This means that \(x<0<y\). So, \(x  y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x  y < 0\), then \(xy=(xy)\), if \(x<0\), then \(x=x\) and if \(y>0\), then \(y=y\).
Hope it's clear. Following this same approach, shouldn't the following part become (xy)? Quote: 1C. \(y\)\(x\)\(0\), \(y<x<0\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\). Which is right, as we consider the range \(y<x<0\). in 2 B you have modified the \(xy\) to (xy) because x is negative and y is positive. So in the same way the 1C should take (xy) as per the sign of x. What am I missing?
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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16 Dec 2017, 11:23
TaN1213 wrote: Bunuel wrote: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\)
(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(xy>xy\):
B. \(x\)\(0\)\(y\).
This means that \(x<0<y\). So, \(x  y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x  y < 0\), then \(xy=(xy)\), if \(x<0\), then \(x=x\) and if \(y>0\), then \(y=y\).
Hope it's clear. Following this same approach, shouldn't the following part become (xy)? Quote: 1C. \(y\)\(x\)\(0\), \(y<x<0\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\). Which is right, as we consider the range \(y<x<0\). in 2 B you have modified the \(xy\) to (xy) because x is negative and y is positive. So in the same way the 1C should take (xy) as per the sign of x. What am I missing? C. \(y\)\(x\)\(0\), \(y<x<0\): x  y > 0, so x  y = x  y; y < 0, so y = y; x < 0, so x = x. \(xy>xy\) > x  y > x  (y) > x  y > x + y
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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17 Dec 2017, 02:16
I am a little bit concerned here. If the question states x and y, can we make an assumption that x ≠ y? For example here, without this assumption, i can let x = y and (2) is NS?



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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17 Dec 2017, 02:22
khiemchii wrote: I am a little bit concerned here. If the question states x and y, can we make an assumption that x ≠ y? For example here, without this assumption, i can let x = y and (2) is NS? Unless it is explicitly stated otherwise, different variables CAN represent the same number. For (2) though x = y is not possible because in this case we'd get x^2 < 0, which is not true for any real number, so x = y does not satisfy xy < 0 and therefore should not be considered for the second statement. Similarly, for (1) it's give that y < x, so x = y cannot be true.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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20 Jan 2018, 21:31
Can anyone tell me the flaw in my logic
x  y > x  y Squaring both sides x^2 +y^22xy> x^2 +y^22xy simplifying, xy<xy,
It will only be possible when product of xy is ve. or xy<0, Hence B.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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22 Jan 2018, 03:27
ManishKM1 wrote: Can anyone tell me the flaw in my logic
x  y > x  y Squaring both sides x^2 +y^22xy> x^2 +y^22xy simplifying, xy<xy,
It will only be possible when product of xy is ve. or xy<0, Hence B. Given: x  y > x  y, you cannot square it. You can square only when you know that both sides are positive. Here, the right hand side may not be positive. For example, if x is 2 and y is 5. 3 > 3 Squaring does not work here since you get 9 > 9 which doesn't hold.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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04 Mar 2018, 00:17
VeritasPrepKarishmaIs it incorrect to say that question is asking is x>y ? Posted from my mobile device



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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05 Mar 2018, 05:23
Mudit27021988 wrote: VeritasPrepKarishmaIs it incorrect to say that question is asking is x>y ? Posted from my mobile device For all real x and y, x – y >= x – y x – y = x – y when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 x – y > x – y in all other cases (i.e. y is not 0 and x < y or x and y have opposite signs) So x – y > x – y is not the same as x > y Check: https://www.veritasprep.com/blog/2014/0 ... thegmat/https://www.veritasprep.com/blog/2014/0 ... tpartii/
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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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12 Mar 2018, 17:51
Bunuel wrote: Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.
Bunuel Xlnt interpretation of the QUESTION. Personally feel if the above interpretation is understood and visualized on the number line the Question becomes easy to solve. Thus the Question holds true if 1) Both x,y are +ve and ONLY if x<y (eg 3,4) 2) Both x,y are ve and ONLY if x>y (eg 3,4) 3) Both x,y are of OPPOSITE sign. Can be x>y OR x<y Stat 1 y<x OR x>y => if both ve, then 'Yes' from (2) => if both +ve then 'No' from (1) Since no UNIQUE outcome so NOT SUFFICIENT Stat 2 xy<0 => so BOTH x,y are of OPPOSITE sign therefore ALWAYS 'Yes' from (3) SUFFICIENT Option B Regards Dinesh



