A and B play a game such that the loser doubles the money of the winner. A loses
the first game and B loses the next game. If both have Rs 120 now, with how much
money did A start?

Let's break down the problem step-by-step.

### Problem Analysis

Let \( A_0 \) and \( B_0 \) be the amounts of money A and B started with, respectively.

### Step 1: First Game

- A loses the first game, so A must double B's money.
- Initially, A has \( A_0 \) and B has \( B_0 \).
- After the first game:
  - A gives B \( B_0 \) amount of money, so A’s new amount is \( A_0 - B_0 \).
  - B receives this money, so B’s new amount is \( B_0 + B_0 = 2B_0 \).

Now, the amounts are:
- A: \( A_0 - B_0 \)
- B: \( 2B_0 \)

### Step 2: Second Game

- B loses the second game, so B must double A's money.
- After the second game:
  - B gives A \( A_0 - B_0 \) amount of money, so B’s new amount is \( 2B_0 - (A_0 - B_0) \).
  - A receives this money, so A’s new amount is \( (A_0 - B_0) + (A_0 - B_0) = 2(A_0 - B_0) \).

Now, the amounts are:
- A: \( 2(A_0 - B_0) \)
- B: \( 3B_0 - A_0 \)

### Step 3: Given Condition

We are told that after these two games, both A and B have Rs 120.

So, we have the equations:
\[
2(A_0 - B_0) = 120 \quad \text{(1)}
\]
\[
3B_0 - A_0 = 120 \quad \text{(2)}
\]

### Step 4: Solve the Equations

From equation (1):
\[
A_0 - B_0 = 60 \quad \Rightarrow \quad A_0 = B_0 + 60 \quad \text{(3)}
\]

Substitute equation (3) into equation (2

Explore how A and B play a game where the loser doubles the winner's money. Solve the problem where A and B both end with Rs 120, starting with different amounts. Detailed step-by-step solution included. Suitable for advanced high school students.

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