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. \(\frac{n+2}{n(n3)}\)
B. \(\frac{2n3}{n(n3)}\)
C. \(\frac{3n3}{n(n3)}\)
D. \(\frac{3n6}{n(n3)}\)
E. \(\frac{3n3}{2n(n3)}\)
A. \(\frac{n+2}{n(n3)}\)
B. \(\frac{2n3}{n(n3)}\)
C. \(\frac{3n3}{n(n3)}\)
D. \(\frac{3n6}{n(n3)}\)
E. \(\frac{3n3}{2n(n3)}\)
Put n = 6.
Sylvia received 1/6 + 1/6 + 1/3 = 2/3 of a cake.
Put n = 6 in options. Since den are same in 4 options, n = 6 gives 6*3 = 18 in the denominator. So we should get 12 in the numerator to get the fraction of 2/3.
When we put n = 6 in numerator of option (D), we get 12.
Answer (D)