Method to obtain 40%

Here's my solution

Let the marks scored in the $i$ th paper be $x_i$ for $i=1,2,3,4,5,6$

Then we have $x_1+x_2+x_3+x_4+x_5+x_6 = 240$ since 240 is 40% of the total marks (600)

We also have the constraints $0 \leq x_i \leq 100$ for $i=1,2,3,4,5,6$

I denote the coefficient of $x^r$ as $((x^r))$ and the binomial coefficient as $C(n,r) = \frac{n!}{r!(n-r)!}$

Now the number of integral solutions of the equations with the constraints is

$((x^{240}))$ in $(1 + x + x^2 + x^3 + \dots + x^{100})^6$

$=((x^{240}))$ in $(1-x^{101})^6 (1 - x)^{-6}$ using the formula for the sum of a geometric sequence

$=((x^{240}))$ in $(C(6,0) - C(6,1)x^{101} + C(6,2) x^{202})(1-x)^{-6}$ . Using the binomial theorem. The powers greater than 240 can be left out since they do not contribute to the coefficient

$=((x^{240}))$ in $C(6,0)(1-x)^{-6} - ((x^{139}))$ in $C(6,1)(1-x)^{-6} + ((x^{38}))$ in $C(6,2)(1-x)^{-6}$

$=C(6,0)C(245,240) - C(6,1)C(144,139) + C(6,2)C(43,38)$ from the binomial theorem for a negative power

Simplifying we get, the number of ways

$= \frac{1!}{5!} (\frac{245!}{240!} - 6 \frac{144!}{139!} + 15 \frac{43!}{38!})$

And we get $A=5,B=5,C=0,D=1,E=9,F=3,G=8$

and $A+B+C+D+E+F+G=\boxed{31}$

This is solution uses a technique of associating the coefficients of a series with a combinatorial meaning.

This problem might be solved using the inclusion and exclusion principle, however, I haven't been able to think of how to do this :( If somebody does, please post the solution as I'd love to know how to use the inclusion and exclusion principle. :)