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

Work this out using real rupee numbers from the start, pinned down by the fact that B lost Rs. 200, then check which option matches.

1. Won 1/12: matches a gain worked out from real numbers below.
2. Lost 1/6: has the wrong direction, J's money actually goes up, not down.
3. Lost 1/3: also the wrong direction and too large a change.
4. Won 1/5: this is T's opening share of the total, not J's change, so it answers a different question.

Using total $60x$: beginning shares are T1=12x, J1=15x, B1=33x; ending shares are T2=15x, J2=20x, B2=25x. B's change is $33x - 25x = 8x$, a loss. Since B lost Rs. 200, $8x = 200$, so $x = 25$.

Then J1 = 15(25) = Rs. 375 and J2 = 20(25) = Rs. 500. J's change is $500 - 375 = Rs.\ 125$, a gain. The total money is $60x = 1500$, so J's gain as a fraction of total is $\frac{125}{1500} = \frac{1}{12}$.

Let's summarize:

• B's loss of Rs. 200 pins down $x = 25$.
• J rises from Rs. 375 to Rs. 500, a Rs. 125 gain.
• $125/1500 = 1/12$, so J won one-twelfth of the total.

This confirms option A using concrete rupee values instead of symbolic fractions alone.