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 T have at the beginning of the game?

• A. \( \dfrac{1}{3} \)
• B. \( \dfrac{1}{8} \)
• C. \( \dfrac{2}{9} \)
• D. \( \dfrac{1}{5} \)

Hint:

Add the given relation J+B = 4T to T itself: total = 4T + T = 5T, so T is one-fifth of the total straight away.

Answer 1

The Correct Option is D

Instead of jumping to the shortcut, build the full picture using a common multiple so every share comes out as a whole number.

1. 1/3: this would mean T has close to as much as J and B combined, which does not match the given 4 times relation.
2. 1/8: too small a share for T given the ratios in the problem.
3. 2/9: close in size but does not come from the actual working of the ratio conditions.
4. 1/5: this is the value that the full working below produces.

Since T and B together are 3 times J, and J and B together are 4 times T, let the total money be a convenient multiple of 5, 4, and 3, say 60x. Because J+B is 4 times T, and J+B+T is the whole 60x, T's share solves as $60x = 4T + T = 5T$, so $T = 12x$. Then J+B = 48x. Using T+B = 3J: substitute B = 48x - J, giving $12x + 48x - J = 3J$, so $60x - J = 3J$, meaning $J = 15x$, and then $B = 48x - 15x = 33x$.

T's share of the total is $\frac{12x}{60x} = \frac{1}{5}$.

Let's summarize:

• Total assumed as 60x for clean whole-number shares.
• T = 12x, J = 15x, B = 33x, all adding to 60x.
• T's fraction of total: $12x/60x = 1/5$.

This confirms option D through the full simultaneous-equation build rather than the shortcut.