Three solid cubes of side lengths 3 cm, 4 cm and 5 cm are melted and recast into a single large cube. What is the ratio of the total surface area of the three original cubes to that of the new cube?

Introduction / Context:
This problem involves volume and surface area relationships for cubes. Three smaller solid cubes are melted and recast into a single larger cube. Because volume is conserved during melting and recasting, the total volume of the smaller cubes equals the volume of the larger cube. Once we find the side length of the new cube, we can compute and compare the total surface areas. This type of question is common in aptitude tests and encourages clear understanding of how linear dimensions, area, and volume scale.

Given Data / Assumptions:
• Side lengths of the three cubes are 3 cm, 4 cm, and 5 cm respectively.
• All cubes are solid and made of the same material.
• The three cubes are melted completely and recast into one large cube.
• Volume is conserved in the process.
• We are asked to find the ratio of total surface area of the three smaller cubes to the surface area of the large cube.

Concept / Approach:
The volume of a cube of side a is a³, and its total surface area is 6a². For the three small cubes, we compute individual volumes and sum them to get the total volume. That total becomes the volume of the large cube, from which we can determine its side length by taking the cube root. Then we compute the total surface area of all three small cubes and the surface area of the large cube. Finally, we express the required ratio as a simplified fraction or ratio of integers.

Step-by-Step Solution:
Step 1: Compute the volumes of the three small cubes: 3³ = 27, 4³ = 64, and 5³ = 125 cubic centimetres. Step 2: Sum the volumes to get total volume: 27 + 64 + 125 = 216 cubic centimetres. Step 3: Let the side of the large cube be A cm. Then its volume A³ equals 216. Step 4: Solve A³ = 216. Since 6³ = 216, we get A = 6 cm. Step 5: Total surface area of the three small cubes is 6(3²) + 6(4²) + 6(5²) = 6(9 + 16 + 25) = 6 * 50 = 300 square centimetres. Step 6: Surface area of the large cube is 6A² = 6 * 6² = 6 * 36 = 216 square centimetres. Step 7: The required ratio is 300 : 216. Divide numerator and denominator by 12 to get 25 : 18.

Verification / Alternative check:
We can verify the side of the large cube by rechecking the volume calculation. The total volume is 216 cubic centimetres, and cube root of 216 is 6 because 6 * 6 * 6 = 216. The surface area calculation is consistent, since 6 * 36 = 216. The ratio 300 / 216 reduces to 25 / 18. This ratio is irreducible because 25 and 18 share no common factor greater than 1. The result is thus stable and unique.

Why Other Options Are Wrong:
2 : 1 would correspond to 300 : 150, which is not equal to 300 : 216.
3 : 2 equals 27 : 18, not 25 : 18, so it does not match the actual ratio.
27 : 20 and 5 : 4 correspond to different fractional values and do not arise from any consistent calculation of volumes and areas in this problem. Only 25 : 18 corresponds to the correct surface area comparison.

Common Pitfalls:
A frequent error is to assume that the ratio of surface areas equals the ratio of volumes directly, which is not correct for solids when dimensions change. Another mistake is miscalculating the total volume by adding side lengths instead of volumes. Some learners also miscompute 6³ or forget the factor 6 in the surface area formula for a cube. Carefully computing each cube's volume and area separately, and then using the cube root correctly, helps avoid these mistakes.

Final Answer:
The ratio of the total surface area of the three smaller cubes to that of the large cube is 25 : 18.