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M31-10

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M31-10  [#permalink]

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New post 07 Jun 2015, 09:27
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

90% (01:00) correct 10% (00:49) wrong based on 30 sessions

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Re M31-10  [#permalink]

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New post 07 Jun 2015, 09:27
1
Official Solution:

If \(0 < x < 1 < y\), which of the following must be true?

A. \(1 < \frac{1}{x} < \frac{1}{y}\)
B. \(\frac{1}{x} < 1 < \frac{1}{y}\)
C. \(\frac{1}{x} < \frac{1}{y} < 1\)
D. \(\frac{1}{y} < 1 < \frac{1}{x}\)
E. \(\frac{1}{y} < \frac{1}{x} < 1\)


Since \(0 < x < 1\), then \(0 < 1 < \frac{1}{x}\) (dividing by \(x\)) and since \(0 < 1 < y\), then \(0 < \frac{1}{y} < 1\) (dividing by \(y\)).

So, we have that \(0 < \frac{1}{y} < 1 < \frac{1}{x}\).


Answer: D
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Re: M31-10  [#permalink]

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New post 07 Sep 2015, 13:13
i picked numbers and this approach is way too time consuming. algebra here is elegant
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Re: M31-10  [#permalink]

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New post 17 Jul 2016, 10:10
I think this is a high-quality question and I agree with explanation. I solved by taking x = 1/2 and y = 2 but the method shown in official solution is superb and can be used in other questions as well to save time.
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Re: M31-10  [#permalink]

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New post 19 Mar 2018, 00:01
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Bunuel wrote:
If \(0 < x < 1 < y\), which of the following must be true?

A. \(1 < \frac{1}{x} < \frac{1}{y}\)
B. \(\frac{1}{x} < 1 < \frac{1}{y}\)
C. \(\frac{1}{x} < \frac{1}{y} < 1\)
D. \(\frac{1}{y} < 1 < \frac{1}{x}\)
E. \(\frac{1}{y} < \frac{1}{x} < 1\)

hey can you please explain what is meant by 1 and can we take value as 0<x<1<y= 0<0.1<1<0.3 and 0<1<1<2
then no situation will give us answer please explain?
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Re: M31-10  [#permalink]

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New post 19 Mar 2018, 00:34
rishabhmishra wrote:
Bunuel wrote:
If \(0 < x < 1 < y\), which of the following must be true?

A. \(1 < \frac{1}{x} < \frac{1}{y}\)
B. \(\frac{1}{x} < 1 < \frac{1}{y}\)
C. \(\frac{1}{x} < \frac{1}{y} < 1\)
D. \(\frac{1}{y} < 1 < \frac{1}{x}\)
E. \(\frac{1}{y} < \frac{1}{x} < 1\)

hey can you please explain what is meant by 1 and can we take value as 0<x<1<y= 0<0.1<1<0.3 and 0<1<1<2
then no situation will give us answer please explain?


It was a formatting error. Edited. Thank you.
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Re: M31-10  [#permalink]

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New post 19 Mar 2018, 05:31
Bunuel wrote:
rishabhmishra wrote:
Bunuel wrote:
If \(0 < x < 1 < y\), which of the following must be true?

A. \(1 < \frac{1}{x} < \frac{1}{y}\)
B. \(\frac{1}{x} < 1 < \frac{1}{y}\)
C. \(\frac{1}{x} < \frac{1}{y} < 1\)
D. \(\frac{1}{y} < 1 < \frac{1}{x}\)
E. \(\frac{1}{y} < \frac{1}{x} < 1\)

hey can you please explain what is meant by 1 and can we take value as 0<x<1<y= 0<0.1<1<0.3 and 0<1<1<2
then no situation will give us answer please explain?


It was a formatting error. Edited. Thank you.

In this question i am not highlighting your error but i was solving this question in practice paper of gmat club and i made it incorrect and i seriously not able to understand from the solution so i asked you and by the way thanks for your kudos.
cheers!!!
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Re: M31-10  [#permalink]

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New post 21 Jul 2019, 11:54
shasadou wrote:
i picked numbers and this approach is way too time consuming. algebra here is elegant


My algebra is not great so by combining number picking + visualizing with a number line I solved in under a minute.

Say x = 1/2 and y=3/2

0 --- x (1/2) --- 1 ---y (3/2)

Now take the reciprocal

--- x (2) --- 1 --- y (2/3) --- 0

So 1/x > 1 > 1/y
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Re: M31-10   [#permalink] 21 Jul 2019, 11:54
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