GMAT Diagnostic Test Question 12
Field: special characters
Difficulty: 750



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}
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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!

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

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

Explanation

Official Answer: D

A: Since X > Y, 1/Y > 1/X.
= (1/X $1/Y) $(1/Y $1/X)
= {1/(1/2) $1/(1/3)} ${1/(1/3) $1/(1/2)}
= (2 $3) $(3 $2)
= (3-2) $(3+2) = 1 $5 = 5 – 1 = 4

B: Since X > Y, 1/Y > 1/X.
= (1/X $1/Y) $(1/Y $1/X)
= {1/(1/3) $1/(1/4)} ${1/(1/4) $1/(1/3)}
= (3 $4) $(4 $3)
= (4-3) $(4+3) = 1 $7 = 7 – 1 = 6

C: Since X > Y, 1/Y > 1/X.
= (1/X $1/Y) $(1/Y $1/X)
= {1/(1/4) $1/(1/5)} ${1/(1/5) $1/(1/4)}
= (4 $5) $(5 $4)
= (5-4) $(5+4) = 1 $9 = 9 – 1 = 8

D: Since X < Y, 1/X > 1/Y
= (1/X $1/Y) $(1/Y $1/X)
= {1/(1/5) $1/(1/4)} ${1/(1/4) $1/(1/5)}
= (5 $4) $(4 $5)
= (5+4) $(5-4) = 9 $1 = 9 + 1 = 10

E: Since X < Y, 1/X > 1/Y
= (1/X $1/Y) $(1/Y $1/X)
= {1/(1/4) $1/(1/2)} ${1/(1/2) $1/(1/4)}
= (4 $2) $(2 $4)
= (4+2) $(4-2) = 6 $2 = 6 + 2 = 8

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.
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Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

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.
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

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

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...

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
_________________

You want somethin', go get it. Period!

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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-DK
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