Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At...

1. Translate the problem requirements: Machine X takes 2 more days than Machine Y to produce \(\mathrm{w}\) widgets. Together they produce \(\frac{5}{4}\) of \(\mathrm{w}\) widgets in 3 days. We need to find how long Machine X alone takes to produce \(2\mathrm{w}\) widgets.
2. Set up rate relationships using simple variables: Define Machine Y's time to produce \(\mathrm{w}\) widgets, then express Machine X's time in terms of this base variable.
3. Use the combined work information: Apply the fact that their combined daily rates over 3 days equals \(\frac{5}{4}\) of \(\mathrm{w}\) widgets to create an equation.
4. Solve for individual machine times: Find how long each machine takes to produce \(\mathrm{w}\) widgets, then scale up to find Machine X's time for \(2\mathrm{w}\) widgets.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• Machine X is slower than Machine Y - it takes 2 more days to produce the same amount (\(\mathrm{w}\) widgets)
• When both machines work together for 3 days, they produce \(\frac{5}{4}\) times \(\mathrm{w}\) widgets (that's 25% more than \(\mathrm{w}\))
• We need to find how long Machine X alone would take to produce \(2\mathrm{w}\) widgets (double the amount)

The key insight is that we're dealing with work rates - how much each machine can produce per day. Since Machine X is slower, it has a lower daily rate.

Process Skill: TRANSLATE - Converting the relationship "2 days longer" into a mathematical framework

2. Set up rate relationships using simple variables

Let's use simple reasoning to set up our variables:

If Machine Y takes some number of days to produce \(\mathrm{w}\) widgets, let's call that number '\(\mathrm{d}\)' days.
Then Machine X takes \((\mathrm{d} + 2)\) days to produce the same \(\mathrm{w}\) widgets.

Now, thinking about daily rates:

• Machine Y's daily rate = \(\mathrm{w}\) widgets ÷ \(\mathrm{d}\) days = \(\frac{\mathrm{w}}{\mathrm{d}}\) widgets per day
• Machine X's daily rate = \(\mathrm{w}\) widgets ÷ \((\mathrm{d} + 2)\) days = \(\frac{\mathrm{w}}{(\mathrm{d} + 2)}\) widgets per day

This makes intuitive sense: since Machine X takes longer, it produces fewer widgets per day.

3. Use the combined work information

When both machines work together, their daily rates add up:

Combined daily rate = Machine Y's rate + Machine X's rate
Combined daily rate = \(\frac{\mathrm{w}}{\mathrm{d}} + \frac{\mathrm{w}}{(\mathrm{d} + 2)}\)

Working together for 3 days, they produce \(\frac{5}{4}\mathrm{w}\) widgets:
\(3 \times \left[\frac{\mathrm{w}}{\mathrm{d}} + \frac{\mathrm{w}}{(\mathrm{d} + 2)}\right] = \frac{5}{4}\mathrm{w}\)

Dividing both sides by \(\mathrm{w}\):
\(3 \times \left[\frac{1}{\mathrm{d}} + \frac{1}{(\mathrm{d} + 2)}\right] = \frac{5}{4}\)

Dividing by 3:
\(\frac{1}{\mathrm{d}} + \frac{1}{(\mathrm{d} + 2)} = \frac{5}{12}\)

Process Skill: SIMPLIFY - Breaking down the combined work equation step by step

4. Solve for individual machine times

To solve \(\frac{1}{\mathrm{d}} + \frac{1}{(\mathrm{d} + 2)} = \frac{5}{12}\), let's find a common denominator on the left side:

\(\frac{(\mathrm{d} + 2)}{\mathrm{d}(\mathrm{d} + 2)} + \frac{\mathrm{d}}{\mathrm{d}(\mathrm{d} + 2)} = \frac{5}{12}\)

\(\frac{(\mathrm{d} + 2 + \mathrm{d})}{\mathrm{d}(\mathrm{d} + 2)} = \frac{5}{12}\)

\(\frac{(2\mathrm{d} + 2)}{\mathrm{d}(\mathrm{d} + 2)} = \frac{5}{12}\)

