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by sanju09 » Thu Aug 26, 2010 10:43 pm
Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by phillybeat » Thu Aug 26, 2010 10:54 pm
sanju09 wrote:Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
(x-11)(x-3) to be negative x has to be 4,5,6,7,8,9,10.

only b and e are correct.
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by sanju09 » Thu Aug 26, 2010 11:03 pm
phillybeat wrote:
sanju09 wrote:Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
(x-11)(x-3) to be negative x has to be 4,5,6,7,8,9,10.

only b and e are correct.
Can we take both to the heavens, do they permit? Is there something wrong with the choices?
The mind is everything. What you think you become. -Lord Buddha



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by phillybeat » Thu Aug 26, 2010 11:06 pm
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
(x-11)(x-3) to be negative x has to be 4,5,6,7,8,9,10.

only b and e are correct.
Can we take both to the heavens, do they permit? Is there something wrong with the choices?
i am not sure .. but what is OA for this Q..
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by narik11 » Thu Aug 26, 2010 11:13 pm
for (x-11)(x-3) to be negative X should be less than 11

option (E) is correct.

@sanju09 option b is not correct, if x takes the value of -2 it evaluates to 0 and is not greater than 0
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by phillybeat » Thu Aug 26, 2010 11:32 pm
narik11 wrote:for (x-11)(x-3) to be negative X should be less than 11

option (E) is correct.

@sanju09 option b is not correct, if x takes the value of -2 it evaluates to 0 and is not greater than 0
if x = -2 then (-2-11)(-2-3) = (-13)(-5) will be positive number..hence x should always positive.and 3<x<11 .

correct me if i am wrong..
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by ymach3 » Thu Aug 26, 2010 11:34 pm
E -- x should be less than 11 for the expression to be negative....
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by phillybeat » Thu Aug 26, 2010 11:37 pm
ymach3 wrote:E -- x should be less than 11 for the expression to be negative....
ok assume that x = 1

then (1-11)(1-3) = (-10)(-2) is positive.

if x=2

(2-11)(2-3) = (-9)(-1) is positive

if x = 3 0r x=11
then (x-11)(x-3) = 0

thats why x>3 and X<11.
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by sanju09 » Thu Aug 26, 2010 11:51 pm
ymach3 wrote:E -- x should be less than 11 for the expression to be negative....
x < 11 is an endless set of values for x, we need to bracket it too, under the fixed constraints only. E is not the right choice.
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by sanju09 » Fri Aug 27, 2010 12:11 am
narik11 wrote:for (x-11)(x-3) to be negative X should be less than 11

option (E) is correct.

@sanju09 option b is not correct, if x takes the value of -2 it evaluates to 0 and is not greater than 0
welcome narik11 and thanks for sharing

We have (x - 11) (x - 3) < 0

I can see two scenarios for this

Scenario 1

(x - 11) < 0 and (x - 3) > 0

x < 11 and x > 3

collectively

3 < x < 11

Scenario 2

(x - 11) > 0 and (x - 3) < 0

x > 11 and x < 3

collectively impossible, isn't it?

Scenario 1 is ok and 3 is the lower bracket now.

In fact, we only need to check A and B now:

A. x^2 + 5 x + 6 < 0

(x + 2) (x + 3) < 0

either x < -2, x > -3 or x > -2, x < -3; which is impossible, hence only this can be concluded from A

-3 < x < -2, which doesn't seem obeying 3 < x < 11

Hence [spoiler]B[/spoiler] without a check
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by diebeatsthegmat » Tue Aug 31, 2010 9:24 am
sanju09 wrote:Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
we have to find a real number x and when we replace x into (x-11)(x-3) the result must give a negative x, correct?
or (x-11)(x-3)<0 or 3<x<11 thus the answer choice must be 4,5,6,7,8,9,10 , all are good enough to fit the requirement.
A/ x^2+5x+6<0 or -3<x<-2 ( nooo)
B. x^2+5x+6>0 or x<-3 and x>-2 .... ( no)
C 5-x<0 or 5<x or x>5 ( no because x could be 14 thus (14-11)*11 >0 not fit the inquirity) and with the condition : 3<X<11 thus 5<X<11 ( maybe)
D x<5 ( noo)
E. x<11 compare with the condition x could be -2 or -5 who knows
the answer must be C
i dont agree that the answer is E
can you give it a clear explanation why its E because if its E i cant understand
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by phillybeat » Tue Aug 31, 2010 9:29 am
diebeatsthegmat wrote:
sanju09 wrote:Which of the following is correct if x is a real number and (x - 11) (x - 3) is negative?
A. x^2 + 5 x + 6 < 0
B. x^2 + 5 x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
we have to find a real number x and when we replace x into (x-11)(x-3) the result must give a negative x, correct?
or (x-11)(x-3)<0 or 3<x<11 thus the answer choice must be 4,5,6,7,8,9,10 , all are good enough to fit the requirement.
A/ x^2+5x+6<0 or -3<x<-2 ( nooo)
B. x^2+5x+6>0 or x<-3 and x>-2 .... ( no)
C 5-x<0 or 5<x or x>5 ( no because x could be 14 thus (14-11)*11 >0 not fit the inquirity) and with the condition : 3<X<11 thus 5<X<11 ( maybe)
D x<5 ( noo)
E. x<11 compare with the condition x could be -2 or -5 who knows
the answer must be C
i dont agree that the answer is E
can you give it a clear explanation why its E because if its E i cant understand



you came up with right values for x = 4,5,6,7,8,9,10.

but when working with your answer choices you assume x = 14 . when x cannot be more than 11.
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by Arcane66 » Tue Aug 31, 2010 12:54 pm
The first thing I did was expand the foiled equation to equal x^2-14x+33<0 Then, I found the smallest answer, or what I thought is the smallest answer, to solve this equation. This happened to be four. I then look at the options and assess them each. Without going into much detail I can see off the bat that A,C,and D are all incorrect. This leaves B and E. I say E is incorrect because I think 11 as x will work which will make this answer false.

The answer, in my opinion, is B.
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by phillybeat » Tue Aug 31, 2010 1:02 pm
Arcane66 wrote:The first thing I did was expand the foiled equation to equal x^2-14x+33<0 Then, I found the smallest answer, or what I thought is the smallest answer, to solve this equation. This happened to be four. I then look at the options and assess them each. Without going into much detail I can see off the bat that A,C,and D are all incorrect. This leaves B and E. I say E is incorrect because I think 11 as x will work which will make this answer false.

The answer, in my opinion, is B.
but if you select x= 11. equation (x - 11) (x - 3) will be '0'.

as per the question (x - 11) (x - 3) should be negative.
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by Arcane66 » Tue Aug 31, 2010 1:25 pm
Hmmm..yeah..this is confusing me a bit. I guess I might have been wrong in selecting B. At second glance, maybe E is correct. Sorry about that I kind of rushed the question a bit and was a bit overconfident and just assumed E couldn't be correct.
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