4.3 Work-Rate, Distance & Mixture Word Problems

4.3 Work-Rate, Distance & Mixture Word Problems

Quick Summary: Technical word problems test an electrician's ability to translate real-world job site scenarios into algebraic equations. Combined work-rate problems require adding individual rates of work ($\frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2}$) to determine how fast multiple technicians complete a project together. Worker-day calculations utilize inverse proportions ($N_1 D_1 = N_2 D_2$) to project schedule adjustments based on crew size. Distance equations ($d = rt$) analyze vehicle dispatch and cable-pulling speeds, while mixture equations ($V_1 C_1 + V_2 C_2 = V_{\text{total}} C_{\text{final}}$) determine exact concentration percentages when combining liquid trade compounds.

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Combined Work-Rate Problems

In electrical contracting, project managers frequently combine workers of differing skill levels and speeds (such as journeymen and apprentices) to complete tasks. A common error made by candidates is adding individual completion times together. Times cannot be added directly; rates of work must be added.

The Work-Rate Formula

Let total work $W$ equal $1$ complete job (such as wiring a panel, pulling wire, or installing fixtures).

• If Person A completes the job in $t_1$ hours, Person A's work rate is $R_1 = \frac{1}{t_1}$ jobs per hour.
• If Person B completes the job in $t_2$ hours, Person B's work rate is $R_2 = \frac{1}{t_2}$ jobs per hour.

When working simultaneously, their combined rate $R_{\text{combined}}$ is the sum of their individual rates:

$R_{\text{combined}} = R_1 + R_2 = \frac{1}{t_1} + \frac{1}{t_2}$

The total time $T$ required for both workers to complete the job together is the reciprocal of the combined rate:

$T = \frac{1}{R_{\text{combined}}} = \frac{1}{\frac{1}{t_1} + \frac{1}{t_2}} = \frac{t_1 t_2}{t_1 + t_2}$

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Step-by-Step Work-Rate Walkthrough

Scenario:

Electrician Dan can wire a commercial distribution panel in $3\text{ hours}$, while apprentice Leo takes $6\text{ hours}$ to complete the exact same panel. Working together, how many hours will it take them to wire the panel?

1. Define Individual Work Rates:

   • Dan's rate: $R_{\text{Dan}} = \frac{1}{3}\text{ panel per hour}$
   • Leo's rate: $R_{\text{Leo}} = \frac{1}{6}\text{ panel per hour}$

2. Sum the Work Rates using a Common Denominator ($6$): $R_{\text{combined}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ panel per hour}$

   Together, Dan and Leo complete $\frac{1}{2}$ of the panel every hour.

3. Calculate Total Combined Time ($T$): $T = \frac{1}{R_{\text{combined}}} = \frac{1}{\frac{1}{2}} = 2.0\text{ hours}$

Alternative Product-over-Sum Formula Check:

$T = \frac{t_1 \cdot t_2}{t_1 + t_2} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2.0\text{ hours}$

Working together, Dan and Leo wire the panel in exactly 2.0 hours.

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Worker-Day & Crew Productivity Proportions

Crew scheduling relies on the concept of total labor volume, measured in worker-days or worker-hours. Assuming each electrician works at a uniform rate, the total amount of labor required to complete a job remains constant.

The Inverse Labor Principle

$\text{Total Labor Volume} = N \times D$

Where:

• $N$ = Number of workers in the crew
• $D$ = Number of days required to complete the project

If the crew size changes from $N_1$ to $N_2$, the days required change inversely from $D_1$ to $D_2$:

$N_1 \cdot D_1 = N_2 \cdot D_2$

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Step-by-Step Worker-Day Walkthrough

Scenario:

A crew of $6\text{ electricians}$ requires $10\text{ days}$ to pull main feeder wire through a high-rise commercial building. Assuming all electricians work at the exact same rate, how many days would it take a crew of $10\text{ electricians}$ to complete the same wire-pulling job?

1. Calculate Total Worker-Days Required: $\text{Total Worker-Days} = N_1 \times D_1 = 6\text{ electricians} \times 10\text{ days} = 60\text{ worker-days}$

2. Set up the Equation for the New Crew Size ($N_2 = 10$): $N_2 \times D_2 = 60$ $10 \times D_2 = 60$

3. Solve for $D_2$: $D_2 = \frac{60}{10} = 6\text{ days}$

Increasing the crew size from $6$ to $10$ electricians reduces the required duration from $10$ days down to 6 days.

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Distance, Speed & Time Relationships

The fundamental equation of motion applies across electrical service dispatches, utility bucket truck travel, and automated cable-pulling tension calculations:

$d = r \cdot t$

Where:

• $d$ = Distance traveled or cable length pulled (miles, feet)
• $r$ = Uniform rate of speed (mph, ft/min)
• $t$ = Elapsed time (hours, minutes)

Algebraic Transformations:

$r = \frac{d}{t} \quad \text{and} \quad t = \frac{d}{r}$

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Application: Motorized Cable-Pulling Rate

A motorized tugger pulls $450\text{ feet}$ of 500 kcmil copper feeder cable through an underground duct bank at a constant speed of $15\text{ feet per minute}$.

$\text{Time required } t = \frac{d}{r} = \frac{450\text{ ft}}{15\text{ ft/min}} = 30\text{ minutes}$

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Chemical & Solution Mixture Problems

Electricians handle liquid chemical compounds such as wire-pulling lubricants (glycol-based friction reducers), battery electrolyte solutions, and anti-freeze mixtures for outdoor snow-melting conduit systems.

The Weighted Mixture Equation

When mixing two solutions of volumes $V_1$ and $V_2$ with concentration percentages $C_1$ and $C_2$, the amount of pure active substance in each solution combines additively:

$\text{Pure Substance in Solution 1} + \text{Pure Substance in Solution 2} = \text{Total Pure Substance}$ $(V_1 \times C_1) + (V_2 \times C_2) = V_{\text{total}} \times C_{\text{final}}$

Where:

• $V_{\text{total}} = V_1 + V_2$
• $C_{\text{final}} = \frac{(V_1 C_1) + (V_2 C_2)}{V_1 + V_2}$

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Step-by-Step Solution Mixture Walkthrough

Scenario:

An electrician mixes $2\text{ liters}$ of an $80%$ glycol conduit lubricant solution with $3\text{ liters}$ of a $30%$ glycol solution. What is the glycol concentration percentage of the resulting $5\text{-liter}$ mixture?

1. Calculate Pure Glycol in Solution 1: $\text{Pure Glycol}_1 = V_1 \times C_1 = 2\text{ L} \times 0.80 = 1.6\text{ Liters}$

2. Calculate Pure Glycol in Solution 2: $\text{Pure Glycol}_2 = V_2 \times C_2 = 3\text{ L} \times 0.30 = 0.9\text{ Liters}$

3. Sum Total Pure Glycol and Total Volume: $\text{Total Pure Glycol} = 1.6 + 0.9 = 2.5\text{ Liters}$ $\text{Total Volume} = 2 + 3 = 5.0\text{ Liters}$

4. Calculate Final Concentration Percentage ($C_{\text{final}}$): $C_{\text{final}} = \frac{\text{Total Pure Glycol}}{\text{Total Volume}} = \frac{2.5\text{ L}}{5.0\text{ L}} = 0.50 = 50\%$

The resulting $5\text{-liter}$ mixture has a glycol concentration of exactly 50%. Mastering work-rate, inverse worker-day proportions, uniform speed, and solution mixture mathematics enables candidates to solve complex applied word problems with confidence.