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

[BTGModeratorVI]

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

(1) a/b > c/d
(2) b/a > d/c

Answer: C
Source: Manhattan Prep

[Brent@GMATPrepNow]

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

(1) a/b > c/d
(2) b/a > d/c

Answer: C
Source: Manhattan Prep

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

Cheers,
Brent

[Scott@TargetTestPrep]

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

(1) a/b > c/d
(2) b/a > d/c

Answer: C

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

Like statement one, 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 inequities 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

[Scott@TargetTestPrep]

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

(1) a/b > c/d
(2) b/a > d/c

Answer: C

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

Like statement one, 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 inequities 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