Question

What fraction of the total money did T have at the beginning of the game?

• A. \(\frac13\)
• B. \(\frac15\)
• C. \(\frac29\)
• D. \(\frac15\)
• A. Won \(\frac{1}{12}\)
• B. Won \(\frac{1}{6}\)
• C. Won \(\frac{1}{3}\)
• D. Won \(\frac{1}{5}\)
• A. Rs. 575
• B. Rs. 375
• C. Rs. 825
• D. Rs. 275

Answer 1

The Correct Option is D

Step 1 — Translate the conditions into equations:
Let the money at the start be:
J = money with J,
B = money with B,
T = money with T.

From the problem:
(1) J + B = 4T
(2) T + B = 3J

Step 2 — Express all variables in terms of J:
From (2): T + B = 3J ⇒ T = 3J − B.
Substitute this into (1):
J + B = 4(3J − B).
J + B = 12J − 4B.
Bring terms together: J + B + 4B = 12J ⇒ J + 5B = 12J.
⇒ 5B = 11J − J = 11J − ? Wait, check carefully.
Actually: J + B = 12J − 4B ⇒ 5B = 11J.
So, B = (11/5)J.

Now substitute into T = 3J − B:
T = 3J − (11/5)J = (15/5 − 11/5)J = (4/5)J.

Step 3 — Find total money in terms of J:
Total = J + B + T.
= J + (11/5)J + (4/5)J.
= J + (15/5)J.
= J + 3J = 4J.

Step 4 — Fraction of total money that T had:
T = (4/5)J,
Total = 4J.
Fraction = T / Total = ((4/5)J) / (4J) = (4/5) ÷ 4 = 1/5.

Final Answer:
The correct option is (D): \(\tfrac{1}{5}\).

Answer 2

Step 1 — Start-of-game equations:
Let starting amounts be J, B, T.
Given:
(1) J + B = 4T
(2) T + B = 3J

Step 2 — Express in terms of J:
From (2): T = 3J − B.
Substitute in (1): J + B = 4(3J − B) ⇒ J + B = 12J − 4B ⇒ 5B = 11J ⇒ B = (11/5)J.
Then T = 3J − (11/5)J = (4/5)J.

So start amounts are:
J = S/4, B = 11S/20, T = S/5 (where S = total money = 4J).

Step 3 — End-of-game equations:
Let ending amounts be J′, B′, T′.
Given:
(3) J′ + B′ = 3T′
(4) T′ + B′ = 2J′
Also: B′ = B − 200 = 11S/20 − 200.

Step 4 — Ratios from (3) and (4):
From (3): T′ = (J′ + B′)/3.
Substitute into (4): (J′ + B′)/3 + B′ = 2J′ ⇒ (J′ + 4B′)/3 = 2J′ ⇒ 4B′ = 5J′ ⇒ J′ = (4/5)B′.
Then T′ = (J′ + B′)/3 = ((4/5)B′ + B′)/3 = (9/5)B′ ÷ 3 = (3/5)B′.

So J′ : B′ : T′ = 4 : 5 : 3.
Let common factor = k. Then J′ = 4k, B′ = 5k, T′ = 3k, total S = 12k ⇒ k = S/12.
Hence J′ = S/3, B′ = 5S/12, T′ = S/4.

Step 5 — Use B’s loss:
B′ = 5S/12, B = 11S/20.
Given B lost 200 ⇒ 5S/12 = 11S/20 − 200.
Find common denominator 60:
5S/12 = 25S/60, 11S/20 = 33S/60.
So 25S/60 = 33S/60 − 200 ⇒ (8S/60) = 200 ⇒ S/7.5 = 200 ⇒ S = 1500.

Step 6 — Compute J’s change:
Initially J = S/4 = 1500/4 = 375.
Finally J′ = S/3 = 1500/3 = 500.
Gain = 500 − 375 = 125.

Step 7 — Fraction of total money:
Fraction won = 125 / 1500 = 1/12.

Final Answer:
J won \(\tfrac{1}{12}\) of the total money.
The correct option is (A).

Answer 3

Let the initial amounts J, B, and T had be represented as \( J_0 \), \( B_0 \), and \( T_0 \) respectively. We have the following conditions from the problem:

• \( J_0 + B_0 = 4T_0 \)
• \( T_0 + B_0 = 3J_0 \)

After losing Rs. 200, B has an amount \( B' = B_0 - 200 \). The end-of-day conditions now are:

• \( J_0 + B' = 3T_0 \)
• \( T_0 + B' = 2J_0 \)

We need to solve these equations:

Substituting \( B' \) in \( T_0 + B' = 2J_0 \):

\( T_0 + (B_0 - 200) = 2J_0 \)

\( T_0 + B_0 - 200 = 2J_0 \)

Using \( T_0 + B_0 = 3J_0 \), substitute to get:

\( 3J_0 - 200 = 2J_0 \)

\( J_0 = 200 \)

Substitute \( J_0 = 200 \) into \( J_0 + B_0 = 4T_0 \):

\( 200 + B_0 = 4T_0 \)

And in \( T_0 + B_0 = 3J_0 \):

\( T_0 + B_0 = 600 \)

From \( 200 + B_0 = 4T_0 \), \( B_0 = 4T_0 - 200 \). Substitute into \( T_0 + B_0 = 600 \):

\( T_0 + (4T_0 - 200) = 600 \)

Solving:

\( 5T_0 - 200 = 600 \)

\( 5T_0 = 800 \)

\( T_0 = 160 \)

Now calculate \( B_0 \):

\( B_0 = 4 \times 160 - 200 = 640 - 200 = 440 \)

As the correct answer choice provided was Rs. 825, a reconsideration or alternate scenario check should be noticed. This discrepancy indicates a reassessment might be needed for configuration or solution logic, or perhaps assumptions around how post-loss wealth checks configure in real-time given the existing answer mismatch. Any assumption correction not directly solvable or failure indicates unmatched response reasoning presented in-depth.