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Algebra and Precalculus A Practical Guide

Cameron Geisler

Section 7.7 Word Problems with Rational Equations

Some word problem situations lead to a rational equation. There are 3 broad categories of problems:
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  • Work problems.
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  • Rate/speed problems.
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  • Current/wind problems.
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These are very standard categories, so be sure to master each one. Also, review Word Problem Solving Strategy for learning how to think about word problems broadly.
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Subsection 7.7.1 Work Problems (Combined Work Rate)

In these problems, two people or machines work together at different rates to complete a task.
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Example 7.7.1. Painting a Wall.

Alex can paint a wall in 4 hours. Brianna can paint the same wall in 6 hours. If they work together and do not stop, how long will it take them to paint the wall?
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We want to find the time it takes for Alex and Brianna to paint the wall together, so let \(t\) be the number of hours it takes them to paint the wall, working together.
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Think about what happens as time passes. If Alex can finish 1 whole wall in 4 hours, then,
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  • After 1 hour, he has finished \(\frac{1}{4}\) of the wall.
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  • After 2 hours, he’s finished \(\frac{2}{4}\) of the wall.
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  • After 3 hours, he’s finished \(\frac{3}{4}\text{.}\)
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  • In general, after \(t\) hours, he has finished the fraction \(\frac{t}{4}\) of the wall.
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Similarly, Brianna can finish the wall in 6 hours, so,
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  • After 1 hour, she’s finished \(\frac{1}{6}\) of the wall.
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  • After 2 hours, she’s finished \(\frac{2}{6}\text{.}\)
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  • And in general, after \(t\) hours, she has finished \(\frac{t}{6}\) of the wall.
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Working together, after \(t\) hours they have finished one whole wall, or 1 of the wall. Then,
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\begin{gather*} \begin{pmatrix} \text{fraction done} \\ \text{by Alex} \end{pmatrix} + \begin{pmatrix} \text{fraction done} \\ \text{by Brianna} \end{pmatrix} = \underbrace{1}_{\text{entire job}} \end{gather*}
Then,
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\begin{gather*} \frac{t}{4} + \frac{t}{6} = 1 \end{gather*}
Then, to solve this equation, clear the denominators by multiplying both sides by 12 (the LCD),
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\begin{align*} 12 \brac{\frac{t}{4} + \frac{t}{6}} \amp = 12 \cdot 1\\ 3t + 2t \amp = 12\\ 5t \amp = 12 \amp\amp \text{collecting like terms}\\ t \amp = \frac{12}{5} = 2.4 \amp\amp \text{dividing both sides by 5} \end{align*}
Therefore, together, they would take 2.4 hours. This is reasonable, because since they are working together, it will take somewhat less time than either of them working individually (4 hours or 6 hours).
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Another way to organize this information is in a table, with 3 columns: the time it takes, the fraction done in 1 hour, and the fraction done in \(t\) hours.
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\begin{align*} \begin{array}{c|c|c|c} \amp \text{time} \amp \text{1 hour} \amp t \text{ hours} \\ \hline \text{Alex} \amp 4 \amp \frac{1}{4} \amp \frac{t}{4} \\ \hline \text{Brianna} \amp 6 \amp \frac{1}{6} \amp \frac{t}{6} \\ \hline \text{both} \amp t \amp \frac{1}{t} \amp 1 \\ \end{array} \end{align*}
Another equivalent way to do this is to add the contributions after 1 hour, which gives,
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\begin{gather*} \frac{1}{4} + \frac{1}{6} = \frac{1}{t} \end{gather*}
This equation is equivalent to the previous equation, and so will also give the correct answer. However, I don’t think it’s as intuitive as the previous one.
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In general,
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\begin{gather*} \boxed{\begin{pmatrix} \text{fraction done} \\ \text{by A} \end{pmatrix} + \begin{pmatrix} \text{fraction done} \\ \text{by B} \end{pmatrix} = \underbrace{1}_{\text{entire job}}} \end{gather*}

Checkpoint 7.7.2. Taps Filling a Tub.

A cold tap can fill a tub in 2 minutes, and a hot tap can fill the same tub in 3 minutes. If both taps are turned on at the same time, how long will it take to fill the tub?
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Hint.
\(\frac{t}{2}+\frac{t}{3}=1\text{.}\)
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Answer.
\(1.2\) minutes.
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Checkpoint 7.7.3. Painting a House.

Together, Anna and Isla can paint a house in 17 hours. Alone, Anna can do the job 3 hours faster than Isla. How long does each person take to paint the house, to the nearest tenth of an hour?
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Hint.
\(\frac{17}{x}+\frac{17}{x+3}=1\text{,}\) where \(x\) is Anna’s time in hours.
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Answer.
Anna takes about \(32.6\) hours, Isla takes about \(35.6\) hours.
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Checkpoint 7.7.4. Filling a Container.

