4.3 Rates, Work-Time & Distance-Speed Word Problems

4.3 Rates, Work-Time & Distance-Speed Word Problems

Rates, work-time allocations, and motion word problems form one of the largest clusters of Quantitative Reasoning questions on the HEC HAT. Whether calculating the time required for combined crews to complete a construction project, determining net filling times for leaking cisterns, or computing train clearance times across moving platforms, candidate success relies on establishing structured rate equations. This section covers the fundamental formulas, relative motion principles, harmonic speed averages, and net work rate systems tested on the HEC HAT.

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Basic Speed-Distance-Time Relations & Unit Conversions

The foundation of all motion word problems rests on the classic tripartite linear formula:

$\text{Distance } (d) = \text{Speed } (r) \times \text{Time } (t)$

$r = \frac{d}{t}, \quad t = \frac{d}{r}$

Essential Unit Conversions

Speed on the HEC HAT is commonly given in kilometers per hour ($\text{km/h}$) or meters per second ($\text{m/s}$). Converting between these units rapidly without scratchwork errors is critical.

$1 \text{ km/h} = \frac{1,000 \text{ meters}}{3,600 \text{ seconds}} = \frac{5}{18} \text{ m/s}$

$1 \text{ m/s} = \frac{3,600 \text{ meters}}{1,000 \text{ seconds}} = \frac{18}{5} \text{ km/h}$

• To convert $\text{km/h} \rightarrow \text{m/s}$: Multiply by $\frac{5}{18}$. (e.g., $90 \text{ km/h} \times \frac{5}{18} = 25 \text{ m/s}$).
• To convert $\text{m/s} \rightarrow \text{km/h}$: Multiply by $\frac{18}{5}$. (e.g., $20 \text{ m/s} \times \frac{18}{5} = 72 \text{ km/h}$).

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Average Speed & Harmonic Mean for Equal Distances

Average speed is defined as total distance traveled divided by total elapsed time:

$S_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$

Equal Distance Traveled at Different Speeds

When an object covers two equal distances $d$ at speeds $v_1$ and $v_2$ (e.g., a round trip from City A to City B and back), taking the arithmetic mean $\frac{v_1 + v_2}{2}$ yields an incorrect result. Because more time is spent traveling at the slower speed, the overall average speed is the Harmonic Mean of the two speeds:

$S_{\text{avg}} = \frac{2 v_1 v_2}{v_1 + v_2}$

Worked Example: A vehicle travels from Lahore to Islamabad at $60 \text{ km/h}$ and returns along the exact same route at $40 \text{ km/h}$. Calculate the average speed for the round trip.

• Using the Harmonic Mean formula: $S_{\text{avg}} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4,800}{100} = 48 \text{ km/h}$.
• Note: The simple average $(60+40)/2 = 50 \text{ km/h}$ is incorrect because the return journey takes longer!

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Relative Speed Concepts (Opposite vs. Same Direction)

Relative speed measures the rate at which the distance between two moving objects is decreasing or increasing.

OPPOSITE DIRECTIONS (Approaching/Diverging):   v_rel = v1 + v2
   [ Object 1 ---> v1 ]       [ v2 <--- Object 2 ]

SAME DIRECTION (Overtaking/Chasing):          v_rel = |v1 - v2|
   [ Object 1 ---> v1 ]   [ Object 2 ---> v2 ]

Scenarios

1. Objects Moving in Opposite Directions: Relative speed is the sum of individual speeds: $v_{\text{rel}} = v_1 + v_2$
2. Objects Moving in the Same Direction: Relative speed is the absolute difference of individual speeds: $v_{\text{rel}} = |v_1 - v_2|$
3. Time to Meet / Overtake: $t = \frac{\text{Initial Distance Between Objects}}{v_{\text{rel}}}$

Worked Example: Two trains start simultaneously from Karachi and Multan, $600 \text{ km}$ apart, traveling toward each other at $70 \text{ km/h}$ and $50 \text{ km/h}$ respectively. How long before they pass each other?

• Relative Speed $v_{\text{rel}} = 70 + 50 = 120 \text{ km/h}$.
• Time $t = \frac{600 \text{ km}}{120 \text{ km/h}} = 5 \text{ hours}$.

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Train Clearance Formulas

Train clearance problems test whether candidates account for the physical length of the train when passing stationary or moving objects.

Worked Example: A train $200$ meters long traveling at $72 \text{ km/h}$ crosses a platform $300$ meters long. Calculate clearance time.

1. Convert speed: $72 \text{ km/h} \times \frac{5}{18} = 20 \text{ m/s}$.
2. Total distance $d = L_{\text{train}} + L_{\text{platform}} = 200 + 300 = 500 \text{ meters}$.
3. Time $t = \frac{500}{20} = 25 \text{ seconds}$.

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Combined & Sequential Work-Time Dynamics

Work problems treat job completion using additive rates. If an agent completes $1$ complete job in $T$ days, their individual work rate $R$ is $R = \frac{1}{T}$ jobs per day.

Combined Work Rate Formula

If Person A completes a task in $A$ days and Person B completes it in $B$ days, their combined rate working together is:

$R_{\text{combined}} = R_A + R_B = \frac{1}{A} + \frac{1}{B} = \frac{A + B}{A \cdot B}$

$\text{Total Combined Time } (T_{\text{total}}) = \frac{A \cdot B}{A + B}$

For three workers $A, B, C$:

$\frac{1}{T_{\text{total}}} = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}$

Sequential Work Schedules (Leaving or Joining Mid-Way)

When workers start together but one leaves before completion, calculate the fraction of work completed during joint operation first, then solve for remaining work.

Worked Example: Worker A can complete a project in $10$ days, while Worker B can complete it in $15$ days. They work together for $4$ days, after which Worker A leaves. How many additional days will Worker B take alone to finish the project?

• Rate $R_A = \frac{1}{10}$, Rate $R_B = \frac{1}{15}$.
• Combined Rate $R_{\text{comb}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$ job/day.
• Work completed in $4$ days $= 4 \times \frac{1}{6} = \frac{2}{3}$ of the project.
• Remaining Work $= 1 - \frac{2}{3} = \frac{1}{3}$.
• Time for B alone $= \frac{\text{Remaining Work}}{R_B} = \frac{1/3}{1/15} = \frac{15}{3} = 5 \text{ days}$.

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Pipes & Cisterns (Inlets & Outlets)

Pipes and cisterns operate under identical logic to work problems, with inlet pipes acting as positive rates and outlet pipes / leaks acting as negative rates.

$\text{Net Rate} = \sum R_{\text{inlets}} - \sum R_{\text{outlets}}$

$\frac{1}{T_{\text{net}}} = \frac{1}{A} + \frac{1}{B} - \frac{1}{C}$

Worked Example: Pipe X can fill a tank in $6$ hours, Pipe Y can fill it in $8$ hours, and Pipe Z can empty the full tank in $12$ hours. If all three pipes are opened simultaneously when the tank is empty, how long will it take to fill the tank?

• Net Rate $= \frac{1}{6} + \frac{1}{8} - \frac{1}{12}$.
• Find common denominator ($24$): Net Rate $= \frac{4 + 3 - 2}{24} = \frac{5}{24}$ tanks per hour.
• Total Time $T_{\text{net}} = \frac{24}{5} = 4.8 \text{ hours}$ (or $4 \text{ hours } 48 \text{ minutes}$).