Question 1108282
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Since the problem asks for the answers as percentages, the radius of the outer sphere is irrelevant.<br>
Let r be the radius of the spherical chocolate.  Then the volume of the chocolate is {{{V(1) = (4/3)(pi)(r^3)}}}.<br>
The radius of the chocolate is half the side length of the cubical cake, so the side length of the cubical cake is 2r.  Then the volume of the cake plus chocolate is {{{V = (2r)^3 = 8r^3}}}. And then the volume of just the cake is {{{V(2) = 8r^3-V(1)}}}.<br>
The radius of the spherical bread is half the space diagonal of the cube, which is the square root of 3 times the radius of the chocolate.  Then the volume of the whole thing is {{{V(3) = (4/3)(pi)(r*sqrt(3))^2 = (4*sqrt(3))(pi)(r^3)}}}.<br>
The fraction of the whole that is chocolate is then {{{V(1)/V(3)}}} and the fraction that is cake is {{{V(2)/V(3)}}}.<br>
Evaluate those fractions using a calculator and convert them to percentages.  Then of course the percentage that is bread is whatever is left to make 100%.