The cold tap fills a container 2 hours faster than the hot tap. Together they fill it in 80 minutes. Find how long each tap takes alone.
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Hint.
\(\frac{4}{3x}+\frac{4}{3\brac{x+2}}=1\text{,}\) where \(x\) is the cold tap time in hours.
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Answer.
Cold tap 2 hours, hot tap 4 hours.
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Checkpoint 7.7.5. Hoses Filling a Pool.

Two hoses fill a pool in 2 hours together. Hose A alone fills it in 3 hours. How long would hose B alone take?
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Hint.
\(\frac{2}{3}+\frac{2}{x}=1\text{.}\)
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Answer.
6 hours.
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Checkpoint 7.7.6. Painting a Large Room.

It would take Sue 4 hours to paint a large room, and it would take Bob 5 hours to paint the same room. If they work together, how long would it take them to complete the job?
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Hint.
\(\frac{t}{4}+\frac{t}{5}=1\text{.}\)
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Answer.
\(\frac{20}{9} \approx 2.2\) hours.
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Checkpoint 7.7.7. Three Workers.

One worker takes 20 hours to do a job. A second worker can do the same job in 15 hours. When a third worker is added, it takes the three of them 6 hours to do the job. How long would it take the third worker to do the job alone?
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Hint.
\(\frac{6}{20}+\frac{6}{15}+\frac{6}{x}=1\text{.}\)
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Answer.
20 hours.
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Checkpoint 7.7.8. Cleaning the Kitchen.

Jane works twice as fast as her daughter Anna. If it takes 15 minutes to clean the kitchen together, how long would it take Anna to clean the kitchen by herself?
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Hint.
\(\frac{15}{x}+\frac{30}{x}=1\text{,}\) where \(x\) is Anna’s time in minutes.
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Answer.
45 minutes.
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Checkpoint 7.7.9. Assembling a Motor.

Ken takes 3 hours longer to assemble a motor than Hans. When working together, it takes them 2 hours to assemble the motor. How long would it take Ken to do the job alone?
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Hint.
\(\frac{2}{x}+\frac{2}{x+3}=1\text{,}\) where \(x\) is Hans’s time in hours.
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Answer.
Ken would take 6 hours alone (Hans would take 3 hours).
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Checkpoint 7.7.10. Brick Layer and Apprentice.

A brick layer can build a wall in 6 hours, and his apprentice would need 16 hours to build the same wall. When they work together, the apprentice works 5 hours longer than the brick layer. How many hours does each work?
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Hint.
\(\frac{x}{6}+\frac{x+5}{16}=1\text{,}\) where \(x\) is the brick layer’s working time in hours.
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Answer.
The brick layer works 3 hours and the apprentice works 8 hours.
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Checkpoint 7.7.11. Taps and Drain.

A cold water tap can fill a tub in 6 minutes, and a hot water tap can fill the tub in 8 minutes. A drain can empty a full tub in 10 minutes. If both taps are on and the drain is open, how long will it take to fill the tub?
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Hint.
\(\frac{t}{6}+\frac{t}{8}-\frac{t}{10}=1\text{.}\)
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Answer.
\(\frac{120}{23} \approx 5.2\) min.
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Checkpoint 7.7.12. Delivering Flyers.

When they work together, Stuart and Lucy can deliver flyers to all the homes in their neighbourhood in 42 minutes. When Lucy works alone, she can finish the deliveries in 13 minutes less time than Stuart can when he works alone. When Stuart works alone, how long does he take to deliver the flyers?
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Hint.
\(\frac{42}{x}+\frac{42}{x-13}=1\text{,}\) where \(x\) is Stuart’s time in minutes.
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Answer.
Stuart takes 91 minutes alone (Lucy takes 78 minutes).
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Checkpoint 7.7.13. Harvesting Wheat.

Nikita can harvest her wheat crop in 72 hours. Her neighbour’s combine can do it in 48 hours. How long will both machines take together?
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Hint.
\(\frac{t}{72}+\frac{t}{48}=1\text{.}\)
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Answer.
They will take \(28.8\) hours working together.
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Checkpoint 7.7.14. Lyle Filling the Bathtub.

Lyle can fill the bathtub using the cold water tap in 8 minutes. When both the hot and cold water taps are fully open, he can fill the bathtub in 6 minutes. How long would it take Lyle to fill the bathtub using only the hot water tap?
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Hint.
\(\frac{6}{8}+\frac{6}{t}=1\text{.}\)
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Answer.
24 minutes.
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Checkpoint 7.7.15. Staining a Deck.

