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GMAT Diagnostic Test Question 13 Field: special characters Difficulty: 750

Rating:

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

A. X = \(\frac{1}{2}\) and Y = \(\frac{1}{3}\) B. X = \(\frac{1}{3}\) and Y = \(\frac{1}{4}\) C. X = \(\frac{1}{4}\) and Y = \(\frac{1}{5}\) D. X = \(\frac{1}{5}\) and Y = \(\frac{1}{4}\) E. X = \(\frac{1}{4}\) and Y = \(\frac{1}{2}\) _________________

Here \((\frac{x+y}{xy})\) must be > \((\frac{y-x}{xy})\). If so, then \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\). Therefore,

Here \((\frac{x+y}{xy})\) must be > \((\frac{x-y}{xy})\). If so, then \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\). Therefore,

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.

We need to maximize the value... the maximum is give by adding everything.

If we want to add because of how the $ is defined A>B. reducing the answers to two (D,E) becuse:

D. 5>4 E. 4>2

However we don't cannot assume $ has the commutative property so in the second part the x switches with y. So now we are interested in minimizing the difference.

Here \((\frac{x+y}{xy})\) must be > \((\frac{y-x}{xy})\). If so, then \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\). Therefore,

Here \((\frac{x+y}{xy})\) must be > \((\frac{x-y}{xy})\). If so, then \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\). Therefore,

By simplification of the expression, the amount of time to solve each question reduces. However, what happens if the expression itself becomes clumsy...or better still if I made an error?

So, i had to go the natural way of the problem set. 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..........Ans D E. (4&2)&(2&4) = 6&2 = 8 _________________

KUDOS me if you feel my contribution has helped you.

Here \((\frac{x+y}{xy})\) must be > \((\frac{y-x}{xy})\). If so, then \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\). Therefore,

Here \((\frac{x+y}{xy})\) must be > \((\frac{x-y}{xy})\). If so, then \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\). Therefore,

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:

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.

Holy Mary Mother of God! How do you accomplish this in less than 2 minutes?

Each question is not strictly solvable in 2 minuets. Most of the questions are solvable under 2 minuets and few may take more than 3 minuets. On an average, each question is solvable in 2 minuets. _________________

This is actually quite straight forward, as these type of "formula" questions should be, but I stumbled enormously on the stylistic differences in the characters. And so, this question looked like a monster until that time. In the question itself the formula in the end uses the lowercase cursive style, while the answer choices use capital non cursive style. I am not sure why I not immediately thought these were one and the same...but I still did. Maybe that is just the limitation of the interface we have to work with...Or is this just me?

One other minute detail is that the symbol in this explanation thread has changed from the greek omega symbol (or for the electrical engineers among us, the OHM symbol) in the original question of the PDF file to a Dollar symbol.

We have assumed that (1/x) = M. So we now have 2M = 2/(1/x)

Substituting for all the answers we have, (A)2/(1/x) = 2/(1/2) = 4 (B)2/(1/x) = 2/(1/3) = 6 (C)2/(1/x) = 2/(1/4) = 8 (D)2/(1/x) = 2/(1/5) = 10 (E)2/(1/x) = 2/(1/4) = 8

Hence the answer is (D)

If we assume that (1/x) = M then 2M = 2/x And taking that assumption into account the answer with the greatest value in A. In my reasoning D is the answer with the lowest value (2/5) Am I missing something?

I solve this one the following way: 1) We know that A=1\x and B=1\y 2) Put the numbers into the equation (without signs) so that: A: (2 3) (3 2) B: (3 4) (4 3) C: (4 5) (5 4) D: (5 4) (4 5) E: (4 2) (2 4)

3. Now we are given that if A>B then A-B and if A<B then B-A. Put the signs and solve the expressions:

lol...i got this one right though i got 600 level one wrong... but it took me around 3.5 minutes i did 3-2$3+2=1$5 as b is high 4 now without looking i did same for 2 and 3 as same pattern is there 1(subtract)$7(add)=6 1$9=8 now as d and e are opposite must be add --subtract and then add 9$1=10 and e 6$2=8

so D out but there were lots of stops 1/x 1/y and x becoming b and y becoming a creating all the confusion but still...

Sorry, I am more than slow apparently. WTH does a $ or Omega mean? Can some kind soul type out the meaning of the question? I have spent half an hour trying to find the meaning for either symbol in math, with no luck. For example, 5! = 5 factorial. So then 5$ = ???? or 5 (omega) = ???.

Sorry, I am more than slow apparently. WTH does a $ or Omega mean? Can some kind soul type out the meaning of the question? I have spent half an hour trying to find the meaning for either symbol in math, with no luck. For example, 5! = 5 factorial. So then 5$ = ???? or 5 (omega) = ???.

The dollar sign and omega are just place holders,same with all the bizzare signs you see on the test. You have to determine which sign to use with the formula.

ie: a>b a$b= a+b then (3$2)=(3+2) if a<b=a$b=(b-a) then (2$3)= (3-2).

I in fact think, the easiest way to solve this is to understand that if A>B then you are adding two positive fractions and if A<B then you are taking a difference of the two. Since you are asked to find the highest value possible as an outcome of that long equation, it is essential to keep in mind that whatever values you choose, try to find it such that A>B. This will off the top help you narrow down your choices.

---------- Kudos? _________________

-DK --------------------------------------------------------- If you like what you read then give a Kudos! Diagnostic Test: 620 The past is a guidepost, not a hitching post. ---------------------------------------------------------

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 $