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M06 #16

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M06 #16 [#permalink] New post 04 Feb 2012, 00:17
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

67% (01:01) correct 33% (00:12) wrong based on 7 sessions
If \frac{t}{u} = \frac{x}{y} and \frac{t}{y} = \frac{u}{x} and t , u , x , and y are non-zero integers, which of the following is true?

A. \frac{t}{u}=1
B. \frac{y}{x}=-1
C. t = u
D. t = \pm u
E. None of the above

A question to Bunuel or someone who actually wrote the test:

From the given equation we get ux=ty (1) and xt=uy (2)
subtracting 2 from 1 this we get x(u-t)= y (t-u) --> this gives us x=-y. This is option B.
So, why is option B wrong?
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Re: M06 #16 [#permalink] New post 04 Feb 2012, 07:52
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mourinhogmat1 wrote:
If \frac{t}{u} = \frac{x}{y} and \frac{t}{y} = \frac{u}{x} and t , u , x , and y are non-zero integers, which of the following is true?

A. \frac{t}{u}=1
B. \frac{y}{x}=-1
C. t = u
D. t = \pm u
E. None of the above

A question to Bunuel or someone who actually wrote the test:

From the given equation we get ux=ty (1) and xt=uy (2)
subtracting 2 from 1 this we get x(u-t)= y (t-u) --> this gives us x=-y. This is option B.
So, why is option B wrong?


Because you can not tell from x(u-t)=y(t-u) that x=-y is necessarily true:
x(u-t)=y(t-u) --> x(u-t)+y(u-t)=(x+y)(t-u) --> either x=-y or t=u.

Complete solution:
Given that: \frac{t}{u} = \frac{x}{y} and \frac{t}{y} = \frac{u}{x}.

So, \frac{t}{u} = \frac{x}{y} and \frac{t}{u} = \frac{y}{x} (from 2), which means that \frac{t}{u} and \frac{x}{y} equal to their reciprocals: \frac{t}{u}=\frac{u}{t} and \frac{x}{y}=\frac{y}{x} --> t^2=u^2 and t^2=u^2 --> |t|=|u| (or which is the same t = \pm u) and |x|=|y| (or which is the same x = \pm y).

Answer: D.

One more thing: notice that answer choices A and C are the same, since we can not have two correct answers than both are wrong (there are some types of questions for which more than one answer can be correct but this is not that type).

Hope it's clear.

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Re: M06 #16 [#permalink] New post 25 Mar 2013, 23:06
This is confusing. I solved this question and immediately got A as the write answer. Thanks for pointing that A and C are the same. But how do we find that A or C is wrong even after getting this answer?
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Re: M06 #16 [#permalink] New post 15 Apr 2013, 16:14
I'm stumped on this one too. Not clear at all.
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Re: M06 #16 [#permalink] New post 16 Apr 2013, 01:46
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Re: M06 #16 [#permalink] New post 16 Apr 2013, 19:32
Bunuel wrote:
youngkacha wrote:
I'm stumped on this one too. Not clear at all.


Can you please tell me which part(s) of the solution didn't you understand?


I'm lost when you take the absolute value of t and u, yet u is +/- and t isn't.

Why is t = +/- u instead of being +/- t = +/- u?
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Re: M06 #16 [#permalink] New post 18 Apr 2013, 15:11
Expert's post
mourinhogmat1 wrote:
If \frac{t}{u} = \frac{x}{y} and \frac{t}{y} = \frac{u}{x} and t , u , x , and y are non-zero integers, which of the following is true?

A. \frac{t}{u}=1
B. \frac{y}{x}=-1
C. t = u
D. t = \pm u
E. None of the above

A question to Bunuel or someone who actually wrote the test:

From the given equation we get ux=ty (1) and xt=uy (2)
subtracting 2 from 1 this we get x(u-t)= y (t-u) --> this gives us x=-y. This is option B.
So, why is option B wrong?


Multiply both the given equalities, we get t/u*t/y = x/y*u/x--> t^2 =u^2-->
D.

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Re: M06 #16 [#permalink] New post 19 Apr 2013, 02:42
Expert's post
youngkacha wrote:
Bunuel wrote:
youngkacha wrote:
I'm stumped on this one too. Not clear at all.


Can you please tell me which part(s) of the solution didn't you understand?


I'm lost when you take the absolute value of t and u, yet u is +/- and t isn't.

Why is t = +/- u instead of being +/- t = +/- u?


|t|=|u| means that t=u (which is the same as -t=-u) or t=-u (which is the same as -t=u), so only two cases.

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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M06 #16 [#permalink] New post 02 Jul 2014, 00:24
mourinhogmat1 wrote:
If \frac{t}{u} = \frac{x}{y} and \frac{t}{y} = \frac{u}{x} and t , u , x , and y are non-zero integers, which of the following is true?

A. \frac{t}{u}=1
B. \frac{y}{x}=-1
C. t = u
D. t = \pm u
E. None of the above

A question to Bunuel or someone who actually wrote the test:

From the given equation we get ux=ty (1) and xt=uy (2)
subtracting 2 from 1 this we get x(u-t)= y (t-u) --> this gives us x=-y. This is option B.
So, why is option B wrong?


This is how I arrived at D:

1 step: we know that t/u = x/y. This could be converted into: t*y = u*x.

2 step: we know that: t/y = u/x.
At the same time: t*y = u*x
From these equations, based on common logic,
we understand that: |t| = |u| and |y| = |x|. For example,
2*3 = 2*3 and 2/3 = 2/3, or -2*3 = 2*-3 and -2/3 = 2/-3
Important to remember about possible negative values

3 step: now lets consider possible answers:
(A) t/u and (C) t =u could be eliminated at once, as they are the same -> impossible in gmat questions of this type
(B) y/x = -1 is possible, but could also be = 1
(D) correct, could be either + or -

As I'm a beginner in gmat, would be greatful if someone can challenge my approach!
Re: M06 #16   [#permalink] 02 Jul 2014, 00:24
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