Instructions

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

Solution

Let the initial amount with them be

J1 , B1 and T1

Now as per given

J1+B1 =4T1 (1)

T1+B1 = 3J1 (2)

Total amount with them will be 4J1 or 5T1

Now Let final amounts with them be J2 , B2 and T2 respectively .

J2 +B2 = 3T2 (3)

T2 +B2 = 2J2 (4)

Total amount with them will be 4T2 or 3J2

Now 5T1 =4T2 =4J1=3J2

From (1)

B1 = 4T1-J1

= 4T1-5/4 T1

= 11/4 T1

B2 = 3T2 -J2

=3T2 -4/3T2

=5/3 T2

5/3 (5/4) T1

= 25/12 T1

Now B1-B2 = 11/4 T1 -25/12T1 =200

so we get 8/12 T1 = 200

we get T1 = 300

so B1 = 11/4 (300) = 825

B2 = 25/12 (300) = 625

T2 = 5/4(300) = 375

J1 = 375

J2 = 4/3 (375) = 500

Total money = 825+375+300 = 1500

J won : 500-375 = 125

So fraction = 125/1500 = 1/12

**Alternative solution:**

It is given that the J and B had 4 times the money as T(in the beginning). Thus, if T had x amount with him, the total amount with them will be 5x.

So, let's assume that the total amount with them is 60x(LCM of 5, 4, and 3).

**In the beginning**: Now, the amount with T will be 12x, and the sum of the amount with J and B will be 48x.

Since the amount with T and B is 3 times that with J, the amount with J will be 15x. Thus, the amount with B will be 33x.

**At the end**:
Since the amount with J and B is three times that of T, T will have
15x at the end. Also, the amount with T and B is twice that of J; the
amount with J will be 20x. Thus, B will have 25x at the end.

Now, the difference between the money with J in the beginning and at the end = $$20x-15x=5x$$

Thus, the required ratio = $$\frac{5x}{60x}=\frac{1}{12}$$

Thus, the correct option is A.

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