A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y...

1. Translate the problem requirements: We need to find what fraction the actual total time represents compared to the hypothetical time if they had maintained the original rate throughout
2. Set up the two time scenarios: Calculate time for the actual scenario (two different rates) and the hypothetical scenario (constant original rate)
3. Express both times in terms of given variables: Use the relationship Time = Work/Rate to write expressions for both scenarios
4. Form the fraction and simplify: Create the ratio of actual time to hypothetical time and algebraically simplify to match answer choices

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in everyday terms. Imagine you're comparing two scenarios:

Scenario 1 (What actually happened): The crew paints \(\mathrm{y}\) houses at their original speed of \(\mathrm{x}\) houses per week, then speeds up and paints the remaining \((80-\mathrm{y})\) houses at \(1.25\mathrm{x}\) houses per week.

Scenario 2 (What could have happened): The crew paints all 80 houses at their original speed of \(\mathrm{x}\) houses per week throughout.

We need to find: What fraction does the actual time represent compared to the hypothetical time?

In plain English, we're asking: "If the actual time is the numerator and the hypothetical time is the denominator, what fraction do we get?"

Process Skill: TRANSLATE - Converting the word problem into a clear mathematical comparison

2. Set up the two time scenarios

Let's think about time using the simple relationship: \(\mathrm{Time} = \frac{\mathrm{Work}}{\mathrm{Rate}}\)

Actual scenario time calculation:

• Time for first \(\mathrm{y}\) houses at rate \(\mathrm{x}\): This takes \(\frac{\mathrm{y}}{\mathrm{x}}\) weeks
• Time for remaining \((80-\mathrm{y})\) houses at rate \(1.25\mathrm{x}\): This takes \(\frac{(80-\mathrm{y})}{1.25\mathrm{x}}\) weeks
• Total actual time = \(\frac{\mathrm{y}}{\mathrm{x}} + \frac{(80-\mathrm{y})}{1.25\mathrm{x}}\) weeks

Hypothetical scenario time calculation:

• Time for all 80 houses at rate \(\mathrm{x}\): This takes \(\frac{80}{\mathrm{x}} = \frac{80}{\mathrm{x}}\) weeks

3. Express both times in terms of given variables

Now let's write these times more clearly using our variables:

Actual Time: \(\frac{\mathrm{y}}{\mathrm{x}} + \frac{(80-\mathrm{y})}{1.25\mathrm{x}}\)

Hypothetical Time: \(\frac{80}{\mathrm{x}}\)

Let's simplify the actual time expression. Since \(1.25\mathrm{x} = \frac{5\mathrm{x}}{4}\), we have:

\(\frac{(80-\mathrm{y})}{1.25\mathrm{x}} = (80-\mathrm{y}) \div \frac{5\mathrm{x}}{4} = (80-\mathrm{y}) \times \frac{4}{5\mathrm{x}} = \frac{4(80-\mathrm{y})}{5\mathrm{x}}\)

So our Actual Time becomes: \(\frac{\mathrm{y}}{\mathrm{x}} + \frac{4(80-\mathrm{y})}{5\mathrm{x}}\)

4. Form the fraction and simplify

The fraction we want is: Actual Time ÷ Hypothetical Time

\(= \left[\frac{\mathrm{y}}{\mathrm{x}} + \frac{4(80-\mathrm{y})}{5\mathrm{x}}\right] \div \left[\frac{80}{\mathrm{x}}\right]\)

When we divide by a fraction, we multiply by its reciprocal:

\(= \left[\frac{\mathrm{y}}{\mathrm{x}} + \frac{4(80-\mathrm{y})}{5\mathrm{x}}\right] \times \frac{\mathrm{x}}{80}\)

\(= \frac{\mathrm{y}}{\mathrm{x}} \times \frac{\mathrm{x}}{80} + \frac{4(80-\mathrm{y})}{5\mathrm{x}} \times \frac{\mathrm{x}}{80}\)

\(= \frac{\mathrm{y}}{80} + \frac{4(80-\mathrm{y})}{5 \times 80}\)

\(= \frac{\mathrm{y}}{80} + \frac{4(80-\mathrm{y})}{400}\)

\(= \frac{\mathrm{y}}{80} + \frac{(80-\mathrm{y})}{100}\)

Let's find a common denominator. The LCD of 80 and 100 is 400:

\(= \frac{5\mathrm{y}}{400} + \frac{4(80-\mathrm{y})}{400}\)

\(= \frac{5\mathrm{y} + 4(80-\mathrm{y})}{400}\)

\(= \frac{5\mathrm{y} + 320 - 4\mathrm{y}}{400}\)

\(= \frac{\mathrm{y} + 320}{400}\)

\(= \frac{320}{400} + \frac{\mathrm{y}}{400}\)

\(= 0.8 + 0.0025\mathrm{y}\)

Process Skill: SIMPLIFY - Systematically reducing the complex fraction to match the answer format

Final Answer

The fraction of time is \(0.8 + 0.0025\mathrm{y}\), which matches answer choice (B).

