Running at their respective constant rate, machine X takes 2 days longer to produce \(w\) widgets than machine Y. At these rates, if the two machines together produce \(\frac{5}{4} w\) widgets in 3 days, how many days would it take machine X alone to produce \(2w\) widgets?
Let’s collect the information in a table. I scaled the production by 4 to get rid of the fraction. The variable \(d\) is the number of days it takes machine X to produce \(w\) widgets (not \(2w\)).
| machine(s) | widgets | days |
|---|---|---|
| X&Y | \(5 w\) | \(12\) |
| X | \(1 w\) | \(d\) |
| Y | \(1w\) | \(d-2\) |
To compare the productivities, we let everything work the same time \(T = 12d(d-2)\), a multiple of the three times in the table.
| machine(s) | widgets | days |
|---|---|---|
| X&Y | \(5d(d-2)\cdot w\) | \(T\) |
| X | \(12(d-2)\cdot w\) | \(T\) |
| Y | \(12d\cdot w\) | \(T\) |
We can infer that \(12d +12(d-2) = 5d(d-2)\) or \[ 5d^2 -34d +24 = 0\;. \] From the answer choices we conclude that \(d\) must be one of the values \(2, 3, 4, 5\), or \(6\). Plugging in odd numbers won’t work (why?), and \(d \ne 2\) because otherwise \(d-2 = 0\). We only have to check 4 and 6, which reveals the correct answer choice (E).