Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

Re: 7)if ab/cd <0, is bc>ad? st1 cd>ab st2 b<c<0 [#permalink]
30 Jan 2012, 13:39

Quote:

7)if ab/cd <0, is bc>ad?

st1 cd>ab

st2 b<c<0 7)if ab/cd <0, is bc>ad?

st1 cd>ab

st2 b<c<0

+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! _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: 7)if ab/cd <0, is bc>ad? st1 cd>ab st2 b<c<0 [#permalink]
30 Jan 2012, 14:59

1

This post received KUDOS

Expert's post

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<ad. Not sufficient.

(2) b < c < 0 --> bc>0 --> from \frac{b}{c}*\frac{a}{d}<0 as bc>0 then ad<0 --> bc>0>ad. Sufficient.

Re: If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0 [#permalink]
28 Mar 2014, 14:12

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.

Re: If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0 [#permalink]
04 Apr 2014, 23:25

yezz wrote:

If ab/cd < 0 , is bc > ad

(1) cd > ab

(2) b < c < 0

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.

gmatclubot

Re: If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0
[#permalink]
04 Apr 2014, 23:25