3 persons A, B and C sit down to play cards. A has Rs. 50 at the start. B has Rs. 40 at the start and C has Rs. 30 at the start. At the beginning of each game, they each bet Rs. 5 and the winner picks up the money at the end of the game. At the end of 4 games, A, B and C have with them Rs. 50, Rs. 40 and Rs. 30 but not necessarily in that order and no person has the same amount he started with. Q1. Who has Rs. 30 at the end? Q2. How much does C end up with ? Q3. If each player ends up with the same amount they started with, find the minimum number of games that would have to be played? __________________
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Answers:
1) (a) A
2) (b) 40
3) (c) 3
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Hi....
Can you tell me the way you have solved this question, because two cases are possible
A-30
B-50
C-40
OR
A-40
B-30
C-50
I am not really sure the about the way to solve it.......
Thanks.......
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3 persons A, B and C sit down to play cards. A has Rs. 50 at the start. B has Rs. 40 at the start and C has Rs. 30 at the start. At the beginning of each game, they each bet Rs. 5 and the winner picks up the money at the end of the game. At the end of 4 games, A, B and C have with them Rs. 50, Rs. 40 and Rs. 30 but not necessarily in that order and no person has the same amount he started with.
Q1. Who has Rs. 30 at the end?
a. A
b. B
c. C
d. Cannot be determined
Q2. How much does C end up with ?
a. Rs. 50
b. Rs. 40
c. (a) or (b)
d. Cannot be determined.
Q3. If each player ends up with the same amount they started with, find the minimum number of games that would have to be played?
a. 4
b. 5
c. 3
d. 6
Sol: As stated in question , A start with Rs. 50,B with Rs. 40 and C with Rs. 30
Also at the end of 4 games, A, B and C have with them Rs. 50, Rs. 40 and Rs. 30 but not necessarily in that order and no person has the same amount he started with.
So, A can have either Rs. 30 or 40,B can have either Rs. 30 or 50 and C can have Rs.40 or 50.
Each winning of the game can fetch the winner Rs. 10 and each loosing can let to a loss of Rs. 5.
Clearly A can not win more than 1 time as then its total amount will be greater than 50 which is not there (as per question).
Now the there can be 4 conditions with A in the four games.
Four win= Rs.90 (not possible)
Three win and one loose=Rs.85(not possible)
Two win and two loose=Rs.60 (not possible)
One win and three loose=Rs.45( not possible)
All loose =Rs.30( only possible condition)
Now as A will have Rs. 30 at the end,B will obviously Rs. 50 at the end.
So now let the number of win be x and the number of loose be (4-x) also 1 win =Rs. 10 and 1 loose =Rs.(-5)
So the equation becomes, -5x + 10 (4-x)=10
Solving, x=2 i.e, number of win = 2 and then obviously the number of loose= 2.(BTW there is no requirement to find the number of loose and win).
Anyways, now the amount at the end with C will be Rs.40
So, again two loose and two win can get him Rs.40.
Now solving the question :
1) A end up with Rs.30 at the end.
So, a) A
2) C end up with Rs.40
b) Rs. 40
3) For each player ending with the same amount as they at the beginning.
The minimum number of games=3(as 2 loose and 1 win will be equal to the beginning amount)
c) 3
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Thanks for the solution
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