- The cars $T$ and $U$ should be parked next to each other.
This means we can park these two cars in two ways, either $TU$ or $UT.$
- The cars $S$ and $V$ also should be parked next to each other.
This means we can park these cars also in two ways, either $SV$ or $VS.$
- The cars $P$ and $Q$ cannot be parked next to each other.
This means we can not park the cars as $PQ$ or $QP.$
- The cars $Q$ and $S$ must be parked next to each other.
This means we can park these two cars in two ways, either $QS$ or $SQ.$
- $R$ is parked to the immediate right of $V$.
This means we can park these two cars in only one way $VR.$
- $T$ is parked to the left of $U$.
This is not “immediately left” but combining with statement $1,$ “immediately left” follows. So, we can park these two cars in only one way $TU.$
Now we can combine the statements $(2)\;\& (4)$ and get two possible ways for cars $Q,S$ and $V.$
But $VSQ$ is not possible because it violates statement $5.$
Now, using statement $5,$ we get the possible car parking sequence $QSVR.$
Now, we can check each and every option.
- There are two cars parked in between $Q$ and $V$, it is incorrect because only $S$ is in between $Q$ and $V.$
- $Q$ and $R$ are not parked together, it is correct.
- $V$ is the only car parked in between $S$ and $R,$ it is correct.
- Car $P$ is parked at the extreme end.
For this option, we can consider all possible combinations of car parking which are only four as given below. $(P$ cannot come near $Q$ due to statement $3)$
- $QSVR\; {\color{DarkBlue} {\color{Red} {P}}\; {TU}}$
- $QSVR\; {\color{DarkBlue} \; {TU}}\;{\color{Red} {P}}$
- ${\color{DarkBlue} {TU}}\; QSVR\; {\color{Red} {P}}$
- ${\color{Red} {P}}\;{\color{DarkBlue} {TU}}\; QSVR\; $
Here, we can see that car $P$ can be parked at the extreme (either left of right) end in three out of the four possible combinations. That is, though this is not a guarantee, it is still a possibility.
So, the correct answer is $(A).$