Question 1

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 an…

Accepted Answer

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.

Question 2

A solid cuboid of dimensions 50 cm × 40 cm × 30 cm is cut into 8 identical smaller cuboids by making 3 cuts, each cut parallel to one of its faces. What is the total surface area, in square centimetres, of all the 8 smaller cuboids together?

Accepted Answer

Introduction / Context: This problem tests understanding of how cutting a solid affects total surface area. A large cuboid is divided into smaller identical cuboids by planes parallel to its faces. Although the total volume remains the same, new internal faces become external surfaces of the smaller pieces, which changes the sum of their surface areas. Such questions are common in aptitude exams because they involve both geometry and spatial reasoning. Given Data / Assumptions: Original cuboid dimensions: 50 cm, 40 cm and 30 cm. The cuboid is cut into 8 identical smaller cuboids. Three cuts are made, each parallel to a face, so that the three dimensions are each divided by 2. No material is lost in cutting. We must find the total surface area of all 8 smaller cuboids combined. Concept / Approach: To obtain 8 equal parts from a cuboid, one cut is made along each dimension, halving length, breadth and height. This yields smaller cuboids with dimensions equal to half of each original dimension. The surface area of a cuboid with dimensions l, b and h is given by 2 * (l * b + b * h + h * l). First, determine the dimensions of each smaller cuboid and compute its surface area. Then multiply by 8 for the total surface area of all parts. Step-by-Step Solution: Original dimensions: length L = 50 cm, breadth B = 40 cm, height H = 30 cm. To get 8 parts, each dimension is halved: L is cut into two segments of 25 cm, B into two segments of 20 cm, and H into two segments of 15 cm. So each small cuboid has dimensions: l = 25 cm, b = 20 cm, h = 15 cm. Surface area of one small cuboid: S_one = 2 * (l * b + b * h + h * l). Compute products: l * b = 25 * 20 = 500; b * h = 20 * 15 = 300; h * l = 15 * 25 = 375. Sum inside brackets: 500 + 300 + 375 = 1175. S_one = 2 * 1175 = 2350 square centimetres. There are 8 identical cuboids, so total surface area S_total = 8 * 2350 = 18800 square centimetres. Verification / Alternative check: We can compare with the surface area of the original cuboid. Original surface area S_original = 2 * (50 * 40 + 40 * 30 + 30 * 50) = 2 * (2000 + 1200 + 1500) = 2 * 4700 = 9400 square centimetres. After cutting, the total surface area 18800 is exactly double. This makes sense because new faces appear inside the original cuboid that become external surfaces of the smaller pieces, increasing the overall surface area. Why Other Options Are Wrong: 11750 and 14100 correspond to miscalculations where one or two face products are omitted or not doubled correctly. 23500 would arise if you mistakenly took the surface area of one small cuboid as the total or miscounted the number of parts. 16400 is another incorrect intermediate value that does not match any consistent application of the surface area formula and the number of pieces. Common Pitfalls: Learners sometimes assume the total surface area remains unchanged after cutting, which is true for volume but not for surface area. Another mistake is to divide the original surface area by 8 instead of recalculating for each smaller cuboid. Misapplying the formula for surface area and confusing l, b and h values also leads to errors. It is important to recompute based on the new dimensions and then multiply by the correct number of identical parts. Final Answer: The total surface area of all the 8 smaller cuboids is 18800 square centimetres.

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