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Director  Joined: 07 Aug 2011
Posts: 510
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27 If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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18 00:00

Difficulty:   55% (hard)

Question Stats: 60% (01:51) correct 40% (01:49) wrong based on 389 sessions

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If y = ||x – 3| – 2|, for how many values of x is y = 4?

A) 1
B) 2
C) 3
D) 4
E) 5

Originally posted by Lucky2783 on 23 Mar 2015, 23:35.
Last edited by Bunuel on 24 Mar 2015, 03:19, edited 1 time in total.
RENAMED THE TOPIC.
Intern  Joined: 20 Mar 2015
Posts: 18
Location: Italy
GMAT 1: 670 Q48 V34 GPA: 3.7
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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1
$$y=||x-3|-2|$$

We can distinguish the two cases:
a) $$x-3\ge 0 \Rightarrow x\ge 3$$
b) $$x-3<0 \Rightarrow x<3$$

case a): $$x\ge 3$$

$$y=||x-3|-2|$$ becomes $$y=|x-3-2|=|x-5|$$
subcase i) $$x \ge 5$$
$$y=x-5 \Rightarrow 4=x-5 \Rightarrow x=9$$
subcase ii) $$3\le x<5$$
$$y=5-x \Rightarrow 4=5-x \Rightarrow x=1$$ ELIMINATE

case b): $$x<3$$

$$y=||x-3|-2|$$ becomes $$y=|3-x-2|=|-x+1|$$
subcase iii) $$x<1$$
$$y=-x+1 \Rightarrow 4=-x+1 \Rightarrow x=-3$$
subcase iv) $$1 \le x<3$$
$$y=x-1 \Rightarrow 4=x-1 \Rightarrow x=5$$ ELIMINATE.

There are only two solutions. B
Director  Joined: 07 Aug 2011
Posts: 510
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27 Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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1
Lucky2783 wrote:
If y = ||x – 3| – 2|, for how many values of x is y = 4?

A) 1
B) 2
C) 3
D) 4
E) 5

easy solution

y = ||x – 3| – 2| can be 4 only and only when X-3= +/-6. so there are 2 values of X . answer B.
Manager  Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 108
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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Lucky2783 wrote:
If y = ||x – 3| – 2|, for how many values of x is y = 4?

A) 1
B) 2
C) 3
D) 4
E) 5

$$||x – 3| – 2|=Y$$..........Eqn (1)

Substitute Y=4 and solve

$$||x – 3| – 2|=4$$

Case I :$$|x – 3| – 2 = 4$$........................................OR.......................................Case II :$$|x – 3| – 2 = -4$$
$$(x – 3)=6$$....OR....$$(x – 3)=-6$$.........................OR....................................... $$(x – 3)=-2$$....OR....$$(x – 3)=2$$

Substituting value of (x-3) in eqn (1),

only (x-3)=6, -6 gives the value of Y=4

Answer is B
Intern  Joined: 02 Mar 2015
Posts: 24
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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The answer is B : 2 values that is 9 and -3
I got there by solving the equation, by putting the value of y = 4.

The problem is I took around 2:36 min. anyway I could have done that in less time (assuming i am a bit quick in arithmetic maths) or the time taken is sufficient for a 700+ level question.

TIA
Director  G
Joined: 23 Jan 2013
Posts: 534
Schools: Cambridge'16
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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||x-3|-2|=4

|x-3|-2=4

has two solutions: x=9, x=-3

|x-3|-2=-4 means |x-3|=-2, but absolute value cannot be negative, NO SOLUTONS

B fits
Manager  B
Joined: 20 Aug 2015
Posts: 94
Location: India
GMAT 1: 710 Q50 V36 GPA: 3
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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Lucky2783 wrote:
If y = ||x – 3| – 2|, for how many values of x is y = 4?

A) 1
B) 2
C) 3
D) 4
E) 5

Let |x – 3| = a

The original equation becomes
|a - 2| = 4
So, a = 6, -2

But a can't be negative, since a is |x-3| and modulus of a number can't be negative.

Substituting, |x-3| = 6, we get 2 values of x.

(B)
Intern  B
Joined: 01 Oct 2017
Posts: 4
Location: India
GMAT 1: 650 Q43 V33 GRE 1: Q150 V150 GPA: 4
Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?  [#permalink]

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1
Take |x-3|=k
So,
|k-2|=4.
So, k=6,-2
Now, |x-3|= -2 which is not possible as modulus is always positive.
|x-3| = 6. So, x=9 and 3.

So, there are 2 solutions. Re: If y = ||x – 3| – 2|, for how many values of x is y = 4?   [#permalink] 11 Aug 2018, 21:12
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