Q6. Sum of a, b, c and d is $120$, such that their HCF is $3$, How many solutions are possible ?
I have tried this ...
I converted the above scene into the following equation as
$ 3a+3b+3c+3d=120$
so a+b+c+d=40
so integral solutions to the equations are ${40 +4-1\choose 4-1}$
Is it correct?
$\newcommand{'}{\ \!'}$ That's a good start, and it tacitly assumes that zero is a permitted value of the components, but you need to also remove solutions with higher common factors than $3$ in the sum to $120$.
So you already have $a\'+b\'+c\'+d\' = 40$ - you need to remove the cases where $\gcd(a\',b\',c\',d\') = 2 $ or $5$ and then add back in the cases where $\gcd(a\',b\',c\',d\') = 10$ (which would be "twice-removed"). This is a typical inclusion-exclusion process.
If zero is not permitted, you will need to modify your initial count to $\binom {40-1}{4-1}$ to allow for each variable to be a positive value, and similarly with the subsequent steps.