Two machines, X and Y, working together at their respective constant rates for 14 hours, and then machine Y, working...

1. Translate the problem requirements: Define variables for the time each machine takes to produce m items alone, then interpret the given working scenario (14 hours together + 4 hours Y alone = m items) and the time relationship (X takes 8 hours less than Y).
2. Set up rate equations from the work scenario: Express the total work done as the sum of combined work for 14 hours plus Y's solo work for 4 hours, setting this equal to m items.
3. Apply the time relationship constraint: Use the fact that X takes 8 hours less than Y to create a second equation relating the two unknown times.
4. Solve the system algebraically: Substitute the time relationship into the work equation to create a single equation in one variable, then solve for Y's time.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for and what information we have.

We want to find: How long it takes machine Y, working alone, to produce m items.

Let's use simple variables:
• Let Y = the time (in hours) for machine Y alone to produce m items
• Let X = the time (in hours) for machine X alone to produce m items

Now let's translate what happened in the work scenario:
• Both machines worked together for 14 hours
• Then machine Y worked alone for 4 more hours
• Together, this produced exactly m items

We also know that machine X is faster than machine Y:
• Machine X takes 8 hours less than machine Y to produce m items
• So: \(\mathrm{X = Y - 8}\)

Process Skill: TRANSLATE - Converting the problem's English description into mathematical relationships

2. Set up rate equations from the work scenario

Now we need to think about work rates. If machine Y takes Y hours to produce m items, then in 1 hour, machine Y produces \(\frac{\mathrm{m}}{\mathrm{Y}}\) items.

Similarly, if machine X takes X hours to produce m items, then in 1 hour, machine X produces \(\frac{\mathrm{m}}{\mathrm{X}}\) items.

Let's break down what happened:
• For 14 hours, both machines worked together
• In those 14 hours, machine X produced: \(14 \times \frac{\mathrm{m}}{\mathrm{X}}\) items
• In those 14 hours, machine Y produced: \(14 \times \frac{\mathrm{m}}{\mathrm{Y}}\) items
• Then machine Y worked alone for 4 more hours, producing: \(4 \times \frac{\mathrm{m}}{\mathrm{Y}}\) items

The total work equals m items:
\(14 \times \frac{\mathrm{m}}{\mathrm{X}} + 14 \times \frac{\mathrm{m}}{\mathrm{Y}} + 4 \times \frac{\mathrm{m}}{\mathrm{Y}} = \mathrm{m}\)

Simplifying:
\(14 \times \frac{\mathrm{m}}{\mathrm{X}} + 18 \times \frac{\mathrm{m}}{\mathrm{Y}} = \mathrm{m}\)

Dividing everything by m:
\(\frac{14}{\mathrm{X}} + \frac{18}{\mathrm{Y}} = 1\)

3. Apply the time relationship constraint

We know that \(\mathrm{X = Y - 8}\), so let's substitute this into our work equation.

Our equation was: \(\frac{14}{\mathrm{X}} + \frac{18}{\mathrm{Y}} = 1\)

Substituting \(\mathrm{X = Y - 8}\):
\(\frac{14}{\mathrm{Y - 8}} + \frac{18}{\mathrm{Y}} = 1\)

Process Skill: APPLY CONSTRAINTS - Using the given relationship to reduce our variables from two to one

4. Solve the system algebraically

Now we have one equation with one unknown. Let's solve:
\(\frac{14}{\mathrm{Y - 8}} + \frac{18}{\mathrm{Y}} = 1\)

To solve this, we'll find a common denominator. The common denominator is \(\mathrm{Y(Y - 8)}\):

\(\frac{14\mathrm{Y} + 18(\mathrm{Y} - 8)}{\mathrm{Y(Y - 8)}} = 1\)

Expanding the numerator:
\(\frac{14\mathrm{Y} + 18\mathrm{Y} - 144}{\mathrm{Y(Y - 8)}} = 1\)
\(\frac{32\mathrm{Y} - 144}{\mathrm{Y(Y - 8)}} = 1\)

Cross-multiplying:
\(32\mathrm{Y} - 144 = \mathrm{Y(Y - 8)}\)
\(32\mathrm{Y} - 144 = \mathrm{Y^2} - 8\mathrm{Y}\)

Rearranging to standard form:
\(\mathrm{Y^2} - 8\mathrm{Y} - 32\mathrm{Y} + 144 = 0\)
\(\mathrm{Y^2} - 40\mathrm{Y} + 144 = 0\)

Let's factor this quadratic. We need two numbers that multiply to 144 and add to -40.
Those numbers are -36 and -4:
\((\mathrm{Y} - 36)(\mathrm{Y} - 4) = 0\)

So \(\mathrm{Y} = 36\) or \(\mathrm{Y} = 4\)

But if \(\mathrm{Y} = 4\), then \(\mathrm{X} = \mathrm{Y} - 8 = 4 - 8 = -4\), which doesn't make sense for time.

Therefore, \(\mathrm{Y} = 36\) hours.

Final Answer

Let's verify: If \(\mathrm{Y} = 36\), then \(\mathrm{X} = 36 - 8 = 28\).

Checking our work equation:
\(\frac{14}{28} + \frac{18}{36} = \frac{14}{28} + \frac{18}{36} = \frac{1}{2} + \frac{1}{2} = 1\) ✓

Therefore, machine Y takes 36 hours to produce m items working alone.

