Pls ans the question ASAP

Dear Student,
Given, the mass of the student, M = 80 kg.
For a weighing machine at rest, the weight of any body of mass M would be:
W = Mg
The weighing machine is kept on a platform which is going down the slope with an acceleration a (say).
The component of acceleration of the platform vertically downwards is 
$a\text{'} = a\sin \left({30}^o\right)=\frac{a}{2}$
As the platform is accelerating down the incline, the net acceleration relative to the platform acting on the person on the balance is
$$\begin{gathered} g\text{'} = g - a\text{'} \\ g\text{'}=g-\frac{a}{2} \end{gathered}$$
Thus, the weight of the student recorded by the weighing balance relative to the accelerating platform is
$$\begin{gathered} W\text{'}=Mg\text{'}=M\left(g-\frac{a}{2}\right) \\ As per question, \\ W=80g \\ W\text{'}=75g \\ W-W\text{'}=5g \\ Also, \\ W-W\text{'}=Mg-M\left(g-\frac{a}{2}\right)=5g \\ M\frac{a}{2}=5g \\ Taking g = 10m{s}^{-2} \\ a=\frac{5g\times 2}{M}=\frac{5\times 10\times 2}{80}=\frac{100}{80}=\frac{5}{4}=1.25m{s}^{-2} \end{gathered}$$
Regards