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Club Requirement

Expert replies
by logitech » Tue Nov 25, 2008 11:06 pm
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?

OA 6
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
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Source: — Problem Solving |

by mals24 » Wed Nov 26, 2008 2:55 am
P(P) = 22
P(H) = 27
P(W) = 28
Suppose the 6 students who sign up for exactly two clubs sign up for Poetry and History.
P(PandH) = 6

Number of persons who sign up for exactly 3 classes:

P(PUHUW) = P(P)+P(H)+P(W)-2(PandH)-(P+H+W)

59 = 22+27+28-2*6-(P+H+W)

(P+H+W) = 6

[U = Union]
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by vivek.kapoor83 » Wed Nov 26, 2008 3:08 am
i also calculated same way mals...but why 2*(P & H) coz...i took total of 6 students who chose for 2clubs and u took 2*6 =12
hwcm pls explain
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by jimmiejaz » Wed Nov 26, 2008 3:14 am
I read this formula here:
https://www.gmathacks.com/gmat-math/thre ... -sets.html

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

But according to mals24,
P(PUHUW) = P(P)+P(H)+P(W)-2(PandH)-(P+H+W)

Both formulas give the same ans as 6.
But, which one is correct?
Can someone throw some light on this please?
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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by jimmiejaz » Wed Nov 26, 2008 3:18 am
mals24 wrote:P(P) = 22
P(H) = 27
P(W) = 28
Suppose the 6 students who sign up for exactly two clubs sign up for Poetry and History.P(PandH) = 6

Number of persons who sign up for exactly 3 classes:

P(PUHUW) = P(P)+P(H)+P(W)-2(PandH)-(P+H+W)

59 = 22+27+28-2*6-(P+H+W)

(P+H+W) = 6

[U = Union]

Mals24,

Cant it be that students took 2 clubs each. then also we get 2+2+2=6?
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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by mals24 » Wed Nov 26, 2008 4:20 am
Ok ill be honest but this is the only formula I know when you have the number of people in exactly 2 sets etc. Everytime I see the word 'exactly' in the question I use this formula. (and I usually get the correct answer)

So I never bothered finding out why this formula works. But I guess now I should.

I got the formula from https://gmat-maths.blocked/2008/05/ ... mulas.html

But I checked BTG forum and this is the formula given by stuart

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) + (total # in none of the groups) - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(# in all 3 groups)

This is the same formula which jimmiejaz you have used. So I guess its time to make some corrections in my flashcards.

Sorry guys for using the previous formula I had no clue it was incorrect.
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by logitech » Wed Nov 26, 2008 8:30 am
True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) + (total # in none of the groups) - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(# in all 3 groups)

That goes to our flashcards. But I would recommend everyone to use a VENN diagram and prove this formula.
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by vivek.kapoor83 » Wed Nov 26, 2008 8:33 am
if u try to get this formula from Venn's diagram ...i think 2(intersections of all three) should not be there.why multiply it by 2.
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by vivek.kapoor83 » Wed Nov 26, 2008 8:39 am
if u try to get this formula from Venn's diagram ...i think 2(intersections of all three) should not be there.why multiply it by 2.
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