If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0

given

ab/cd<0


case 1 a,b are positive and either c or d is negative

case 2 c,d are positive and either a or b is negative

case 3 a,b are negative and either c or d negative

case 4 c,d are negative and either a or b negative



st1 - cd>ab

we have to consider case 2 and case 4

i.e c,d both are postive or negative and either a or b postive

case 2:
consider cd postive and a negative,b positive

so will get bc>ad

consider cd postive and b negative,a positive

so will get bc<ad

Not sufficient



st2-b<c<0


we have to consider case 3 and 4

case 3
consider a,b negative and c negative,d positive

bc>ad

case 4

consider c,d negative and b negative,a positive

bc>ad

sufficient.

My answer is B

Good question.
1)CD>AB;
In this case CD should pe positive and AB negative in order to satisfy the condition that ab/cd<0; So Eiher two numbers from CD should be positive or negative and one number from AB should be positive and another should be negative or vice versa. If we put different signs of numbers in equation bc>ad we will have different results, so insuff.
2) B<C<0
ab<0 and cd>0
so a - positive, and d - negative, let's put them into equation
bc>ad; on the left side we have positive number, on the right side we have negative number, Suff.
ab>0 and cd<0
so a - negative, and d - positive, and again
bc>ad; on the left side we have positive number, on the right side we have negative number, Suff.
So I choose B.

Totally stumped with this question

we have either ab is <0 or cd < 0

stmnt1) cd> ab so ab is <0 but we cannot deduce that bc > ad. hence insuff
stmnt2) b<c<0 so bc is +ve and for ab/cd to be <0 either a or d should be -ve. hence bc > ad. So suff

B....

Given \(\frac{ab}{cd}<0\)
This means either ab is -ve or cd is -ve
Case 1: ab is -ve & cd is +ve
this means
(a<0 and b>0) or (a>0 and b<0) AND
(c<0 and d<0) or (c>0 and d>0)

Case 2: ab is +ve & cd is -ve
this means
(a<0 and b<0) or (a>0 and b>0) AND
(c<0 and d>0) or (c>0 and d<0)

Ques: Is bc > ad?

S1: cd>ab. This means cd is +ve and ab is -ve. Therefore we have Case 1:
(a<0 and b>0) or (a>0 and b<0) AND
(c<0 and d<0) or (c>0 and d>0)

With this when bc is negative, ad is +ve. Therefore Is bc > ad? is False
But when bc is +ve, ad is -ve. Therefore Is bc > ad? is True. Hence NOT SUFF

S2: b<c<0. gives c as -ve and b as -ve.
Therefore Case 1 can only be:
(a>0 and b<0) AND (c<0 and d<0). Which gives Is bc > ad? as True
Also Case 2 can only be:
(a<0 and b<0) AND (c<0 and d>0). Which gives Is bc > ad? as True
Hence SUFF.....

+1 B, good question

If \(\frac{ab}{cd} <0\), ab > 0 and cd < 0, OR ab<0 and cd > 0.

Statement (1): cd > ab
So, cd > 0 and ab < 0
This could mean that c,d > 0 OR c,d<0
AND
a>0, b<0, OR a<0, b>0
If c,d >0 , a>0, and b<0, the result is bc<0 and ad>0, then answer is NO
If c,d <0 , a>0, and b<0, the result is bc>0 and ad<0, then answer is YES
INSUFF

Statement (2): b < c < 0, this means that bc >0
Then,
if a<0, d>0, so bc >0 AND ad<0, the answer is YES.
If a>0, d<0, so bc>0 AND ad<0, the answer is also YES.
SUFFICIENT

ANSWER : B

If someone has a faster method, that would be a great help!

So we know that ac and cd have different signs. Now question is is bc>ad? Let's begin.

Staetment 1 cd>ab. Cd is positive and ab is negative. But still we don't know if bc>ad.

Statement 2 tells us that b and c are both negative hence bc is positivei. Now, what do we know about ad? Well, if ab/cd is negative and we already now that bc is positive then we have one negative more between a and d, anyways, ad will be negative and so bc>ad.

Therefore B stands

Hope this helps

Sol:

either AB or CD is -ve & remaining one will be positive:

consider option 2 which is b<c<0 which means B and C are negative. negative*negative= positive. so combining the earlier given and this statement proves that ad should be negative.
i.e. either A or D is negative so -ve * +ve = -ve

+ve > -ve i.e. BC> AD
Hence B.

Ans is B...I will explain as follows Statement 2 tells us that b<c<0 implies both b and c are negative and it is given that ab<0 or cd is negative now between a and d one has to be positive then only expression ab/cd would be <0 so bc positive and ad is negative implies bc>ad. hence proved. Hope this explanation helps

stmt-1: cd > ab, here we do not know the signs and hence this will not be sufficient to find out whether bc > ad. Insuff.

stmt-2: b<c<0 , it says both b and c are negative. Then bc will be positive.
It is given that ab/cd < 0, either a or d has to be –ve for this to be true as both b and c are –ve. This way ad will be negative.
So we know bc +ve and ad –ve. Definitely bc > ad. Sufficient.
Answer is B.

Assumption : \(\frac{ab}{cd}\)<0

So, 2 cases : ab < 0 and cd > 0 OR ab > 0 and cd < 0

Question : bc > ad ?

(1) : cd > ab Not sufficient
Many cases possible.

(2) b<c<0, b and c are both negative
so, bc >0

Case 1 from assumption : ab < 0 and cd > 0
so, a>0 and d<0 => ad<0

Case 2 from assumption : ab > 0 and cd < 0
so, a<0 and d>0 => ad<0

So, in every case, bc > 0 and ad < 0, so bc>ad

Sufficient

We can modify the original condition and the question. If we multiply (cd)^2 to each side of ab/cd<0, since squared number is always positive, the sign of inequality does not change.
Then we get abcd<0. From bc>ad?, if bc is positive, we get ad<0 from abcd<0. Hence, since the condition 2) gives bc>0, we get ad<0. The answer is yes and the condition is sufficient. Thus, the correct answer is B.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

(ab/cd) < 0

This means that

1) ab is (+), cd is (-)
2) ab is (-), cd is (+)

Statement 1

cd > ab

Then this is only possible when ab is (-) and cd is (+)

Refer below:



Thus Statement 1 is insufficient

Statement 2

This means that we have two cases

1) ab is (+), cd is (-)
2) ab is (-), cd is (+)

Both b and c are negative. Infact b < c < 0

Case 1
Refer below


Case 2
Refer below



Statement 2 is sufficient

B is the answer.

answer: Option B

Check solution as attached

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