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04 Sep 2011, 08:03
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sl750
Can someone explain this step
(1) Suppose if 1/x > 1/y:
(1/x $ 1/y) = 1/x + 1/y = (x+y)/(xy).
(1/y $ 1/x) = 1/x - 1/y = (y-x)/(xy). . Why does the sign change here? We are given that, for A>B A$B = A+B. So how does the sign change when we flip the quantities around the operator $
Thanks
(1) Suppose if 1/x > 1/y:
(1/x $ 1/y) = 1/x + 1/y = (x+y)/(xy).
(1/y $ 1/x) = 1/x - 1/y = (y-x)/(xy). . Why does the sign change here? We are given that, for A>B A$B = A+B. So how does the sign change when we flip the quantities around the operator $
Thanks
Signs change because the stem presents a scenario for LHS<RHS as well. See the enlarged text below from the question stem.
If A>B, A$B = A+B and if A<B, A$B = B-A, then which of the followings is the highest for \((\frac{1}{x}$\frac{1}{y})$(\frac{1}{y}$\frac{1}{x})\)
BTW, We get the maximum value for the expression when,
\(\frac{1}{x}>\frac{1}{y}\);-------Condition 1
For all the options in which Condition 1 hold true, we need to find the highest value for \(\frac{1}{x}\).
D satisfies both of those conditions.
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04 Sep 2011, 11:18
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A) (2$3)$(3$2) = 1$5 = 4
B) (3$4)$(4$3) = 1$7 = 6
c) (4$5)$(5$4) = 1$9 = 8
D) (5$4)$(4$5) = 9$1 = 10
E) (4$2)$(2$4) = 6$2 = 8
hence D
B) (3$4)$(4$3) = 1$7 = 6
c) (4$5)$(5$4) = 1$9 = 8
D) (5$4)$(4$5) = 9$1 = 10
E) (4$2)$(2$4) = 6$2 = 8
hence D
29 Jun 2012, 07:52
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LUGO
gmatoverduegirl
LUGO
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.
