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Difficulty: MediumIntegers, Absolute Value, and Number Lines

On a vertical number line representing elevation relative to sea level (00 meters), Submersible XX is located at a coordinate of −45-45 meters. Submersible YY is located at a depth such that the distance between Submersible XX and Submersible YY is 3030 meters. If Submersible YY is closer to sea level than Submersible XX, what is the value of ∣2Y−X∣|2Y - X|?

  1. 1515Answer
  2. B
    6060
  3. C
    7575
  4. D
    105105
  5. E
    135135

Answer

The value of ∣2Y−X∣|2Y - X| is 1515.
Submersible XX is at −45-45. Being 3030 meters away means Submersible YY is at either −15-15 or −75-75. Since Submersible YY is closer to sea level (00), its coordinate is −15-15. Substituting X=−45X = -45 and Y=−15Y = -15 into ∣2Y−X∣|2Y - X| gives ∣2(−15)−(−45)∣=∣−30+45∣=∣15∣=15|2(-15) - (-45)| = |-30 + 45| = |15| = 15.

Step-by-Step Solution

1
Determine the possible coordinates for Submersible YY.
Since X=−45X = -45 and the distance between XX and YY is 3030, Y=−45+30=−15Y = -45 + 30 = -15 or Y=−45−30=−75Y = -45 - 30 = -75.
Distance on a number line from coordinate xx is given by x±dx \pm d.
2
Select the correct coordinate for YY based on the given constraint.
Y=−15Y = -15, because ∣−15∣=15<45=∣−45∣|-15| = 15 < 45 = |-45|, meaning YY is closer to sea level (00).
The problem states that Submersible YY is closer to sea level than Submersible XX.
3
Substitute X=−45X = -45 and Y=−15Y = -15 into the expression ∣2Y−X∣|2Y - X| and evaluate.
∣2(−15)−(−45)∣=∣−30+45∣=∣15∣=15|2(-15) - (-45)| = |-30 + 45| = |15| = 15.
Multiplying 22 by −15-15 yields −30-30, and subtracting −45-45 is equivalent to adding 4545.

Key Concept

Distance on a Number Line and Absolute Value Evaluation
Estimated Time:1m 15s
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Topics

Integers, Absolute Value, and Number Lines

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