Three trains T1, T2 and T3 leaves station X for station Y in equal time intervals, when three trains reach station Y simultaneously and then they leave for station Z, which is 480 km away from station Y. T1 reach station Z two hours after T2 and T3 turns back immediately towards station Y after reaching station Z. T1 and T3 meet a point O, which is 160 km away from station Z and the difference between speed of T1 & T3 is V km/hr. Find the time (in minutes) taken by T3 to reach point O from station Y.

The correct answer is 160.

Let's break down the problem step-by-step:

Step 1: Relationship between the speeds of the trains
Let the distance between station X and station Y be $d_{\mathrm{XY}}$.
The three trains T1, T2, and T3 leave station X at equal time intervals $t$. Let T1 leave at time $0$, T2 at time $t$, and T3 at time $2$$t$.
Since they reach station Y simultaneously at some time $T$, the time taken by each train to travel from X to Y is:
$t_1=T$
$t_2=T-t$
$t_3=T-2t$

Since the departure times are in arithmetic progression (AP), the times taken by the trains to reach station Y also form an AP. Therefore, the reciprocals of their speeds $v_1$, $v_2$, and $v_3$ are in AP:
$\frac{1}{v_1}+\frac{1}{v_3}=\frac{2}{v_2}$ --- (Equation 1)

Step 2: Analysis of the journey from Y to Z
The trains leave station Y simultaneously for station Z, which is $480$ km away.
T1 and T3 meet at point O, which is $160$ km away from station Z. This means:
Distance of point O from station Y = $480-160=320$ km.

By the time they meet at point O:
T1 has traveled directly from Y to O: Distance = $320$ km.
T3 has reached Z and turned back to meet T1 at O: Distance = $480+160=640$ km.
Since they started from Y at the same time and met at O, the time taken by both trains is equal:
$\frac{320}{v_1}=\frac{640}{v_3}\Rightarrow v_3=2v_1$

Step 3: Calculating individual speeds
Substituting $v_3=2v_1$ into Equation 1:
$\frac{1}{v_1}+\frac{1}{2v_1}=\frac{2}{v_2}$
$\frac{3}{2v_1}=\frac{2}{v_2}\Rightarrow v_2=\frac{4}{3}$

We are given that T1 reaches station Z two hours after T2:
$\frac{480}{v_1}-\frac{480}{v_2}=2$

Substitute $v_2=\frac{4}{3}$:
$\frac{480}{v_1}-\frac{480}{\frac{4}{3}}=2$
$\frac{480}{v_1}-\frac{360}{v_1}=2$
$\frac{120}{v_1}=2\Rightarrow v_1=60\text{km/hr}$

Thus, the speed of T3 is:
$v_3=2\times 60=120\text{km/hr}$

Step 4: Finding the time taken by T3 to reach point O from station Y
On its journey to Z, the distance from station Y to point O is $320$ km.
The time taken by T3 to reach point O is:
$\text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{320}{120}\text{hours}=\frac{8}{3}\text{hours}$

Converting this time into minutes:
$\text{Time}=\frac{8}{3}\times 60=160\text{minutes}$