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12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh

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Bunuel
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12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:00
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12 Days of Christmas GMAT Competition with Lots of Fun

If \(|2x – 2| < x + 1\), which of the following represents the correct range of values of x ?

A. \(\frac{1}{3} < x\)
B. \(x < 3\)
C. \(3 < x < 11\)
D. \(\frac{1}{3} < x < 3\)
E. \(3 < x < 9\)


 


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Bunuel
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   15 Dec 2022, 06:45
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Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

If \(|2x – 2| < x + 1\), which of the following represents the correct range of values of x ?

A. \(\frac{1}{3} < x\)
B. \(x < 3\)
C. \(3 < x < 11\)
D. \(\frac{1}{3} < x < 3\)
E. \(3 < x < 9\)


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

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    \(|2x – 2| < x + 1\);

    \(2|x – 1| < x + 1\).

When x < 1, then x - 1 (the expression in the modulus) is negative and thus |x - 1| = -(x - 1), so for this case we'd have:

    \(2(-(x – 1)) < x + 1\);

    \(2(-(x – 1)) < x + 1\);

    \(2 - 2x < x + 1\);

    \(1 < 3x\);

    \(x > \frac{1}{3}\).

When x >= 1, then x - 1 (the expression in the modulus) is nonnegative and thus |x - 1| = x - 1, so for this case we'd have:

    \(2(x – 1) < x + 1\);

    \(2x - 2 < x + 1\);

    \(x < 3\).

So, when x < 1, we have \(x > \frac{1}{3}\) and when x >= 1, we have \(x < 3\). Therefore, the complete range is \(\frac{1}{3} < x < 3\).

Answer: D.
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:10
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Q) [2x-2] < x+1

open the mods

2x-2 < x+1 and -2x+2< x+1

taking same units to 1 side (keeping x positive)

X<3 and x>\frac{1}{3}

SO,

x is greater than \frac{1}{3} and less than 3.
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:11
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|2x–2|<x+1
Case 1 (2x-2)<x+1 {when 2x-2>0)
2x-x<3
x<3

Case 2 : -(2x-2)<x+1
3x>1
x>1/3

OA:D
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:16
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Option c since we will remove the mod and get 2 conditions for (2x-2)> 0 and <0
Solving them we will get range of x

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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:18
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\(|2x – 2| < x + 1\),

possible range of x
2x-2<x+1
x<3
and
-2x+2<x+1
1<3x
1/3<x
1/3<x<3
OPTION D


Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

If \(|2x – 2| < x + 1\), which of the following represents the correct range of values of x ?

A. \(\frac{1}{3} < x\)
B. \(x < 3\)
C. \(3 < x < 11\)
D. \(\frac{1}{3} < x < 3\)
E. \(3 < x < 9\)


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:19
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Asked: If \(|2x – 2| < x + 1\), which of the following represents the correct range of values of x ?

|2x-2| < x + 1

Case 1: x>=1
|2x-2| = 2x-2 < x + 1
x < 3
1<=x<3

Case 2: x<1
|2x-2| = 2-2x < x + 1
3x > 1
x > 1/3
1/3 < x < 1

Combining 1<=x<3 & 1/3 < x < 1
1/3<x<3

IMO D
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:21
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So when you see |x|<a, it implies =>-a < x < a.
Now using the above result, we try to remove the modulus from the question

=> |2x-2| < x+1,
=> -(x+1) < 2x-2 < x+1
=> -x-1 < 2x-2 and 2x-2 < x+1 (just split the above inequality into two)
=> x>1/3 and x<3 (Solve individually)
=> 1/3 < x < 3 (answer choice D)
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:23
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|2x 2| < x + 1

=> -(2x+2)<x+1
2x-2<x+1 => x<3

=> (2x+2)<x+1
-3x<1 => x>1/3

so 1/3<x<3
ans D
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12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   Updated on: 15 Dec 2022, 07:03
Method 1

Let X=0, we get
|-2|<1 which is not possible, Hence eliminate choice A and B.

Now, Let X=4, we get
|6|<5, not possible, hence eliminate choice C and E.
Correct Option Choice D.

Method 2

When |2X-2|>0, we get X>1 and
2X-2<X+1 = X<3
on Combining above 2 we get 1<X<3

when |2X-2|<0 , we get X<1 and
-2X+2>X+1 =1>3X =X<1/3
On combining we get 1/3<X<3 choice D.

Originally posted by Catman on 14 Dec 2022, 07:29.
Last edited by Catman on 15 Dec 2022, 07:03, edited 1 time in total.
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:33
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|2x–2| < x+1

1. 2x-2 < x+1
2x-x<1+2
x<3

2. -2x+2 < x+1
-2x-x<1-2
-3x<-1
Now, multiply both sides by -1 and we get

3x > 1
x>1/3

hence, 1/3<x<3

Answer, D
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:38
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Opening the mod.

2x-2<x+1 AND 2x-2>-x-1
x<3 AND 3x>3
X<3 and X>1/3

Hence, 1/3<X<3
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:38
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|2x-2| < x+1
2x-2 < x+1 and -(2x-2) < x+1
Simplifying both sides we get
x<3 and X> 1/3
Thus the answer is D. 1/3<x<3
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:39
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|2x-2|<x+1, to find the range of x we need to solve 2 cases:

1) if 2x-2>0, then the expression is 2x-2<x+1
x>1/3
2) if 2x-2<0, then the expression is -2x+2>x+1
x<3

From these 2 cases we can see that 1/3<x<3

Answer: D
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 07:49
Quote:
If |2x–2|<x+1, which of the following represents the correct range of values of x ?


|2x-2| < x +1
|2x|-2 < x+1
2|x| < x+3
Case A: When x>0
2x < x+3
=> x<3
Case B: When x<0
-2x < x+3
=> -3x<3
=> -x < 1
=> x > -1

IMO B
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 08:02
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Concept:
< sign in modulus means RANGE of values

solving the modulus on both sides:
-x-1 <2x-2<x+1

it gives x>1/3 and x<3 so ans id D.
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 08:03
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|2x–2|<x+1

2x-2<x+1
x<3

-2x+2<x+1
x>1/3

D is the answer
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 08:08
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By opening the modulus we recieve 2 cases and if we track those on no. line we get
x<3 and x>1/3
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 08:25
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To solve the question, let us substitute values and test the ranges.

for (A),(C) and (E):

Let x = 4.
|2*4 - 2| < 4 +1
|6| < 5
6 <5 which is incorrect, so options a,c,e are incorrect.

Let x = 0.
|2*0-2| < 0+1
|-2| < 1
2 < 1 which is incorrect, so option B is incorrect.

therefore only option D is left and is our solution
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink] New post   14 Dec 2022, 08:28
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Bunuel wrote:
12 Days of Christmas GMAT Competition with Lots of Fun

If \(|2x – 2| < x + 1\), which of the following represents the correct range of values of x ?

A. \(\frac{1}{3} < x\)
B. \(x < 3\)
C. \(3 < x < 11\)
D. \(\frac{1}{3} < x < 3\)
E. \(3 < x < 9\)


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

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This is an inequality question. We can either solve it directly or by POE.

We will apply POE as the options are not closed.

Case 1: x=0;
Inequality won't hold good. So B can be eliminated.


Case 2: x=8;
Inequality won't hold good. So A, C, and E can be eliminated.

Hence IMO D
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Re: 12 Days of Christmas GMAT Competition - Day 1: If |2x 2| < x + 1, wh [#permalink]
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