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Intern  Joined: 26 Sep 2012
Posts: 23
Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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Question Stats: 61% (02:12) correct 39% (02:29) wrong based on 137 sessions

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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) $$\frac{N-X-Y+Z}{N}$$
Senior Manager  Joined: 13 Jun 2013
Posts: 266
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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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
Intern  Joined: 02 Feb 2016
Posts: 15
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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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!
Math Expert V
Joined: 02 Aug 2009
Posts: 7984
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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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

_________________
Intern  Joined: 02 Feb 2016
Posts: 15
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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Got it, thank you! I should read questions more carefully.
Manager  B
Joined: 05 Dec 2015
Posts: 100
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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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
Non-Human User Joined: 09 Sep 2013
Posts: 13266
Re: Of the N students in a certain class, X took a course in biology and Y  [#permalink]

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_________________ Re: Of the N students in a certain class, X took a course in biology and Y   [#permalink] 13 Mar 2018, 11:02
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