GMAT Diagnostic Test Question 13

Two scenerios: (1) A > B and (2) A < B.

(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).

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{1}{x} + \frac{1}{y}) $ (\frac{1}{x} - \frac{1}{y})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) $ (\frac{y-x}{xy})\)

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,

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) + (\frac{y-x}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y+y-x}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2y}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2}{x})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = 2A if we suppose 1/x = A.

(1) Suppose if 1/x < 1/y:
(1/x $ 1/y) = 1/y - 1/x = (x-y)/(xy).
(1/y $ 1/x) = 1/y + 1/x = (x+y)/(xy).

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{1}{y} - \frac{1}{x}) $ (\frac{1}{y} + \frac{1}{y})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x-y}{xy}) $ (\frac{x+y}{xy})\)

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,

\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y}{xy}) - (\frac{x-y}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{x+y-x+y}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2y}{xy})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = \((\frac{2}{x})\)
\((\frac{1}{x} $ \frac{1}{y}) $ (\frac{1}{y} $ \frac{1}{x})\) = 2A if we suppose 1/x = A.

Therefore, in short, the value of the operation is equal to 2A. A = 1/X in each case and highest A is 1/(1/5) = 5 in D. So 2A in D is 10.

Hope it is clear.

I think this explanation might be too much (although it leads to the correct answer).

I seem to have a big problem with looking to "simplify expressions and so on. In this problem, I simply wrote out each possibility:

A| 2 3 3 2 = (2-3) 1 (3+2) 5 = 4
B| 3 4 3 2 = " 1 " 7 = 6
C| 4 5 4 3 = " 1 " 9 = 8
D| 5 4 4 5 = " 9 " 1 = 10
E| 4 2 2 4 = " 6 " 2 = 8

Answer = D

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

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.

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

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.

I solved this in the following way -

[ (1/x) $ (1/y)] $[ (1/y) $ (1/x)]

Assume (1/x) = M and (1/y) = N.

So now we have, (M $ N) $ (N $ M)

Step 1:
Let M > N.

Then, (M + N) $ ( M - N) and now since (M + N) > ( M - N)
=> M + N + M - N = 2M

Step 2:
Let N > M.

Then, (N - M) $ ( N + M ) and now since ( N + M ) > (N - M)
=> N + M - (N - M)
= N + M - N + M
= 2M

So either way we get 2M as the answer.

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)

I did it a simpler way:

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.

D. 5-4=1
E. 4-2=2

Leaving us with answer D!

Hope it helps!

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

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?

Any consideration is welcome

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:

A: (2-3)-(3+2)=-6
B: (3-4)-(4+3)=-8
C: (4-5)-(5+4)=-10
D: (5+4)+(4-5)=8
E: (4+2)+(2-4)=4

The highest value is 8, which is D.

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) = ???.

interesting is that even when I was comptuting wrong with A-B instead of B-A in ... A<B, A$B = B-A i came to correct answer

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.

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Kudos?

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