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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by sukh » Sat Jul 23, 2011 12:39 am
Let M be the set of all students on the mathematics teamand C be the set of all members on the chess team . Exactly 17 students are on atleast one of the teams and exactly 7 students are on both teams , If M has twice as many members as C , exactly how many students are members of M, but not members of C ? a) 1 b)7 c) 8 d) 9 e) 16
why it should not be solved using conventional set formulas
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by cans » Sat Jul 23, 2011 12:50 am
atleast one team = 17
on both = 7
thus only on M + only on C = 10 (=17-7)
M=2C. To Find: only M
M = only M + both & C = only C + both ->
M = only M + 7 & C = only 7 + 7
only M + 7 = 2* (only C) + 14 -> only M = 2*(only C) + 7 = 10 - (only C)
=> only C = 1 and thus only M = 9
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