What is the value of x?
(1) 4 < x < 6
(2) |x| = 4x − 15
The OA is B.
When we solve the statement (2) it give us two options for x (x=5 and x=3). Why is sufficient statement (2)?
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What is the value of x?
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Jay@ManhattanReview
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Let's take Statement 2: |x| = 4x − 15Vincen wrote:What is the value of x?
(1) 4 < x < 6
(2) |x| = 4x − 15
The OA is B.
When we solve the statement (2) it give us two options for x (x=5 and x=3). Why is sufficient statement (2)?
1. Taking x as positive
x = 4x − 15 => x = 5
3. Taking x as negative
-x = 4x − 15 => x = 3
At this point, it does seem that Statement 2 is insufficient; however, it is not so, let's relook at |x| = 4x − 15.
Since the left-hand side, |x| is positive or 0, the right-hand side must also be non-negative; thus, 4x − 15 ≥ 0 => x ≥ 3.75.
Since means that the derived value x = 3 is not an eligible value, thus x = 5: a unique value. Sufficient.
The correct answer: B
Hope this helps!
-Jay
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Brent@GMATPrepNow
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Target question: What is the value of x?Vincen wrote:What is the value of x?
(1) 4 < x < 6
(2) |x| = 4x − 15
Statement 1: 4 < x < 6
ASIDE: Some students will assume that x is an integer and incorrectly conclude that x must equal 5. This is a common mistake on the GMAT.
The truth of the matter is that there are infinitely many values of x that satisfy statement 1.
For example x could equal 4.1, or x could equal 4.132 or 4.54 or 5 or 5.4211 etc
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: |x| = 4x − 15
There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
Given: |x| = 4x − 15
So, there are two possible cases to examine:
Case a: x = 4x − 15. When we solve this equation, we get: x = 5
Case b: x = -(4x − 15). When we solve this equation, we get: x = 3
At this point, it LOOKS LIKE there are two possible values of x. However, we haven't performed step 3 yet: Plug solutions into original equation to check for extraneous roots
Let's do that.
Case a: Plug x = 5 into original equation to get: |5| = 4(5) − 15
Simplify right side of equation to get: |5| = 5
This works, so x = 5 IS a possible value of x
Case b: Plug x = 3 into original equation to get: |3| = 4(3) − 15
Simplify right side of equation to get: |3| = -3
This DOES NOT work, so x = 3 is NOT a possible value of x
Since there is only ONE possible value of x (x = 5), statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
