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There are 30 students in Mr. Peterson’s gym class. 14 of the

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sandy
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There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink] New post   26 Mar 2018, 15:53
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There are 30 students in Mr. Peterson’s gym class. 14 of them play basketball, 13 play baseball, and 9 play neither basketball nor baseball.

Quantity A
Quantity B
The number of students who play both basketball and baseball
6


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

Drill 2
Question: 4
Page: 549-550
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Official Answer
C
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HEcom
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink] New post   29 Mar 2018, 07:02
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...
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink] New post   29 Mar 2018, 13:50
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sandy wrote:
There are 30 students in Mr. Peterson’s gym class. 14 of them play basketball, 13 play baseball, and 9 play neither basketball nor baseball.

Quantity A
Quantity B
The number of students who play both basketball and baseball
6


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

Drill 2
Question: 4
Page: 549-550


There are 30 sttudents of whom 9 play neither sports so we have a pool of 21 students (30-9) who play basketbal and baseball. Adding the number of students who play basketball and baseball is 27 (13+14). Based on this, we can conclude 6 students (27-21) are playing both sports.

Answer D
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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink] New post   29 Mar 2018, 13:58
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] New post   30 Mar 2018, 01:05
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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] New post   04 Apr 2018, 16:00
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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] New post   16 Aug 2021, 18:26
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: There are 30 students in Mr. Peterson’s gym class. 14 of the [#permalink]
16 Aug 2021, 18:26
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