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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]
31 Aug 2012, 05:01
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ziko wrote:
If ab > cd and a, b, c and d are all greater than zero, which of the following CANNOT be true?
A. c > b B. d > a C. b/c < d/a D. a/c > d/b E. (cd)2 < (ab)2
Could someone explain how to cross multiply such enequalities, what is the general rule?
When you multiply/divide both parts of an inequality by a positive value you should keep the sign. When you multiply/divide both parts of an inequality by a negative value you should flip the sign.
Now, if we multiply both parts of the given inequality by 1/(ac) (which according to the stem must be positive), then we'll get: b/c>d/a, so option C which says that b/c < d/a cannot be true.
Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]
13 Feb 2015, 13:45
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]
13 Feb 2015, 18:48
Expert's post
Hi All,
This question can be solved by TESTing VALUES. Since there are so many variables, and the answers look a "little crazy", the key is to keep your numbers small and simple.
We're told that (A)(B) > (C)(D) and that A, B, C and D are all POSITIVE. We're asked which of the following CANNOT be true. If we can prove that an answer COULD be true, even once, then we can eliminate it.
Let's TEST VALUES: Starting with Answer A, is there a way for C to be > B?
IF.... A = 3 B = 2 C = 3 D = 1 Here, C is greater than B. Eliminate Answer A.
We can ALSO eliminate a couple of other answers with this example: Answer D (3/3 > 1/2) and Answer E (3^2 < 6^2) can be eliminated TOO.
With 2 answers remaining, I'm going to deal with the easier-looking option:
With Answer B, is there a way for D to be > A?
IF... A = 3 B = 2 C = 1 D = 4 Here, D is greater than A. Eliminate Answer B.
Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]
14 Nov 2015, 18:28
if you take a=3 b=4 c=2 d=1 even option c will be true 3*4>2*1 and b/c>d/a 4/2>1/3 so even option C is true Can you correct me if i am wrong. Some thing wrong in this question
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