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# overlapping sets

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Joined: 25 Aug 2009
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10 Mar 2010, 05:28
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Question Stats:

63% (04:02) correct 38% (00:58) wrong based on 11 sessions

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1) All of the students of Music High School are in the band, the orchestra, or both.80% of the students are in only one group.There are 119 students in the band.If 50% of the students are in the band only, how many students are in the orchestra only?
A)30 B)51 C)60 D)85 E)119

OA-(B)

2) Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females.If the ratio of females speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there?
A)192 B)195 C)200 D)205 E)208

OA- (D)
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11 Mar 2010, 04:03
for the
indolent wrote:
1) All of the students of Music High School are in the band, the orchestra, or both.80% of the students are in only one group.There are 119 students in the band.If 50% of the students are in the band only, how many students are in the orchestra only?
A)30 B)51 C)60 D)85 E)119

OA-(B)

2) Of the 645 speckled trout in a certain fishery that contains only speckled and rainbow trout, the number of males is 45 more than twice the number of females.If the ratio of females speckled trout to male rainbow trout is 4:3 and the ratio of male rainbow trout to all trout is 3:20, how many female rainbow trout are there?
A)192 B)195 C)200 D)205 E)208

OA- (D)

First question

let's suppose X is the no of total students

so no of students who are in one group only = 0.8X
and no of students who are only band = 0.5x
so, no of students who are orchestra only = 0.8x-0.5x=0.3x

now the no of students who are in both groups = x - 0.8x = 0.2x
so, no of students who are in band = 0.5x+0.2x
=>119 = 0.5x + 0.2x
=>119 = 0.7x
=>x = 170

no of students in orchestra only = 0.3 x = 170 * 0.3 = 51

For the Second question, I m getting following answer

lets suppose trouts are
r(m) are male rainbow trouts and r(f) are female rainbow trouts
s(m) are male rainbow trouts and s(f) are female rainbow trouts

since, of 645 speckled trout
s(m)+s(f) = 645,

the number of males is 45 more than twice the number of females
so s(m) = 2s(f)+45 =>3s(f)+45 = 645 => S(f) = 200

means s(f) is 200 and s(m) is 445

ratio of females speckled trout to male rainbow trout is 4:3
so, s(f)/r(m) = 4/3=> 200/r(m) = 4/3=>r(m) = 150

and the ratio of male rainbow trout to all trout is 3:20
so, r(m)/alltrouts = 3/20 => 150/alltrouts = 3/20 => all trouts = 1000

now no of all trouts = s(f) + s(m) + r(m)+r(f)

1000 = 200+445+150+r(f)=>r(f) = 1000 - 795 => 205

hence no of female rainbow trouts is 205 D
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Manager
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Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
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21 May 2011, 23:38
B NOT B TOTAL

O S 0.3

NOT O S 0.5 zero

TOTAL 119 S 0.3 S

0rchestra/not band)+(band/not orchestra)=0.8 )
b/not o=0.5 therefore o/not b=0.3
not o/not b is zero

S0.7=119

S=170

0.3S=51
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23 May 2011, 23:01
1. 0.7x = 119 gives x = 170 = total
thus 0.3x * 170 = 51

2. 445+200 = 645

200:150 = 4:3
150: 1000 = 3:20

thus 1000-645 = 355
355-150 = 205 D
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Re: overlapping sets   [#permalink] 23 May 2011, 23:01
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