GMAT Advanced Quant

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by GMATGuruNY » Sun May 29, 2016 2:54 am
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 Matt@VeritasPrep » Tue Jun 07, 2016 11:51 pm
S1::

a/b > c/d

a/b - c/d > 0

(ad - bc)/bd > 0

if ac > 0, we have

ac(ad - bc)/(abcd) > 0

in which case (abcd) shares a sign with (ad - bc). If ac < 0, the opposite holds, so NOT SUFFICIENT to determine the sign of abcd. Similar logic for S2.

Together, however

(ad - bc)/bd > 0

and

(bc - ad)/ac > 0

Since the left hand side of each equation is positive, we can multiply them together to find

((ad - bc)(bc - ad)) / abcd > 0

(abcd - (bc)² - (ad)² - abcd)/abcd > 0

(-(bc)² - (ad)²) / abcd > 0

-((bc)² + (ad)²) / abcd > 0

(bc)² + (ad)² must be positive, so the numerator MUST be negative. abcd shares the sign of the numerator, so it must also be negative; SUFFICIENT.

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by 800_or_bust » Wed Jun 08, 2016 12:41 pm
gmat_for_life wrote:If abcd ≠ 0, is abcd < 0?

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

[spoiler]OA is C. Could anyone please show some examples to prove that C is sufficient?[/spoiler]

Regards,
Amit
Hmm, deceptively tricky. It's fairly easy to prove by counterexample that each of the two statements alone are not sufficient. However, together they are sufficient.

The key is to recognize that the second inequality is comparing the reciprocals of the first. The only way both of these inequalities can be true is if a/b is positive and c/d is negative (and hence b/a would also be positive, and c/d negative). This means a and b must either be both positive or both negative, and that c and d must have opposite signs from one another (one positive and one negative). This means we have an odd number of negative terms (either 1 or 3) which guarantees that the resulting product will be negative.
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by Matt@VeritasPrep » Fri Jun 10, 2016 10:06 am
800_or_bust wrote: The only way both of these inequalities can be true is if a/b is positive and c/d is negative (and hence b/a would also be positive, and c/d negative).
This is true, but it might be worth unpacking, since it isn't self-evident, I don't think.

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by BalysLTU » Sat Oct 21, 2017 4:21 am
gmat_for_life wrote:If abcd ≠ 0, is abcd < 0?

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

[spoiler]OA is C. Could anyone please show some examples to prove that C is sufficient?[/spoiler]

Regards,
Amit
Here's a quite lengthy solution .. I feel that the previously mentioned method w/ regards to recognizing that the comparison is made between reciprocal numbers is the fastest way to solve this.

Statement 1: a/b > c/d
a/b - c/d > 0
(ad - cb)/(bd) > 0

Can't do anything more here.. Let's check Statement 2

Statement 2: b/a>d/c
b/a - d/c > 0
(bc - da)/ac > 0

Can't do anything more here.. Let's check both statements together:

Statement 1&2:

(ad - cb)/(bd) > 0 and (bc - da)/ac > 0
(ad - cb)/(bd) > 0 and (cb - ad)/ac > 0
(ad - cb)/(bd) > 0 and (-ad + cb)/ac > 0

Multiply (-ad + cb)/ac > 0 by -1 to get a common nominator

(ad - cb)/(bd) > 0 and (ad - cb)/ac < 0

This shows us that the products of b&d and a&c must be of different signs. therefore the product of both products will be -ve

Answer: C

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by Brent@GMATPrepNow » Wed Nov 01, 2017 3:03 pm
gmat_for_life wrote:If abcd ≠ 0, is abcd < 0?

1. a/b>c/d
2. b/a>d/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 reverses.
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 c/d and d/c are both NEGATIVE, then the product cd is also NEGATIVE
If 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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by Jay@ManhattanReview » Wed Nov 01, 2017 9:09 pm
gmat_for_life wrote:If abcd ≠ 0, is abcd < 0?

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

[spoiler]OA is C. Could anyone please show some examples to prove that C is sufficient?[/spoiler]

Regards,
Amit
Since abcd ≠ 0, it implies that none among a, b, c, and d is 0.

For abcd < 0, either ONLY one among a, b, c, and d is negative OR three among a, b, c, and d are negative.

(1) a/b > c/d

Case 1 : 1/2 > 1/3, then abcd > 0. The answer is No.
Case 2 : 1/2 > -1/3, then abcd < 0. The answer is Yes. Insufficient.

(2) b/a > d/c

Case 1 : 2/1 > 1/1, then abcd > 0. The answer is No.
Case 2 : 2/1 > -1/1, then abcd < 0. The answer is Yes. Insufficient.

(1) and (2) combined:

From (1), we have a/b > c/d and from (2), we have b/a > d/c, Since the LHS (b/a) and the RHS (d/c) of Statement (2) is the raciprocal of the LHS (a/b) and the RHS (c/d) of Statement (1), we can anticipate that the LHS (b/a) of Statement (2) should be less than RHS (d/c) of Statement (2); however, it is not so!!!

We see that the LHS (b/a) of Statement (2) is still greater than RHS (d/c) of Statement (2). This is possible if both c/d and d/c are negative and both a/b and b/a are positive.

Case 1: One between c and d is negative and both a and b are positive. Thus, ONLY one among a, b, c, and d is negative, making abcd < 0. The answer is Yes.

Case 2: One between c and d is negative and both a and b are negative too, making a/b and b/a positive. Thus, three among a, b, c, and d are negative, making abcd < 0. The answer is Yes.

Sufficient.

The correct answer: C

Hope this helps!

-Jay
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