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josemarioamaya
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Is ad > bc ? Updated on: 20 Sep 2013, 00:39
Originally posted by josemarioamaya on 19 Sep 2013, 19:32.
Last edited by Bunuel on 20 Sep 2013, 00:39, edited 1 time in total.
RENAMED THE TOPIC.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

Question Stats:

56% (01:35) correct 44% (01:22) wrong based on 232 sessions

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Is ad > bc ?

(1) a/b > c/d
(2) b/d >0
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C
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Bunuel
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Is ad > bc ? 20 Sep 2013, 00:43
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Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.
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CAMANISHPARMAR
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Is ad > bc ? 28 Jul 2018, 10:39
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So this is a very interesting DS question!!

Statement 2 is easier to evaluate. All we can infer from this statement is either both b & d are positive or negative. Clearly insuff

Statement 1 is difficult to evaluate. Considering many scenarios could be time consuming. All we need to come up with two scenerios which lead to both a "yes" and a "no". So if b & d are having same sign then the answer is "yes" but if they have apposite sign then the answer is "no" to the question Is ad > bc ?

If we combine Statement 1 & 2 we get a definite yes hence the correct answer is C.
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Is ad > bc ? 28 Jul 2018, 14:29
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CAMANISHPARMAR
So this is a very interesting DS question!!
Statement 1 is difficult to evaluate. Considering many scenarios could be time consuming. All we need to come up with two scenerios which lead to both a "yes" and a "no". So if b & d are having same sign then the answer is "yes" but if they have apposite sign then the answer is "no" to the question Is ad > bc ?


Hi CAMANISHPARMAR

Sometimes plugging-in values is time consuming but we can easily fix numbers and play with others, for example in statement 1

\(\frac{a}{b} > \frac{c}{d}\)

Let a=0, b=1, this will make LHS =0.......So choose any small numbers c= -1 & d=2.......0 > -1.......Answer is Yes

Let a=0, b=1, this will make LHS =0.......So reverse numbers above c= 2 & d=-1.......0 > 2.......Answer is No

Insufficient

Sometimes we tricked to find nice numbers and spend long time to make it fit.

I hope it helps yo maybe in other questions.
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Is ad > bc ? 29 Jul 2018, 08:14
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Bunuel
Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Can you explain where am i going wrong, i am getting E.

Is ad>bc?

1) a/b > c/d
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.

2) b/d > 0
insuff.

(1) + (2)
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.
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Is ad > bc ? 29 Jul 2018, 11:56
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thefibonacci
Bunuel
Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Can you explain where am i going wrong, i am getting E.

Is ad>bc?

1) a/b > c/d
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.

2) b/d > 0
insuff.

(1) + (2)
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.

Hi,

I'm happy to help.

When b =2 & d=4 , then the highlighted past above does not satisfy the condition in statement 1 because -1/2 < -1/4.
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Is ad > bc ? 29 Jul 2018, 21:04
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thefibonacci
Bunuel
Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Can you explain where am i going wrong, i am getting E.

Is ad>bc?

1) a/b > c/d
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.

2) b/d > 0
insuff.

(1) + (2)
if (a,b,c,d) = (2,1,1,1) --> ans is yes
if (a,b,c,d) = (-1,2,-1,4) --> ans is no
insuff.

If (a,b,c,d) = (-1,2,-1,4), then (a/b = -1/2) < (c/d = -1/4), so this set violates (1).
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Is ad > bc ? 29 Jul 2018, 22:39
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Bunuel
Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.


Hi Buenel,

Why cant A be sufficient for this question:

please find my explanation below and rectify me if I am going wrong any where:

a/b>c/d;
a/b-c/d>0;
(ad-bc)/bd>0;
ad/bd-bc/bd>0;
ad/bd>bc/bd

since,denominator in both the cases are same ,we can say that the numerators decides everything and hence A alone is sufficient.

Regards,
Parth


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Is ad > bc ? 29 Jul 2018, 23:16
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parthkar
Bunuel
Is ad > bc ?

(1) a/b > c/d. If b and d have the same sign, then when cross-multiplying we'll get ad > bc but if they have the opposite signs, then when cross-multiplying we'll get ad < bc. Not sufficient.

(2) b/d >0. b and d have the same sign but no info about a and c. Not sufficient.

(1)+(2) Since from (2) we have that b and d have the same sign, then from (1) we have the firs case: ad > bc. Sufficient.

Answer: C.

Hope it's clear.


Hi Buenel,

Why cant A be sufficient for this question:

please find my explanation below and rectify me if I am going wrong any where:

a/b>c/d;
a/b-c/d>0;
(ad-bc)/bd>0;
ad/bd-bc/bd>0;
ad/bd>bc/bd

since,denominator in both the cases are same ,we can say that the numerators decides everything and hence A alone is sufficient.

Regards,
Parth


Sent from my iPhone using GMAT Club Forum mobile app

Consider the following: \(\frac{2}{(-2)} > \frac{3}{(-2)}\). The denominators are the same but is 2 > 3? No.

The same in \(\frac{ad}{bd}>\frac{bc}{bd}\). ad would be greater than bc if and only bd is positive. In this case we could reduce the inequality by it and get ad > bc. But if bd is negative, then by reducing by it, since it's negative, we should flip the sign and we;d get ad < bc.

Hope it's clear.
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Is ad > bc ? 30 Jul 2018, 13:40
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parthkar
please find my explanation below and rectify me if I am going wrong any where:

a/b>c/d;
a/b-c/d>0;
(ad-bc)/bd>0;
ad/bd-bc/bd>0;
ad/bd>bc/bd

since,denominator in both the cases are same ,we can say that the numerators decides everything and hence A alone is sufficient.

The last statement you made (in red) is where you went wrong. All of your math is correct, but you can't make that assumption.

A good guideline is to think through the math you're doing, rather than making assumptions about how the math will work out. (That's how you avoid careless mistakes.)

ad/bd > bc/bd

If you want to 'ignore' the denominators, what you're actually doing is multiplying both sides of the inequality by bd.

However, you can't do that, because you don't know (without using statement 2) whether bd is positive or negative. So, you don't know whether or not to flip the inequality. You wouldn't know whether it would come out to ad > bc, or ad < bc!

Since you don't know, the statement you used is insufficient.
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