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

Getting B ...

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



then AB/CD < 0

B is SUFF

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

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

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.

(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.

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.

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.

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

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.

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

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 .

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VeritasKarishma plz solve this

To solve this, you only need to understand one concept - the sign of the answer doesn't change whether you multiply or divide factors. It is something we discuss when working with inequalities with factors. We say that

(x+a)(x+b)(x+c) < 0 has the same solution as (x+a)(x+b) / (x+c) < 0 because it doesn't matter whether the factors are getting multiplied or divided.

So considering 4 factors a, b, c and d, whether they are multiplied or divided, the sign of the result will stay the same.

(2) abcd = -4

So when all are multiplied together, we get a negative result. Then when two are multiplied and other two divide them i.e. with ab/cd, the result will be negative too.
Sufficient

(1) ad + bc = 0

This means ad = - bc
So ad and bc have opposite signs. So a/d and b/c will have opposite signs too.

Then ab/cd = (a/d) * (b/c) will be negative since one of them is positive and the other negative.
Sufficient

Answer (D)

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