Casting — How many spherical bullets (diameter 2 cm) can be made from a lead cube of edge 22 cm?

Introduction / Context:
When recasting, volume is conserved. Number of bullets = (volume of cube) / (volume of one bullet). Use exact π = 22/7 for clean integers here.

Given Data / Assumptions:

• Cube edge a = 22 cm ⇒ V_cube = a^3 = 10648 cm^3
• Bullet diameter = 2 cm ⇒ radius = 1 cm
• Sphere volume V_s = (4/3)πr^3

Concept / Approach:
Compute V_s with r = 1 using π = 22/7, then divide the cube volume by V_s and take the integer count.

Step-by-Step Solution:
V_s = (4/3) * (22/7) * 1^3 = 88/21 cm^3N = 10648 / (88/21) = (10648 * 21) / 88 = 121 * 21 = 2541

Verification / Alternative check:
88 * 121 = 10648, so cancellation is exact.

Why Other Options Are Wrong:
2662 and 1347 come from rounding with π = 3.14; 5324 doubles the correct count.

Common Pitfalls:
Using diameter instead of radius in sphere volume; forgetting exact fraction 22/7.

Final Answer:
2541