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23 Jan 2019, 02:39
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Question Stats:
64% (01:55) correct
36%
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wrong
based on 603
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If \(xy < 0\) and \(x + y < 0\), then which of the following must be true?
I. \(x - y > 0\)
II. \(x > 0\) or \(y > 0\)
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
I. \(x - y > 0\)
II. \(x > 0\) or \(y > 0\)
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
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25 Apr 2023, 05:44
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Official Solution:
If \(xy < 0\) and \(x + y < 0\), then which of the following must be true?
I. \(x - y > 0\)
II. \(x > 0\) or \(y > 0\)
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
\(xy < 0\) implies that \(x\) and \(y\) have opposite signs. Hence, in conjunction with \(x + y < 0\), either \(x > 0 > y\) and \(y\) is further from 0 than \(x\) (to make \(x + y\) negative, for example \(x = 1\) and \(y = -2\)), or \(y > 0 > x\) and \(x\) is further from 0 than \(y\) (to make \(x + y\) negative, for example \(x = -2\) and \(y = 1\)).
Let's examine the options:
I. \(x - y > 0\)
The above implies that \(x > y\). This option is not always true because we know that either \(x > 0 > y\) or \(y > 0 > x\). Discard.
II. \(x > 0\) or \(y > 0\)
This option must be true. Since from \(xy < 0\), it follows that one of them must be positive.
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
This one is indeed tricky. If \(x > 0 > y\) and \(y\) is further from 0 than \(x\), then \(\frac{x}{y} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(y\), and then flipping the inequality sign to get \(\frac{x}{y} + 1 > 0\), which leads to \(\frac{x}{y} > -1\). On the other hand, if \(y > 0 > x\) and \(x\) is further from 0 than \(y\), then \(\frac{y}{x} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(x\), and then flipping the inequality sign to get \(1 + \frac{y}{x} > 0\), which leads to \(\frac{y}{x} > -1\). Therefore, in either case, either \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\) is always true.
Hence, only options II and III are always true.
Answer: D
If \(xy < 0\) and \(x + y < 0\), then which of the following must be true?
I. \(x - y > 0\)
II. \(x > 0\) or \(y > 0\)
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
\(xy < 0\) implies that \(x\) and \(y\) have opposite signs. Hence, in conjunction with \(x + y < 0\), either \(x > 0 > y\) and \(y\) is further from 0 than \(x\) (to make \(x + y\) negative, for example \(x = 1\) and \(y = -2\)), or \(y > 0 > x\) and \(x\) is further from 0 than \(y\) (to make \(x + y\) negative, for example \(x = -2\) and \(y = 1\)).
Let's examine the options:
I. \(x - y > 0\)
The above implies that \(x > y\). This option is not always true because we know that either \(x > 0 > y\) or \(y > 0 > x\). Discard.
II. \(x > 0\) or \(y > 0\)
This option must be true. Since from \(xy < 0\), it follows that one of them must be positive.
III. \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\)
This one is indeed tricky. If \(x > 0 > y\) and \(y\) is further from 0 than \(x\), then \(\frac{x}{y} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(y\), and then flipping the inequality sign to get \(\frac{x}{y} + 1 > 0\), which leads to \(\frac{x}{y} > -1\). On the other hand, if \(y > 0 > x\) and \(x\) is further from 0 than \(y\), then \(\frac{y}{x} > -1\) must be true. We can derive this by dividing the inequality \(x + y < 0\) by the negative value \(x\), and then flipping the inequality sign to get \(1 + \frac{y}{x} > 0\), which leads to \(\frac{y}{x} > -1\). Therefore, in either case, either \(\frac{x}{y} > -1\) or \(\frac{y}{x} > -1\) is always true.
Hence, only options II and III are always true.
Answer: D
