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Nina1987
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 26 Oct 2021, 13:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

44% (02:30) correct 56% (02:36) wrong based on 255 sessions

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If y = |||x| – 3| – x| , how many values are possible for y=1?

A) 0
B) 1
C) 2
D) 4
E) 6

This Qs is from VeritasKarishma blog post. While I absolutely love her graphical approach in general, for this one particularly I find it way to complex and so looking for an alternate approach.
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C
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KarishmaB
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 26 Oct 2021, 21:17
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Nina1987
If y = |||x| – 3| – x|| , how many values are possible for y=1?

A) 0
B) 1
C) 2
D) 4
E) 6

This Qs is from VeritasKarishma blog post. While I absolutely love her graphical approach in general, for this one particularly I find it way to complex and so looking for an alternate approach.
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The algebraic approach would be worse. This question is not GMAT type and is discussed for intellectual purposes only. Take some time to understand why the graph changes at each step as it does and then walk away from it. If what is done at each step is clear to you, that is all you need to take away from this question. You would not be required to solve such a question in the test.
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 19 May 2023, 13:04
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Could someone please explain how the answer is C?
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Regor60
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 19 May 2023, 13:26
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Appears to be an absolute value sign missing in the question since there are only 7 appearing ?

Posted from my mobile device
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Avneeshyadav01
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 25 May 2023, 01:56
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The algebraic approach is lengthier than the graphical solution. We need to solve the modulus from outside to inside.
Given, Y = 1
Hence |||x|-3|-x| = 1

Case 1: ||x|-3|-x = 1
||x|-3| = x+1

Subcase 1.1:
|x|-3 = x+1
|x| = x+4

Subcase 1.1.1:
X = x+4 (no solution)

Subcase 1.1.2:
X= -x-4
X= -2------------------------------(1)

Subcase 1.2:
|x|-3 = -x-1
|x| = 2-x

Subcase 1.2.1:
x=2-x
x=1---------------------------------(2)

Subcase 1.2.2:
X=x-2 (no solution)

Case 2: ||x|-3|-x = -1
||x|-3| = x-1

Subcase: 2.1
|x|-3 = x-1
|x| = x+2

Subcase: 2.1.1:
X = x+2 (no solution)

Subcase: 2.1.2:
X= -x-2
X= -1 -----------------------------------------(3)

Subcase 2.2:
|x|-3 = -x+1
|x| = -x+4

Subcase 2.2.1:
X= -x+4
X= 2 --------------------------------------------(4)

Subcase 2.2.2:
X = x-4 (no solution)

Now, we have the solutions above. Since solving for modulus might give some extraneous solutions, we need to verify all possible solutions by putting values in the question stem
x y
1 1
-1 3
2 1
-2 3
Hence, for y to be 1, x can have only two values i.e. x=1 & x=2. So option C is the right choice.
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 30 Oct 2023, 02:33
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KarishmaB
Nina1987
If y = |||x| – 3| – x|| , how many values are possible for y=1?

A) 0
B) 1
C) 2
D) 4
E) 6

This Qs is from VeritasKarishma blog post. While I absolutely love her graphical approach in general, for this one particularly I find it way to complex and so looking for an alternate approach.

The algebraic approach would be worse. This question is not GMAT type and is discussed for intellectual purposes only. Take some time to understand why the graph changes at each step as it does and then walk away from it. If what is done at each step is clear to you, that is all you need to take away from this question. You would not be required to solve such a question in the test.
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KarishmaB can you please post a link to the original question in this thread so that we can take a look at the graphical approach.
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 30 Oct 2023, 08:47
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modify the eq:

\(||x|-3|=x+1 , x-1\)

plot the graphs, we get 2 pts of intersection
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If y = |||x| 3| x|| , how many values are possible for y=1? 1) 0 06 Jul 2025, 19:18
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