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 amount did B start with?

• A. Rs. 575
• B. Rs. 375
• C. Rs. 825
• D. Rs. 275

Hint:

Express B's beginning and ending share in terms of T1 using the given ratios, then use B's Rs. 200 loss to solve for T1.

Answer 1

The Correct Option is C

Solve this directly with algebra on the four given conditions, using T1 as the base variable instead of assuming a total of 60x.

1. Rs. 575: does not satisfy the beginning and ending equations together with the Rs. 200 loss.
2. Rs. 375: this is actually J's starting amount, not B's, so it answers the wrong quantity.
3. Rs. 825: this is confirmed by the direct algebraic solve below.
4. Rs. 275: too small to satisfy $J1+B1=4T1$ once T1 is solved.

From $J1+B1=4T1$, we get $J1 = 4T1 - B1$. Plug into $T1+B1=3J1$: $T1+B1 = 3(4T1-B1) = 12T1 - 3B1$, so $4B1 = 11T1$, giving $B1 = \frac{11}{4}T1$.

Similarly at the end, from $J2+B2=3T2$ and $T2+B2=2J2$: $J2=3T2-B2$, so $T2+B2=2(3T2-B2)=6T2-2B2$, giving $3B2=5T2$, so $B2=\frac{5}{3}T2$.

Since total money is conserved, $5T1 = 4T2$ (both routes give the same fixed total M), so $T2 = \frac{5}{4}T1$. Then $B2 = \frac{5}{3} \times \frac{5}{4}T1 = \frac{25}{12}T1$.

B's loss: $B1-B2 = \frac{11}{4}T1 - \frac{25}{12}T1 = \frac{33-25}{12}T1 = \frac{8}{12}T1 = \frac{2}{3}T1 = 200$, so $T1 = 300$.

Then $B1 = \frac{11}{4}(300) = Rs.\ 825$.

Let's summarize:

• Beginning: $B1 = \frac{11}{4}T1$.
• Ending: $B2 = \frac{25}{12}T1$.
• Loss condition gives $T1=300$, so $B1 = 825$.

This confirms option C through direct substitution instead of the 60x shortcut.