three tests contain maximum marks of 50 each and the fourth test contains maximum of 100 marks.how many ways can a student can get total of 60% aggregate????

Let the tests with a maximum of 50 marks be A, B and C and the test with a maximum of 100 marks be D.
Let the marks lost in A, B and C be a, b and c respectively and the marks lost in D be d.
Total marks = 250
Aggregate = 60% => Marks scored = 150
=> a + b + c + d = 250 - 150 = 100 such that $$0 \leq a, b, c \leq 50$$ and $$0 \leq d \leq 100$$
Number of integral solutions of a + b + c + d = 100 is $$^{103}C_3$$
But $$0 \leq a, b, c \leq 50$$
So, let's remove the cases in which a, b or c is greater than 50.
On giving 51 to a, we get the equation as a + b + c + d = 49.
Number of integral solutions = $$^{52}C_3$$
a, b or c can be greater than 51 => Number of integral solutions to be removed = $$3 * ^{52}C_3$$
=> Number of ways possible = $$^{103}C_3$$ - $$3 * ^{52}C_3$$