Let total distance = x
Now, according to the question,
\( 40 \Large \times 3 + 60 \times 4.5 = \frac{3x}{5} \)
= > \( \Large 120 +270 = \frac{3x}{5} \)
x = \(\Large \frac{390 \times 5}{3}\)= 650km
Remaining distance = \( \Large \left( 650 - \frac{3 \times 650}{5} \right) \)
= 650 - 390 = 260 km.
Required average speed = 260/4 = 65 Km/hr