$2$

Say a cone weighs $x$, a sphere weighs $y$ and a cube weighs $z$.

Then we have $$2x + y = z \quad \text{(1)}$$since two cones and a sphere together weigh the same as a cube.

Also $$y + z = 3 x \quad \text{(2)}$$since a sphere and a cube together weigh the as three cones.

Taking $y$ from each side of (1) (or taking a sphere from each side of the first scales) gives $$2x=z-y\quad \text{(3)}$$

Thento find out how much one cone weighs, we take equation (3) from equation (2) to give

\begin{eqnarray} 3x - 2x &=& (y+z) - (z-y) \\ x &=& y+z-z+y \\ &=& 2y \end{eqnarray}

So a cone weighs the same as two spheres.

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Say a cone weighs $x$, a sphere weighs $y$ and a cube weighs $z$.

Then we have $$2x + y = z \quad \text{(1)}$$since two cones and a sphere together weigh the same as a cube.

Also $$y + z = 3 x \quad \text{(2)}$$since a sphere and a cube together weigh the as three cones.

Taking $y$ from each side of (1) (or taking a sphere from each side of the first scales) gives $$2x=z-y\quad \text{(3)}$$

Thento find out how much one cone weighs, we take equation (3) from equation (2) to give

\begin{eqnarray} 3x - 2x &=& (y+z) - (z-y) \\ x &=& y+z-z+y \\ &=& 2y \end{eqnarray}

So a cone weighs the same as two spheres.

*This problem is taken from the UKMT Mathematical Challenges.*

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