7.2 Distance-Rate-Time and Work Problems

Distance, Rate, and Time Relationships

One of the most common applications of rates on the SHSAT is the relationship between distance ($d$), rate of speed ($r$), and time ($t$). The fundamental formula governing these problems is: $\text{Distance} = \text{Rate} \times \text{Time}$ $d = r \times t$

Depending on what the problem asks you to find, you can rearrange this formula algebraically:

• To find rate: $r = \frac{d}{t}$
• To find time: $t = \frac{d}{r}$

Because the SHSAT does not permit calculators, many problems are designed with easily divisible numbers or require you to set up algebraic relationships rather than do heavy computation. A common challenge is ensuring that all units of measurement are consistent. If a speed is given in miles per hour, and a time is given in minutes, you must convert the time to hours before applying the formula.

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Opposite Direction and Same Direction Travel

Many SHSAT problems involve two moving objects. These typically fall into two categories:

Case 1: Objects Moving in Opposite Directions

When two objects move toward each other or away from each other, their relative speed is the sum of their individual speeds. $\text{Combined Speed} = r_1 + r_2$ The total distance between them is the sum of the distances traveled by each object: $d_{\text{total}} = (r_1 \times t) + (r_2 \times t) = (r_1 + r_2)t$

Let's look at an example: Two cyclists start at the same time from towns 84 miles apart and ride toward each other. Cyclist A travels at a constant speed of 12 miles per hour, and Cyclist B travels at a constant speed of 16 miles per hour. How long will it take them to meet? Since they are traveling toward each other, their rates add up: $\text{Combined Speed} = 12 \text{ mph} + 16 \text{ mph} = 28 \text{ mph}$ Now use the time formula: $t = \frac{d_{\text{total}}}{\text{Combined Speed}} = \frac{84 \text{ miles}}{28 \text{ mph}}$ Simplify by dividing both numerator and denominator by 7: $t = \frac{12}{4} = 3 \text{ hours}$ It will take them 3 hours to meet.

Case 2: Objects Moving in the Same Direction (Chase or Catch-Up)

When one object chases another, the distance between them decreases at a rate equal to the difference in their speeds. $\text{Relative Speed of Overtaking} = r_{\text{faster}} - r_{\text{slower}}$ Alternatively, you can set their distances equal to each other at the point of overtaking: $d_{\text{faster}} = d_{\text{slower}}$

Example: A pedestrian leaves a transit station walking at a rate of 4 miles per hour. Three hours later, a runner leaves the same station and travels along the same route at a rate of 10 miles per hour. How many hours after leaving the station will the runner catch up to the pedestrian? Let $t$ represent the travel time of the runner (in hours). Since the pedestrian started 3 hours earlier, the pedestrian's travel time is $t + 3$ hours. At the point where the runner catches up, both have traveled the exact same distance from the station. Set their distances equal: $d_{\text{runner}} = d_{\text{pedestrian}}$ $10t = 4(t + 3)$ Distribute the 4: $10t = 4t + 12$ Subtract $4t$ from both sides: $6t = 12$ Divide by 6: $t = 2 \text{ hours}$ The runner will catch up to the pedestrian 2 hours after starting.

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The Average Speed Trap

A classic trap on the SHSAT is asking for the average speed of a round trip. If you drive to a location at 40 miles per hour and return along the same route at 60 miles per hour, the average speed for the entire trip is not the simple average of the speeds (which would be $\frac{40 + 60}{2} = 50$ mph). This is because you spend more time traveling at the slower speed than at the faster speed.

To find the average speed, you must always use the definition: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Let's work through this scenario: A truck travels 120 miles from a warehouse to a delivery site at an average speed of 30 miles per hour. The truck then returns along the same 120-mile route at an average speed of 60 miles per hour. What is the average speed of the truck for the entire round trip? First, calculate the time spent on each leg of the journey: $\text{Time going} = \frac{120 \text{ miles}}{30 \text{ mph}} = 4 \text{ hours}$ $\text{Time returning} = \frac{120 \text{ miles}}{60 \text{ mph}} = 2 \text{ hours}$ Second, calculate the total distance and total time: $\text{Total Distance} = 120 \text{ miles} + 120 \text{ miles} = 240 \text{ miles}$ $\text{Total Time} = 4 \text{ hours} + 2 \text{ hours} = 6 \text{ hours}$ Third, divide total distance by total time: $\text{Average Speed} = \frac{240 \text{ miles}}{6 \text{ hours}} = 40 \text{ mph}$ The average speed for the round trip is 40 miles per hour. Notice that 40 is lower than the simple arithmetic average of 45 mph, reflecting the fact that the truck spent twice as much time (4 hours vs. 2 hours) traveling at the slower speed.

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Rate-of-Work Problems

Work problems involve one or more people (or machines) completing a task. The key to these problems is to work with rates rather than times.

If a worker can complete a job in $H$ hours, their work rate is: $\text{Rate} = \frac{1}{H} \text{ of the job per hour}$

When multiple workers work together, their individual rates of work add up to a combined rate: $\text{Combined Rate} = R_1 + R_2 + R_3 + \dots$ Once you have the combined rate, the time required to complete 1 full job together is: $\text{Combined Time} = \frac{1}{\text{Combined Rate}}$

Let's solve a representative work problem: Working alone, Alicia can paint a room in 6 hours, while Beatrice can paint the same room in 3 hours. How many hours will it take them to paint the room if they work together? First, find their individual hourly rates:

• Alicia's rate = $\frac{1}{6}$ of the room per hour.
• Beatrice's rate = $\frac{1}{3}$ of the room per hour. Second, add their rates to find the combined rate: $\text{Combined Rate} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \text{ of the room per hour}$ Third, calculate the time to complete the job by taking the reciprocal of the combined rate: $\text{Combined Time} = \frac{1}{\frac{1}{2}} = 2 \text{ hours}$ Working together, they will finish in 2 hours.

For two workers with individual times $a$ and $b$, you can also use the shortcut formula for combined time: $\text{Combined Time} = \frac{a \times b}{a + b}$ Substituting our values: $\text{Combined Time} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \text{ hours}$

Advanced Work Problems: Different Start Times

A more challenging variation of work problems occurs when one worker starts before another. To solve these, calculate the fraction of the job completed by the first worker before the second worker joins. Then, divide the remaining fraction of the job by their combined work rate.

For example, if Alicia paints alone for 2 hours, she completes $2 \times \frac{1}{6} = \frac{1}{3}$ of the room. The remaining fraction of the job is $1 - \frac{1}{3} = \frac{2}{3}$. When Beatrice joins, they work at their combined rate of $\frac{1}{2}$ room per hour. The time they spend working together is: $\text{Time together} = \frac{\text{Remaining Job}}{\text{Combined Rate}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{\text{Remaining Job}}{\text{Combined Rate}} = \frac{2}{3} \times 2 = \frac{4}{3} \text{ hours}$ This translates to 1 hour and 20 minutes. Breaking the job down into chronological phases prevents algebra mistakes when rates are variable.