In the kitchen, there was a bowl of berries.
When Billy walked by, he took 1/2 of the berries and 2 extra.
Later Bob walked by, and he took 1/3 of what Billy took and 2 extra berries.
Soon Joe walked by, and took 1/3 of what Bob took and then put 1 of those berries back into the bowl.
Sometime afterward, their mom walked by and saw only 1 berry remaining in the bowl.
How many berries were originally in the bowl?
x = original number of berries
x - 1/2x -2 - 1/6 - 2 - 1/18 x = 1
x - 13/18 x - 4 = 1
5/18 x = 5
x = 5 * 18/5 = 18 berries originally
See Chris answer below..... I am not sure I interpreted the Q correctly
I interpreted 'he took 1/3 of what Billy took and 2 extra berries.' to exclude the 'extras' etc
+130071 What all 4 took must be all the berries in the bowl....call this N
Billy took (1/2)N + 2
Bob took (1/3) [ (1/2)N + 2 ] + 2
Joe took (1/3) [ (1/3) [ (1/2)N + 2 ] + 2 ] -1
Mom took 1
So we have this rather convoluted equation
[ (1/2) N + 2 ]+ (1/3) [ (1/2)N + 2 ] + 2 + (1/3) [ (1/3) [ (1/2)N + 2 ] + 2 ] -1 + 1 = N
Because this is a little tedious ( but not impossible) to solve, I'll get a little help from WolframAlpha
N = 20
Billy took (1/2) (20) + 2 = 10 + 2 = 12
Bob took (1/3) of what Billy took + 2 = (1/3)12+ 2 = 4 + 2 = 6
Joe took (1/3) of what Bob took and put one back =(1/3)(6) - 1 = 2 - 1 = 1
And Mom took 1
And 12 + 6 + 1 + 1 = 20 berries in the bowl originally
