If A$B=A+B, if A>B and A$B=B-A, if A<B, then which of

[gmatter2012]

If A$B=A+B, if A>B and A$B=B-A, if A<B, then which of the followings is highest for (1/x$1/y)$(1/y$1/x)?

A. x=1/2 and y=1/3
B. x=1/3 and y=1/4
C. x=1/4 and y=1/5
D. x=1/5 and y=1/4
E. x=1/4 and y=1/2

[sathishkumarjva9888]

Ans: D.

Can you tel the OA pls??

[gmatter2012]

Ans: D.

Can you tel the OA pls??

Hi thanks for attempting, can you please show your reasoning for ans D

P.S - Since no expert has provided solution to this question, there would be something wrong with this question, if you would show your working then may be it would help us all.

[Anurag@Gurome]

If A$B = A+B, if A > B and A$B = B-A, if A < B, then which of the followings is highest for (1/x$1/y)$(1/y$1/x)?

For positive numbers A and B and A > B, (A$B)$(B$A) = (A + B)$(A - B)
Now, (A + B) > (A - B)
Hence, (A + B)$(A - B) = (A + B) + (A - B) = 2A

And for positive numbers A and B and A < B, (A$B)$(B$A) = (B - A)$(B + A)
Now, (B - A) < (B + A)
Hence, (B - A)$(B + A) = (B + A) - (B - A) = 2A

Hence, the expression will attain highest value for highest value of A, i.e. highest value of 1/x, i.e. least value of x.

The correct answer is D.

[gmatter2012]

If A$B = A+B, if A > B and A$B = B-A, if A < B, then which of the followings is highest for (1/x$1/y)$(1/y$1/x)?

For positive numbers A and B and A > B, (A$B)$(B$A) = (A + B)$(A - B)
Now, (A + B) > (A - B)
Hence, (A + B)$(A - B) = (A + B) + (A - B) = 2A

And for positive numbers A and B and A < B, (A$B)$(B$A) = (B - A)$(B + A)
Now, (B - A) < (B + A)
Hence, (B - A)$(B + A) = (B + A) - (B - A) = 2A

Hence, the expression will attain highest value for highest value of A, i.e. highest value of 1/x, i.e. least value of x.

The correct answer is D.

Thank you Anurag, but some areas I couldn't understand, kindly bear with me.

For First case when A>B then

function is defined as A$B= A+B

but we do not have any definition of B$A but since A>B then I would assume B$A= A+B
How are we getting B$A= A-B when A>B

similarly when A<B then function is defined as A$B=B-A
but there is no definition for B$A but since A<B i would assume B$A = B-A

proceeding further for the first case:
(A+B)$(A+B) obviously I would get stuck here, as A+B = A+B and there is no definition for this condition.

same for the second case (B-A)$(B-A)..Dead end as both are equal

so my question is when A>B and A$B is defined as A+B how are we getting B$A= A-B

similarly when A<B and A$B= B-A then how are we deducing B$A = B+A

It seems that we are trying to keep the expression inside the brackets positive

example
For the first case when A>B then A$B = A+B( This of course is according to definition) and B$A= A-B( This will be positive as A>B)

when A<B
then A$B= B-A ( This again is according to definition ) but B$A is not defined but we are taking
B$A= B+A ( as this will be positive A<B)

but no where has it been mentioned that B$A is positive we know that A is positive and B is positive .
So please tell me how when A>B B$A= A -B

and when A<B then B$A = A+B

Am I missing something? or is my query justified, Kindly Bear with me.

[Anurag@Gurome]

but we do not have any definition of B$A but since A>B then I would assume B$A= A+B
How are we getting B$A= A-B when A>B

We don't need to have definition of B$A.
Just replace A with B and B with A in the definition of A$B.
And we have, the definition of B$A as follows,

• B$A = B + A, if B > A and B$A = A - B, if B < A

Hope that helps.