Question

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 Rs. 200.

What fraction of the total money did J win/lose?

• A. Won \( \dfrac{1}{12} \)
• B. Lost \( \dfrac{1}{6} \)
• C. Lost \( \dfrac{1}{3} \)
• D. Won \( \dfrac{1}{5} \)

Hint:

Find J's starting and ending share in terms of a common total, then see whether the share rises or falls.

Answer 1

The Correct Option is A

Step 1: Express the starting shares using a common multiple.
Let the total money be \(60x\) (a common multiple of 5, 4, 3). Starting shares work out to \(T1 = 12x\), \(J1 = 15x\), \(B1 = 33x\) (from \(J+B=4T\) and \(T+B=3J\)).

Step 2: Express the ending shares the same way.
At the end, \(J2+B2 = 3T2\) and \(T2+B2=2J2\), and the total is still \(60x\). This gives \(T2 = 15x\), \(J2 = 20x\), \(B2 = 25x\).

Step 3: Find J's change.
J went from \(15x\) to \(20x\), a gain of \(5x\). As a fraction of the total: \( \dfrac{5x}{60x} = \dfrac{1}{12}\). Since this is positive, J won money.

Step 4: Verify with actual rupee values.
Using \(x = 25\) (found from B's Rs. 200 loss), J1 = 375, J2 = 500, gain = Rs. 125, and \( \dfrac{125}{1500} = \dfrac{1}{12}\), matching Step 3.

Step 5: Rule out the other options.
Lost 1/6 and Lost 1/3 have the wrong sign, since J actually gained. Won 1/5 is T's starting fraction from the earlier question, not J's change.

Final Answer:
J won one-twelfth of the total money. \[ \boxed{\text{Won } \dfrac{1}{12}} \]