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Is x – 3 > y – 3? [#permalink]
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Updated on: 25 Jul 2013, 02:13
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70% (01:10) correct 30% (00:57) wrong based on 522 sessions
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Is x – 3 > y – 3? (1) x > y. (2) xу is not equal to 0.
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Originally posted by banksy on 15 Feb 2011, 13:28.
Last edited by Bunuel on 25 Jul 2013, 02:13, 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, 13:59
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, 14:05



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Re: Is x – 3 > y – 3? [#permalink]
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07 May 2014, 21:14
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, 21:06
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, 03:01
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, 08:46
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, 09:42
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, 09:54
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, 10:43
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, 11:00
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, 06:01
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, 21:54
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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Is x – 3 > y – 3? [#permalink]
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Updated on: 13 Mar 2018, 13:57
banksy wrote: Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0. For DS AbsValue questions I try to read the question and spend about 10 seconds to see if I can understand what it is asking before I even look at the statements. Is x – 3 > y – 3? When would this be true and when would it be not true? Well, if they are both positive and x>y, then yes. If they are both negative and x>y, then no. Read statement (1) and (2), neither account for the option of a Negative Y. Example: x=1 y=900, satisfies both statements, answer to question is no. Yes/No= E
Originally posted by msurls on 13 Mar 2018, 05:38.
Last edited by msurls on 13 Mar 2018, 13:57, edited 1 time in total.



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Re: Is x – 3 > y – 3? [#permalink]
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13 Mar 2018, 07:48
msurls wrote: banksy wrote: Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0. For DS AbsValue questions I try to read the question and spend about 10 seconds to see if I can understand what it is asking before I even look at the statements. Is x – 3 > y – 3? When would this be true and when would it be not true? This is really the same as "Is x>y?"Well, if they are both positive and x>y, then yes. If x>y but Y has a greater magnitude than X, then no. Read statement (1) and (2), neither account for the option of a Negative Y. Example: x=1 y=900, satisfies both statements, answer to question is no. Yes/No= E Hello You are correct that answer is E. However, I would like to point out the highlighted part in your analysis. (I have highlighted) x3 > y3 is NOT the same as x>y. Eg., if x=1 and y=5, then x3 > y3 BUT x < y Actually x3 > y3 means that the distance of 'x' and '3' is more than the distance of 'y' and '3'.



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Re: Is x – 3 > y – 3? [#permalink]
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15 Mar 2018, 13:10
banksy wrote: Is x – 3 > y – 3?
(1) x > y. (2) xу is not equal to 0. Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2): x = 4, y = 3: Yes x = 3, y = 2: No Since we have two answer, "yes" and "no", both conditions together are not sufficient. Therefore, the answer is E. Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is x – 3 > y – 3? [#permalink]
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26 Mar 2018, 09:17
fluke wrote: 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" I got the solution explained above. My doubt is shouldn't be we checking the other 2 scenarios also which are (x3)>(y3) and (x3) > (y3) my thinking is as there are two modulus involved we should check all the 4 possible scenario x3 > y3 x3 > (y3) (x3)>(y3) (x3) > (y3)




Re: Is x – 3 > y – 3?
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