In an arithmetic reasoning puzzle, two examination rooms P and Q contain some students. If 10 students are sent from room P to room Q, the numbers in both rooms become equal. If instead 20 students are sent from room Q to room P, then the number of students in P becomes double the number of students in Q. How many students are originally in room Q?

Introduction / Context:
This problem is a typical linear equation word problem involving two groups of students and transfers between them. We are given two different hypothetical movements of students between the rooms and told how the numbers change in each scenario. The goal is to set up algebraic equations describing each situation and then solve the system to find the original number of students in room Q.

Given Data / Assumptions:
- Let P be the original number of students in room P.
- Let Q be the original number of students in room Q.
- Scenario 1: When 10 students move from P to Q, the numbers become equal in both rooms.
- Scenario 2: When 20 students move from Q to P, the number in P becomes double the number in Q.
- No students enter or leave the building; only movement between the two rooms is considered.

Concept / Approach:
We express each scenario as an equation. In the first scenario, the new population of P is P - 10 and the new population of Q is Q + 10, and these two are equal. In the second scenario, the new population of P is P + 20 and the new population of Q is Q - 20, and now P + 20 is twice Q - 20. Solving these two equations simultaneously gives us P and Q. Finally, we select the option that matches the value of Q.

Step-by-Step Solution:
Step 1: From scenario 1, when 10 students move from P to Q, we have P - 10 = Q + 10. Step 2: Rearranging gives P - Q = 20. This is our first equation. Step 3: From scenario 2, when 20 students move from Q to P, we have P + 20 = 2(Q - 20), because the number in P becomes double the number in Q. Step 4: Expand the second equation: P + 20 = 2Q - 40, so P - 2Q = -60. Step 5: Now we solve the system: (1) P - Q = 20 and (2) P - 2Q = -60. Step 6: Subtract equation (2) from equation (1): (P - Q) - (P - 2Q) = 20 - (-60) gives Q = 80. Step 7: Substitute Q = 80 into P - Q = 20 to get P - 80 = 20, so P = 100.

Verification / Alternative check:
Check scenario 1 with P = 100 and Q = 80: after sending 10 from P to Q, room P has 90 and room Q has 90, so they are equal. Check scenario 2: after sending 20 from Q to P, room P has 120 and room Q has 60. Indeed, 120 is twice 60, so the second condition is satisfied as well. This confirms that Q = 80 is correct.

Why Other Options Are Wrong:
Option 100 or 70 or 120 or 90: Substituting any of these values for Q into the equations does not satisfy both scenarios simultaneously. Either the equal-student condition fails or the double-condition fails, so these values cannot be correct.

Common Pitfalls:
A common error is mixing up which room gains and which loses students in each scenario, leading to equations with wrong signs. Another mistake is misinterpreting double as P = Q + 20 instead of P = 2Q. Carefully writing updated counts for each room and explicitly forming the equations before solving helps avoid these mistakes.

Final Answer:
The number of students originally in room Q is 80.