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150 students at seward high school. 66 play baseball, 45 [#permalink]
24 Jul 2007, 20:36

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Difficulty:

5% (low)

Question Stats:

0% (00:00) correct
100% (00:03) wrong based on 1 sessions

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Re: PS: high school sports [#permalink]
25 Jul 2007, 00:15

ArvGMAT wrote:

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Why have you added the triple? I think you should minus two times tripple. What do you think?
In the condition we have:
"27 play exactly 2 sports and 3 play all 3 sports"
I think that in 27 we don't have triple.

Re: PS: high school sports [#permalink]
25 Jul 2007, 17:34

ArvGMAT wrote:

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.

Re: PS: high school sports [#permalink]
25 Jul 2007, 20:16

Robin in NC wrote:

ArvGMAT wrote:

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.

OMG. This looks correct too. I think the number I found in the first post is actually the no. of people who play just one sport. I should have added the rest too.

Re: PS: high school sports [#permalink]
26 Jul 2007, 15:09

Robin in NC wrote:

ArvGMAT wrote:

150 students at seward high school. 66 play baseball, 45 basketball and 42 soccer. 27 play exactly 2 sports and 3 play all 3 sports. How many of the 150 play none of the 3 sports?

Using a Venn diagram, there are 3 areas that overlap 2 sports, and 1 area that overlaps all three sports. Put 9 into each of the "double" areas (for the 27 that play 2 sports), and 3 into the center of the diagram. (Assuming that the 27 are equally divided among the sports, but that really doesn't matter). Then you can see that for each sport, you need to subtract 21. That means that, of the students that only play one sport, you have 45 in baseball, 24 in basketball, and 21 in soccer. That, plus the 27 (2 sports) and 3(3 sports)= 120 students that play sports. So 30 don't.

you minus the doubles cause you have counted them twice, once when we are given the number of basket ball player, football etc..you add the triple cause triple +single -double will give you the total number of players...

try it with a simple example..

Andrey2010 wrote:

Hi fresinha12

Why have you added the triple? I think you should minus two times tripple. What do you think? In the condition we have: "27 play exactly 2 sports and 3 play all 3 sports" I think that in 27 we don't have triple.

[quote="ywilfred"]Set up a venn diagram with the following: #(all three) = 3 #(basketball and baseball) = x #(basketball and soccer) = y #(soccer and baseball) = z

So x+y+z = 27

#(none) = n

63-x-z + 42-x-y + 39-y-z + x + y + z + 3 + n= 150 147-x-y-z+n= 150 147 - (x+y+z) + n = 150 147 - (27) + n = 150 n = 30

Sorry, too tired and lazy to draw it out =([/quote]

Great explanation, was stuck in this ques for a while. Thanks.

ioiio, you are correct except you don't multiply the 27 by 2. The 27 figure encompasses all 2-sport individuals and thus no manipulation is needed. The formula is: A + B + C -AB - AC - BC - 2ABC + Neither = Total

In this problem, it states that AB + AC + BC = 27 --> don't need to multiply by 2.

Plus, the 3 figure is multiplied by 2, not 3.

Answer is 30.
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