Question 1096800
 Rahim averaged 70 in his first m number of exams.
 After taking n more exams, he had an overall averaged of 75.
 In terms of m & n, his averaged for the last n exams was.
:
let a = the average of the n exams
{{{((70m+an))/((m+n))}}} = 75
multiply both sides aby (m+n)
70m + an = 75(m+n)
70m + an = 75m + 75n
solve for a
an = 75m - 70m + 75n
an = 5m + 75n
a = {{{((5m+75n))/n}}} is the average for the last n exams