To solve this, we can set up a linear relationship between the actual Celsius scale (0ºC to 100ºC) and the thermometer's faulty scale (20ºC to 150ºC).

1. Set up the linear equation:

The faulty thermometer's scale can be represented as:
\[
T_{\text{faulty}} = a \cdot T_{\text{actual}} + b
\]

Using the freezing point:
\[
20 = a \cdot 0 + b \Rightarrow b = 20
\]

Using the boiling point:
\[
150 = a \cdot 100 + 20
\]
\[
130 = 100a \Rightarrow a = 1.3
\]

So, the relation is:
\[
T_{\text{faulty}} = 1.3 \cdot T_{\text{actual}} + 20
\]

2. Find the faulty reading at 60ºC actual temperature:
\[
T_{\text{faulty}} = 1.3 \cdot 60 + 20 = 78 + 20 = 98
\]

Therefore, the thermometer will read 98ºC at an actual temperature of 60ºC.

Using the relation $\frac{T - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{constant}$, we write $\frac{T - 0}{100 - 0} = \frac{T - 40}{80 - 40}$. This simplifies to $\frac{T}{100} = \frac{T - 40}{40}$, which gives $40T = 100T - 4000$ or $60T = 4000$, hence $T = \frac{200}{3}\,^\circ\text{C}$.