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mm007
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Agree with D.

S1: One of ad or bc has to be negative for them to add to zero.

S2: One or three of abcd has to be negative for their product to be negative.
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Getting B ...

if ABCD <0 (either one or three of them is negetive)



then AB/CD < 0

B is SUFF
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My answer is D

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

statement 2 is sufficient


Thus, the answer is D
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From 1 we get the expression as -(b^2/d^2) hence negative

from 2 we get (-a^2b^2)/4

Both are negative

Answer is D
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mm007
For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4


should be D.

i. one of the integer is -ve so ab/cd is -ve. sufficient
ii. either one or three of the integers is/are -ve, so ab/cd is again -ve. sufficient.
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(1) SUFFICIENT: This statement can be rephrased as ad = <bc. For the signs of ad and bc to be opposite one
another, either precisely one or three of the four integers must be negative. The answer to our rephrased question is
"no," and, therefore, we have achieved sufficiency.
(2) SUFFICIENT: For the product abcd to be negative, either precisely one or three of the four integers must be
negative. The answer to our rephrased question is "no," and, therefore, we have achieved sufficiency.
The correct answer is D.
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1) ad + bc = 0
ad=-bc
therefore a=-(bc/d) -->replacing in the question
For ab/cd to be +ve
we get -(b^2/d^2) --> -ve

(2) abcd = –4
here we can have either one integer to be -ve or three integer to be -ve
therefore (ab/cd ) will be -ve in any case.

Hence D as both are sufficient to prove the statement to be false.
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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?
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mohitvarshney
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.
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mm007
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

Hence answer is (D)
Hope it helps
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Q: a, b, c, d are non-zero integers. Is ab/cd positive?

St1: ad + cb = 0, ad = -cb. One of the four elements is of an opposite sign, therefore either one or three elements are negative and ab/cd = -ve. Sufficient.

St2: abcd = -4, ab = -4/cd, ab/cd = -4/cd^2, therefore ab/cd is negative.
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For non-zero integers a, b, c and d, is ab/cd positive?

(1) ad + bc = 0

(2) abcd = -4

1) a=1 b=-1 c=1 d=-1 ad+bc=0 ab/cd is +ve
a=2 b=-3 c=-2 d=-3 ad+bc=0 ab/cd is -ve two solutions so wrong
not sufficient
2) a=1 b=1 c=1 d=-4 abcd=-4 ab/cd is -ve
a=-1 b=-1 c=-1 d=4 abcd=-4 ab/cd is -ve
sufficient
so answer is B not D
Please correct me if i am wrong
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If ad+bc =0 , square of both sides should also be zero which gives us
ad^2 + bc^2 + 2abcd = 0
Since ad^2 and bc^2 can’t be zero ,
Hence abcd<0 , which means three of them have same sign and one opposite
Condition A will suffice ,
Condition B directly states the above
Both are sufficient .

Posted from my mobile device
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VeritasKarishma plz solve this
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Concept:
If xy > 0, then x and y are of the same sign. Either both positive or both negative.
If x/y > 0, then x and y are of the same sign. Either both positive or both negative.
Hence, So, xy > 0 means x/y > 0

Question:
Is ab/cd >0 we can simplify this to let ab=x, cd=y, x/y>0 or xy>0. (Is abcd>0?)

Statement 2: abcd=-4 Sufficient as per above.

Statement 1: ad+bc=0. ad=-bc Both are having opposite signs and we are multiplying, Hence sufficient.

Answer D.

Bunuel: Please confirm if my approach is ok.
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