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Is x – 3 > y – 3? [#permalink]
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15 Feb 2011, 12:28
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Is x – 3 > y – 3? (1) x > y. (2) xу is not equal to 0.
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Last edited by Bunuel on 25 Jul 2013, 01:13, edited 2 times in total.
Added the OA.



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Re: 180. Is Ix – 3I > Iy – 3I? (1) x > y. (2) xу is not equal [#permalink]
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15 Feb 2011, 12:59
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Is x – 3 > y – 3? x3 > (y3) x3 > y+3 x+y > 6 or x3 > y3 xy > 0 x>y so; if x>y and x+y>6; we can be sure that x3 > y3 (1) x > y. But we don't know whether x+y>6. Not sufficient. x=2 y=1 1<2 x=5 y=2 2>1 (2) xу is not equal to 0. We don't know whether x>y or x+y>6. Not sufficient. Same sample set from 1 can be used; Together; We don't know whether x+y>6. Not sufficient. Same sample set from 1 can be used. Ans: "E"
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Re: 180. Is Ix – 3I > Iy – 3I? (1) x > y. (2) xу is not equal [#permalink]
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15 Feb 2011, 13:05



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Re: Is x – 3 > y – 3? [#permalink]
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23 Feb 2014, 04:05



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Re: Is x – 3 > y – 3? [#permalink]
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07 May 2014, 20:14
Is x – 3 > y – 3?
To make true the statement we have to alternatives: x>y (both x and y positives) x<y (both x and y negatives)
(1) x > y. If both are positive the answer will be YES but if they are negatives the answer will be NO (2) xу is not equal to 0. This statement is irrelevant.
(1)+(2) We can not determine. Not sufficient.



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Re: Is x – 3 > y – 3? [#permalink]
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04 Oct 2014, 20:06
E
Picking smart numbers is the best approach for this problem: (1) N: x=6, y=6 Y: x=6, y=5 NS
(2) Same numbers as (1) NS
(1)+(2) NS



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Re: Is x – 3 > y – 3? [#permalink]
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24 May 2016, 02:01
Hello Experts, Since both sides of the eqn are positive I assumed we can approach the problem as below: (x3)^2  (y3)^2 >0 which leads to (x3y3)(x3y+3)>0 and further (xy6)(xy)>0 ie, we get 2 solutions : x>y or (xy)>6. Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.
Thanks & Regards, Nab



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Re: Is x – 3 > y – 3? [#permalink]
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24 May 2016, 07:46
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Nab77 wrote: Hello Experts, Since both sides of the eqn are positive I assumed we can approach the problem as below: (x3)^2  (y3)^2 >0 which leads to (x3y3)(x3y+3)>0 and further (xy6)(xy)>0 ie, we get 2 solutions : x>y or (xy)>6. Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.
Thanks & Regards, Nab Hi Nab, you have done two mistakes in the highlighted portion.. 1) \((x3)^2  (y3)^2 >0.............. (x3+y3)(x3(y3))>0...... (x+y6)(xy)>0.....\) and NOT (xy6)(xy)>0 2) It is NOT x>y or (xy)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6.. so two cases a) BOTH (x+y6) and (xy) are +ive or BOTh are ive
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Is x – 3 > y – 3? [#permalink]
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24 May 2016, 08:42
chetan2u wrote: Nab77 wrote: Hello Experts, Since both sides of the eqn are positive I assumed we can approach the problem as below: (x3)^2  (y3)^2 >0 which leads to (x3y3)(x3y+3)>0 and further (xy6)(xy)>0 ie, we get 2 solutions : x>y or (xy)>6. Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.
Thanks & Regards, Nab Hi Nab, you have done two mistakes in the highlighted portion.. 1) \((x3)^2  (y3)^2 >0.............. (x3+y3)(x3(y3))>0...... (x+y6)(xy)>0.....\) and NOT (xy6)(xy)>0 2) It is NOT x>y or (xy)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6.. so two cases a) BOTH (x+y6) and (xy) are +ive or BOTh are ive Hi Chetan, Oops yes I'm sorry that was a typo, i did get (x+y6)(xy)>0. From 2) Do you mean that it is not an OR condition but an AND condition? I didn't understand that quite well, can you please explain again. Especially the part BUT x>y and (x+y)>6 or x<y and (x+y)<6. I didn't understand how the equality signs changed. Thanks for your reply. Regards, Nab



