On an X temperature scale, water freezes at `-125.0^(@) X` and boils at `375.0^(@) X`. On a Y temperature scale, water freezes at `-70.0^(@)Y` and boils at `-30.0^(@)Y`. The value of temperature on X-scale equal to the temperature of `50.0^(@)Y` on Y-scale is

Options

A. `455.0^(@) X`
B. `-125.0^(@) X`
C. `1375.0^(@) X`
D. `1500.0^(@) X`

Explanation

To solve the problem, we need to find the temperature on the X scale that corresponds to 50.0 degrees on the Y scale. We will use the reference points provided for both temperature scales.

### Step-by-Step Solution:

1. **Identify the reference points for both scales:**
- On the X scale:
- Freezing point of water: \( T_{1x} = -125.0^\circ X \)
- Boiling point of water: \( T_{2x} = 375.0^\circ X \)
- On the Y scale:
- Freezing point of water: \( T_{1y} = -70.0^\circ Y \)
- Boiling point of water: \( T_{2y} = -30.0^\circ Y \)

2. **Set up the relationship between the two temperature scales:**
We can use the linear relationship between the two scales. The formula for converting temperatures between two scales is:
\[ \frac{x - T_{1x}}{T_{2x} - T_{1x}} = \frac{y - T_{1y}}{T_{2y} - T_{1y}} \]

3. **Substitute the known values into the equation:**
\[ \frac{x + 125}{375 - (-125)} = \frac{y + 70}{-30 - (-70)} \]
Simplifying the denominators:
- \( T_{2x} - T_{1x} = 375 + 125 = 500 \)
- \( T_{2y} - T_{1y} = -30 + 70 = 40 \)

Thus, the equation becomes:
\[ \frac{x + 125}{500} = \frac{y + 70}{40} \]

4. **Substitute \( y = 50.0 \) into the equation:**
\[ \frac{x + 125}{500} = \frac{50 + 70}{40} \]
Simplifying the right side:
\[ \frac{50 + 70}{40} = \frac{120}{40} = 3 \]

5. **Set up the equation:**
\[ \frac{x + 125}{500} = 3 \]

6. **Solve for \( x \):**
Multiply both sides by 500:
\[ x + 125 = 1500 \]
Subtract 125 from both sides:
\[ x = 1500 - 125 = 1375 \]

7. **Conclusion:**
The value of temperature on the X scale that corresponds to 50.0 degrees on the Y scale is:
\[ x = 1375.0^\circ X \]