If xФy = (2x - y)/(2y - x), where x ≠ 2y, then is a≠b > b≠a ?

@jaiminjani "condition 1 is included in condition 2" is wrong to assume.

e.g., let a =2, b = 3

2 < 3

but 2*2 > 3

@subhashghosh I did not say condition 2 is included in condition1.
I meant condition 1 is included in condition 2.

i.e. say a=1 and b =3
1. a<b => 1<3
2. 2a<b => 2*1 <3

In condition 2, a is multiplied by possitive integer and hemce if condition 2 is true. condition 1 will always be true.

If I am wrong please correct me with example which satisfies condition 2 and does not satisfy condition 1

Hi

When both 'a' and 'b' are negative then even if condition 2 is true, condition 1 might fail -

let a = -3 and b = -4

1. -3 < -4 not true
2. -6 < -4 true

Hope this helps

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If xФy = (2x - y)/(2y - x), where x ≠ 2y, then is aФb > bФa ?

(1) a < b
(2) 2a < b

If we modify the original condition and the question, we can see that the question is asking if (2a-b)/(2b-a)>(2b-a)/(2a-b). If we multiply both sides by (2b-a)^2(2a-b)^2, the question can be modified to(2a-b)^3(2b-a)>(2b-a)^3(2a-b)? Then, the question changes to whether (2a-b)^3(2b-a)-(2b-a)^3(2a-b)>0 is true. Then, we get (2a-b)(2b-a)[(2b-a)^2-(2a-b)^2]>0? Essentially, if we further modify the question, we can see that the question is asking 3(2a-b)(2b-a)(3b^2+3a^2)>0? There are 2 variables (a and b), and in order to match the number of variables and the number of equations, we need 2 equations. Since the condition 1) and 2) each has 1 equation, there is high chance C is the answer. However, using both the condition 1) and 2), we can see that there is no way we can find out the relationship between 2b and a. Therefore, the correct answer is E.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Can someone explain what am i doing wrong!!!
Thanks

Official answer from Kaplan -

What did i do -


Now i know what some of you will say - that we need to prove exp1 > exp2 so we can't take a > b or via-versa a < b. But what thought was that as long as we are getting firm a < b or a > b answer based on condition in option 1 and 2 we will get a sufficient Yes/No response.
I hope i was able to convey what i was thinking!!!
So please let me know where/what am i doing wrong.
Thanks

I followed same approach so can say u have done silly error in step 4

(2a-b+2b-a) (2a-b-2b+a)> 0
(a+b)(3a-3b) > 0
3(a+b)(a-b) > 0
3(a^2 - b^2) > 0
3a^2 > 3b^2
a^2 > b^2

Condtn 1 : a < b
To check value if it satisfies
So assume a=-1 and b=1 (-1 < 1)
But (-1)^2 > (1)^2
1 > 1 ---> NO
So does not satisfy

Condtn 2 : 2a < b
Take same values a=-1 b=1 (-2<1)
But to check a^2 > b^2
(-1)^2 > (1)^2
So does not satisfy

Together same result
Hence E

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