If abcd is not equal to zero, is abcd <0 ?

I got E somehow ...

If we see all the posibilities of a, b, c and d: a could be > 0 and a could be < 0. All b, c, d must be > 0. (Because the number like 1-111, one thousand minus one hundred eleven and one, doesnt exist)
(1) If a/b > c/d

c/d must be > 0. Then a/b must be > 0. Then a and b must be both > 0.

abcd >0 (suff)

(2) b/a > d/c
d/c must be > 0. Then b/a must be > 0. Then a and b must be both > 0.
abcd >0 (suff)

Pls correct me if Im wrong.

Hi,
abcd is not a 4-digit number ..
abcd= a*b*c*d...

Now rework your solution

Great solution.

However I think it should be
1) If bc>ad, then ac>0 and bd<0 .. so ac*bd<0 or abcd<0
2) If bc<ad, then ac<0 and bd>0.. so ac*bd<0 or abcd<0

Yeah but the answer still is C

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c


In the original condition, there are 4 variables(a,b,c,d), which should match with the number of equations. So you need 4 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. When 1) & 2), a/b>c/d, a/b-c/d>0, (ad-bc)/bd > 0 is derived and becomes b/a>d/c, b/a-d/c >0, (cb- da) /ca >0. Since they both are positive integers, multiply the both equations. Then, (ad-bc)(cb-da)/abcd>0, (ad-bc)^2/abcd<0 is derived. As (ad-bc)^2>0 is always valid, abcd<0, which is yes. Thus, the answer is C.


 For cases where we need 3 more equations, such as original conditions with“3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

Hi chetan2u,

I understand why (1) and (2) by themselves are insufficient, but I do not understand why combining them is sufficient. Can you please elaborate?

Another approach.
In inequalities when we take the reciprocals of the two sides we flip the inequality sign
5 > 3 => 1/5 < 1/3 or -5 <- 3 => -1/5 > -1/3

However, when 1 side is positive and the other negative we do not flip the inequality – the positive side is always greater than the negative side.

5 > -3 => 1/5 > - 1/3

In this case, statement 2 is the reciprocal of statement 1, and the sign is not flipped; this means that a*b is positive and c*d is negative, so a*b*c*d < 0

Answer C

EACH clearly insuff

both

(ad-bc)/bd >0 and -(ad-bc)/ac > 0 , now let (ad-bc) = x , bd = y , ac = z

thus we have 2 inequalities x/y>0 and -x/z > 0 this means that x and y has same sign but y and z has opposite signs i.e. yz<0 , since y = bd , z = ac

therefore abdc<0

C

APPROACH #1

Target question: Is abcd <0

Statement 1: a/b > c/d
There are several values of a, b, c and d that satisfy statement 1. Here are two:
Case a: a = 1, b = 1, c = 1, d = 2. Plugging these values into the statement 1 inequality, we get 1/1 > 1/2, which works. In this case, abcd = (1)(1)(1)(2) = 2. So, abcd > 0
Case b: a = 1, b = 1, c = -1, d = 2. Plugging these values into the statement 1 inequality, we get 1/1 > -1/2, which works. In this case, abcd = (1)(1)(-1)(2) = -2. So, abcd < 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b/a > d/c
There are several values of a, b, c and d that satisfy statement 2. Here are two:
Case a: a = 1, b = 1, c = 2, d = 1. Plugging these values into the statement 2 inequality, we get 1/1 > 1/2, which works. In this case, abcd = (1)(1)(2)(1) = 2. So, abcd > 0
Case b: a = 1, b = 1, c = 2, d = -1. Plugging these values into the statement 2 inequality, we get 1/1 > -1/2, which works. In this case, abcd = (1)(1)(2)(-1) = -2. So, abcd < 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that a/b > c/d
Statement 2 tells us that b/a > d/c
So, the fraction a/b is greater than the fraction c/d
When we invert the two fractions, b/a is also greater than d/c
This should strike us as odd.
In MOST cases, when we invert two fractions in an inequality, the inequality symbol should reverse.
For example, 2/3 < 7/6 and 3/2 > 6/7
Likewise, 1/30 > 1/50 and 30/1 < 50/1
Notice that in my above examples, both fractions are POSITIVE

The same holds true when both fractions are NEGATIVE
For example, -2/3 < -1/6 and -3/2 > -6/1
Likewise, -5/7 > -10/3 and -7/5 < -3/10

The COMBINED statements tell a different story. Here, when we invert the fractions, the inequality symbol does NOT reverse.
This means that it is not the case that both fractions are POSITIVE, and it is not the case that both fractions are NEGATIVE
So, one fraction must be positive and one fraction must be negative.
In both inequalities, the fractions with the c and d are LESS THAN the fractions with a and b
So, it must be the case that the fractions c/d and d/c are both NEGATIVE
And the fractions a/b and b/a are both POSITIVE

If fractions c/d and d/c are both NEGATIVE, then the product cd is also NEGATIVE
If fractions a/b and b/a are both POSITIVE, then the product ab is POSITIVE
So, abcd = (POSITIVE)(NEGATIVE) = SOME NEGATIVE VALUE
In other words, abcd < 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C
-------------------------------------------------------
APPROACH #2

Target question: Is abcd <0
We can test values (as I did above) to show that statement 1 alone and statement 2 alone are insufficient.

