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Given : a,b,c and d are all non-zero ints.
asked: ab/cd > 0 ?

(1) ad + bc = 0
---------------------
ad = - bc --> one of abcd has a different sign that the others
ex: all +ve and one is -ve or all -ve and one is +ve
So, is ab/cd > 0 ? NO

statement 1 is sufficient

(2) abcd = -4
------------------
abcd = -ve # --> one abcd also has a different sign than the others
So, is ab/cd > 0 ? NO

For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]
03 Aug 2013, 08:26

For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0 (2) abcd = -4

I dont agree with the OA.

IMO B.

Statement 1 : ad = -bc 1) If all integers a,b,c & d are -ve/+ve, then statement 1 holds true and ab/bc is positive. 2) If any one integer is -ve & other 3 integers are positive, statement 1 holds true and ab/bc is negative.

Let me know where my thinking is wrong?

Last edited by Zarrolou on 03 Aug 2013, 08:30, edited 1 time in total.

Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]
03 Aug 2013, 08:34

mohitvarshney wrote:

For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0 (2) abcd = -4

I dont agree with the OA.

IMO B.

Statement 1 : ad = -bc 1) If all integers a,b,c & d are -ve/+ve, then statement 1 holds true and ab/bc is positive. 2) If any one integer is -ve & other 3 integers are positive, statement 1 holds true and ab/bc is negative.

Let me know where my thinking is wrong?

If \(a,b,c,d\) are all positive or negative (1) does not hold true,as \(positive+positive>0\) and \(negative + negative < 0\) ( and not equal 0). Your second point is correct.

Hope it's clear. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]
17 Aug 2013, 00:26

1

This post was BOOKMARKED

From statement 1, we know that \(ad+bc=0\) \(=>\) \(ad=-bc\) \(=>\) \(-\frac{a}{b}=\frac{c}{d}\) \(=>\) \(-\frac{a*b}{b*b}=\frac{c*d}{d*d}\) \(=>\) \(-\frac{ab}{b^2}=\frac{cd}{d^2}\) \(=>\) \(\frac{ab}{cd}=-\frac{b^2}{d^2}\) Thus from 1, we come to know that ab/cd is negative since \(b^2\) and \(d^2\) are positive

From statement 2, \(abcd=-4\) Now the only way this can happen is if one of the term is negative. And if one of the term is negative, then ab/cd has to be negative.

Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]
10 Oct 2013, 13:57

mm007 wrote:

For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4

So this question is basically testing negatives and positives (Remember >0).

First Statement

ad = -bc. Now we could rearrange this to be a/c = -b/d. Now replacing in the original statement we would have (-b/d)(b/d) . Since this is basically the same fraction but with different signs then the result HAS to be negative. Therefore this statement is Sufficient

Second Statement

abcd = -4. Now here, we see that the result is -ve. So actually, we can either have 1 negative or 3 negatives. But either choice will give us ab/cd <0. Because the only thing we need is to have an odd number of negative signed numbers. I suggest to try it with different combinations and see it for yourself

Re: For non-zero integers a, b, c and d, is ab/cd positive? [#permalink]
25 Nov 2014, 03:53

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