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correct if x is a real

Expert replies
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



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Source: — Problem Solving |

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



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

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.
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

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
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion

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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