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Difficulty: MediumIntegers, Absolute Value, and Number Lines

A scientist monitors the temperature of two research chambers. Chamber A is kept at −5∘C-5^\circ\text{C} and Chamber B is kept at 7∘C7^\circ\text{C}. The scientist sets a third chamber, Chamber C, to a temperature of T∘CT^\circ\text{C} such that the distance between TT and the temperature of Chamber A on the Celsius scale is exactly 33 times the distance between TT and the temperature of Chamber B. If the temperature of Chamber C is warmer than Chamber A but colder than Chamber B, what is the value of TT?

Answer: 4

Answer

The correct temperature value of Chamber C is 4.
The temperature of Chamber C must be 4∘C4^\circ\text{C} because the distance from 44 to −5-5 is ∣4−(−5)∣=9|4 - (-5)| = 9 and the distance from 44 to 77 is ∣4−7∣=3|4 - 7| = 3. The distance of 99 is exactly 33 times the distance of 33. Furthermore, 44 lies between −5-5 and 77, satisfying the condition that Chamber C is warmer than Chamber A but colder than Chamber B.

Step-by-Step Solution

1
Represent the distances on the number line using absolute value expressions.
The distance to Chamber A is ∣T−(−5)∣=∣T+5∣|T - (-5)| = |T + 5| and the distance to Chamber B is ∣T−7∣|T - 7|.
Distance between two points xx and yy on a number line is represented by ∣x−y∣|x - y|.
2
Set up the algebraic equation reflecting the relationship between the distances.
∣T+5∣=3∣T−7∣|T + 5| = 3|T - 7|
The problem states the distance to Chamber A is exactly 3 times the distance to Chamber B.
3
Solve the absolute value equation by considering both positive and negative cases.
Case 1: T+5=3(T−7)⇒T=13T + 5 = 3(T - 7) \Rightarrow T = 13. Case 2: T+5=−3(T−7)⇒T=4T + 5 = -3(T - 7) \Rightarrow T = 4.
The equation ∣x∣=∣y∣|x| = |y| implies x=yx = y or x=−yx = -y.
4
Verify which solution satisfies the temperature boundary condition.
Since Chamber C must be warmer than −5∘C-5^\circ\text{C} but colder than 7∘C7^\circ\text{C}, the only valid value is T=4T = 4.
The value T=13T = 13 is warmer than both chambers and does not lie between them.

Key Concept

Using absolute value to represent distance on a number line and solving absolute value equations with boundary conditions.

Alternative Method

Alternatively, visualize this on a number line. The total distance between Chamber A (−5-5) and Chamber B (77) is 1212 units. Since Chamber C lies between them and the distance from C to A is 33 times the distance from C to B, we can divide the 1212-unit interval into 3+1=43 + 1 = 4 equal parts. Each part is 12÷4=312 \div 4 = 3 units. Chamber C is located 11 part away from Chamber B (towards Chamber A), which places it at 7−3=47 - 3 = 4.
Estimated Time:1m 30s
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Topics

Integers, Absolute Value, and Number Lines

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