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On the number line, which of the following specifies the set of all
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06 Aug 2023, 09:00
06 Aug 2023, 09:00
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On the number line, which of the following specifies the set of all numbers x such that |x - 3| + |x - 4| < 2?
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
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Re: On the number line, which of the following specifies the set of all
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06 Aug 2023, 10:37
06 Aug 2023, 10:37
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Bunuel wrote:
On the number line, which of the following specifies the set of all numbers x such that |x - 3| + |x - 4| < 2?
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
We can divide our analysis into three regions and open the modulus operator according to the positive-negative nature in that region
- Region 1: x < 3
- Region 2: 3 < x < 4
- Region 3: x > 4
Region 1: x < 3
\(-(x-3) - (x-4) < 2\)
\(-2x < -5\)
\(x > 2.5\)
Region 2: 3 < x < 4
\((x-3) - (x-4) < 2\)
\(1 < 2\)
Hence, the entire region satisfies the inequality.
\(3 < x < 4\)
Region 3: x > 4
\(x - 3 + x - 4 > 2\)
\(2x < 9\)
\(x < 4.5\)
The inequality satisfies at the critical points, i.e. at x = 3 and at x = 4
Combined: 2.5 < x < 4.5
Option D
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Re: On the number line, which of the following specifies the set of all
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12 Feb 2024, 00:57
12 Feb 2024, 00:57
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Laracbo wrote:
Bunuel wrote:
On the number line, which of the following specifies the set of all numbers x such that |x - 3| + |x - 4| < 2?
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
A. 1 < x <6
B. 1.5 < x < 5.5
C. 2 < x < 5
D. 2.5 < x < 4.5
E. 3 < x < 4
Would it be possible that you further elaborate on your solution? I somehow cannot really follow.It would be very helpful for me to know:
1. How did you come up with these 3 cases? I understand that at x=3 and x=4, the absolute value expressions become zero. However, I do not understand why you set the inequality signs in that particular direction e.g. why do we test for x<3 but then for 4 ≤ x? What's the reasoning that once x is less than and the other times is greater than?
2. Furthermore, I would greatly appreciate if you could explain why you changed the order of operations whilst testing the differnet cases (at least for case 1 and 2). For example here:
|x – 3| + |x – 4| < 2
(-x + 3) + (-x + 4) < 2 --> why do we have neg. x now and+ 3 / 4?
Needless to say I am not familiar with this underlying concept of determining the interval of all possible values of x for absolute value expressions, so if there is any blog posts or other ressources explaining this, I would greatly appreciate it (I could not find anything myself upon my inital reserach).
Many thanks for your support.
Kind wishes
Lara
The critical points (aka key points or transition points) for \(|x - 3| + |x - 4| < 2\) are 3 and 4 (those are the values of x for which the expressions in the absolute values become 0). This gives us the following three ranges to consider:
- \(--------(3)--------(4)--------\)
\(x < 3\)
\(3 ≤ x < 4\)
\(4 ≤ x\)
When x moves from one range to another, one of the absolute values we have, |x - 3| or |x - 4|, changes sign when the absolute value is dropped.
1. When \(x < 3\), then:
- \(x - 3 < 0 \), and thus \(|x - 3| = -(x - 3)\) (recall that \(|a| = -a\), when \(a ≤ 0\)).
\(x - 4 < 0 \), and thus \(|x - 4| = -(x - 4)\).
Hence, in this range \(|x - 3| + |x - 4| < 2\) becomes \(-(x - 3) - (x - 4) < 2\). This gives \(x > 2.5\). Since we consider \(x < 3\) range, then combining \(x > 2.5\) with \(x < 3\) gives a set for which the inequality holds for the considered range: \(2.5 < x < 3\).
2. When \(3 ≤ x < 4\), we are moving to the second range, then:
- \(x - 3 ≥ 0 \), and thus \(|x - 3| = x - 3\) (recall that \(|a| = a\), when \(a ≥ 0\)). Notice how |x - 3| gets expanded with plus sign here in contrast with minus sign in the previous range.
\(x - 4 < 0 \), and thus \(|x - 4| = -(x - 4)\).
Hence, in this range \(|x - 3| + |x - 4| < 2\) becomes \((x - 3) - (x - 4) < 2\). This gives \(1 < 2\)m which is true. This implies that the inequality holds for the entire range we consider: \(3 ≤ x < 4\).
3. When \(4 ≤ x\), we are moving to the thrid range, then:
- \(x - 3 > 0 \), and thus \(|x - 3| = x - 3\).
\(x - 4 ≥ 0 \), and thus \(|x - 4| = x - 4\). Notice how |x - 4| gets expanded with plus sign here in contrast with minus sign in the previous range.
Hence, in this range \(|x - 3| + |x - 4| < 2\) becomes \((x - 3) + (x - 4) < 2\). This gives \(x < 4.5\). Since we consider \(4 ≤ x\) range, then combining \(x < 4.5\) with \(4 ≤ x\) gives a set for which the inequality holds for the considered range: \(4 ≤ x < 4.5\).
Therefore, \(|x - 3| + |x - 4| < 2\) holds for the following three ranges:
- \(2.5 < x < 3\)
\(3 ≤ x < 4\)
\(4 ≤ x < 4.5\)
Observe that these three ranges comprise one continuous range \(2.5 < x < 4.5\), so that's our final answer.
Answer: D
10. Absolute Value
- Theory
Math: Absolute value (Modulus)
Absolute Value: Tips and hints
Properties of Absolute Values on the GMAT
Absolute modulus : A better understanding
GMAT Question Patterns: Abolute Value
The Reason Behind Absolute Value Questions on the GMAT
For more check Ultimate GMAT Quantitative Megathread
Hope it helps.
