Statement 1: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. 

Statement 2: By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline. 

In case of pure rolling on inclined plane acceleration of a body
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}$ 
For hollow cylinder  $\frac{I}{M{R}^2}=\frac{M{R}^2}{M{R}^2}=1$
For solid cylinder  $\frac{I}{M{R}^2}=\frac{\frac{1}{2}M{R}^2}{M{R}^2}=\frac{1}{2}$
Hence, acceleration of solid cylinder is more than hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first. 
In the case of rolling there will be no work done by friction.
Therefore, total mechanical energy remains conserved.
$\therefore {\left(KE\right)}_{solid}={\left(KE\right)}_{hollow}=$ decrease in $PE=mgh$