LUGO wrote:

gmatoverduegirl wrote:

LUGO wrote:

Actually something that you could do for sure under two minutes is the following:

Possible combinations are:

+ + - = 2B

- + + = 2B

+ - - = 2A

- - + = - 2A

Clearly you can not have a negative answer for greatest? nor two choices with the same answer 2B? Hence the right answer has to be of the type 2A meaning => [(A>B) = A+ B] - [(A<B) = B - A] = [5 + 4] - [4 - 5] = 9 + 1 = 10.

It took me 18 seconds

Lugo can you please tell me how you came with the answers 2B, 2B, 2A and -2A? sorry but i still don't get it

All answers are positive real numbers, right? So there is no need to worry about adding/subtracting negative numbers. Also all answers can be converted to integer numbers since the formula to solve is made of inverse number of reals which are actually integer numbers i.e. 1 / (1/4) is 4 and 1 / (1/5) is 5 right?

Next, the formula to solve is (bracket_1) +/- (bracket_2) where either bracket is a combination of adding/subtracting 2 integer numbers and given to you in answers A,B,C,D and E. Think about all possible combinations of bracket_1 and bracket_2 in terms of sign. Given that you have two brackets and two possible signs (2^2) = 4 X 2 cases = 8 cases. However eliminate the ones where both bracket_1 and bracket_2 are equal, i.e. A>B and A>B or A<B and A<B since, from the answers provided and formula written in the question, no two real numbers can be the same, right?

This means that you are left with only 4 possible cases:

Bracket_1 +/- Bracket_2

+ + - = (A>B) + (A<B) = A+B+

B-A =

2B -

+ + = (A<B)

+ (A>B) = B-A

+A+B =

2B + - - = (A>B) - (A<B) = A+B-B+A = 2A

- - + = (A<B) - (A>B) = B-A-A-B = -2A Next the question ask you for greatest. Notice than from the cases above 2A and 2B can be greater than any other combination, correct? However 2B is duplicated twice so which one do you choose? It has to be 2A since each question can only have one possible answer. Besides try few cases from the answers provided and see what happens.

I hope it helps....

LUGO

Lugo, not sure everything is right here.

Some cases can never happen and you have to switch the order of the brackets when the second is negative.

Thus we are only left with (only possible cases as

"ra011y" explained it the same way in page 1) :

A>B (i.e. B<A) : ==> Bracket_1 (A+B) > Bracket_2 (A-B) ==> Bracket_1

+ Bracket_2

==> (+) + (-) = (A>B) $ (B<A) = (A>B) + (B<A) = (A+B) + (A-B) = A+B+A-B = 2A

A<B (i.e. B>A) : ==> Bracket_1 (B-A) < Bracket_2 (A+B) ==> Bracket_1

- Bracket_2

==> (-) - (+) = (A<B) $ (B>A) = (B>A) - (A<B) = (B+A) - (B-A) = B+A-B+A = 2A

Hope it's clear.