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# M31-10

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Math Expert
Joined: 02 Sep 2009
Posts: 58402

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07 Jun 2015, 09:27
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Difficulty:

15% (low)

Question Stats:

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

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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. $$\frac{1}{y} < \frac{1}{x} < 1$$

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Joined: 02 Sep 2009
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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 $$0 < \frac{1}{y} < 1 < \frac{1}{x}$$.

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Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
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07 Sep 2015, 13:13
i picked numbers and this approach is way too time consuming. algebra here is elegant
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KUDO me plenty
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Joined: 23 Apr 2014
Posts: 57
Location: United States
GMAT 1: 680 Q50 V31
GPA: 2.75

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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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Joined: 23 Sep 2016
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19 Mar 2018, 00:01
1
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
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Joined: 02 Sep 2009
Posts: 58402

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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

It was a formatting error. Edited. Thank you.
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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

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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21 Jul 2019, 11:54
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
Re: M31-10   [#permalink] 21 Jul 2019, 11:54
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# M31-10

Moderators: chetan2u, Bunuel