Question 771740
n fishermen in the original group of fishermen.  Each would pay {{{2400/n}}} PhP.

"Two men withdrew", then n-2 are in the adjusted group.  Each would pay {{{2400/(n-2)}}} PhP.


The adjusted group forces each member to pay MORE than the each member of the original group; 100 PhP difference and {{{2400/(n-2)>2400/n}}}


{{{highlight(2400/(n-2)-2400/n=100)}}}
{{{24/(n-2)-24/n=1}}}
Multiply both sides by n(n-2)
{{{24n-24(n-2)=n(n-2)}}}
*{{{24n-24n+48=n^2-2n}}}__________(mistake WAS found but now fixed)
{{{n^2-2n=24n-24n+48}}}
{{{n^2-2n-48=0}}}


Resorting to General Solution to Quadratic Formula but omitting some steps here,... {{{n=(2+sqrt(4-4*(-48)))/2}}} .
{{{n=1+-sqrt(49)}}},  
{{{highlight(n=8)}}}
EIGHT fishermen originally.