Statements 1 and 2 combined
Statement 1 says: \(\frac{a}{b} > \frac{c}{d}\)
Rewrite as: \(\frac{a}{b} - \frac{c}{d} > 0\)
Find common denominators and combine to get: \(\frac{ad - bc}{bd} > 0\)

Statement 2 says: \(\frac{b}{a} > \frac{d}{c}\)
Rewrite as: \(\frac{d}{c} - \frac{b}{a} < 0\)
Find common denominators and combine to get: \(\frac{ad - bc}{ac} < 0\)

This tells us that \(\frac{ad - bc}{bd}\) is POSITIVE, and \(\frac{ad - bc}{ac} \) is NEGATIVE

So, \((\frac{ad - bc}{bd})(\frac{ad - bc}{ac}) = \) (POSITIVE)(NEGATIVE) \(=\) NEGATIVE

In other words: \(\frac{(ad - bc)^2}{abcd}\) is NEGATIVE

Since \((ad - bc)^2\) is POSITIVE, it must be the case that abcd is NEGATIVE
In other words, abcd < 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

chetan2u

Hi,

I don't understand the following parts:

If ad-bc>0, then bd>0 how is bd>0 here?

If bc-ad>0, then ac>0 how is ac>0 here?


1) If bc>ad, then ac>0 and bd<0 .. so ac*bd<0 or abcd<0
2) If bc<ad, then ac<0 and bd>0.. so ac*bd>0 or abcd<0 how is ac*bd>0 here?


Thanks!

Hi..
(ad-bc)/bd>0..
Thus means either both numerator and denominator are POSITIVE or both are NEGATIVE..
They both have to be same sign.. so ad-bc and BD will have SAME sign

hi,

can we take this approach?:
AD-BC/BD>0
BC-DA/AC>O

MEANS THAT IF AD-BC IS NEGATIVE BC-AD IS POSITIVE AND V.V
===AD-BC===BD===BC-DA===AC
===+======+====(-)=====(-) SO ACBD IS (-)*+ WHICH IS NEGATIVE
===(-)=====(-)====+=====+ SO BDAC IS +*(-) WHICH IS NEGATIVE.

Solution:

Statement One Only:

a/b > c/d

If a, b, c, and d are all positive (e.g., a = 2, b = 1, c = 3, d = 4), then abcd > 0. However, if a, b, and c are positive and d is negative (e.g., a = 2, b = 1, c = 3, d = -4), then abcd < 0. Statement one alone is not sufficient.

Statement Two Only:

b/a > d/c

Similar to the logic used in statement one, we see that if a, b, c, and d are all positive (e.g., a = 1, b = 2, c = 3, d = 4), then abcd > 0. However, if a, b, and c are positive and d is negative (e.g., a = 1, b = 2, c = 3, d = -4), then abcd < 0. Statement two alone is not sufficient.

Statements One and Two Together:

If both inequalities are true, it’s impossible for the fractions on both sides of the inequalities to be positive. That is because the fractions in statement two are reciprocals of their counterparts in statement one. For example, if 3 > 2 (or 3/1 > 2/1), it’s impossible to have 1/3 > 1/2. Therefore, we know the fractions are not all positive. Now, can they be all negative? The answer is also no. For example, if -2 > -3 (or -2/1 > -3/1), it’s impossible to have -1/2 > -1/3. However, since both inequalities have to be true, it’s only possible that the fractions on one side of the inequalities are positive, while the fractions on the other side are negative. In particular, the fractions on the left hand side of the inequalities (a/b and b/a) have to be positive, and the fractions on the right hand side (c/d and d/c) have to be negative. In other words, three of the four values, a, b, c, and d have to be positive and the last one (c or d) has to be negative. This makes the product abcd negative, which is less than 0. Both statements together are sufficient.

Answer: C

You have omitted the possibility that both a and b are negative. In short, a and b must have the same sign, while c and d must have opposite signs.

If abcd is not equal to zero, is abcd < 0?
1) a/b > c/d
2) b/a > d/c

1) a/b > c/d
Here we cannot re-frame statement as ad > bc as we don't know the signs of b and d.
So what we can do here is ,
a/b - c/d >0
(ad-bc)/bd >0
Statement 1 alone is not sufficient to answer.

2) b/a > d/c
Here also we cannot write the statement as bd > ac as we don't know whether a and c are positive or not.
b/a -d/c >0
(bc - ad)/ac >0
Statement 2 alone is not sufficient to answer.

On combining both statements, (ad-bc)/bd and (bc-ad)/ac are positive since each term are greater than 0.

Multiplying both,
(ad-bc)/bd * (bc-ad)/ac >0
-(ad-bc)^2/abcd >0

(ad-bc)^2 will be always positive , So in order to satisfy the equation, abcd should be negative i.e abcd <0 . Its enough to answer the question stem.

Option C is the answer.

Thanks,
Clifin J Francis,
GMAT SME.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.