We need to find the total time of the journey, given the distances traveled and the speeds of X (on a bike), Y (switching between car and cycle), and Z (in the car).

Step-by-Step Solution:

1. Key Variables:

   • Total distance from $P$ to $Q = 126 \, \text{km}$

   • Speeds:

     • Car: $48 \, \text{km/hr}$
     • Bike: $16 \, \text{km/hr}$
     • Cycle: $16 \, \text{km/hr}$

   • Unknowns:

     • Distance $d_1$: Distance traveled by the car before Y gets out to cycle.
     • Time $t_1$: Time taken by the car before Y gets out to cycle.

2. Car Travel (Y and Z):

   • In the first phase, both Y and Z travel together in the car at $48 \, \text{km/hr}$.
   • Distance covered by the car: $d_1 = 48 \cdot t_1$, where $t_1$ is the time of this phase.

3. Switching Phase:

   • At $d_1$, Y gets out of the car to cycle. Now:
     • Y continues the remaining distance $126 - d_1$ on a cycle at $16 \, \text{km/hr}$.
     • Z returns in the car to pick up X, covering a distance $d_2$.

4. Return Trip for Z:

   • Z takes time $t_2$ to return and meet X on the bike:
     • $d_2 = 16 \cdot t_2$ (distance covered by X on the bike while Z drives back to meet X).
     • $d_1 - d_2 = 48 \cdot t_2$ (distance Z travels back in the car).

5. Meeting Phase:

   • After Z meets X, they travel together in the car for the remaining distance $126 - d_1 - d_2$ at $48 \, \text{km/hr}$.

6. Total Time for Journey:

   • Total time $T$ must be the same for X, Y, and Z: $T = \text{Time taken by X on the bike} + \text{Time taken by Z to return and pick up X} + \text{Time for the final phase}$ $T = \frac{d_2}{16} + t_2 + \frac{126 - d_1 - d_2}{48}$ and $T = \text{Time taken by Y to cycle from } d_1 \text{ to Q}.$ $T = \frac{126 - d_1}{16}.$

Solving the Equations:

1. From the equations: $d_1 = 48 \cdot t_1, \quad d_2 = 16 \cdot t_2, \quad d_1 - d_2 = 48 \cdot t_2.$

   Solve for $d_1$, $d_2$, and $t_2$ in terms of $t_1$.

2. Substituting back into the total time equation, compute $T$.

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