Solution
- In order to find the shapes that will balance out scale C, we need to analyze scales A and B.
- Let the weight of a cylinder be C, the weight of a sphere be S, and the weight of a cube be B.
On Scale A:
[tex]\begin{gathered} \text{ 1 cylinder, 1 cube, and 1 sphere balances 3 cylinders and 1 sphere} \\ \text{ Written as an equation of their weights, we have:} \\ \\ C+S+B=C+C+C+S \\ C+S+B=3C+S \\ \text{ Subtract S and C from both sides} \\ B=3C-C+S-S \\ B=2C \end{gathered}[/tex]
- This implies that 2 cylinders weigh just as much as 1 Cube.
On Scale B:
[tex]\begin{gathered} \text{ Two cubes balance out 1 cube, 3 cylinders, and 1 sphere.} \\ \text{ Thus, we can write:} \\ \\ B+B=B+C+S+S+S \\ 2B=B+C+3S \\ \text{ Subtract B from both sides} \\ B=3S+C \\ \text{ But we know that} \\ B=2C \\ \\ 2C=3S+C \\ 2C=C+3S \\ \text{ Subtract C from both sides} \\ \therefore C=3S \\ \\ \\ \text{ Thus further implies that:} \\ B=2C=2(3S)=6S \end{gathered}[/tex]
- This implies that 3 spheres weigh as much as 1 cylinder and 6 spheres weigh as much as 1 Cube.
On Scale C:
[tex]\begin{gathered} \text{ On the left hand side:} \\ 1\text{ cylinder and 3 spheres: }C+3S \\ \\ \text{ On the right hand side:} \\ 1\text{ cylinder and 1 cube: }C+B \\ \\ \text{ Let the last shape to be added to the right-hand side be X} \\ \\ C+3S=C+B+X \\ \text{ Subtract C from both sides} \\ 3S=B+X \\ \text{ But }B=6S \\ 3S=6S+X \\ \therefore X=-6S+3S=-3S \\ \\ \text{ This implies that we actually need 3 spheres or 1 Cylinder on the left scale} \end{gathered}[/tex]
Final Answer
The answer is 1 CYLINDER on the left platform of Scale C
