96. Incorrect scales. When the pudding was placed on one pan of the scales, they showed 4 g more than $\frac{9}{11}$ of its true weight. When it was placed on the other pan, the scales showed 48 g more than in the first case. What is the true weight of the pudding?

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-031.jpg?height=371&width=508&top_left_y=2079&top_left_x=774)
96. If the scales were incorrect due to the different weights of their pans, the true weight of the pudding would be 154 g; the first indication of the scales would give 130, and the second 178 g. Half the sum of the scale readings (the arithmetic mean) is 154. However, from the diagram in the problem, it is clear that the pans weigh the same and that the error is due to the difference in the lengths of the lever arms*. Therefore, the scale readings were 121 and 169 g, and the true weight is 143 g. Taking the square root of the product of the scale readings, we get 143 (the geometric mean). The lengths of the lever arms are in the ratio of 11 to 13.

If we denote the true weight by \(x\), then for the cases discussed, we get the following equations:

\[
\begin{gathered}
\frac{\left(\frac{9}{11} x+4\right)+\left(\frac{9}{11} x+52\right)}{2}=x, \quad \text { from which } x=154 \\
\sqrt{\left(\frac{9}{11} x+4\right)+\left(\frac{9}{11} x+52\right)}=x \quad \text { from which } x=143
\end{gathered}
\]