On a liner X temperature scale , water freezes at `-125.0^@X` and boils at `360.0^@X`. On a linear Y temperature scale water freezes at `-70.00^@Y` and boils at `-30.00^@Y` . A temperature of `50.00^@Y` corresponds to what temperature on the X scale ?

Answer

To solve the problem, we need to convert the temperature from the Y scale to the X scale using the given freezing and boiling points of water on both scales.

### Step-by-Step Solution:

1. **Identify the Freezing and Boiling Points:**
- For the X scale:
- Freezing point of water: \( T_{fX} = -125.0^\circ X \)
- Boiling point of water: \( T_{bX} = 360.0^\circ X \)
- For the Y scale:
- Freezing point of water: \( T_{fY} = -70.0^\circ Y \)
- Boiling point of water: \( T_{bY} = -30.0^\circ Y \)

2. **Calculate the Temperature Range for Each Scale:**
- For the X scale:
\[ \Delta T_X = T_{bX} - T_{fX} = 360.0 - (-125.0) = 360.0 + 125.0 = 485.0^\circ X \]
- For the Y scale:
\[ \Delta T_Y = T_{bY} - T_{fY} = -30.0 - (-70.0) = -30.0 + 70.0 = 40.0^\circ Y \]

3. **Identify the Temperature to Convert:**
- We need to convert \( T_Y = 50.0^\circ Y \) to the X scale.

4. **Calculate the Difference from Freezing Point on the Y Scale:**
- The difference from the freezing point on the Y scale:
\[ \Delta T_{Y, 50} = T_Y - T_{fY} = 50.0 - (-70.0) = 50.0 + 70.0 = 120.0^\circ Y \]

5. **Set Up the Ratio of Temperature Changes:**
- Since the scales are linear, the ratio of the temperature changes can be set up as:
\[ \frac{\Delta T_X}{\Delta T_Y} = \frac{T - T_{fX}}{\Delta T_{Y, 50}} \]
- Substituting the known values:
\[ \frac{485.0}{40.0} = \frac{T - (-125.0)}{120.0} \]

6. **Cross-Multiply and Solve for T:**
\[ 485.0 \times 120.0 = 40.0 \times (T + 125.0) \]
\[ 58200 = 40.0T + 5000 \]
\[ 58200 - 5000 = 40.0T \]
\[ 53200 = 40.0T \]
\[ T = \frac{53200}{40.0} = 1330.0^\circ X \]

### Final Answer:
The temperature of \( 50.0^\circ Y \) corresponds to \( 1330.0^\circ X \).

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