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Difficulty: MediumIntegers, Absolute Value, and Number Lines

A highway has exits at mile markers −7-7 and 99. A new rest stop is to be built at mile marker cc such that its distance from the exit at mile marker −7-7 is exactly 33 times its distance from the exit at mile marker 99. Which of the following is a possible value of cc?

  1. A
    -3
  2. B
    1
  3. 5Answer
  4. D
    8
  5. E
    10

Answer

The value 5 is a possible mile marker for the rest stop.
The correct answer is the value 5. The distance between the rest stop at mile marker 5 and the exit at -7 is |5 - (-7)| = 12 miles. The distance between the rest stop at mile marker 5 and the exit at 9 is |5 - 9| = 4 miles. Since 12 is exactly 3 times 4, this satisfies the given condition.

Step-by-Step Solution

1
Express the distance from the rest stop cc to each exit using absolute value.
Distance to the exit at −7-7 is ∣c−(−7)∣=∣c+7∣|c - (-7)| = |c + 7|; distance to the exit at 99 is ∣c−9∣|c - 9|.
Absolute value represents the non-negative distance between two points on a number line.
2
Set up the algebraic equation using the given relationship.
∣c+7∣=3∣c−9∣|c + 7| = 3|c - 9|
The problem states that the distance to the exit at −7-7 is 3 times the distance to the exit at 99.
3
Solve the absolute value equation by separating it into two cases.
Case 1: c+7=3(c−9)c + 7 = 3(c - 9) or Case 2: c+7=−3(c−9)c + 7 = -3(c - 9)
An equation of the form ∣x∣=∣y∣|x| = |y| implies x=yx = y or x=−yx = -y.
4
Solve Case 1 for cc.
c+7=3c−27  ⟹  2c=34  ⟹  c=17c + 7 = 3c - 27 \implies 2c = 34 \implies c = 17
Distribute the 3, subtract cc from both sides, add 27 to both sides, and divide by 2.
5
Solve Case 2 for cc.
c+7=−3c+27  ⟹  4c=20  ⟹  c=5c + 7 = -3c + 27 \implies 4c = 20 \implies c = 5
Distribute the -3, add 3c3c to both sides, subtract 7 from both sides, and divide by 4.
6
Compare the solutions to the given options.
The two mathematically valid positions are 1717 and 55. Only 55 is listed among the options.
To identify the correct choice from the multiple-choice options.

Key Concept

Calculating distances and solving equations involving absolute values on a number line.

Alternative Method

Test the given answer choices directly by calculating the distances for each option. For the value 5, the distance to the exit at -7 is |5 - (-7)| = 12, and the distance to the exit at 9 is |5 - 9| = 4. Since 12 is 3 times 4, this value is correct.
Estimated Time:1m 0s
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Topics

Integers, Absolute Value, and Number Lines

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