Question 935409
Working forward is easier  (for me, at least).


Let p be the original amount of money.


Bought vegetables.  Money remaining, {{{(3/4)p}}}.


Bought fruit.  Money remaining, {{{(1/2)(3/4)p}}}.


Bought cheese, last, using {{{1/3}}} of {{{(1/2)(3/4)p}}}, so what remains is  {{{blue((2/3))(1/2)(3/4)p}}}, which
is the quantity, $12.



{{{highlight_green((2/3)(1/2)(3/4)p=12)}}}
-
{{{((2*1*3)/(3*2*4))p=12}}}

{{{(1/4)p=12}}}

{{{p=4*12}}}

{{{highlight(p=48)}}},  original amount of money.