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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink]
HEcom wrote:
Whaat.... The answer should be "B"....The number of students who play both basketball and baseball should be 3... these three students get counted with 10 students who play baseball only and again with 11 students who play basketball only...


You are neglecting the fact that 9 students dont play either sports. You cannot add 13 to 14 because inside the pool of 13 students who play baseball there are a few who play basketball also and vice and versa.
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink]
1
total o of students n(U)=30
no of students playing basketball n(Bk)=14
no of students playing baseball n(Bb)=13
playing neither of them n(Bk U BB)'=9

let x be the no of students playing both,
then
n(U)= n(Bk)+n(Bb)-N(Bk & Bb)+ n(Bk U Bb)' & means intersection of Bk and Bb
30=14+13-x+9
x=6

ANS: C
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink]
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Explanation

Use the group formula and fill in what you know.

So Total = Group1 + Group2 – Both + Neither becomes 30 = 14 + 13 – Both + 9. So Both = 6, and the answer is choice (C).
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink]
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