How many spherical bullets of diameter 4 cm can be cast by melting a solid cube of edge 44 cm? (Use π = 22/7 where needed.)

Introduction / Context:Volume is conserved in melting/recasting. We divide the cube’s volume by the volume of one sphere (bullet) to get the count. The standard approximation π = 22/7 is commonly used in such problems for an exact integer result.

Given Data / Assumptions:

• Cube edge a = 44 cm → V_cube = a^3.
• Bullet sphere diameter = 4 cm → radius r = 2 cm.
• Use V_sphere = (4/3)πr^3 with π = 22/7.

Concept / Approach:n = V_cube / V_sphere; compute carefully to preserve integer results with π = 22/7.

Step-by-Step Solution:

Verification / Alternative check:Reduce 85184/704 = 121 exactly; integer multiplication by 21 yields 2541.

Why Other Options Are Wrong:Near misses result from rounding π or arithmetic slips; only 2541 is exact with 22/7.

Common Pitfalls:Using π ≈ 3.14 leads to a non-integer; forgetting radius is half the diameter.

Final Answer:2541