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In the Reagan High School swim club, 120 members swim the backstroke o

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In the Reagan High School swim club, 120 members swim the backstroke o [#permalink]

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New post 15 May 2017, 12:01
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In the Reagan High School swim club, 120 members swim the backstroke or the crawl or both. If 30 of these members do not swim the backstroke, how many members swim both the crawl and the backstroke?

(1) Of the 120 members, 72 do not swim the crawl.
(2) A total of 48 members swim the crawl.
[Reveal] Spoiler: OA

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Re: In the Reagan High School swim club, 120 members swim the backstroke o [#permalink]

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New post 16 May 2017, 08:46
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Let no. of students who do only backstroke be B.
Let no. of students who do only crawl be C.
Let no. of students who do both be X.

Given,
B+C+X=120
C = 30

And B+X = 120 - C = 120 - 30 = 90

S1. Of the 120 members, 72 do not swim the crawl. ---> B=72.

So, X = 90 - B = 90 - 72 = 18

Sufficient!

S2. A total of 48 members swim the crawl.

So, C+X = 48

X = 48 - C = 48 - 30 = 18

Sufficient!

Hence, answer will be D.
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Re: In the Reagan High School swim club, 120 members swim the backstroke o [#permalink]

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New post 16 May 2017, 10:08
120 members here can be divided into three types:

Those who swim only the backstroke (and not crawl) - lets call them B
Those who swim only the crawl (and not backstroke) - already given to be as 30
Those who can swim both backstroke and crawl - lets call them S (we have to find this value)

So we have B+30+S = 120 Or B+S = 90 Or S = 90-B. Now lets look at the statements

Statement 1. 72 do not swim the crawl, thus they swim only backstroke. B = 72. This gives S = 90-72 = 18.
Sufficient

Statement 2. In total, 48 members swim the crawl. This includes those who can only crawl (and not backstroke) and also those who can crawl as well as backstroke.
This means 30+S = 48
S = 18. Sufficient.

Hence D
Re: In the Reagan High School swim club, 120 members swim the backstroke o   [#permalink] 16 May 2017, 10:08
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In the Reagan High School swim club, 120 members swim the backstroke o

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