Question

a student has completed 1st day 1/3 sums, 2nd day 2/5 sums solved, he cant solved 8 sums. total how many sums he has to solve ? 

Answer 1

aaa
Let there be  N sums in total.
First day solved  =  N * 1/3 = N/3
2nd day  solved = N *2/5  =  2N/5
Now,    total N =  N/3 + 2N/5 + 8         as  total = 1st day solved + 2nd day solved
                                                                           + remaining sums
N =  N (1/3 + 2/5) + 8    =>  N = N * 11/15 + 8    => 
N ( 1 - 11/15) = 8      =>  N  =  8 * 15  / 4    = 30