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

Step 1: Set up the beginning relations.
Let J1, B1, T1 be the starting amounts. Given: \(J1 + B1 = 4T1\) and \(T1 + B1 = 3J1\). Since the total money \(M = J1+B1+T1\), the first relation gives \(M = 4T1 + T1 = 5T1\), so total money is always 5 times T's starting amount.

Step 2: Solve for T1's fraction directly.
From \(M = 5T1\), we get \( \dfrac{T1}{M} = \dfrac{1}{5}\) right away, without even needing the second equation for this particular question.

Step 3: Cross-check with actual numbers.
Using the full solved system (from the other two conditions and B losing Rs. 200), the actual values turn out to be T1 = 300, J1 = 375, B1 = 825, giving total \(M = 1500\). Then \( \dfrac{T1}{M} = \dfrac{300}{1500} = \dfrac{1}{5}\), matching Step 2.

Step 4: Rule out the other options.
1/3, 1/8, and 2/9 do not equal 300/1500; they come from mixing up which pair sums to which multiple.

Final Answer:
T started with one-fifth of the total money. \[ \boxed{\dfrac{1}{5}} \]