On a hypothetical scale `A` the ice point is `42^(@)` and the steam points is `182^(@)` For another scale `B`. The ice points is `-10^(@)` and steam point in `90^(@)`. If `B` reads `60^(@)`. The reading of `A` is.

Options

A. `160^(@)`
B. `140^(@)`
C. `120^(@)`
D. `11^(@)`

Explanation

To solve the problem, we need to establish the relationship between the two temperature scales (A and B) using the given ice points and steam points.

### Step-by-Step Solution:

1. **Identify the Fixed Points:**
- For scale A:
- Ice point = 42°A
- Steam point = 182°A
- For scale B:
- Ice point = -10°B
- Steam point = 90°B

2. **Set Up the Formula:**
The relationship between the two scales can be expressed as:
\[ \frac{T_A - T_{A_{ice}}}{T_{A_{steam}} - T_{A_{ice}}} = \frac{T_B - T_{B_{ice}}}{T_{B_{steam}} - T_{B_{ice}}} \]
Where:
- \(T_A\) = temperature on scale A
- \(T_{A_{ice}} = 42\)
- \(T_{A_{steam}} = 182\)
- \(T_B = 60\) (given)
- \(T_{B_{ice}} = -10\)
- \(T_{B_{steam}} = 90\)

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ \frac{T_A - 42}{182 - 42} = \frac{60 - (-10)}{90 - (-10)} \]

4. **Simplify the Equation:**
Calculate the denominators:
- For scale A: \(182 - 42 = 140\)
- For scale B: \(60 + 10 = 70\) and \(90 + 10 = 100\)

The equation now becomes:
\[ \frac{T_A - 42}{140} = \frac{70}{100} \]

5. **Cross-Multiply to Solve for \(T_A\):**
Cross-multiplying gives:
\[ 100(T_A - 42) = 70 \cdot 140 \]

6. **Calculate the Right Side:**
Calculate \(70 \cdot 140 = 9800\):
\[ 100(T_A - 42) = 9800 \]

7. **Divide by 100:**
\[ T_A - 42 = 98 \]

8. **Solve for \(T_A\):**
\[ T_A = 98 + 42 = 140 \]

### Final Answer:
The reading of scale A when scale B reads 60° is **140°A**.

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