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 establish a relationship between the two temperature scales (X and Y) based on the freezing and boiling points of water on each scale.

### Step-by-Step Solution:

1. **Identify the freezing and boiling points on 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 linear relationship between the two scales:**
The relationship can be expressed as:
\[ \frac{T - T_{1x}}{T_{2x} - T_{1x}} = \frac{T_y - T_{1y}}{T_{2y} - T_{1y}} \]
where \( T \) is the temperature on the X scale corresponding to \( T_y \) on the Y scale.

3. **Substitute the known values into the equation:**
We want to find the temperature on the X scale when \( T_y = 50.0^\circ Y \):
\[ \frac{T - (-125)}{375 - (-125)} = \frac{50 - (-70)}{-30 - (-70)} \]
Simplifying the denominators:
\[ \frac{T + 125}{500} = \frac{120}{40} \]

4. **Simplify the right-hand side:**
\[ \frac{120}{40} = 3 \]
So, we have:
\[ \frac{T + 125}{500} = 3 \]

5. **Cross-multiply to solve for \( T \):**
\[ T + 125 = 3 \times 500 \]
\[ T + 125 = 1500 \]

6. **Isolate \( T \):**
\[ T = 1500 - 125 \]
\[ T = 1375^\circ X \]

### Final Answer:
The value of temperature on the X scale equal to the temperature of \( 50.0^\circ Y \) on the Y scale is \( 1375.0^\circ X \).