Question 1105576
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There was an error in the algebra in the solution provided by the other tutor; the given answer does not satisfy the conditions in the problem.<br>
Let b and t represent the numbers Ben and Tom start out with, respectively:
Start:
Ben = b;
Tom = t.<br>
Ben gives 1/5 of the number he has to Tom; after that:
Ben = (4/5)b;
Tom = t+(1/5)b.<br>
Tom now gives 1/4 of what he has  -- which is (1/4)t+(1/20)b -- to Ben; after that:
Ben = (4/5)b+(1/4)t+(1/20)b = (17/20)b+(1/4)t;
Tom = (3/4)t+(3/20)b.<br>
At this point the two of them have the same number of sweets, 72.<br>
Ben: {{{(17/20)b+(1/4)t = 72}}}
(1) {{{17b+5t=1440}}}
Tom: {{{(3/4)t+(3/20)b = 72}}}
(2) {{{15t+3b=1440}}}<br>
Solve (1) and (2) by elimination:
{{{15t+51b = 4320}}}
{{{15t+3b = 1440}}}
{{{48b = 2880}}}
{{{b = 60}}}<br>
Ben started with 60 sweets; Tom with 84.<br>
Check:
Start: Ben 60, Tom 84
After Ben gives 1/5 of his to Tom: Ben 60-12 = 48; Tom 84+12 = 96
After Tom gives 1/4 of his to Ben: Ben 48+24 = 72; Tom 96-24 = 72