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by karthikpandian19 » Wed Jan 04, 2012 9:50 pm
Among 60 members of a club, 6p players
play soccer, 11p players play tennis, 8p
players play golf and 2p players play
none of the games. If p players play all of
the games, how many players play only
one game?
(1) The number of players who play
soccer and golf but not tennis is half the
number of players who play any other
combination of two sports
(2) p = 3
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Source: — Data Sufficiency |

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by Anurag@Gurome » Wed Jan 04, 2012 11:11 pm
karthikpandian19 wrote:Among 60 members of a club, 6p players play soccer, 11p players play tennis, 8p players play golf and 2p players play none of the games. If p players play all of the games, how many players play only one game?
(1) The number of players who play soccer and golf but not tennis is half the number of players who play any other combination of two sports
(2) p = 3
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(5p - x - y) + (10p - x - z) + (7p - y - z) + x + y + z + p = 60
23p - x - y - z = 60
We have to find (5p - x - y) + (10p - x - z) + (7p - y - z).

(1) y = (1/2) * (x + z); NOT sufficient.
(2) p = 3 implies 23(3) - x - y - z = 60 or x + y + z = 9
So, we can find the value of (5p - x - y) + (10p - x - z) + (7p - y - z); SUFFICIENT.

The correct answer is B.
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by karthikpandian19 » Thu Jan 05, 2012 9:30 pm
Anurag@Gurome wrote:
(5p - x - y) + (10p - x - z) + (7p - y - z) + x + y + z + p = 60
23p - x - y - z = 60
We have to find (5p - x - y) + (10p - x - z) + (7p - y - z).
Can you explain this portion of the equation?
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by amit.lohchab » Fri Jan 06, 2012 10:26 am
Participants who play only one game is asked in question so check out the venn diagram which will provide you this equation.
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