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clubs problem

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clubs problem

by pradeepkaushal9518 » Fri Jul 02, 2010 3:53 am
is the number of members of club X is greater than the numbers of members of club Y?

1. of the members of club X, 20 % are also member of club Y
2. of the members of club y,30 % are also members of club X

plz explain.
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by mj78ind » Fri Jul 02, 2010 3:57 am
Stmt 1 we do not know anything about y

Stmt 2 do not know anything about x

Combine the two, let the members of club X only be x, only club Y be y and common be z.

Then per 1, 0.2(x+z) = z and per 2, 0.3(y+z) = z, hence solving x = 1.5y +0.5z, thus even if z = 0, still y <x. Hence C
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by albatross86 » Fri Jul 02, 2010 3:59 am
Is X > Y?

(Please refer to the venn diagram)

1. 20% of the members in club X are also members of club Y.

This says XY = 0.2*(X + XY) = 0.2X + 0.2XY
=> 0.8XY = 0.2X
=> XY = X/4

Nothing about the value of X or Y.

INSUFFICIENT

2. 30% of the members of club Y are also members of club X

This says XY = 0.3*(Y + XY),
=>XY = 0.3Y + 0.3XY
=>0.7XY = 0.3Y
=> XY = 3/7*Y

Nothing about the value of X or Y.

Both 1 and 2:

XY = X/4 = 3/7Y

=> X = 12/7*Y

Thus X > Y

SUFFICIENT

Pick C
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Last edited by albatross86 on Fri Jul 02, 2010 9:15 am, edited 1 time in total.
~Abhay

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by mj78ind » Fri Jul 02, 2010 7:25 am
albatross86 wrote:Is X > Y?

(Please refer to the venn diagram)

1. 20% of the members in club X are also members of club Y.

This says XY = 0.2*X, but nothing about the value of X or Y.

INSUFFICIENT

2. 30% of the members of club Y are also members of club X

This says XY = 0.3*Y, but nothing about the value of X or Y.

Both 1 and 2:

XY = 0.2*X = 0.3*Y

=> X/Y = 3/2

Thus X > Y

SUFFICIENT

Pick C
@abhay
I think we differ slightly in our interpretations, according to me 0.2 (x+xy) = xy and 0.3(y + xy) = xy

I think this question is in the OG12 but just do not have the urge to go check it again after being recently massacred at the GMAT altar ........ what do you think?
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by albatross86 » Fri Jul 02, 2010 9:08 am
You're totally right mj78ind! Thanks for pointing that out, will correct it :)
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by ankurmit » Sun Jul 04, 2010 11:31 pm
I am not getting how you both got this equation... can u please explain
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by san2009 » Mon Jul 05, 2010 1:28 am
using a double set matrix:
statement 1: insuff b/c we know nothing abt the other
statement 2: insuff b/c we know nothing abt the other

statement 1 + statement 2:
let number of members in Y be Y
let number of members in x by X
if you put the info from both statement into a double set matrix...the common box can be expressed as
1/5Y=3/10x
simplifiying, you get y=3/2x ---- thus y wll be greater than x (note x will always be positive...can't have negative number of members) --- sufficient to answer
thus, answer is C

What's the OA?
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