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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26 Apr 2018, 09:03
VeritasPrepKarishma wrote: Mudit27021988 wrote: VeritasPrepKarishmaIs it incorrect to say that question is asking is x>y ? Posted from my mobile device For all real x and y, x – y >= x – y x – y = x – y when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 x – y > x – y in all other cases (i.e. y is not 0 and x < y or x and y have opposite signs) So x – y > x – y is not the same as x > y Check: https://www.veritasprep.com/blog/2014/0 ... thegmat/https://www.veritasprep.com/blog/2014/0 ... tpartii/Hi Karishma I understand that either of the above mentioned two conditions should be satisfied to exclude the '=' from the inequality. In our case, the 1st condition is considered but how do we check that the absolute value of x is greater than y in this particular question. Thank you for your help. AS



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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02 Jun 2018, 05:06
Bunuel wrote: Is x  y > x  y?
(1) \(y<x\)
Try two positive number \(x=3>y=1\) > is \(31>31\)? > is \(2>2\)? Answer NO.
Try ANY other case but both positive: \(x=5>y=7\) > is \(5(7)>57\)? > is \(2>2\)? Answer YES.
Two different answers. Not sufficient.
(2) \(xy<0\), means \(x\) and \(y\) have different signs.
Now we can spot here that when \(x\) and \(y\) have different signs \(xy\) always "contribute" to each other so that its absolute value will increase: \(x=3\), \(y=1\) > \(xy=3+1=4=4\) or \(x=3\), \(y=1\) > \(xy=31=4=4\).
But \(xy\) is difference (thus not "contributing") of two positive values (as neither equals to zero). \(x=3\), \(y=1\) > \(xy=31=2\) or \(x=3\), \(y=1\) > \(xy=31=2\).
So \(xy<0\) means \(xy>xy\) is always true.
Sufficient.
Answer: B.
Hope it helps. Bunuel, Followed your Algebraic Approach and Plugin the number approach. How do we decide, which one to use during exam? in such questions, considering 2 min/question estimate.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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21 Jun 2018, 06:34
It would be easier by plugging in the neumerical values. Answer is B .



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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21 Jun 2018, 22:51
Hi Bunuel,
could you check the following approach:
Absolute value property: If \(x+y > x+y\) then \(xy<0\)
Rewrite the original statement: \(xy > xy\) to \(xy+y > x+yy\)
According to the property above the final question becomes: \((xy)(y) < 0\)? > Is \(xy < y^2\) ?
Stm 2: \(xy<0\) > sufficient.
Is that a valid approach?



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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21 Jun 2018, 23:15



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Re: Is x  y > x  y? (1) y < x (2) xy < 0
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22 Jun 2018, 01:06
study wrote: Is x  y > x  y?
(1) y < x (2) xy < 0
For x  y we have (i) if x  y > 0, x > y Hence we have if x > 0, then y >= 0 or y < 0 if x < 0, then y < 0 & (ii) if x  y < 0, x < y, we have if y > 0, then x >=0 or x < 0 if y < 0, then x < 0 Statement 1: y < x , hence we have case 1, however there are multiple possibilities, since we do not know signs of x & y. Hence Statement 1 is Not Sufficient. Statement 2: xy < 0 Hence either x > 0, y < 0 or x < 0 , y > 0, lets take x > 0, y < 0 case & plug in numbers x = 1, y = 1, we have for x  y > x  y LHS = 2 RHS = 0, hence LHS > RHS lets take the case where x < 0 , y > 0 x = 1, y = 1, we get LHS = 2 RHS = 0, hence LHS > RHS Statement 2 is Sufficient. Answer B. Thanks, GyM
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