Asked by Anonymous

Adam, Bryant and Clara shared some pies. Adam ate 2/3 of the number of pies 
  and 2/3 of a pie. Bryant then ate 2/3 of the remaining number of pies and 2/3 of a 
pie. Finally, after Bryant had eaten his pies. Clara ate 2/3 of the remaining 
number of pies and 2/3 of a pie.  There was no pie left.
(a) How many pies were there at first?
(b) How many pies did Bryant eat?

Answers

Answered by mathhelper
very confusing pie eating event.

Assume all the pies are the same , let x represent the number of pies
Adam ate: (2/3)(x) + (2/3) = (2/3)(x + 1)
Pies left = x - (2/3)(x+1)
= x - 2/3 x - 2/3 = 1/3 x - 2/3 = (x-2)/3

Bryant ate: (2/3)(x-2)/3 + 2/3
= 2/9 x - 4/9 + 2/3
= 2/9 x + 2/9
pies left over after that:
= (1/3 x - 2/3) - (2/9 x + 2/9)
= 1/9 x - 8/9

Clara ate: (2/3)((1/9 x - 8/9) + 2/3
= 2/27 x - 16/27 + 2/3
= 2/27 x + 2/27

pies left over after that
= 1/9 x - 8/9 - 2/27 x - 2/27
= 1/27 x - 26/27
this is supposed to be equal to zero

1/27 x = 26/27
x = 26

so we started with 26 pies, and Bryant ate
2/9 x + 2/9
= (2/9)(26) + 2/9 = 6 pies
Answered by Anonymous
Thank u very much
Answered by Anonymous
Let number of pies = X
Adam : 2/3X + 2/3
Bryant : 2/5 (X-Adam) + 2/3
= 2/3 (1/3x - 2/3) + 2/3
= 2/9x - 4/9 + 6/9 = 2/9x + 2/9
Clara = 2/3 (X-Adam-Bryant) + 2/3
= 2/3 (1/3x - 2/3 - 2/9x - 2/9) + 2/3 = 2/3 (1/9x - 8/9) + 2/3
= 2/27x + 2/27
A + B + C = 2/27x + 2/27 + 2/9x + 2/9 + 2/3x + 2/3 = x
26/27x + 26/27 = x
1/17x = 26/27
X = 26 (a)

(2/9)(26)+2/9 = 6 (b)
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