Question 1078361
Let {{{a}}}, {{{b}}}, and {{{c }}} = the
number of apples in the respective boxes
at the 1st step
-----------------------
step 1
A has {{{ a = b + 128 }}}
B has {{{ b }}}
C has {{{ c = 0 }}}
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step 2
A has {{{ a - .25a = .75a }}}
{{{ .75a = .75*( b + 128 ) }}}
B has {{{ b + .25*( b + 128 ) }}}
{{{ 1.25b + 32 }}}
C has {{{ c = 0 }}}
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step 3
A has {{{ .75a = .75*( b + 128 ) }}}
B has {{{ (7/8 )*( 1.25b + 32 ) }}}
C has {{{ (1/8)*( 1.25b + 32 ) }}}
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After this step,
{{{ (1/8)*( 1.25b + 32 ) = (3/4)*( b + 128 ) -149 }}}
Multiply both sides by {{{ 8 }}}
{{{ 1.25b + 32 = 6*( b + 128 ) - 1192 }}}
{{{ 1.25b + 32 = 6b + 768 - 1192 }}}
{{{ 4.75b = 1192 + 32 - 768 }}}
{{{ 4.75b = 456 }}}
{{{ b = 96 }}}
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A.
The number of apples in box C is
C has {{{ (1/8)*( 1.25b + 32 ) }}}
{{{ (1/8)*( 1.25*96 + 32 ) }}}
{{{ (1/8)*( 120 + 32 ) }}}
{{{ (1/8)*152 = 19 }}}
C has 19 apples at the end
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B.
Find {{{a}}}, the original number of apples in box A
step 3
A has {{{ .75a = .75*( b + 128 ) }}}
{{{ .75a = .75*( 96 + 128 ) }}}
{{{ .75a = .75*224 }}}
{{{ a = 224 }}}
There were 224 apples in box A to start with
===================================
check the answers:
1st step:
{{{ a = b + 128 }}}
{{{ a = 96 + 128 }}}
{{{ a = 224 }}}
{{{ b = 96 }}}
{{{ c = 0 }}}
--------------------
2nd step:
{{{ .75a = .75*224 }}}
{{{ .75a = 168 }}}
and
{{{ 1.25b = 1.25*96 }}}
{{{ 1.25b = 120 }}}
and
{{{ c = 0 }}}
---------------------
3rd step:
A has:
{{{ .75a = 168 }}}
and
B has:
{{{ (7/8 )*( 1.25b + 32 ) }}}
{{{ (7/8 )*( 1.25*96 + 32 ) }}}
{{{ (7/8)*( 120 + 32 ) }}}
{{{ (7/8)*152 = 133 }}}
and
C has:
{{{ (1/8)*( 1.25b + 32 ) }}}
{{{ (1/8)*( 1.25*96 + 32 ) }}}
{{{ (1/8)*152 = 19 }}}
------------------------------
The totals must be the same at
start and finish
{{{ a = 224 }}}
{{{ b = 96 }}}
{{{ c = 0 }}}
Total = 320
-----------------
After step C:
{{{ a = 168 }}}
{{{ b = 133 }}}
{{{ c = 19 }}}
Total = 320 
OK