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Re: Is x – 3 > y – 3? [#permalink]
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24 May 2016, 08:54
Nab77 wrote: chetan2u wrote: Nab77 wrote: Hello Experts, Since both sides of the eqn are positive I assumed we can approach the problem as below: (x3)^2  (y3)^2 >0 which leads to (x3y3)(x3y+3)>0 and further (xy6)(xy)>0 ie, we get 2 solutions : x>y or (xy)>6. Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.
Thanks & Regards, Nab Hi Nab, you have done two mistakes in the highlighted portion.. 1) \((x3)^2  (y3)^2 >0.............. (x3+y3)(x3(y3))>0...... (x+y6)(xy)>0.....\) and NOT (xy6)(xy)>0 2) It is NOT x>y or (xy)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6.. so two cases a) BOTH (x+y6) and (xy) are +ive or BOTh are ive Hi Chetan, Oops yes I'm sorry that was a typo, i did get (x+y6)(xy)>0. From 2) Do you mean that it is not an OR condition but an AND condition? I didn't understand that quite well, can you please explain again. Especially the part BUT x>y and (x+y)>6 or x<y and (x+y)<6. I didn't understand how the equality signs changed. Thanks for your reply. Regards, Nab Hi, the equation is \((x+y6)(xy)>0\)... The Left Hand Side can be >0 under two cases.. 1) when both (x+y6) and (xy) are greater than 0... since Positive * Positive = Positive..so xy>0 or x>y................and x+y6>0 or x+y>0......... Example x = 5, y=2.. x>y and x+y>6 ... so (5+26)(52)>0.....1*3>0...YES 2) when both (x+y6) and (xy) are lesser than 0... since Negative * Negative = Positive..so xy<0 or x<y................and x+y6<0 or x+y<6......... x= 2 and y =3.....x<y and x+y<6 ... so (3+26)(23)>0.....(1)(1)>0..........1>0......YES
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Re: Is x – 3 > y – 3? [#permalink]
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24 May 2016, 09:43
Quote: Hi, the equation is \((x+y6)(xy)>0\)... The Left Hand Side can be >0 under two cases.. 1) when both (x+y6) and (xy) are greater than 0... since Positive * Positive = Positive..so xy>0 or x>y................and x+y6>0 or x+y>0......... Example x = 5, y=2.. x>y and x+y>6 ... so (5+26)(52)>0.....1*3>0...YES 2) when both (x+y6) and (xy) are lesser than 0... since Negative * Negative = Positive..so xy<0 or x<y................and x+y6<0 or x+y<6......... x= 2 and y =3.....x<y and x+y<6 ... so (3+26)(23)>0.....(1)(1)>0..........1>0......YES Yes i realize where I went wrong! Thanks a tonne for the explanation. Thanks & Regards, Nab



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Re: Is x – 3 > y – 3? [#permalink]
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24 May 2016, 10:00
banksy wrote: Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0. (1) x > y. Let's say x=4 and y=3, then x – 3 > y – 3 but it x=3 and y=4, then the aboveineqzuality will not hold true. Insufficient (2) xу is not equal to 0 This statement means either x or y is not equal to 0. It can be +ve or ve. not sufficient. Combining both statements doesn't give a unique answer. Hence E is the answer
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Re: Is x – 3 > y – 3? [#permalink]
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04 Oct 2017, 05:01
banksy wrote: Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0. DS: Is x – 3 > y – 3? Statement 1 : x>y Lets say x =6 , y = 4 x3 = 3>y3 = 1 Lets say x = 1, y = 4 x3= 2<y3 = 7 NOT SUFFICIENT Statement 2 : xy is not equal to 0 . So, neither x nor y is not equal to 0 . Lets say x =6 , y = 4 x3 = 3>y3 = 1 Lets say x = 1, y = 4 x3= 2<y3 = 7 NOT SUFFICIENT Combined : same examples NOT SUFFICIENT Answer E
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Re: Is x – 3 > y – 3? [#permalink]
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05 Oct 2017, 20:54
Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0
2. XY not equal to 0 we don't know the value of X and Y so Insufficient
2. If x =10 and Y = 100 then Y3 is greater if x=10 and Y= 5 then X3 is greater so insuffiecient
together the same example holds good, not sufficient hence Answer should be E




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