A solid right circular cylinder of radius 7 cm and height 21 cm is melted and recast into small identical bullets. Each bullet consists of a right circular cylinder surmounted by a hemisphere on one base. The total height of each bullet is 3.5 cm and the radius of its base is 2.1 cm. Approximately how many complete bullets can be made from the original solid?

Introduction / Context:
This question combines volume of a cylinder and hemisphere and the idea of melting and recasting a solid. The original solid is a large cylinder, and it is converted into smaller composite solids shaped like bullets. Each bullet consists of a cylindrical part plus a hemispherical cap. To find how many bullets can be formed, we compare the total volume of the original cylinder with the volume of one bullet and then take the integer number of complete bullets.

Given Data / Assumptions:

Original solid is a right circular cylinder with radius R = 7 cm and height H = 21 cm.
Each bullet has radius r = 2.1 cm at the base.
Each bullet has total height 3.5 cm and consists of a cylinder plus a hemisphere.
The hemispherical part has radius 2.1 cm.
Height of cylindrical part of each bullet = 3.5 cm minus 2.1 cm = 1.4 cm.
Take pi = 22 / 7 for calculations.

Concept / Approach:
First, we find the volume of the large cylinder using V = pi * R^2 * H. Next, we compute the volume of one bullet as the sum of the volume of its cylindrical part and the volume of its hemispherical part. That means V_bullet = pi * r^2 * h_cyl + (2 / 3) * pi * r^3. Dividing the total original volume by the volume of one bullet gives the number of bullets. Because only complete bullets are counted, we take the integer part of this ratio.

Step-by-Step Solution:
Volume of original cylinder: V_big = pi * R^2 * H = (22 / 7) * 7^2 * 21. Compute V_big: (22 / 7) * 49 * 21 = (22 / 7) * 1029 = 22 * 147 = 3234 cubic centimetres. For each bullet, height of cylindrical part h_cyl = 3.5 cm - 2.1 cm = 1.4 cm. Volume of cylindrical part: V_cyl = pi * r^2 * h_cyl = (22 / 7) * 2.1^2 * 1.4. Volume of hemispherical part: V_hemi = (2 / 3) * pi * r^3 = (2 / 3) * (22 / 7) * 2.1^3. Performing the calculation gives V_bullet = V_cyl + V_hemi = approximately 38.808 cubic centimetres. Number of bullets = V_big / V_bullet = 3234 / 38.808 which is about 83.33. Therefore, the number of complete bullets that can be formed is 83.

Verification / Alternative check:
We can quickly check by multiplying 83 by the bullet volume: 83 * 38.808 is slightly less than 3234, so the material is sufficient for 83 complete bullets. If we tried 84 bullets, the required volume would exceed the original 3234 cubic centimetres, so 84 is not possible. This confirms that 83 is the maximum number of complete bullets that can be formed.

Why Other Options Are Wrong:
89 and 92 would require more volume than available in the original cylinder and so are impossible.
74 and 79 are smaller than the correct integer count and would leave some usable volume unused, which contradicts the idea of forming as many complete bullets as possible.

Common Pitfalls:
A common mistake is to ignore the hemispherical part and treat the bullet as a simple cylinder, which underestimates its volume and leads to a larger incorrect count. Another error is to take the total height of the bullet as the height of the cylindrical part and forget to subtract the radius reserved for the hemisphere. Finally, rounding pi or intermediate values too early can slightly distort the final answer, so it is better to keep enough precision until the end and then choose the closest valid integer.

Final Answer:
Approximately 83 complete bullets can be made from the original cylinder.