Inequalities from Manhattan Advanced Quant

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If abcd ≠ 0, is abcd < 0?

1. a/b > c/d
2. b/a > d/c

Manhattan Prep (2015-05-17). GMAT Advanced Quant (Kindle Locations 2464-2466). Manhattan Prep Publishing. Kindle Edition.

OA C
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by GMATGuruNY » Thu Aug 27, 2015 3:32 am
bubbliiiiiiii wrote:If abcd ≠ 0, is abcd < 0?

1. a/b > c/d
2. b/a > d/c
Statement 1:
Case 1: a/b > c/d --> 2/1 > 1/1
In this case, abcd = 2*1*1*1 = 2, so the answer to the question stem is NO.
Case 2: a/b > c/d --> 2/1 > -1/1
In this case, abcd = (2)(1)(-1)(1) = -2, so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2:
Case 3: b/a > d/c --> 2/1 > 1/1
In this case, abcd = 1*2*1*1 = 2, so the answer to the question stem is NO.
Case 4: b/a > d/c --> 2/1 > -1/1
In this case, abcd = (1)(2)(1)(-1) = -2, so the answer to the question stem is YES.
INSUFFICIENT.

RULE:
If x>y, then 1/x > 1/y only if x is positive and y is negative.

Statements combined:
According to the rule above:
If a/b > c/d, then b/a > d/c only if a/b is positive and c/d is negative.
Thus, the two statements combined require that a/b > 0 and that c/d < 0.
Since a/b > 0, ab > 0.
Since c/d < 0, cd < 0.
Thus, abcd < 0.
SUFFICIENT.

The correct answer is C.
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by Brent@GMATPrepNow » Tue Sep 17, 2019 8:11 am
bubbliiiiiiii wrote:If abcd ≠ 0, is abcd < 0?

1. a/b > c/d
2. b/a > d/c

Manhattan Prep (2015-05-17). GMAT Advanced Quant (Kindle Locations 2464-2466). Manhattan Prep Publishing. Kindle Edition.

OA C
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

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