GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Sep 2019, 02:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58176
If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

24 Apr 2018, 23:10
14
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:29) correct 38% (02:44) wrong based on 124 sessions

### HideShow timer Statistics

If $$0 < a < b < c < d$$ and $$y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd)$$, in which of the following intervals is y < 0?

I. $$a < x < b$$
II. $$b < x < c$$
III. $$c < x < d$$

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only

_________________
Senior Manager
Joined: 21 Jan 2015
Posts: 458
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28
GMAT 2: 690 Q49 V35
WE: Sales (Consumer Products)
If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

25 Apr 2018, 00:07
1
1
Ans: D
given that a,b,c,d are >0 and y = (x-a)(x-b)(x-c)(x-d) we need to find out is Y<0 or y is negative
now by given conditions:
I. y<0
II. y >0
III. y<0

I and III satisfy y<0
Ans : D
_________________
--------------------------------------------------------------------
The Mind is Everything, What we Think we Become.
Intern
Joined: 11 Jan 2018
Posts: 19
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

25 Apr 2018, 01:51
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c
Senior Manager
Joined: 21 Jan 2015
Posts: 458
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28
GMAT 2: 690 Q49 V35
WE: Sales (Consumer Products)
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

25 Apr 2018, 02:14
2
1
minatminat wrote:
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c

I think the problem is in the steps you are folowing..
for example
Statement 1 says x lies between a and b; a<x<b and we already know that a<b<c<d this mean b,c and d are greater than x. now by putting them in resolved given value of y = (x-a)(x-b)(x-c)(x-d) we get only one +ve and three -ve signs so y is negative.
Statement 2 says: b<x<c means c and d are both greater than x. so now we get two -ve and two +ve sings so y is +ve.
Statement 3 says: c<x<d means only d is bigger than x so one -ve and three +ve signs. which gives us y -ve.

_________________
--------------------------------------------------------------------
The Mind is Everything, What we Think we Become.
Intern
Joined: 18 Jul 2016
Posts: 35
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

25 Apr 2018, 15:58
2
IMO D

I used wavy line method by egmat to reach to conclusion.

Thanks.
Intern
Joined: 11 Jan 2018
Posts: 19
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

25 Apr 2018, 21:06
dkumar2012 wrote:
minatminat wrote:
B?

Im probably wrong but I will still give it a go...

y=(x^2−ax−bx+ab)(x^2−cx−dx+cd) y<0

(x^2−ax−bx+ab) = (x-a)(x-b)
(x^2−cx−dx+cd) = (x-c)(x-d)

For y <0
(x-a)(x-b) is +ve , then (x-c)(x-d) is - ve
So, x>a or x>b and x<c or x<d
So, the condition that matches y<0 is b<x<c

I think the problem is in the steps you are folowing..
for example
Statement 1 says x lies between a and b; a<x<b and we already know that a<b<c<d this mean b,c and d are greater than x. now by putting them in resolved given value of y = (x-a)(x-b)(x-c)(x-d) we get only one +ve and three -ve signs so y is negative.
Statement 2 says: b<x<c means c and d are both greater than x. so now we get two -ve and two +ve sings so y is +ve.
Statement 3 says: c<x<d means only d is bigger than x so one -ve and three +ve signs. which gives us y -ve.

Oh right! I see it now!!! thanks !!!!
Manager
Joined: 07 May 2018
Posts: 62
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),  [#permalink]

### Show Tags

20 Jun 2019, 11:08
1
D3N0 wrote:
Ans: D
given that a,b,c,d are >0 and y = (x-a)(x-b)(x-c)(x-d) we need to find out is Y<0 or y is negative
now by given conditions:
I. y<0
II. y >0
III. y<0

I and III satisfy y<0
Ans : D

Hi Could you explain how you got

1. y <0 for the first statement
2. y > 0 for the second statement
3. y < 0 for the third statement
Re: If 0 < a < b < c < d and y = (x^2 - ax - bx + ab)(x^2 - cx - dx + cd),   [#permalink] 20 Jun 2019, 11:08
Display posts from previous: Sort by