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A group of n college students bought three identical round cakes to sh

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A group of n college students bought three identical round cakes to sh  [#permalink]

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New post 26 Nov 2018, 23:08
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Question Stats:

57% (02:23) correct 43% (02:37) wrong based on 45 sessions

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A group of n college students bought three identical round cakes to share. They divided the first cake into equal-sized pieces, one piece for each of them. They did the same with the second cake. After 3 of the students decided they did not want any more cake, the remaining students divided the third cake into equal-sized pieces, one piece for each of them. If Silvia received 1 piece from each of the three cakes, then, in terms of n, the amount of cake that she received was the same as what fraction of 1 cake?


A. \(\frac{n+2}{n(n−3)}\)

B. \(\frac{2n−3}{n(n−3)}\)

C. \(\frac{3n−3}{n(n−3)}\)

D. \(\frac{3n−6}{n(n−3)}\)

E. \(\frac{3n−3}{2n(n−3)}\)

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A group of n college students bought three identical round cakes to sh  [#permalink]

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New post Updated on: 27 Nov 2018, 10:47
Bunuel wrote:
A group of n college students bought three identical round cakes to share. They divided the first cake into equal-sized pieces, one piece for each of them. They did the same with the second cake. After 3 of the students decided they did not want any more cake, the remaining students divided the third cake into equal-sized pieces, one piece for each of them. If Silvia received 1 piece from each of the three cakes, then, in terms of n, the amount of cake that she received was the same as what fraction of 1 cake?


A. \(\frac{n+2}{n(n−3)}\)

B. \(\frac{2n−3}{n(n−3)}\)

C. \(\frac{3n−3}{n(n−3)}\)

D. \(\frac{3n−6}{n(n−3)}\)

E. \(\frac{3n−3}{2n(n−3)}\)


GMATinsight :

Sir I did not understand the last line of the question "in terms of n, the amount of cake that she received was the same as what fraction of 1 cake?

I did the following way let n be 9
so cake 1 ; 9 pieces
Cake 2 : 9 pieces
Cake 3 : 6 pieces

total pieces received by Silvia is 3

total cake pieces : 9+9+6= 24

1/9+1/9+1/6= 7/18

i.e. option D suffices this relation

Since I am not able to understand what is exactly to be derived so I am not able to find the correct answer option.. please help ....
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Originally posted by Archit3110 on 27 Nov 2018, 08:02.
Last edited by Archit3110 on 27 Nov 2018, 10:47, edited 1 time in total.
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A group of n college students bought three identical round cakes to sh  [#permalink]

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New post 27 Nov 2018, 10:43
So what I would do here is basically to work with the numbers.
Let's sat that there are 10 students, so Silvia eats
1/10 + 1/10 + 1/7 = 24/70

I would plug the numbers in and check for the correct answer.
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Re: A group of n college students bought three identical round cakes to sh  [#permalink]

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New post 27 Nov 2018, 19:09
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(1/n + 1/n + 1/(n-3)) = 2/n + 1/(n-3) = 3n-6/n(n-3)
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Re: A group of n college students bought three identical round cakes to sh   [#permalink] 27 Nov 2018, 19:09
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