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:
After Game 1: A1 = 4k/3, B1 = 5k/3.Game 2 transfer: T = f * B1 = f * (5k/3).Final amounts: A2 = A1 + T; B2 = B1 − T.Set equality: A1 + T = B1 − T ⇒ A1 = B1 − 2T.Substitute: 4k/3 = 5k/3 − 2f*(5k/3) ⇒ 4/3 = 5/3 − (10/3)f.Solve: (10/3)f = 5/3 − 4/3 = 1/3 ⇒ f = (1/3) * (3/10) = 1/10.
Verification / Alternative check:
With f = 1/10, T = (1/10)*(5k/3) = k/6. A2 = 4k/3 + k/6 = 9k/6 = 3k/2; B2 = 5k/3 − k/6 = 9k/6 = 3k/2: equal.
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:
1/10
Discussion & Comments