A solid sphere of radius 21 cm is melted and recast into a solid cube. In this process 20 percent of the material is wasted. The cube is then melted and recast into a solid hemisphere, again with 20 percent material wasted. The hemisphere is finally melted and recast into two solid spheres of equal radius, with 20 percent material wasted in this process as well. What is the radius of each of the two new spheres, in centimetres?

Introduction / Context:
This is a multistep mensuration problem involving repeated melting and recasting of solids with loss of material at each stage. The initial solid is a sphere, which is converted to a cube, then to a hemisphere, and finally to two spheres of equal radius. At every melting stage, 20 percent of the material is lost. The challenge is to track the volume through each step and finally compute the radius of the last pair of spheres.

Given Data / Assumptions:

• Initial sphere radius R = 21 cm.
• Volume of a sphere = (4/3) * pi * r^3.
• Each recasting stage wastes 20 percent of the available material.
• Sequence: sphere → cube → hemisphere → two equal spheres.
• At each stage, the remaining 80 percent volume forms the new solid.
• We need the radius of each final sphere.

Concept / Approach:
If the original volume is V0, then after a 20 percent wastage, only 80 percent, that is (4/5) * V0, remains and is used for the next solid. Every subsequent melting also multiplies the available volume by 4/5. Therefore, after three wastage steps, the final total volume of metal is (4/5)^3 * V0. This total final volume equals the combined volume of the two identical spheres. Once the final total volume is known in terms of the original volume and radius, we can equate it to 2 * (4/3) * pi * r^3 and solve for the new radius r.

Step-by-Step Solution:
Step 1: Initial sphere volume V0 = (4/3) * pi * 21^3.Step 2: After first melting to form the cube, volume of cube V1 = (4/5) * V0 because 20 percent is wasted.Step 3: After second melting from cube to hemisphere, volume V2 = (4/5) * V1 = (4/5)^2 * V0.Step 4: After third melting from hemisphere to two spheres, volume V3 = (4/5) * V2 = (4/5)^3 * V0.Step 5: Let r be the radius of each final sphere. Combined final volume of two spheres = 2 * (4/3) * pi * r^3.Step 6: Equate final volume to reduced original volume: 2 * (4/3) * pi * r^3 = (4/5)^3 * (4/3) * pi * 21^3.Step 7: Cancel common factor (4/3) * pi on both sides to get 2r^3 = (4/5)^3 * 21^3. Hence r^3 = (1/2) * (4/5)^3 * 21^3 = (32/125) * 21^3.Step 8: Since 21^3 = 9261, r^3 = (32/125) * 9261 = 296352 / 125 ≈ 2370.816, so r ≈ cube root of 2370.816 ≈ 13.3 cm.

Verification / Alternative check:
A rough check is to compare the final radius with the original radius. There is wastage at every step and the final volume is split into two spheres, so the final radius must be much smaller than 21 cm. The factor (32/125) under the cube root is around 0.256, and cube root of 0.256 is a bit more than 0.6. Multiplying 21 by about 0.63 gives a value close to 13.2 or 13.3 cm, which supports the computed answer of approximately 13.3 cm.

Why Other Options Are Wrong:

• 10.5 cm is too small and would imply a final volume much less than the reduced volume after wastage.
• 12.6 cm still gives a combined volume lower than the required reduced volume.
• 14.0 cm and 9.8 cm are far from the cube root based estimate and do not satisfy the volume relation derived from (4/5)^3.

Common Pitfalls:

• Forgetting that wastage happens three times, which leads to using (4/5) only once or twice instead of cubing it.
• Not dividing by 2 when equating the final volume to two spheres of equal radius.
• Trying to keep pi in decimal form too early and losing accuracy in the intermediate steps.

Final Answer:
The radius of each of the two final spheres is approximately 13.3 cm.