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Difficulty: HardSolving Linear Equations

A newly designed temperature scale, Scale X, is related to the Celsius scale (∘C^{\circ}\text{C}) by a linear equation. Water freezes at 0∘C0^{\circ}\text{C}, which corresponds to −15∘X-15^{\circ}\text{X}, and water boils at 100∘C100^{\circ}\text{C}, which corresponds to 135∘X135^{\circ}\text{X}. If a chemical reaction must be maintained at a temperature where the reading on Scale X is exactly 2.52.5 times the reading on the Celsius scale, what is this temperature in degrees Celsius?

  1. A
    −11.5-11.5
  2. B
    −8.2-8.2
  3. −15.0-15.0Answer
  4. D
    13.613.6
  5. E
    15.015.0

Answer

−15.0-15.0 degrees Celsius
By writing the linear relationship between Scale X (XX) and Celsius (CC) as X=mC+kX = mC + k, we determine the constants using the given coordinates: (0,−15)(0, -15) yields k=−15k = -15, and (100,135)(100, 135) yields m=1.5m = 1.5. The resulting equation is X=1.5C−15X = 1.5C - 15. We then substitute the given condition X=2.5CX = 2.5C, resulting in 2.5C=1.5C−152.5C = 1.5C - 15. Subtracting 1.5C1.5C from both sides gives C=−15C = -15.

Step-by-Step Solution

1
Set up the general linear equation relating Scale X (XX) and Celsius (CC).
X=mC+kX = mC + k
Since the relationship is linear, it can be represented by a slope-intercept linear model.
2
Use the freezing point of water to find the y-intercept kk.
When C=0C = 0, X=−15X = -15, so −15=m(0)+k  ⟹  k=−15-15 = m(0) + k \implies k = -15.
The freezing point of water provides the point (0,−15)(0, -15) on the linear graph.
3
Use the boiling point of water to find the slope mm.
When C=100C = 100, X=135X = 135, so 135=m(100)−15  ⟹  150=100m  ⟹  m=1.5135 = m(100) - 15 \implies 150 = 100m \implies m = 1.5.
The boiling point of water provides the second point (100,135)(100, 135) to determine the rate of change.
4
Substitute the condition X=2.5CX = 2.5C into the linear equation and solve for CC.
2.5C=1.5C−15  ⟹  1.0C=−15  ⟹  C=−152.5C = 1.5C - 15 \implies 1.0C = -15 \implies C = -15.
This isolates the Celsius variable to find the temperature where the Scale X value is exactly 2.52.5 times the Celsius value.

Key Concept

Formulating and solving a linear equation from word-problem constraints and coordinate pairs.

Alternative Method

Instead of deriving the full equation, you can test the options directly. For example, check −15∘C-15^{\circ}\text{C}. The distance from freezing (0∘C0^{\circ}\text{C}) to −15∘C-15^{\circ}\text{C} is −15-15 units. Since Scale X changes by 1.51.5 units for every 11 unit of Celsius (calculated from a change of 150150 on Scale X for 100100 on Celsius), Scale X will change by 1.5×(−15)=−22.51.5 \times (-15) = -22.5 units from its freezing point value of −15-15. This yields −15−22.5=−37.5∘X-15 - 22.5 = -37.5^{\circ}\text{X}. Checking the ratio: −37.5−15=2.5\frac{-37.5}{-15} = 2.5, which matches the given condition.
Estimated Time:2m 0s
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Topics

Solving Linear Equations

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