What time after 3:00 o’clock would the hour-hand and the minute-hand of a continuously driven clock subtends an angle of 45° for the second time?

$x = 15 + \dfrac{x}{12} + \dfrac{45}{6}$

$\frac{11}{12}x = 22.5$

$x = 24.54 ~ \text{min}$
 

Sirang-relo Method

$x = \dfrac{2}{11} \cdot \theta$

$x = \dfrac{2}{11} \cdot (90 + 45)$

$x = 24.54$
 

Hence, required time = 3:24.54