Cross-multiplying:
\(12(2\mathrm{d} + 2) = 5\mathrm{d}(\mathrm{d} + 2)\)
\(24\mathrm{d} + 24 = 5\mathrm{d}^2 + 10\mathrm{d}\)
\(24 = 5\mathrm{d}^2 + 10\mathrm{d} - 24\mathrm{d}\)
\(24 = 5\mathrm{d}^2 - 14\mathrm{d}\)
\(5\mathrm{d}^2 - 14\mathrm{d} - 24 = 0\)

Using the quadratic formula or factoring:
\((5\mathrm{d} + 6)(\mathrm{d} - 4) = 0\)

Since \(\mathrm{d}\) must be positive: \(\mathrm{d} = 4\)

Therefore:

• Machine Y takes 4 days to produce \(\mathrm{w}\) widgets
• Machine X takes \(4 + 2 = 6\) days to produce \(\mathrm{w}\) widgets

To produce \(2\mathrm{w}\) widgets, Machine X would take: \(6 \times 2 = 12\) days

5. Final Answer

Let's verify our answer makes sense:

• Machine Y rate: \(\frac{\mathrm{w}}{4}\) widgets per day
• Machine X rate: \(\frac{\mathrm{w}}{6}\) widgets per day
• Combined rate: \(\frac{\mathrm{w}}{4} + \frac{\mathrm{w}}{6} = \frac{3\mathrm{w}}{12} + \frac{2\mathrm{w}}{12} = \frac{5\mathrm{w}}{12}\) widgets per day
• In 3 days together: \(3 \times \frac{5\mathrm{w}}{12} = \frac{15\mathrm{w}}{12} = \frac{5\mathrm{w}}{4}\) ✓

Therefore, Machine X alone would take 12 days to produce \(2\mathrm{w}\) widgets.

The answer is (E) 12.

Common Faltering Points

Errors while devising the approach

• Misinterpreting the relationship between machines: Students might incorrectly assume that Machine X takes 2 days total (instead of 2 days longer than Machine Y) to produce \(\mathrm{w}\) widgets, leading to completely wrong rate calculations.
• Confusing individual vs. combined work rates: Students may struggle to understand that when machines work together, their individual rates add up. They might try to average the rates or use other incorrect combinations.
• Missing the connection between time and rate: Students often fail to recognize that if a machine takes more time to complete the same work, its rate (work per unit time) is lower. This leads to setting up inverse relationships incorrectly.

Errors while executing the approach

• Algebraic manipulation errors when combining fractions: When solving \(\frac{1}{\mathrm{d}} + \frac{1}{(\mathrm{d}+2)} = \frac{5}{12}\), students frequently make errors finding common denominators or cross-multiplying, leading to incorrect quadratic equations.
• Quadratic equation solving mistakes: Students may incorrectly factor \(5\mathrm{d}^2 - 14\mathrm{d} - 24 = 0\) or make arithmetic errors when using the quadratic formula, potentially getting \(\mathrm{d} = -\frac{6}{5}\) instead of \(\mathrm{d} = 4\).
• Sign errors in equation setup: When rearranging the equation \(24\mathrm{d} + 24 = 5\mathrm{d}^2 + 10\mathrm{d}\), students might incorrectly handle the signs, writing \(5\mathrm{d}^2 + 14\mathrm{d} - 24 = 0\) instead of \(5\mathrm{d}^2 - 14\mathrm{d} - 24 = 0\).

Errors while selecting the answer

• Forgetting to scale up for the final question: After correctly finding that Machine X takes 6 days to produce \(\mathrm{w}\) widgets, students might select answer choice (B) 6, forgetting that the question asks for time to produce \(2\mathrm{w}\) widgets (which requires doubling to get 12 days).
• Selecting the wrong machine's time: Students might calculate both machines' times correctly (Y: 4 days, X: 6 days for \(\mathrm{w}\) widgets) but then use Machine Y's time in their final calculation, leading to \(4 \times 2 = 8\) days instead of \(6 \times 2 = 12\) days.