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23 Feb 2012, 20:27
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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:
15%
(low)
Question Stats:
79% (01:22) correct
21%
(01:44)
wrong
based on 363
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If (a – b)c < 0, which of the following cannot be true?
A. a < b
B. c < 0
C. |c| < 1
D. ac > bc
E. a^2 – b^2 > 0
A. a < b
B. c < 0
C. |c| < 1
D. ac > bc
E. a^2 – b^2 > 0
23 Feb 2012, 22:22
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dvinoth86
Time saving approach:
Given: \((a-b)c<0\).
Now, since only option D has all three unknowns in it, I'd analyze this answer choice first:
D. \(ac > bc\) --> \(ac-bc>0\) --> \((a-b)c>0\). As you can see this option says directly opposite of what is given in the stem, hence it must be false.
There is no need even to check other options, as you cannot have two correct answers.
Answer: D.
P.S. Having all three unknowns does not mean that D is automatically a correct answer. An option could have two or even one unknown and still be a correct answer. For example if one of the options were \(c=0\) or \(a-b=0\) (a=b) then it would mean that it's false, since in this case \((a-b)c=0\), which means that this option would have been a correct answer.
Hope it helps.
23 Feb 2012, 21:05
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Question: If \((a-b)*c<0\) , which of the following cannot be true?
1. \(a<b\)
2. \(c<0\)
3. \(|c|<1\)
4. \(ac>bc\)
5. \(a^2-b^2>0\)
We know that if \((a-b)*c<0\) , either \(c<0\) or \((a-b)<0\) which means \(a<b\)
1. \(a<b\) true.
2. \(c<0\) true.
3. \(|c|<1\) Could be true since \(c\) could be \(-ve\) or \(+ve\).
4. \(ac>bc\) Now if \((a-b)\) is \(+ve\) then \(a>b\) but at the same time \(c\) is \(-ve\) so \(ac\) is less than \(bc\). On the other hand if \((a-b)\) is \(-ve\) than \(a<b\) and \(ac\) cannot be greater than \(bc\) since \(c\) is \(+ve\).
5. \(a^2-b^2>0\) Now from the original statement if \(c<0\) then \((a-b)>0\) so \(a>b\) and \(a^2\) and \(b^2\) are both \(+ve\) so this could be possible.
Hence Answer D
1. \(a<b\)
2. \(c<0\)
3. \(|c|<1\)
4. \(ac>bc\)
5. \(a^2-b^2>0\)
We know that if \((a-b)*c<0\) , either \(c<0\) or \((a-b)<0\) which means \(a<b\)
1. \(a<b\) true.
2. \(c<0\) true.
3. \(|c|<1\) Could be true since \(c\) could be \(-ve\) or \(+ve\).
4. \(ac>bc\) Now if \((a-b)\) is \(+ve\) then \(a>b\) but at the same time \(c\) is \(-ve\) so \(ac\) is less than \(bc\). On the other hand if \((a-b)\) is \(-ve\) than \(a<b\) and \(ac\) cannot be greater than \(bc\) since \(c\) is \(+ve\).
5. \(a^2-b^2>0\) Now from the original statement if \(c<0\) then \((a-b)>0\) so \(a>b\) and \(a^2\) and \(b^2\) are both \(+ve\) so this could be possible.
Hence Answer D
