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M61-07

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Math Revolution GMAT Instructor
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M61-07  [#permalink]

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New post 18 Jun 2018, 00:25
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

40% (01:09) correct 60% (03:47) wrong based on 5 sessions

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If \(0 < 2x + 3y < 10\) and \(-10 < 3x+2y < 0\), then which of the following must be true?

I. \(x < 0\)

II. \(y < 0\)

III. \(x < y\)


A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

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Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
Posts: 6625
GMAT 1: 760 Q51 V42
GPA: 3.82
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M61-07  [#permalink]

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New post Updated on: 05 Aug 2018, 22:05
Official Solution:


If \(0 < 2x + 3y < 10\) and \(-10 < 3x+2y < 0\), then which of the following must be true?

I. \(x < 0\)

II. \(y < 0\)

III. \(x < y\)


A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


Label the inequalities as follows:

\(0 < 2x+3y < 10\) --- (1)

\(-10 < 3x+2y < 0\) --- (2)

We consider each statement individually.

Statement I:

Multiplying (1) by -2 yields \(-20 < -4x – 6y < 0\), and multiplying (2) by 3 yields \(-30 < 9x + 6y < 0\). Adding these inequalities gives \(-50 < 5x < 0\) or \(-10 < x < 0\).

This statement is true.

Statement II:

Multiplying (1) by 3 yields \(0 < 6x + 9y < 30\), and multiplying (2) by -2 yields \(0 < -6x – 4y < 20\). Adding these inequalities gives \(0 < 5y < 50\) or \(0 < y < 10\).

This statement is false.

Statement III:

Multiplying (1) by – 1 yields \(-10 < -2x – 3y < 0\). Adding this to inequality (2) yields \(-20 < x – y < 0\).
This implies that \(x < y\), and statement III is true.

Answer: D
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Originally posted by MathRevolution on 18 Jun 2018, 00:26.
Last edited by MathRevolution on 05 Aug 2018, 22:05, edited 1 time in total.
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Re: M61-07  [#permalink]

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New post 01 Aug 2018, 03:16
Must do before exam. It teaches a lot of concepts. This is an easy question. If someone has command over inequalities can surely do it correctly.
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Re: M61-07 &nbs [#permalink] 01 Aug 2018, 03:16
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