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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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03 Jul 2014, 23:57
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62% (01:36) correct 38% (02:01) wrong based on 123 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) \(\frac{X+YZN}{N}\) (D) \(\frac{N+X+YZ}{N}\) (E) \(\frac{NXY+Z}{N}\)
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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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04 Jul 2014, 02:50
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+YZN) / N (D) (N+X+YZ) / N (E) (NXY+Z) / N N= x+yboth + z both= x+y+zN thus fraction becomes, (x+y+zN)/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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07 Mar 2016, 13:42
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+YZN) / N (D) (N+X+YZ) / N (E) (NXY+Z) / N N= x+yboth + z both= x+y+zN thus fraction becomes, (x+y+zN)/N = ((x+y+z)/N)1 B Hi! It's absolutely clear how you got both= x+y+zN, but what did you do to get the next equation? And where does N1 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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07 Mar 2016, 19:45
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+YZN) / N (D) (N+X+YZ) / N (E) (NXY+Z) / N N= x+yboth + z both= x+y+zN thus fraction becomes, (x+y+zN)/N = ((x+y+z)/N)1 B Hi! It's absolutely clear how you got both= x+y+zN, but what did you do to get the next equation? And where does N1 in denominator come from? Thanks a lot! Hi,
let me take it from where you left.. both= x+y+zN.. But we have to find the fraction.. \(Fraction = \frac{Both}{total}= \frac{both}{N}\) => \(\frac{(x+y+zN)}{N}= \frac{(x+y+z)}{N}\frac{N}{N}\) => \(\frac{(x+y+z)}{N}1\)
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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08 Mar 2016, 05:39
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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29 Jun 2016, 23:59
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)/101 > gives you 13/10  1 or 13/9 depending on whether the denominator is (N1) 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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13 Mar 2018, 11:02
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