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

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If abcd is not equal to zero, is abcd <0 ? [#permalink]

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29 Mar 2016, 09:52
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If abcd is not equal to zero, is abcd <0 ?

(1) a/b > c/d

(2) b/a > d/c
[Reveal] Spoiler: OA

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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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29 Mar 2016, 10:18
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If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

getting c/d to LHS
a/b- c/d > 0

statement is not sufficient

2) b/a > d/c

getting d/c to LHS
b/a- d/c >0
(cb- da) /ca >0

statement is not sufficient

Combining both, multiplying together

it means

1/abcd <0
implies abcd <0

Hence C is the right option

PS : I do mind Kudos
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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30 Mar 2016, 08:42
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.
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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30 Mar 2016, 09:09
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Expert's post
gianghoang217 wrote:
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...

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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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30 Mar 2016, 09:20
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samichange wrote:
If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c

I donot mind Kudos

Hi,

The Q tells us that $$abcd \neq{0}$$...
so none of a,b,c or d is 0..

lets see the statements

1)$$\frac{a}{b} > \frac{c}{d}$$
=> $$\frac{a}{b} - \frac{c}{d}>0$$
$$\frac{(ad-bc)}{bd}>0$$
so two cases..
this means if ad>bc, then bd>0
this means if ad<bc, then bd<0

Insuff

2) $$\frac{b}{a} > \frac{d}{c}$$
=> $$\frac{b}{a} - \frac{d}{c}>0$$
$$\frac{(bc-ad)}{ac}>0$$
so two cases..
this means if bc>ad, then ac>0
this means if bc<ad, then ac<0

Insuff

combined-
two cases
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

Suff
C
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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30 Mar 2016, 09:28
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chetan2u wrote:
samichange wrote:
If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c

I donot mind Kudos

Hi,

The Q tells us that $$abcd \neq{0}$$...
so none of a,b,c or d is 0..

lets see the statements

1)$$\frac{a}{b} > \frac{c}{d}$$
=> $$\frac{a}{b} - \frac{c}{d}>0$$
$$\frac{(ad-bc)}{bd}>0$$
so two cases..
this means if ad>bc, then bd>0
this means if ad<bc, then bd<0

Insuff

2) $$\frac{b}{a} > \frac{d}{c}$$
=> $$\frac{b}{a} - \frac{d}{c}>0$$
$$\frac{(bc-ad)}{ac}>0$$
so two cases..
this means if bc>ad, then ac>0
this means if bc<ad, then ac<0

Insuff

combined-
two cases
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

Suff
C

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
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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30 Mar 2016, 23:47
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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.
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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26 Jun 2016, 19:06
chetan2u wrote:
samichange wrote:
If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c

I donot mind Kudos

Hi,

The Q tells us that $$abcd \neq{0}$$...
so none of a,b,c or d is 0..

lets see the statements

1)$$\frac{a}{b} > \frac{c}{d}$$
=> $$\frac{a}{b} - \frac{c}{d}>0$$
$$\frac{(ad-bc)}{bd}>0$$
so two cases..
this means if ad>bc, then bd>0
this means if ad<bc, then bd<0

Insuff

2) $$\frac{b}{a} > \frac{d}{c}$$
=> $$\frac{b}{a} - \frac{d}{c}>0$$
$$\frac{(bc-ad)}{ac}>0$$
so two cases..
this means if bc>ad, then ac>0
this means if bc<ad, then ac<0

Insuff

combined-
two cases
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

Suff
C

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?
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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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25 Oct 2016, 13:13
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samichange wrote:
If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c

I donot mind Kudos

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

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Re: If abcd is not equal to zero, is abcd <0 ? [#permalink]

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29 Oct 2016, 01:55
samichange wrote:
If abcd is not equal to zero ,is abcd <0 ?

1) a/b > c/d

2) b/a > d/c

I donot mind Kudos

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
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If abcd is not equal to zero, is abcd <0 ? [#permalink]

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01 Nov 2017, 16:03
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samichange wrote:
If abcd is not equal to zero, 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 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

Cheers,
Brent
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