The answer is E. 36

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the work scenario

Students often struggle to correctly parse that machine Y works for the ENTIRE duration (14 + 4 = 18 total hours) while machine X only works for the first 14 hours. They might incorrectly assume both machines work for 14 hours, then there's a separate 4-hour period, leading to wrong rate equations.

2. Confusing the time relationship constraint

The problem states "machine X takes 8 hours less than machine Y," but students frequently mix this up and write \(\mathrm{Y} = \mathrm{X} - 8\) instead of the correct \(\mathrm{X} = \mathrm{Y} - 8\). This fundamental error in constraint setup leads to completely wrong equations.

3. Incorrectly setting up rate equations

Students often confuse rate (items per hour) with time (hours per batch). They might write the work equation as \(14\mathrm{X} + 18\mathrm{Y} = \mathrm{m}\) instead of the correct \(\frac{14}{\mathrm{X}} + \frac{18}{\mathrm{Y}} = 1\), failing to properly convert between time-to-complete and rate-per-hour.

Errors while executing the approach

1. Algebraic manipulation errors when solving rational equations

When solving \(\frac{14}{\mathrm{Y}-8} + \frac{18}{\mathrm{Y}} = 1\), students frequently make errors in finding the common denominator or expanding terms. Common mistakes include incorrectly distributing \(18(\mathrm{Y}-8)\) or making sign errors when cross-multiplying.

2. Factoring the quadratic incorrectly

After reaching \(\mathrm{Y^2} - 40\mathrm{Y} + 144 = 0\), students often struggle to find the correct factor pairs. They need two numbers that multiply to +144 and add to -40 (which are -36 and -4), but may incorrectly identify these or make arithmetic errors in the factoring process.

3. Arithmetic errors in verification

When checking their answer by substituting back into the original work equation, students may make calculation errors with fractions like \(\frac{14}{28} + \frac{18}{36}\), potentially leading them to doubt their correct answer.

Errors while selecting the answer

1. Selecting the wrong variable value

After solving and finding both \(\mathrm{X} = 28\) and \(\mathrm{Y} = 36\), students might accidentally select \(\mathrm{X} = 28\) instead of \(\mathrm{Y} = 36\). Since the question specifically asks for machine Y's time, this represents a critical final-step error despite correct mathematical work.

2. Not rejecting the impossible solution

When the quadratic gives solutions \(\mathrm{Y} = 36\) and \(\mathrm{Y} = 4\), students might not properly check that \(\mathrm{Y} = 4\) leads to \(\mathrm{X} = -4\) (impossible for time) and could randomly pick the wrong root, especially if they don't perform the constraint check.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a smart value for m

Instead of keeping m as a variable, let's choose \(\mathrm{m} = 180\) items. This number is selected because it works well with the time relationships we'll discover and makes rate calculations clean (180 has many factors).

Step 2: Set up the time relationship

Let Y take t hours to produce 180 items alone
Then X takes \((\mathrm{t} - 8)\) hours to produce 180 items alone

Step 3: Express rates in terms of items per hour

Y's rate = \(\frac{180}{\mathrm{t}}\) items per hour
X's rate = \(\frac{180}{\mathrm{t} - 8}\) items per hour

Step 4: Set up the work equation

Work done in 14 hours together + Work done by Y alone for 4 hours = 180 items

\(14 \times \left[\frac{180}{\mathrm{t}} + \frac{180}{\mathrm{t}-8}\right] + 4 \times \frac{180}{\mathrm{t}} = 180\)

Step 5: Simplify by dividing by 180

\(14 \times \left[\frac{1}{\mathrm{t}} + \frac{1}{\mathrm{t}-8}\right] + \frac{4}{\mathrm{t}} = 1\)

\(\frac{14}{\mathrm{t}} + \frac{14}{\mathrm{t}-8} + \frac{4}{\mathrm{t}} = 1\)

\(\frac{18}{\mathrm{t}} + \frac{14}{\mathrm{t}-8} = 1\)

Step 6: Solve for t

Multiply through by \(\mathrm{t(t-8)}\):

\(18(\mathrm{t}-8) + 14\mathrm{t} = \mathrm{t(t-8)}\)

\(18\mathrm{t} - 144 + 14\mathrm{t} = \mathrm{t^2} - 8\mathrm{t}\)

\(32\mathrm{t} - 144 = \mathrm{t^2} - 8\mathrm{t}\)

\(40\mathrm{t} - 144 = \mathrm{t^2}\)

\(\mathrm{t^2} - 40\mathrm{t} + 144 = 0\)

Step 7: Factor the quadratic

\((\mathrm{t} - 4)(\mathrm{t} - 36) = 0\)

So \(\mathrm{t} = 4\) or \(\mathrm{t} = 36\)

Step 8: Check validity

If \(\mathrm{t} = 4\), then X would take \((4 - 8) = -4\) hours, which is impossible.

Therefore, \(\mathrm{t} = 36\) hours.

Verification: Y takes 36 hours, X takes 28 hours. Y's rate = 5 items/hour, X's rate = \(\frac{180}{28} = \frac{45}{7}\) items/hour. In 14 hours together: \(14\left(5 + \frac{45}{7}\right) = 14\left(\frac{80}{7}\right) = 160\) items. Plus 4 hours of Y alone: \(4(5) = 20\) items. Total: \(160 + 20 = 180\) ✓