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# GMAT Diagnostic Test Question 13

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GMAT Diagnostic Test Question 13 [#permalink]

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06 Jun 2009, 22:14
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GMAT Diagnostic Test Question 13
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Difficulty: 750
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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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Last edited by bb on 28 Sep 2013, 21:25, edited 2 times in total.
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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18 Oct 2009, 09:13
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stilite wrote:
GMAT TIGER wrote:
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

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
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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04 Feb 2010, 16:22
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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)
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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01 Mar 2010, 18:34
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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!
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Joined: 26 Aug 2009
Posts: 22
Re: GMAT Diagnostic Test Question 12 [#permalink]

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28 Aug 2009, 06:21
2
KUDOS
GMAT TIGER wrote:
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

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Re: GMAT Diagnostic Test Question 12 [#permalink]

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02 Apr 2010, 10:42
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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
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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16 Jul 2009, 22:42
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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.
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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28 Oct 2009, 13:21
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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
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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07 Jun 2009, 16:54
Explanation

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. _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Last edited by bb on 28 Sep 2013, 12:42, edited 1 time in total. Updated Intern Joined: 24 Jun 2009 Posts: 40 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 16 Jul 2009, 13:39 could you please explain how did you get 2A? thanks SVP Joined: 29 Aug 2007 Posts: 2473 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 18 Oct 2009, 09:57 linfongyu wrote: 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. _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Intern Joined: 24 Oct 2009 Posts: 2 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 03 Feb 2010, 20:59 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. Intern Joined: 19 Jul 2010 Posts: 1 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 26 Aug 2010, 18:42 ra011y wrote: I solved this in the following way - 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? Any consideration is welcome Intern Joined: 27 Aug 2010 Posts: 23 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 27 Aug 2010, 11:10 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. Manager Joined: 24 Apr 2010 Posts: 62 Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 03 Sep 2010, 10:15 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$2=8 so D out but there were lots of stops 1/x 1/y and x becoming b and y becoming a creating all the confusion but still... Intern Joined: 24 Sep 2010 Posts: 39 Schools: CEIBS Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 30 Sep 2010, 21:47 Sorry, I am more than slow apparently. WTH does a$ 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$= ???? or 5 (omega) = ???. Manager Joined: 27 Jul 2010 Posts: 194 Location: Prague Schools: University of Economics Prague Re: GMAT Diagnostic Test Question 12 [#permalink] ### Show Tags 09 Oct 2010, 05:44 interesting is that even when I was comptuting wrong with A-B instead of B-A in ... A<B, A$B = B-A i came to correct answer
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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14 Nov 2010, 00:16
bterwilliger wrote:
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).
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Re: GMAT Diagnostic Test Question 12 [#permalink]

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15 Dec 2010, 19:39
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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Re: GMAT Diagnostic Test Question 12 [#permalink]

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04 Sep 2011, 05:26
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
Re: GMAT Diagnostic Test Question 12   [#permalink] 04 Sep 2011, 05:26

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