The weight of a drilled out cylinder Cylinder is $2.71$kg or $2710$g, diameter is $5$cm and $7.5$cm high. The volume is $147.18$cm$^3$ and the density is just the mass divided by the volume, I'm not sure if I need to convert it into grams then calculate it. So I just left it to $2.71$kg / $147.18$cm$^3$.
Anyways, the question is

If a $15$mm diameter hole is drilled through the centre of this cylinder to a depth of $6$cm from one end, and $6.4$ diameter through the remainder what would be the modified cylinders reduce weight?

This question confuses me and I do not understand it. Please help, there is another question part of this which is

In $\%$ how much material has been removed by the drilling of the 2 holes?

Thanks for helping!
 A: The density can be computed but is not really needed. If the original weight is given in kg, your answer should presumably be in kg. 
The old volume is $(\pi)(5/2)^2(7.5)$  cm$^3$. You may want to calculate this using a calculator, but it will turn out it is not necessary to do so.
Now we calculate the new volume, which is the old volume minus the combined volume of the holes. The two holes have volume $(\pi)(1.5/2)^2(6)$ and $(\pi)(0.32)^2(1.5)$ respectively. I have assumed the second hole has diameter $6.4$ mm, the units were not specified. It has depth $1.5$ cm, since the original cylinder has height $7.5$, and the first hole is $6$ cm deep. We had to convert the hole diameters to cm from the given information, which was in mm. 
So the new volume is
$$(\pi)(5/2)^2(7.5)- (\pi)(1.5/2)^2(6)-(\pi)(0.32)^2(1.5).$$
Thus the fraction of the original material remaining is 
$$\frac{(\pi)(5/2)^2(7.5)- (\pi)(1.5/2)^2(6)-(\pi)(0.32)^2(1.5)}{(\pi)(5/2)^2(7.5)}.\tag{1}$$
Cancel the $\pi$ and calculate. For the weight remaining, multiply (1) by the original weight. 
The percentage of material removed can be calculated from the information above. 
