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# If ab > cd and a, b, c and d are all greater than zero

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If ab > cd and a, b, c and d are all greater than zero [#permalink]

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31 Aug 2012, 05:50
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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?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Aug 2012, 06:01, edited 1 time in total.
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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31 Aug 2012, 06:01
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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.

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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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24 Sep 2012, 22:03
Hi All,

Is there a way to verify the validity of options a and b. If answer c was also a valid inequality, how could we have verified the first two options?

any inputs on this anyone? Thanks in advance.
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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31 Oct 2012, 15:12
ab > cd
ab/abcd > cd/abcd
1/cd > 1/ab
cd < ab ==> d/a < b/c (b/c CANNOT BE < d/a)
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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13 Feb 2015, 14:45
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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13 Feb 2015, 19:48
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.

The final answer is the only one left....

[Reveal] Spoiler:
C

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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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14 Nov 2015, 19:28
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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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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14 Nov 2015, 21:27
Hi ashokingmat,

If you compare your work to Answer C in the original prompt, you'll notice that you've used the WRONG inequality.

You wrote it as B/C > D/A. Your example is a good one though (it does prove that C is not true), you just made a little mistake in your work.

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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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20 Jun 2017, 18:08
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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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31 Jul 2017, 05:04
Its C..

I was stuck between B & C only. B can be sometimes true, but C can never be true. Hence, C.

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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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31 Jul 2017, 06:34
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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?

a,b,c,d > 0
ab>cd

A. INCORRECT. Can/Cannot be TRUE: as c can be greater than b or less than b. We can't say which is greater
B. INCORRECT . Can/Cannot be TRUE: Again similar to option A, here also d can be greater than or less than a. We can't say which is greater.
C. CORRECT. Cannot be TRUE: ab>cd -> b/c > d/a. So, b/c < d/a is totally incorrect.
D. INCORRECT. a/c > d/b This is totally possible.
E. INCORRECT. (ab)^2 > (cd)^2. This is also true.

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Re: If ab > cd and a, b, c and d are all greater than zero [#permalink]

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01 Aug 2017, 17:05
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

We can divide the given inequality by ac since we know that all of our variables are greater than zero:

ab/(ac) > cd/(ac)

b/c > d/a

Since b/c will always be greater than d/a, b/c < d/a can NEVER be true.

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Re: If ab > cd and a, b, c and d are all greater than zero   [#permalink] 01 Aug 2017, 17:05
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