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# OG 12.problem 130

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Manager
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OG 12.problem 130 [#permalink]

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05 Dec 2009, 15:20
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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]

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05 Dec 2009, 15:34
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]

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06 Dec 2009, 07:29
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
Manager
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Re: OG 12.problem 130 [#permalink]

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22 Feb 2010, 01:16
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]

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22 Feb 2010, 07:49
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.
Manager
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Re: OG 12.problem 130 [#permalink]

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22 Feb 2010, 08:39
|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
Intern
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Re: OG 12.problem 130 [#permalink]

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26 Feb 2010, 14:40
|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]

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01 Mar 2010, 01:28

E is correct.
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Re: OG 12.problem 130 [#permalink]

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21 Nov 2011, 10:06
1
KUDOS
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]

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23 Nov 2011, 06:07
1
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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 08 Sep 2011 Posts: 75 Concentration: Finance, Strategy Followers: 3 Kudos [?]: 2 [0], given: 5 Re: OG 12.problem 130 [#permalink] ### Show Tags 23 Nov 2011, 09:13 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, 09:13 Similar topics Replies Last post Similar Topics: 7 If the time is currently 1:30 pm, what time will it be in exactly 647 8 21 Jul 2015, 03:19 1 The sum of 5 different positive 2-digit integers is 130. What is the 4 22 Jan 2015, 08:01 If the time is currently 1:30pm, what time will it be in 1 23 Apr 2012, 06:55 Bottle R contains 250 capsules and costs$6.25. Bottle T contains 130 5 07 Jan 2008, 03:37
OG Problem 2 30 Aug 2010, 08:28
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# OG 12.problem 130

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