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150 students at seward high school. 66 play baseball, 45

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150 students at seward high school. 66 play baseball, 45 [#permalink]

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24 Jul 2007, 20:36
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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?

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Re: PS: high school sports [#permalink]

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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?

I'm getting 57 by using venn diagram method.

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25 Jul 2007, 02:23
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.

Ans : 57.

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25 Jul 2007, 10:35
I get 21 ...

65+45+42-27+3=129

150-129=21 dont play anything

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25 Jul 2007, 10:44
fresinha12 wrote:
I get 21 ...

65+45+42-27+3=129

150-129=21 dont play anything

I agree with the method and the answer. I made a calculation error. it has to be 21

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25 Jul 2007, 10:47
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.

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25 Jul 2007, 11:08
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.

Ans : 57.

I agree with this rationale, but I get 60.

66+45+42=153

2*27+3*3=63

153-63=90 that play one sport.

Therefore 150-90 = 60 that play no sport.

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25 Jul 2007, 14:40
emoryhopeful wrote:
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.

Ans : 57.

I agree with this rationale, but I get 60.

66+45+42=153

2*27+3*3=63

153-63=90 that play one sport.

Therefore 150-90 = 60 that play no sport.

That was a stupid one. I guess I got influenced by the previous post. Yeah... 60 looks correct.

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25 Jul 2007, 15:19
23 play no sports at all.

each of the given sports have a total of 10 players that play that and at least one other sport. these add up to 127 sports participants.

150 - 127 = 23 play nothing.

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Re: PS: high school sports [#permalink]

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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.

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Re: PS: high school sports [#permalink]

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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.

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25 Jul 2007, 20:55
Set up a venn diagram with the following:
#(all three) = 3
#(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 =(

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26 Jul 2007, 00:11
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(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 =(

Great explanation. Thanks a lot. Helped me correct my mistake with the venn diagram.

Using the formula for 3 overlapping sets we get 30 as well.

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Re: PS: high school sports [#permalink]

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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.

duh. what a dumb mistake I made. great write-up.

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27 Jul 2007, 12:19
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.

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27 Jul 2007, 12:27
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(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 =(

nothing to add. but i do differently

total = x + y + z - (xy + yz + xz) - 2(xyz) + n
150 = 66 + 45 + 42 - (27) - 2(3) + n
n = 30

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31 Jul 2007, 02:12
Himalayan wrote:
ywilfred wrote:
Set up a venn diagram with the following:
#(all three) = 3
#(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 =(

nothing to add. but i do differently

total = x + y + z - (xy + yz + xz) - 2(xyz) + n
150 = 66 + 45 + 42 - (27) - 2(3) + n
n = 30

Hi Himalayan,

However, could you plz tell me if it is the standard formula which you have used to solve the problem.

Regards

Nikhil

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02 Aug 2007, 09:14
I am still sticking with 21.....

whats the OA??

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16 Jun 2008, 04:35
[quote="ywilfred"]Set up a venn diagram with the following:
#(all three) = 3
#(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.

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16 Jun 2008, 13:02
ioiio wrote:
66 + 45 + 42 - 2*27 - 3*3 = 153 - 60 = 93 play atleast one.

Ans : 57.

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.

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Re:   [#permalink] 16 Jun 2008, 13:02

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