We learn that there are six animals that will be assigned to six stalls, with one animal per stall.
There are four lions: F, G, H, and J
There are two tigers: K and M
The stalls are arranged as follows with each one facing exactly one other stall (i.e. 1 faces 4, 2 faces 5, and 3 faces 6).
_ _ _ 1 2 3 _ _ _ 4 5 6
Now let's go through our conditions.
Rule 1 - the tigers' stalls cannot face each other
Therefore, K and M cannot be assigned to 1 and 4, 2 and 5, or 3 and 6.
Rule 2 - a lion must be assigned to Stall 1
Rule 3 - H must be assigned to Stall 6
Rule 4 - J must be assigned to a stall numbered one higher than K's stall
This means that K cannot be in Stall 5 because we know that H is in Stall 6. This also means that J cannot be assigned to Stall 2 because a lion is in Stall 1.
Rule 5 - K cannot be assigned to the stall that faces H's stall
This means that K cannot be assigned to Stall 3. This means that K can be assigned to either Stall 2 or Stall 4. If K is assigned to Stall 2, then M must be assigned to Stall 4.
So, what do we know so far?
L _ _ 1 2 3 _ _ H 4 5 6
Let's go through all the possible scenarios to see if answer choice (A) could be true.
We know that J cannot be in Stall 2.
What if J is in Stall 3?
K must be in Stall 2. If F is in Stall 4, then M would have to be in Stall 5. This violates Rule 1 because we know that the tigers' stalls cannot face each other.
What if J is in Stall 4? K would have to be in Stall 3. However, we know from our deduction made under Rule 5 that K cannot be assigned to Stall 3 because K cannot be assigned to the stall that faces H's stall.
Finally, what if J is in Stall 5? We can see immediately that this will not work because Rule 3 tells us that H must be assigned to Stall 6.
Therefore, it is impossible for F's stall to be numbered one higher than J's stall, making (A) incorrect.
Hope this helps! Please let us know if you have any further questions.
