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What is the value of x?

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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Thu Jun 08, 2017 7:09 am
jack0997 wrote:What is the value of x?

(1) |y| <= 3x
(2) |5x - 1| ƒ = x ‚ + 7

OA C
Statement 1:

|y| ≤ 3x

Since |y| ≥ 0, we have:

-3x ≥ 0
ƒ=> x ≤ 0
However, the value of x cannot be determined. - Insufficient

Statement 2: |5x - 1| ƒ = x ‚ + 7

ƒ=> 5x - 1 ƒ=+/-( „x ‚ + 7)…
ƒ=> 5x + 1 =ƒ x ‚ + 7; taking positive value
=ƒ> 4x ƒ = 8
ƒ=> x =ƒ 2

OR

5x - 1 ƒ= -x - 7; taking negative value
=ƒ> 6x ƒ = -6
=ƒ> x ƒ = -1
Thus, the value of x cannot be uniquely determined. - Insufficient

Statement1 & 2 together:

Since x ≤ 0, the only possible value of x ƒ = -1. - Sufficient

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by [email protected] » Thu Jun 08, 2017 1:08 pm
Hi jack0997,

This question is built around Absolute Value rules, so you can either approach it algebraically or with a bit of 'brute force' (meaning that most Absolute Value equations have more than one solution, so you just have to 'find' them all).

We're asked for the value of X.

1) |Y| <= 3X

While we don't know the value of Y, we do know that |Y| cannot be negative. Thus, |Y| is greater than, or equal to, 0. By extension, this means that 3X CANNOT be negative, so X CANNOT be negative. X is >= 0, but we don't know the exact value of X.
Fact 1 is INSUFFICIENT.

2) |5X - 1| ƒ = X ‚ + 7

When an equation contains an Absolute Value, there are likely to be two different possible values for X (depending on whether the value inside the absolute value is positive or negative)....

5X - 1= X + 7
4X = 8
X = 2

-(5X - 1) = X + 7
-5X + 1 = X + 7
-6 = 6X
X = -1

Thus, there are two solutions: -1 and 2.
Fact 2 is INSUFFICIENT

Combined, we know...
X >= 0
X = -1 or +2

There's only one solution that 'fits' both Facts: +2
Combined, SUFFICIENT

Final Answer: C

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by Matt@VeritasPrep » Thu Jun 08, 2017 5:22 pm
S1:

3x ≥ |y|

Since |y| ≥ 0, we know that 3x ≥ 0, and that x ≥ 0. That's helpful, but not sufficient.

S2:

|5x - 1| = x + 7

This has two solutions, either

5x - 1 = x + 7 (which gives x = 2)

or

-(5x - 1) = x + 7 (which gives x = -1)

Since x could be 2 or -1, this is also not sufficient.

Together, I know that x can only be 2, since S1 tells me x ≥ 0. With one only solution left, the answer is C.
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by Matt@VeritasPrep » Thu Jun 08, 2017 5:23 pm
Jay@ManhattanReview wrote: Since |y| ≥ 0, we have:

-3x ≥ 0
ƒ=> x ≤ 0
Other way round.

We've got

3x ≥ |y|

and

|y| ≥ 0

So we combine the two inequalities

3x ≥ |y| ≥ 0

then drop the middle term

3x ≥ 0

and divide

x ≥ 0
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