Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?
A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0
PS: I do not have the OA for this question.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Inequalities / Quadratic Equation
This topic has expert replies
I would guess that the answer is E, i.e. 11-x>0
Here is my reasoning:
If (x-11)(x-3)<0 then at least one of them must be negative. Intuitively then, x-11 is the negative part and x-3 is positive because in the former case you subtract a larger number.
x-11<0 | *(-1)
11-x>0
Any other takers?
Here is my reasoning:
If (x-11)(x-3)<0 then at least one of them must be negative. Intuitively then, x-11 is the negative part and x-3 is positive because in the former case you subtract a larger number.
x-11<0 | *(-1)
11-x>0
Any other takers?
-
sunil_snath
- Senior | Next Rank: 100 Posts
- Posts: 39
- Joined: Wed Nov 18, 2009 8:41 am
- Thanked: 3 times
Even B looks right tho 
From the question X is greater than 3 and less than 11.
So B, (X+3)(X+2)>0 is true. But E is obvious
From the question X is greater than 3 and less than 11.
So B, (X+3)(X+2)>0 is true. But E is obvious
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Here's my solution. It includes a systematic way to solve quadratic inequalities.Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?
A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0
PS: I do not have the OA for this question.
The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
Brent Hanneson - Creator of GMATPrepNow.com
-
rohan_vus
- Master | Next Rank: 500 Posts
- Posts: 189
- Joined: Thu Apr 03, 2008 2:03 pm
- Location: USA
- Thanked: 21 times
With respect to the logic behind bold statement , one solution is x = -4 which satisfies option B but its not a solution to original inequality . This question seems kinda odd to meBrent Hanneson wrote:Here's my solution. It includes a systematic way to solve quadratic inequalities.Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?
A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0
PS: I do not have the OA for this question.
The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
In our answer choices, we are looking for an inequality such that every possible value for x from the original inequality [ (x-11)(x-3) < 0 ] must also be a solution to the inequality among the answer choices. You are ready the question the other way around.rohan_vus wrote:With respect to the logic behind bold statement , one solution is x = -4 which satisfies option B but its not a solution to original inequality . This question seems kinda odd to meBrent Hanneson wrote:Here's my solution. It includes a systematic way to solve quadratic inequalities.Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?
A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0
PS: I do not have the OA for this question.
The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
x = -4 is not a solution to the original inequality (x-11)(x-3) < 0, so we need not consider that value for x when checking the answer choices. We need only consider values of x between 3 and 11.
When we plug these allowable values into the inquality in answer choice B, EVERY value of x is such that x^2 + 5x + 6 > 0
The same cannot be said for the other answer choices.
Brent Hanneson - Creator of GMATPrepNow.com
-
rohan_vus
- Master | Next Rank: 500 Posts
- Posts: 189
- Joined: Thu Apr 03, 2008 2:03 pm
- Location: USA
- Thanked: 21 times
We need only consider values of x between 3 and 11
This is what i am driving at!.
All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .
This is what i am driving at!.
All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .
GMAT/MBA Expert
-
Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Sorry, my bad. I was reading something completely different from what you were saying.rohan_vus wrote:We need only consider values of x between 3 and 11
This is what i am driving at!.
All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .
All possible values for x (3<x<11) make E valid as well.
Looks like the answer is B and E
Brent Hanneson - Creator of GMATPrepNow.com