This makes intuitive sense: the 0.8 represents the time savings from the increased speed on most of the houses, while the \(0.0025\mathrm{y}\) term accounts for the houses painted at the original slower rate.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what fraction is being asked for

Students often confuse the direction of the fraction. The question asks for "what fraction of the time it would have taken if they had painted all houses at the original rate." This means we need (actual time)/(hypothetical time). Students might incorrectly set up (hypothetical time)/(actual time), leading to a completely wrong approach.

2. Misunderstanding the two-phase work scenario

Students may incorrectly assume that the crew works at both rates simultaneously rather than sequentially. The problem clearly states they paint the first \(\mathrm{y}\) houses at rate \(\mathrm{x}\), then paint the remaining \((80-\mathrm{y})\) houses at rate \(1.25\mathrm{x}\). Missing this sequential nature leads to incorrect time calculations.

3. Confusing rate and time relationships

Students often struggle with the inverse relationship between rate and time (\(\mathrm{Time} = \frac{\mathrm{Work}}{\mathrm{Rate}}\)). They might incorrectly multiply work by rate instead of dividing, or forget that when the rate increases by a factor of 1.25, the time decreases by that same factor.

Errors while executing the approach

1. Arithmetic errors when handling the fraction \(1.25\mathrm{x}\)

When converting \(1.25\mathrm{x}\) to a fraction \(\frac{5\mathrm{x}}{4}\), students often make mistakes in the division step. Specifically, when calculating \(\frac{(80-\mathrm{y})}{1.25\mathrm{x}}\), they might incorrectly compute this as \((80-\mathrm{y}) \times \frac{4}{5\mathrm{x}}\) instead of the correct \(\frac{(80-\mathrm{y}) \times 4}{5\mathrm{x}}\), leading to wrong coefficients in their final expression.

2. Common denominator calculation errors

When combining fractions like \(\frac{\mathrm{y}}{80} + \frac{(80-\mathrm{y})}{100}\), students frequently make errors finding the LCD (400) or incorrectly converting the fractions. For example, they might write \(\frac{\mathrm{y}}{80}\) as \(\frac{4\mathrm{y}}{400}\) instead of the correct \(\frac{5\mathrm{y}}{400}\).

3. Algebraic simplification mistakes

In the final steps when expanding \(\frac{5\mathrm{y} + 4(80-\mathrm{y})}{400}\), students often make distribution errors, writing \(5\mathrm{y} + 4(80-\mathrm{y})\) as \(5\mathrm{y} + 320 + 4\mathrm{y}\) instead of the correct \(5\mathrm{y} + 320 - 4\mathrm{y}\), which leads to completely different final coefficients.

Errors while selecting the answer

1. Decimal conversion errors

Students may correctly arrive at \(\frac{(\mathrm{y} + 320)}{400}\) but then incorrectly convert this to decimal form. They might calculate \(\frac{320}{400} = 0.8\) correctly but then compute \(\frac{\mathrm{y}}{400} = 0.025\mathrm{y}\) instead of the correct \(0.0025\mathrm{y}\), leading them to select an answer that doesn't exist among the choices.

2. Rearrangement confusion

Even with the correct expression \(0.8 + 0.0025\mathrm{y}\), students might try to rearrange it to match other answer choices, not recognizing that answer choice (B) is already in the correct form. They might waste time trying to force their answer to look like choices (A), (C), (D), or (E).

Alternate Solutions

Smart Numbers Approach

Instead of working with variables algebraically, we can choose specific values for \(\mathrm{x}\) and \(\mathrm{y}\) that make calculations manageable.

Step 1: Choose convenient values

Let's select:

• \(\mathrm{x} = 4\) houses per week (original rate)
• \(\mathrm{y} = 20\) houses (first batch painted at original rate)

These values are chosen because:

• \(1.25 \times 4 = 5\), giving us a clean increased rate
• 20 and 60 (remaining houses) divide nicely

Step 2: Calculate actual time scenario

• Time for first 20 houses at 4 houses/week = \(20 \div 4 = 5\) weeks
• Remaining houses = \(80 - 20 = 60\) houses
• New rate = \(1.25 \times 4 = 5\) houses/week
• Time for remaining 60 houses at 5 houses/week = \(60 \div 5 = 12\) weeks
• Total actual time = \(5 + 12 = 17\) weeks

Step 3: Calculate hypothetical time (all at original rate)

Time to paint all 80 houses at 4 houses/week = \(80 \div 4 = 20\) weeks

Step 4: Find the fraction

Fraction = Actual time ÷ Hypothetical time = \(17 \div 20 = 0.85\)

Step 5: Verify with answer choices

Testing answer choice (B): \(0.8 + 0.0025\mathrm{y}\)

\(= 0.8 + 0.0025(20)\)

\(= 0.8 + 0.05\)

\(= 0.85\) ✓

The answer is (B).