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Clearly if the expression is true, then x must be between 3 and 11.
If I look at expression E, it says x< 11. I am not sure why this must be true as the questions asks. What if x< 3. Take x= 2.
Then expression equals (-9)*(-1) = +9. This means that we cant say that x must be less than 11.
What is going on. Someone please help with the question
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Clearly if the expression is true, then x must be between 3 and 11.
If I look at expression E, it says x< 11. I am not sure why this must be true as the questions asks. What if x< 3. Take x= 2.
Then expression equals (-9)*(-1) = +9. This means that we cant say that x must be less than 11.
What is going on. Someone please help with the question
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Clearly if the expression is true, then x must be between 3 and 11.
If I look at expression E, it says x< 11. I am not sure why this must be true as the questions asks. What if x< 3. Take x= 2.
Then expression equals (-9)*(-1) = +9. This means that we cant say that x must be less than 11.
What is going on. Someone please help with the question

Hi,

the reason is -
we are not saying that - x can be ANY value < 11..... But WHATEVER be the value of x, it will be <11.... and 3<x<11 lies in the range given by x<11

similarily had you got a choice say x>0, this also would be true, as the range 3<x<11 lies in the range of x given by x>0..
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Bunuel
If (x-11)(x-3) < 0, which of the following must be true?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

(x-11)(x-3) < 0; 3<x<11 --> Given condition
If we review options, E is always true.

Let us review others:

x^2 + 5x + 6 = (x+3)(x+2)

(x+3)(x+2)<0, then -3<x<-2 --> This will not be true with the given conditions (3<x<11). A is out


x^2 + 5x + 6 = (x+3)(x+2)

(x+3)(x+2)>0, then x> -2 and x < -3 --> This will not be true with the given conditions (3<x<11). B is out


(C)

5 - x < 0; 5<x; x> 5. But if x = 12 , then given condition is not satisfied. So C is also out.

D. x - 5 < 0

x<5

But if x = 1, (which is less than 3) then given condition is not satisfied.

E is the answer.
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x lies between 3 and 11
3<x<11
so x-11<0 === 11-x>0 (E)
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Bunuel
If (x-11)(x-3) < 0, which of the following must be true?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


(x-11)(x-3)<0 can be written as (11-x)(x-3)>0

So,to inequality hold true multiple of (11-x) and (x-3) must be positive and if I let both elements positive/negative then the range becomes 3<x<11

Answer Choice E matched which says 11-x>0 or x<11,which must be true (I started to look from E to A for making easy decision first)

Correct Answer E
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Bunuel
If (x-11)(x-3) < 0, which of the following must be true?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

<----(-1)----0---1---2---3---4---5---6---7---8---9---10---11---12--->
Actually, the question stem says the green part only.
E says:
11-x>0
---> 11>x
---> for the value of x in the green part 11 always must be greater than that green value.
So, the correct choice is E to me.
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