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OG 12.problem 130

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OG 12.problem 130 [#permalink] New post 05 Dec 2009, 14:20
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A
B
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D
E

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–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4


The answer is E, why not B?
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Re: OG 12.problem 130 [#permalink] New post 05 Dec 2009, 14:34
ISBtarget wrote:
–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4


The answer is E, why not B?


b is saying -5≤x≤5
e is saying -5≤x≤3
x+1≤4 = x≤3
-x -1 ≤ 4
-x ≤ 5
x>=-5
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Re: OG 12.problem 130 [#permalink] New post 06 Dec 2009, 06:29
take both conditions
if x>0
ie. if x+1>0
or x>-1

then |x+1|<=4
x+1<= 4
x<= 3

If x<0
i.e if x+1<0
or x<-1

then -|x+1|<=4
-x-1<= 4
-x<= 5
x>= -5

combining the limits we get
-5<x<3 therefore E
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Re: OG 12.problem 130 [#permalink] New post 22 Feb 2010, 00:16
ISB target ....try and put x=4 in eqn 2. It still holds good but doesnt satisfies your number options for the line. Eqn 5 does that for x=4 and is justified only for the numbers given in the quesn.

Hope its clear now.
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Re: OG 12.problem 130 [#permalink] New post 22 Feb 2010, 06:49
ISBtarget wrote:
–5 –4 –3 –2 –1 0 1 2 3 (number line shaded portion and X can take any value from -5 until 3)
130. Which of the following inequalities is an algebraic
expression for the shaded part of the number line
above?
(A) |x| ≤ 3
(B) |x| ≤ 5
(C) |x − 2| ≤ 3
(D) |x − 1| ≤ 4
(E) |x + 1| ≤ 4


The answer is E, why not B?


Length = 3-(-5) = 8
center = (-5)+8/2 = -1
Now only D & E has 8/2 = 4 on right side.
Also at x=-5 only E is satisfied and not D so answer is E.
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Re: OG 12.problem 130 [#permalink] New post 22 Feb 2010, 07:39
|x+a| <= b

you can take the above case and simplify as below
if the origin is at (-a,0) then x will have as many values to the right of -a as to the left of -a. Bacially the values of x are symetric at the value (-a,0)

for the expression given in the question, the values of x are -5<=x<=3 .. the values are symetric at x=-1

so the expression would be a = 1 and b =4 ... |x+1|<=4
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Re: OG 12.problem 130 [#permalink] New post 26 Feb 2010, 13:40
|x + 1| ≤ 4

remove the absolute value by

-4 ≤ x+1 ≤ 4
add -1 to all side

-1 -4 ≤ x+1 -1 ≤ 4 -1

-5 ≤ x ≤ 3
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Re: OG 12.problem 130 [#permalink] New post 01 Mar 2010, 00:28
:-D :-D :-D
E is correct.
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Re: OG 12.problem 130 [#permalink] New post 21 Nov 2011, 09:06
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All the answers above are back solving from answer E.

But if we were to solve directly from question to get the answer,
We need to find the center point in the shaded part of the number line.

Here it is easy to identify that -1 is the center point and the region is 4 units from it on either side.

So x is less than 4 units on either side of -1
this gives |x+1|<=4.
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Re: OG 12.problem 130 [#permalink] New post 23 Nov 2011, 05:07
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Check out this post. It discusses the visual approach of handling mods. Once you understand it, you can answer this question in 10 secs.
http://www.veritasprep.com/blog/2011/01 ... edore-did/
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Re: OG 12.problem 130 [#permalink] New post 23 Nov 2011, 08:13
E. It is not B because B implies that it could be 4 or 5 as well when we know that it 3 is the highest positive number
Re: OG 12.problem 130   [#permalink] 23 Nov 2011, 08:13
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OG 12.problem 130

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