Two-game money transfer scenario: A has twice as much money as B. After game 1, B wins one-third of A’s initial money from A. In game 2, what fraction of B’s then-current money must A win back so that they end up with equal amounts?

Introduction / Context:This problem requires careful tracking of money transfers in two stages. It tests proportional reasoning and equation setup: first, compute each person’s holdings after the first transfer; then, determine the fraction of B’s new amount that must be transferred back to A for equality after the second game.

Given Data / Assumptions:

• Initially A = 2k, B = k.
• Game 1: B wins (1/3) of A’s initial money, which is (1/3)*2k = 2k/3.
• After Game 1: A = 2k − 2k/3 = 4k/3; B = k + 2k/3 = 5k/3.
• Game 2: A wins back a fraction f of B’s current money (5k/3).

Concept / Approach:Let the transfer in Game 2 be f*(5k/3) from B to A. Set the final amounts equal and solve for f. The ratio relationship 4k/3 vs 5k/3 makes the algebra neat and the result a simple fraction.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• 1/3, 1/4, 1/5: Lead to unequal final amounts when substituted.

Common Pitfalls:

• Taking one-third of A’s current (post-loss) money instead of A’s initial money.
• Forgetting that the second transfer depends on B’s post–Game-1 amount.

Final Answer: