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18 Jun 2018, 01:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:57) correct
28%
(02:28)
wrong
based on 46
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
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
Updated on: 05 Aug 2018, 23:05
Originally posted by MathRevolution on 18 Jun 2018, 01:26.
Last edited by MathRevolution on 05 Aug 2018, 23:05, edited 1 time in total.
Last edited by MathRevolution on 05 Aug 2018, 23:05, edited 1 time in total.
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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
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
01 Aug 2018, 04:16
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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.
