How many bullets can be made out of a cube of lead whose edge measures $22$ cm, each bullet being $2$ cm in diameter?

### Concept & Volume Equivalence When melting a solid shape to form multiple smaller solid shapes, the total volume remains conserved. The number of smaller objects is calculated by dividing the total volume of the original object by the volume of one smaller object. $$\text{Number of bullets} = \frac{\text{Volume of Cube}}{\text{Volume of one spherical bullet}}$$ Volume of a cube $= a^3$ and Volume of a sphere $= \frac{4}{3}\pi r^3$. ### Step-by-Step Solution * **Given:** Edge of the lead cube, $a = 22$ cm. * Volume of the cube $= 22^3 = 22 \times 22 \times 22 = 10648 \text{ cm}^3$. * Diameter of each spherical bullet $= 2$ cm, so its radius $r = 1$ cm. * Volume of one bullet $= \frac{4}{3} \times \pi \times (1)^3 = \frac{4}{3} \times \frac{22}{7} \times 1 = \frac{88}{21} \text{ cm}^3$. * Number of bullets $= \frac{10648}{\frac{88}{21}} = 10648 \times \frac{21}{88}$. * Simplify the division: $10648 \div 88$. We know $88 = 8 \times 11$. * $10648 \div 11 = 968$. * $968 \div 8 = 121$. * Number of bullets $= 121 \times 21 = 2541$. ### Exam Strategy & Shortcut Instead of computing the large cube volume ($10648$), keep factors separated: $\frac{22 \times 22 \times 22 \times 3 \times 7}{4 \times 22}$. Cancel one $22$ immediately. The $4$ in the denominator cancels with factors of $2$ from the remaining $22$s, leaving $11 \times 11 \times 3 \times 7 = 121 \times 21 = 2541$. ### Common Pitfall Using the diameter ($2$ cm) instead of the radius ($1$ cm) in the sphere volume formula, which would yield an answer $8$ times smaller than the correct value. ### Final Answer Therefore, the correct answer is **2541**.