The stake (in Rs) in the bet between B and C is
The stake (in Rs) in the bet between B and C is
The correct option is B 5
From the question, none of them win / loss equal number of bets. Thus, the number of wins are 4,3,2,1,0 in some order . It is also known that The number of bets won by D is equal to the number of bets lost by A. thus the only possibilities for this are
A D Wins (case 1)13Wins (case 2)04 But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0
A wins 1 bet and D wins 3. Thus C wins 0. Also,
It is also known that
PersonNo. of WinsNo. of LossesA13B40C04D31E22
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows
In the table above every + means a gain and every - means a loss.
For example: +17 in the last row means, E won 17 from A and similarly in the first row -17 means A lost 17 to E.
Taking into account the following conditions:
(vii) E won Rs.17 against A and the stake involved in the bet between C and D is Rs.4. The gain of A is equal to the loss of D.
And we already know that A has only one gain (against C) and D only one loss (against B). Lets assume that value to be x.
Now, we know that
(iii)The stake of any bet is an integer amount (in Rs) and is neither less than Rs.4 nor more than Rs.21
(iv)Neither the gain of any person who won atleast one bet nor the loss of any person who lost at least one bet isless than Rs.20 or more than Rs.40.
Hence we can write
Eqn1: 4 ≤ x ≤ 21
Eqn2: 20 ≤ x ≤40
From the above 2 eqns we get: 20 ≤ x ≤ 21
So x can have a value of either 20 or 21.
Step 2
Now take stake (A vs D) = y
In the first row, the sum of all the stakes (neglecting them as gains or losses) should be equal to the total stake i.e. 55. So I get stake(A vs B) = (38-y). Similarly we can get, stake(D vs E) = 51-(x+y)
Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).
Now we have all the cells filled in the table
ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53 Step 3
Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53
⇒ 103 - 2 (x + z) = 53
⇒ 2 (x + z) = 50
⇒ x + z = 25
Now z can have two values 4 or 5. But, (ii)For any person the stakes of any 2 bets were different. And since C has 2 stakes - both “z” and “4”, therefore z is not equal to 4.
z = 5, and hence x = 20.
This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5
Now total loss of A = 38 - (x + y) + y + 17
=55 - x
=55 - 20 = 35
From the question, none of them win / loss equal number of bets. Thus, the number of wins are 4,3,2,1,0 in some order . It is also known that The number of bets won by D is equal to the number of bets lost by A. thus the only possibilities for this are
A D Wins (case 1)13Wins (case 2)04But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0
A wins 1 bet and D wins 3. Thus C wins 0. Also,
It is also known that
PersonNo. of WinsNo. of LossesA13B40C04D31E22
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows
In the table above every + means a gain and every - means a loss.
For example: +17 in the last row means, E won 17 from A and similarly in the first row -17 means A lost 17 to E.
Taking into account the following conditions:
(vii) E won Rs.17 against A and the stake involved in the bet between C and D is Rs.4. The gain of A is equal to the loss of D.
And we already know that A has only one gain (against C) and D only one loss (against B). Lets assume that value to be x.
Now, we know that
(iii)The stake of any bet is an integer amount (in Rs) and is neither less than Rs.4 nor more than Rs.21
(iv)Neither the gain of any person who won atleast one bet nor the loss of any person who lost at least one bet isless than Rs.20 or more than Rs.40.
Hence we can write
Eqn1: 4 ≤ x ≤ 21
Eqn2: 20 ≤ x ≤40
From the above 2 eqns we get: 20 ≤ x ≤ 21
So x can have a value of either 20 or 21.
Step 2
Now take stake (A vs D) = y
In the first row, the sum of all the stakes (neglecting them as gains or losses) should be equal to the total stake i.e. 55. So I get stake(A vs B) = (38-y). Similarly we can get, stake(D vs E) = 51-(x+y)
Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).
Now we have all the cells filled in the table
ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53Step 3
Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53
⇒ 103 - 2 (x + z) = 53
⇒ 2 (x + z) = 50
⇒ x + z = 25
Now z can have two values 4 or 5. But, (ii)For any person the stakes of any 2 bets were different. And since C has 2 stakes - both “z” and “4”, therefore z is not equal to 4.
z = 5, and hence x = 20.
This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5
Now total loss of A = 38 - (x + y) + y + 17
=55 - x
=55 - 20 = 35
