Math Problem Statement

A, B, C, and D play dice games where the loser doubles the amount of money of all other players. Four games are played in all, and A, B, C, and D lose these games in this sequence only. Each player has Rs. 96 at the end of the fourth game. What is the amount with C at the end of the third game?

Solution

Let's solve the problem step by step.

Understanding the Problem

  • Initial Condition: All players have an equal amount of money initially. Let's denote this amount by xx.
  • Game Rules:
    • The loser doubles the amount of money of all other players.
    • Four games are played, and the losers are A, B, C, and D, respectively.
  • Final Condition: Each player has Rs. 96 at the end of the fourth game.

Step-by-Step Calculation

After Game 1:

  • Loser: A
  • Other Players' Money Doubles:
    • If A loses, B, C, and D's money doubles.
  • Equations: 2(B+C+D)=3x(since B, C, and D’s money doubles)2(B + C + D) = 3x \quad \text{(since B, C, and D's money doubles)} A's money is x3x=2xx - 3x = -2x after doubling others' money, but A had to match the others' gain, so A would end up with the same amount as each other player.

So after the first game, the amount with each player will be:

A=x3x+x=x(due to matching others’ gain),B=2x,C=2x,D=2x.A = x - 3x + x = x \quad \text{(due to matching others' gain)}, \quad B = 2x, \quad C = 2x, \quad D = 2x.

After Game 2:

  • Loser: B
  • Other Players' Money Doubles:
    • A, C, and D’s money doubles.
  • Equations: A=2x,C=4x,D=4x,B=x2A = 2x, \quad C = 4x, \quad D = 4x, \quad B = \frac{x}{2}

After Game 3:

  • Loser: C
  • Other Players' Money Doubles:
    • A, B, and D’s money doubles.
  • Equations: A=4x,B=2x,D=8x,C=x4A = 4x, \quad B = 2x, \quad D = 8x, \quad C = \frac{x}{4}

After Game 4:

  • Loser: D
  • Other Players' Money Doubles:
    • A, B, and C’s money doubles.
  • Equations: A=8x,B=4x,C=x2,D=x8A = 8x, \quad B = 4x, \quad C = \frac{x}{2}, \quad D = \frac{x}{8}

Given that after the fourth game, each player has Rs. 96:

8x=96    x=128x = 96 \implies x = 12

Calculation of Amount with C After the Third Game

After the third game, the amount with C was:

x4=124=3\frac{x}{4} = \frac{12}{4} = 3

Thus, after the third game, C had Rs. 3.

Answer:

The amount with C at the end of the third game was Rs. 3.


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Related Questions:

  1. What was the amount with B at the end of the second game?
  2. How much money did D have after losing the second game?
  3. What was the initial amount each player had before the games started?
  4. How much money did each player have after the first game?
  5. What would have been the result if B lost the first game instead of A?

Tip: To solve problems involving sequential changes like this, always track the changes step by step for each participant after every round or event.

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Math Problem Analysis

Mathematical Concepts

Arithmetic
Game Theory
Equations

Formulas

Doubling principle

Theorems

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Suitable Grade Level

Grades 7-9