· Each person has initially 10 rupees.

· We can mark the possible values to be used in the blanks, these possible values are between minimum and maximum, and excluding value 10 rupees.

· Since, one loses the bet and another gains the same amount within the pair, or the pair does not participate. This implies that each round’s total of the 4 persons will be 40 only.

· Round 1: Q had (-2), R and S remained same. So, P must have had (+2) = 12.

· Round 2: P had (+1) from 12, Q had (+2), S had (-2). So, R must have (-1) = 9.

· Round 4: P and Q had same money in this round. Only possible amount is 11.

· Round 5: Total of the 4 persons must be 40. R must have value = 40 – 10 – 10 – 13 = 7.

	Pulak	Qasim	Ritesh	Suresh	Pairs
Round 1	12	8	10	10	P,Q
Round 2	13	10	9	8	(P,R) (Q,S)
Round 3				10
Round 4	11	11
Round 5	10	10	7	13
Round 6
Round 7		12	4
Round 8	13			10
Possible Values	11/12	9/11	5/6/7/8/9	9/11/12

· Round 5: P and Q, both had (-1). So, R and S must have (+1), with values in round 4: (R=6, S=12).

· Round 3: R must have value=8 or 7, so that in round 4, R can have value=6, after losing bet.

To find the values of P, Q, and R, we can make 2 cases for Q:

§ Case 1: Q’s value=9. To get the sum=40 in round 3, even if we chose the max possible for P and R (P=12, R=8), then the sum will be = 12+9+8+10 = 39. Therefore, this case is not possible.

§ Case 2: Q’s value=11, after (+1). Q’s amount is unchanged from round 3 to round 4, so there must be another person whose amount remains same from round 3 to round 4.

This person can only be P with value=11, after (-2), as R’s value cannot be 6 in round 3, and we know that S’s value has changed.

S had (+2). Q had (+1). P had (-2). Therefore, R’s value in this case must be 8, after (-1).

· Round 6: Q cannot have value=9, as after winning or losing a bet, he can only lose/win 1 or 2 rupees, not 3 rupees.

Therefore, Q’s value in round 6 must be 11.

	Pulak	Qasim	Ritesh	Suresh	Pairs
Round 1	12	8	10	10	P,Q
Round 2	13	10	9	8	(P,R) (Q,S)
Round 3	11	11	8	10
Round 4	11	11	6	12
Round 5	10	10	7	13
Round 6		11
Round 7		12	4
Round 8	13			10
Possible Values	11/12	9/11	5/6/7/8/9	9/11/12

· Round 8: Q cannot have value=9, from 12 in round 7, as bet cannot be of amount=3.

Therefore, Q’s value in round 8 = 11. This implies that R’s value must be =6 (40-13-11-10).

· Round 7: Q had (-1) from round 7 to round 8, and R had (+2). Therefore, there must be (+1) and (-2) for other persons. Since, P cannot have amount=15, in round 7 for (-2) to happen in round 8. Therefore, (-2) must have happened to S, with amount in round 7 = 12. P’s value in round 7 = 12, so that (+1) can be done in round 8.

	Pulak	Qasim	Ritesh	Suresh	Pairs
Round 1	12	8	10	10	P,Q
Round 2	13	10	9	8	(P,R) (Q,S)
Round 3	11	11	8	10	(P,S) (Q,R)
Round 4	11	11	6	12	(R,S)
Round 5	10	10	7	13
Round 6		11
Round 7	12	12	4	12
Round 8	13	11	6	10
Possible Values	11/12	9/11	5/6/7/8/9	9/11/12

· Round 6: From round 6 to round 7, Q had (+1). For the opposite (-1) to happen, it is only possible for R with value in round 6 = 5. This is because in round 6, P and S cannot have value=13, for (-1) to happen.

· Round 5 to round 6: Q had (+1), and R had (-2). Other persons must have (-1) and (+2). Since P cannot have value in round 6 = 9, for (-1) to happen, therefore, P must have (+2) for value in round 6 = 12. S must have (-1) for value in round 6 = 12.

	Pulak	Qasim	Ritesh	Suresh	Pairs
Round 1	12	8	10	10	P,Q
Round 2	13	10	9	8	(P,R) (Q,S)
Round 3	11	11	8	10	(P,S) (Q,R)
Round 4	11	11	6	12	(R,S)
Round 5	10	10	7	13
Round 6	12	11	5	12	(P,R) (Q,S)
Round 7	12	12	4	12
Round 8	13	11	6	10
Possible Values	11/12	9/11	5/6/7/8/9	9/11/12

For Round 5: In round 6, pairs are (P,R) (Q,S). Therefore, these cannot be repeated in round 5, as it is given in the set info.

Therefore, P (-1 in round 5) can only pair with S (+1 in round 5). Answer is option(c): Pulak and Suresh

CAT