In a termperature scale `X` ice point of water is assigned a value of `20^(@) X` and the boiling point of water is assiggned a value of `220^(@) X`. In another scale `Y` the ice poiny of water is assigned a value of `-20^(@) Y` and the boiling point is given a value of `380^(@)Y`. Then temperature on both the scales will be same is

Options

A. `30^(@)`
B. `40^(@)`
C. `50^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to find the temperature at which the readings on the two temperature scales \(X\) and \(Y\) are the same. We have the following information:

1. For scale \(X\):
- Ice point of water: \(20^\circ X\)
- Boiling point of water: \(220^\circ X\)

2. For scale \(Y\):
- Ice point of water: \(-20^\circ Y\)
- Boiling point of water: \(380^\circ Y\)

### Step 1: Establish the linear relationship
We can set up a linear relationship between the two scales based on the given points. The relationship can be expressed as:

\[ \frac{X - X_1}{X_2 - X_1} = \frac{Y - Y_1}{Y_2 - Y_1} \]

Where:
- \(X_1 = 20\), \(X_2 = 220\)
- \(Y_1 = -20\), \(Y_2 = 380\)

### Step 2: Substitute the values into the equation
Substituting the known values into the equation gives:

\[ \frac{X - 20}{220 - 20} = \frac{Y + 20}{380 + 20} \]

This simplifies to:

\[ \frac{X - 20}{200} = \frac{Y + 20}{400} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives:

\[ 400(X - 20) = 200(Y + 20) \]

### Step 4: Expand both sides
Expanding both sides results in:

\[ 400X - 8000 = 200Y + 4000 \]

### Step 5: Rearranging the equation
Rearranging the equation gives:

\[ 400X - 200Y = 8000 + 4000 \]

This simplifies to:

\[ 400X - 200Y = 12000 \]

### Step 6: Simplify the equation
Dividing the entire equation by 200 gives:

\[ 2X - Y = 60 \]

### Step 7: Set \(X = Y\) to find the temperature at which both scales are equal
To find the temperature where both scales are the same, we set \(X = Y\):

\[ 2X - X = 60 \]

This simplifies to:

\[ X = 60 \]

### Conclusion
Thus, the temperature at which the readings on both scales are the same is:

\[ \boxed{60^\circ} \]