If ab/cd ad (1) cd ab (2) b

Question

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

Bunuel · Accepted Answer

If \frac{ab}{cd}<0 , is bc>ad ? \frac{ab}{cd}<0 means that ab and cd have the opposite signs. (1) cd > ab --> as they have the opposite signs then cd>0>ab --> c and d have the same sign, and a and b have the opposite signs. Now, if c , d and b are positive and a is negative then: bc>0>ad BUT if c , d and a are positive and b is negative then: bc<0 bc>0 --> from \frac{b}{c}*\frac{a}{d}<0 as bc>0 then ad<0 --> bc>0>ad . Sufficient. Answer: B.

crejoc · Answer

IMO B
given ab/cd < 0 , so two cases
1) ab < 0, cd >0 or
2) ab >0 , cd <0

stmt 1 cd > ab
so case 1 satisfy, qstn is bc > ad? not sufficnt

stmt 2
b<c<0
applying in above two cases

1) ab<0 , cd> 0
b< 0 , so for ab<0, a must be >0
c<0 , so for cd>0, d must be <0

so a>0, b<0, c<0, d<0 , apply to check bc > ad?
defintely bc>ad

2) ab >0 , cd < 0

b<0 , so for ab>0 , a must be <0
c<0 , so for cd<0 , d must be >0

so a<0, b<0, c<0, d>0 , apply to check bc>ad?
defintely bc>ad.
so stmt2 suffint

hrish88 · Answer

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

over2u · Answer

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.

raghavs · Answer

Totally stumped with this question

kp1811 · Answer

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 ad. So suff

jeeteshsingh · Answer

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: b0 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.....

metallicafan · Answer

+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!

jlgdr · Answer

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

email2vm · Answer

Sol: either AB or CD is -ve & remaining one will be positive: consider option 2 which is b -ve i.e. BC> AD Hence B.

mridupawandas · Answer

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

HKD1710 · Answer

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 ad. Sufficient. Answer is B.

Alex75PAris · Answer

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) b0 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

MathRevolution · Answer

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.

adiagr · Answer

(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: plus and minus.JPG 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 abcd1.JPG Case 2 Refer below abcd2.JPG Statement 2 is sufficient B is the answer.

GMATinsight · Answer

answer: Option B

Check solution as attached

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

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