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Of the N students in a certain class, X took a course in biology and Y

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Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 03 Jul 2014, 22:57
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Of the N students in a certain class, X took a course in biology and Y took a course in history, and Z took neither of the courses. If there were students who took both two courses, in terms of N, X, Y and Z, what fraction of the students took both two courses?


(A) \(1 - \frac{X+Y+Z}{N}\)

(B) \(\frac{X+Y+Z}{N} - 1\)

(C) \(\frac{X+Y-Z-N}{N}\)

(D) \(\frac{N+X+Y-Z}{N}\)

(E) \(\frac{N-X-Y+Z}{N}\)
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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 04 Jul 2014, 01:50
Maksym wrote:
Of the N students in a certain class, X took a course in biology and Y took a course in history, and Z took neither of the courses. If there were students who took both two courses, in terms of N, X, Y and Z, what fraction of the students took both two courses?

(A) 1 - (X+Y+Z) / N
(B) (X+Y+Z) / N - 1
(C) (X+Y-Z-N) / N
(D) (N+X+Y-Z) / N
(E) (N-X-Y+Z) / N


N= x+y-both + z

both= x+y+z-N

thus fraction becomes, (x+y+z-N)/N
= ((x+y+z)/N)-1
B
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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 07 Mar 2016, 12:42
manpreetsingh86 wrote:
Maksym wrote:
Of the N students in a certain class, X took a course in biology and Y took a course in history, and Z took neither of the courses. If there were students who took both two courses, in terms of N, X, Y and Z, what fraction of the students took both two courses?

(A) 1 - (X+Y+Z) / N
(B) (X+Y+Z) / N - 1
(C) (X+Y-Z-N) / N
(D) (N+X+Y-Z) / N
(E) (N-X-Y+Z) / N


N= x+y-both + z

both= x+y+z-N

thus fraction becomes, (x+y+z-N)/N
= ((x+y+z)/N)-1
B


Hi!

It's absolutely clear how you got both= x+y+z-N, but what did you do to get the next equation? And where does N-1 in denominator come from?

Thanks a lot!
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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 07 Mar 2016, 18:45
1
1
Viktoriaa wrote:
manpreetsingh86 wrote:
Maksym wrote:
Of the N students in a certain class, X took a course in biology and Y took a course in history, and Z took neither of the courses. If there were students who took both two courses, in terms of N, X, Y and Z, what fraction of the students took both two courses?

(A) 1 - (X+Y+Z) / N
(B) (X+Y+Z) / N - 1
(C) (X+Y-Z-N) / N
(D) (N+X+Y-Z) / N
(E) (N-X-Y+Z) / N


N= x+y-both + z

both= x+y+z-N

thus fraction becomes, (x+y+z-N)/N
= ((x+y+z)/N)-1
B


Hi!

It's absolutely clear how you got both= x+y+z-N, but what did you do to get the next equation? And where does N-1 in denominator come from?

Thanks a lot!


Hi,

let me take it from where you left..
both= x+y+z-N..
But we have to find the fraction..
\(Fraction = \frac{Both}{total}= \frac{both}{N}\)
=> \(\frac{(x+y+z-N)}{N}= \frac{(x+y+z)}{N}-\frac{N}{N}\)
=> \(\frac{(x+y+z)}{N}-1\)

Hope it helps you

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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 08 Mar 2016, 04:39
Got it, thank you! I should read questions more carefully.
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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 29 Jun 2016, 22:59
Hello - I'm having trouble understanding why using smart numbers isn't working, can someone assess my method please?

N students: 10
X Bio: 3
Y History: 5
Z Neither: 5

Formula: N = X + Y + Z - Both --> 10 = 3 + 5 + 5 - both --> both = 3

Plug variables into answer B gives you: (X+Y+Z) / N - 1 --> (3 + 5 + 5)/10-1 --> gives you 13/10 - 1 or 13/9 depending on whether the denominator is (N-1) or only /N, but either way does not equal 30% (3/10) as I would expect
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Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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New post 13 Mar 2018, 10:02
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Re: Of the N students in a certain class, X took a course in biology and Y &nbs [#permalink] 13 Mar 2018, 10:02
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