At the start of a game of cards, J and B together had four times as much money as T, while T and B together had three times as much as J. At the end of the evening, J and B together had three times as much money as T, while T and B together had twice as much as J, B lost ₹ 200. What fraction of the total money did J win/lose?

### Concept & Tracking Shares of a Total To find a win/loss fraction, calculate the player's initial fraction of the total pool and their final fraction of the total pool, then find the difference. $$\text{Difference} = \text{Final Fraction} - \text{Initial Fraction}$$ ### Step-by-Step Solution * Let the total money be $M$. * **Find J's starting fraction:** T and B together had three times as much as J. * $T + B = 3J$. Since $J + B + T = M$, substituting gives $3J + J = M \Rightarrow 4J = M \Rightarrow J_{start} = \frac{M}{4}$. * **Find J's ending fraction:** At the end, T and B together had twice as much as J. * $T + B = 2J$. Substituting into the total gives $2J + J = M \Rightarrow 3J = M \Rightarrow J_{end} = \frac{M}{3}$. * **Calculate the difference:** J's money changed from $\frac{1}{4}$ of the total to $\frac{1}{3}$ of the total. * Change $= J_{end} - J_{start} = \frac{M}{3} - \frac{M}{4}$. * Find a common denominator (12): $\frac{4M}{12} - \frac{3M}{12} = \frac{M}{12}$. * Because the ending fraction is larger than the starting fraction, J won $\frac{1}{12}$ of the total money. ### Exam Strategy & Shortcut Use the rule $(A+B) = k \times C \Rightarrow C = \frac{1}{k+1}$ of the total. Start: $T+B = 3J \Rightarrow J$ is $\frac{1}{4}$. End: $T+B = 2J \Rightarrow J$ is $\frac{1}{3}$. The difference is $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$. Since $\frac{1}{3} > \frac{1}{4}$, it is a win. ### Common Pitfall Confusing the ratios and calculating for the wrong person, or assuming the total money $M$ changed during the game (it remains constant in closed-system card games). ### Final Answer Therefore, the correct answer is **Won 1/12**.