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?

Guest Feb 10, 2021

#1**0 **

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

Guest Feb 10, 2021

edited by
Guest
Feb 10, 2021

#2**+1 **

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

CPhill Feb 10, 2021