Ten identical solid spherical balls each of radius 3 cm are melted and recast to form a single solid sphere. During the process 20% of the material is wasted. What is the radius, in centimetres, of the new larger sphere?

Introduction / Context:
This problem checks understanding of how volume scales for spheres and how to handle melting and recasting of solid material. When multiple small spheres are melted and formed into a single larger sphere, the total volume available determines the radius of the new sphere. The complication here is that a fixed percentage of the material is wasted, so only a fraction of the original volume is used to form the new sphere.

Given Data / Assumptions:

Number of small spheres = 10.
Radius of each small sphere r_small = 3 cm.
Wastage of solid material = 20%, so usable volume is 80% of the total initial volume.
The new solid is one larger sphere of radius R.
Volumes are computed using V = (4 / 3) * pi * r^3.

Concept / Approach:
Total initial volume is the sum of the volumes of the 10 small spheres. Because 20% is wasted, only 80% of this total volume is available to form the big sphere. Set the volume of the big sphere equal to the usable volume and solve for its radius R. Since the formula involves r^3, the radius scales with the cube root of the volume factor, which is a common pattern in such recasting problems.

Step-by-Step Solution:
Volume of one small sphere = (4 / 3) * pi * 3^3 = (4 / 3) * pi * 27 = 36 * pi cubic centimetres. Total volume of 10 small spheres = 10 * 36 * pi = 360 * pi cubic centimetres. Given 20% of the solid is wasted, the usable volume is 80% of 360 * pi. Usable volume = 0.8 * 360 * pi = 288 * pi cubic centimetres. Let R be the radius of the new sphere. Its volume is V_big = (4 / 3) * pi * R^3. Set this equal to the usable volume: (4 / 3) * pi * R^3 = 288 * pi. Cancel pi on both sides to get (4 / 3) * R^3 = 288. Multiply both sides by 3 / 4: R^3 = 288 * 3 / 4 = 216. Therefore, R = cube root of 216 = 6 cm.

Verification / Alternative check:
We can verify by checking volumes numerically. For R = 6 cm, V_big = (4 / 3) * pi * 6^3 = (4 / 3) * pi * 216 = 288 * pi cubic centimetres, which matches the usable volume computed earlier. This confirms that our algebra and interpretation of the 20% wastage are correct. No other option, when converted back to volume, will match 288 * pi exactly.

Why Other Options Are Wrong:
A radius of 24 cm would give a volume far larger than the original total volume of all ten small spheres and is therefore impossible.
A radius of 12 cm also yields a volume much greater than 360 * pi, even before accounting for wastage.
A radius of 8 cm results in a volume smaller than the usable volume of 288 * pi and thus does not use all available material.
A radius of 10 cm similarly produces a volume that does not match the exact usable volume and arises from an incorrect volume ratio calculation.

Common Pitfalls:
Many students forget to subtract the wasted volume and erroneously use the full sum of the initial volumes. Others add radii instead of volumes, which is incorrect because volume of a sphere depends on the cube of the radius. Some learners also make algebraic mistakes when solving R^3 from the equation, such as taking square roots instead of cube roots. Keeping track of the percentage and the cubic relationship is essential.

Final Answer:
The radius of the new larger sphere is 6 cm.