Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
23 Feb 2012, 20:27
Kudos
Bookmarks
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:
76% (01:23) correct
24%
(01:39)
wrong
based on 344
sessions
History
Date
Time
Result
Not Attempted Yet
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
Kudos
Bookmarks
dvinoth86
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
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
Kudos
Bookmarks
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
