13.428 A batch of identical parts was processed on three machines of different designs in the following sequence: first, only the first machine worked for as many hours as it would take for the second and third machines to complete the entire job together; then, only the second machine worked for as many hours as it would take for the first and third machines to complete the entire job together. The remaining part of the batch was processed on the third machine for as many hours as it would take for the first and second machines to complete the entire job together.

How many times faster would this work be completed if all three machines worked together?
Solution. Let $x, y, z$ be the efficiencies of the machines, and $V$ be the volume of work. According to the problem, $\frac{V}{y+z}$ is the time taken by the first machine, $\frac{V}{x+z}$ is the time taken by the second machine, and $\frac{V}{x+y}$ is the time taken by the third machine.

From this, $\frac{V x}{y+z}+\frac{V y}{x+z}+\frac{V z}{x+y}=V \Rightarrow \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1$.

Furthermore, $\frac{V}{x+y+z}$ is the time it would take to complete the work if all three machines worked simultaneously.

According to the question, we need to determine the following quantity $t$:

$t=\frac{\frac{V}{x+y}+\frac{V}{x+z}+\frac{V}{z+y}}{\frac{V}{x+y+z}}=\frac{x+y+z}{x+y}+\frac{x+y+z}{x+z}+\frac{x+y+z}{y+z}=$
$=\left(1+\frac{z}{x+y}\right)+\left(1+\frac{y}{x+z}\right)+\left(1+\frac{x}{y+z}\right)=3+\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)=4$.

Answer: 4 times.