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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?

 Feb 10, 2021
 #1
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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

 Feb 10, 2021
edited by Guest  Feb 10, 2021
 #2
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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

 

 

cool cool cool

 Feb 10, 2021
edited by CPhill  Feb 10, 2021
edited by CPhill  Feb 10, 2021

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