Mixing aluminum and carbon nanotubes to achieve target density

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I am trying to do a simple calculation to figure out how much of a certain material must be added to another to produce a specific volume and weight.

Sadly, such a simple problem i cannot figure out! Naturally i thought i would ask.

In essence i wanted to know how much single walled carbon nanotubes need to placed into a pure aluminum structure to produce a bulk density of 1.5g/cm^3, where the density of aluminum is 2.7g/cm^3 and the single wall nanotubes are .15g/cm^3. The volume must be 31.4cm^3, and i had done a rough calculation for a theoretical nanotube which was 2nm in diameter and 10cm in length and found it to be 3.14 *10^-13 cm^3 per tube.

Thus, we get close to the needed density when we split the volumes in half (half aluminum, half carbon nanotube). Regardless of the possibility of such a structure i am left speechless in that i cannot figure out how to do this simple calculation to determine what volume % of each is needed to get 47grams, or a density of 1.5g/cm^3

Would anyone mind jumpstarting my brain really quick?
 
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I have difficulties to follow what has to be done, especially what the carbon nanotubes are meant to do. I would list all data you have, not in text, but in a list: material, density, mass, volume, container, volume. And do not forget the units. Then say what you want: a specific target density, mass or volume. Or just fill an Al cylinder with carbon? Your problem is a bit lost within your essay above: "split volume", "single walled" etc. are unexplained. Furthermore, does the density of the nanotubes differ from the density if packed in a container, i.e. do they behave like a fluid? Another general idea is: if in doubt, draw it out!
 
Let x be the volume of carbon nanotubes, then the total mass is (31.4cm^3 - x)*2.7g/cm^3 + x*.15g/cm^3. Set that equal to the total volume multiplied with the target density and solve for x, it is a linear equation.

Divide by the volume of one carbon nanotube to get the number of nanotubes.