Question 596067
Let c = cost per person (before the increase of 50 people) and let n = number of people (originally -- before the increase of 50 people)



Since the total cost is 1680, and this cost is distributed evenly among 'n' people, this means that each person will pay {{{c=1680/n}}} dollars.


Adding 50 people to the group increases the count to n+50 and it decreases the cost per person by $10. So the cost per person is now c-10



The same cost of 1680 is now distributed amongst n+50 people and the cost per person is now c-10



This means that {{{c-10=1680/(n+50)}}}



Let's use this equation and {{{c=1680/n}}} to solve for n



{{{c-10=1680/(n+50)}}}


{{{1680/n-10=1680/(n+50)}}}


{{{n(n+50)(1680/n)-10n(n+50)=n(n+50)(1680/(n+50))}}}


{{{1680(n+50)-10n(n+50)=1680n}}}


{{{1680n+84000-10n^2-500n=1680n}}}


{{{1680n+84000-10n^2-500n-1680n=0}}}


{{{-10n^2-500n+84000=0}}}


{{{10n^2+500n-84000=0}}}


Now use the quadratic formula to solve for n


{{{n = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{n = (-(500)+-sqrt((500)^2-4(10)(-84000)))/(2(10))}}}


{{{n = (-500+-sqrt(250000-(-3360000)))/(20)}}}


{{{n = (-500+-sqrt(3610000))/20}}}


{{{n = (-500+sqrt(3610000))/20}}} or {{{n = (-500-sqrt(3610000))/20}}}


{{{n = (-500+1900)/20}}} or {{{n = (-500-1900)/20}}}


{{{n = 1400/20}}} or {{{n = -2400/20}}}


{{{n = 70}}} or {{{n = -120}}}


Toss out the negative solution since you can't have a negative number of people.


Therefore, the only solution is {{{n = 70}}}


So there were originally <font color="red">70</font> people.