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Of all the students in a certain dormitory, 1/2 are first-ye

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Of all the students in a certain dormitory, 1/2 are first-ye [#permalink] New post 27 Dec 2009, 09:47
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Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jan 2013, 06:28, edited 2 times in total.
Renamed the topic and edited the question.
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Re: FDP problem [#permalink] New post 27 Dec 2009, 13:31
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I think you dropped a fraction when you typed/copied this in. Working backward from the spoiler answer, I'm thinking the second sentence should read as follows? Please confirm.

"If 4/5 of the first-year students have not declared a major ..."
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Re: FDP problem [#permalink] New post 27 Dec 2009, 17:09
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Expert's post
msand wrote:
Of all the students in a certain dormitory, 1/2 are first-year students and the rest are second-year students. If 4/5 of the first-year students have not declared a major and if the fraction of second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major, what fraction of all the students in the dormitory are second-year students who have not declared a major?

A. 1/15
B. 1/5
C. 4/15
D. 1/3
E. 2/5

OA :
[Reveal] Spoiler:
B


First-year=Second-year=x --> all students 2x;

\frac{4}{5}x of the first-year students have not declared a major --> \frac{1}{5}x declared;

Second-year students who have declared a major is 3 times the fraction of first-year students who have declared a major =\frac{3}{5}x --> who have not \frac{2}{5}x;

Hence: \frac{\frac{2}{5}x}{2x}=\frac{1}{5}

Answer: B.
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Re: FDP problem [#permalink] New post 27 Dec 2009, 21:10
tot students=x

1st year student=x/2----> non majaor=4/5(x/2)-----> maj=1/5(x/2)

2nd year student=x/2---->maj=3(1/5(x/2))=3/10(x)--->non major=x/2-3/10(x)=1/5(x)

hence 1/5
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Re: FDP problem [#permalink] New post 27 Dec 2009, 23:05
1/5...dude form the table as given i manhattan series for sets..this would be a cake walk
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Re: FDP problem [#permalink] New post 30 Jan 2013, 06:25
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You can also do this problem sans algebra (in your head):

100 Students Total:

1st yr = 50 students
4/5x50 = 40 = No Major
Therefore: 10 = Major

2nd yr = 50 students
10x3 = 30 = Major
Therefore: 20 = No Major
20/100 = 1/5 = B
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Re: Of all the students in a certain dormitory, 1/2 are first-ye [#permalink] New post 08 Jun 2013, 06:34
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In this kind of problem table structure is the weapon for me...

Let 2X would be total student which revealed that 1 yrs student is X and 2nd Yr is X as well.

No form the table




MAJOR NO MAJOR Total Student
1st yr Student x/5 4x/5 x
2nd yr Student 3x/5 2x/5 x
Total Student 4x/5 6x/5 2x

From the table 2nd yr student with major is 2x/5 so the answer would be =2x/5/2x =>1/5 Ans

Rgds
Prasannajeet
Re: Of all the students in a certain dormitory, 1/2 are first-ye   [#permalink] 08 Jun 2013, 06:34
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