Three solid spheres of radii 3 cm, 4 cm and 5 cm are melted and recast into a single solid sphere. What percentage decrease occurs in the total surface area?

Introduction / Context:
This question combines volume conservation with surface area comparison, which is a classic theme in aptitude and engineering exams. When multiple solids are melted and recast into a new solid, volume is conserved but surface area generally changes. You are asked to compute how much the total surface area decreases when three smaller spheres are melted into a single larger sphere.

Given Data / Assumptions:
• Radii of three original spheres: 3 cm, 4 cm and 5 cm. • They are melted and recast into one solid sphere. • Volume is conserved during melting and recasting. • We are asked to find the percentage decrease in total surface area.

Concept / Approach:
The volume V of a sphere of radius r is given by V = (4 / 3) * π * r^3. When multiple spheres are melted and recast into a single sphere, the total initial volume equals the final volume. So we first compute the sum of volumes of the three small spheres and equate it to the volume of the new sphere to find its radius R. Then we compute the initial total surface area, which is the sum of 4 * π * r^2 for each sphere, and compare it with the surface area of the new sphere 4 * π * R^2. Finally, we use the percentage decrease formula.

Step-by-Step Solution:
1. Volume of first sphere: V1 = (4 / 3) * π * 3^3 = (4 / 3) * π * 27. 2. Volume of second sphere: V2 = (4 / 3) * π * 4^3 = (4 / 3) * π * 64. 3. Volume of third sphere: V3 = (4 / 3) * π * 5^3 = (4 / 3) * π * 125. 4. Sum the volumes: V_total = (4 / 3) * π * (27 + 64 + 125) = (4 / 3) * π * 216. 5. Simplify: (4 / 3) * 216 = 288, so V_total = 288π. 6. Let the radius of the new sphere be R. Then (4 / 3) * π * R^3 = 288π. 7. Cancel π and multiply both sides by 3 / 4: R^3 = 288 * (3 / 4) = 216. 8. Thus R = cube root of 216 = 6 cm. 9. Total initial surface area: S_initial = 4 * π * (3^2 + 4^2 + 5^2) = 4 * π * (9 + 16 + 25) = 4 * π * 50 = 200π. 10. Final surface area: S_final = 4 * π * R^2 = 4 * π * 6^2 = 4 * π * 36 = 144π. 11. Decrease in surface area: ΔS = S_initial - S_final = 200π - 144π = 56π. 12. Percentage decrease = (ΔS / S_initial) * 100 = (56π / 200π) * 100 = 28 %.

Verification / Alternative check:
As a quick reasonableness check, notice that the final radius 6 cm is larger than any of the original radii 3 cm, 4 cm and 5 cm. Because surface area grows with the square of radius and volume grows with the cube, combining smaller spheres into one larger sphere almost always decreases the total surface area. The drop from 200π to 144π is significant but not extreme, making a decrease around 25 to 30 percent reasonable. Our calculation of 28 percent fits nicely in this expected range.

Why Other Options Are Wrong:
• 12 %: This would give a very small change, inconsistent with the large difference between 200π and 144π. • 14 %: Still too small given that 56π is more than a quarter of 200π. • 16 %: This corresponds to a much smaller surface area drop than actually computed. • 32 %: This would require an even larger drop than 56π, which does not match the exact values.

Common Pitfalls:
A common mistake is to add radii directly instead of adding volumes, which violates volume conservation. Another error is to compute only one of the surface areas and forget to compare initial and final values. Also, some students forget to include π consistently or cancel it improperly. Being systematic about using r^3 for volume and r^2 for surface area helps avoid these traps.

Final Answer:
The percentage decrease in total surface area is 28 %.