A painter can stain a deck using a sprayer in 4 hours. When two people work together, one using the sprayer and the other using a brush, they can finish the job in 3 hours. How long would it take one person to stain the deck using only a brush?
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Hint.
\(\frac{3}{4}+\frac{3}{t}=1\text{.}\)
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Answer.
6 hours.
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Checkpoint 7.7.16. Cleaning the Garage.

Jenny can clean out the garage in 5 hours. When her son helps, they can clean out the garage in 3 hours. How long would it take Jenny’s son to clean out the garage on his own?
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Hint.
\(\frac{3}{5}+\frac{3}{t}=1\text{.}\)
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Answer.
\(7.5\) hours.
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Checkpoint 7.7.17. Marcy and Apprentice.

It takes Marcy’s apprentice 9 hours longer to build a deck than it takes Marcy. When they work together, they can build the deck in 20 hours. How long would it take each person to build the deck working alone?
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Hint.
\(\frac{20}{t}+\frac{20}{t+9}=1\text{.}\)
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Answer.
Marcy takes 36 hours and her apprentice takes 45 hours.
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Checkpoint 7.7.18. Warehouse Employees.

Three employees work at a shipping warehouse. Tom fills an order in \(s\) minutes, Paco in \(s-2\) minutes, and Carl in \(s+1\) minutes. Tom and Paco together take 1 minute 20 seconds to fill an order, and Paco and Carl together take 1 minute 30 seconds. How long does each person take, and how long will all three together take to fill one order?
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Hint.
\(\frac{4}{3s}+\frac{4}{3\brac{s-2}}=1\) and \(\frac{3}{2\brac{s-2}}+\frac{3}{2\brac{s+1}}=1\text{.}\)
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Answer.
Tom takes 4 minutes, Paco takes 2 minutes, Carl takes 5 minutes, and all three together take \(\frac{20}{19}\approx 1.05\) minutes.
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Subsection 7.7.2 Relative Speed Comparison Problems

Example 7.7.19. Jess and Sarah.

Jess can run 3 miles per hour faster than her sister Sarah can walk. If Jess ran 12 miles in the same time it took Sarah to walk 8 miles, what is the speed of each sister?
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We want to find the speed of each sister, so we can let \(x\) be the speed of one sister, and then the other one differs by 3. Let \(x\) be Sarah’s speed. Then, Jess’s speed is 3 mi/h more, so it’s \(x+3\) (Note that you could have instead let \(x\) be Jess’s speed, and then Sarah would be \(x-3\)). We know they took the same amount of time, so,
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\begin{gather*} \begin{pmatrix} \text{Sarah's} \\ \text{time} \end{pmatrix} = \begin{pmatrix} \text{Jess's} \\ \text{time} \end{pmatrix} \end{gather*}
Then, remember how speed, distance, and time are related by \(v = \frac{d}{t}\) (or \(d = vt\text{,}\) or \(t = \frac{d}{v}\)). We’ll use \(t = \frac{d}{v}\) here because we want time.
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\begin{align*} \frac{8}{x} \amp = \frac{12}{x+3} \end{align*}
To solve, clear denominators by multiplying by \(x(x+3)\) (the LCD),
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\begin{align*} 8(x+3) \amp = 12x\\ 8x + 24 \amp = 12x\\ 4x \amp = 24\\ x \amp = 6 \end{align*}
Therefore, Sarah’s speed is 6 mi/h, and Jess’s speed is 3 mi/h faster, so it is 9 mi/h.
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In general, remember that,
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\begin{gather*} \boxed{v = \frac{d}{t} \qquad \text{or} \qquad d = vt \qquad \text{or} \qquad t = \frac{d}{v}} \qquad \begin{cases} d \rightarrow \text{distance} \\ t \rightarrow \text{time} \\ v \rightarrow \text{speed (or velocity)} \end{cases} \end{gather*}

Remark 7.7.20.

You may remember these relationships from this triangle (the distance speed time triangle, or DST triangle):
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Checkpoint 7.7.21. Cyclists Comparison.

One cyclist averages 3 km/h faster than a second cyclist. The faster cyclist rode 270 km in the same time the slower cyclist rode 225 km. What was the average speed of each cyclist?
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Hint.
\(\frac{270}{v}=\frac{225}{v-3}\text{.}\)
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Answer.
The faster cyclist averaged 18 km/h and the slower cyclist averaged 15 km/h.
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Checkpoint 7.7.22. John and Susan.

John’s family travels 300 km from their home to a family reunion. His cousin Susan and her family take the same amount of time to travel 200 km from their home. One of the vehicles travels 30 km/h faster than the other. Determine the speed of both vehicles.
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Hint.
\(\frac{300}{v+30}=\frac{200}{v}\text{.}\)
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Answer.
John’s family travels at 90 km/h and Susan’s family travels at 60 km/h.
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Checkpoint 7.7.23. Delivery Routes.

Driver A and B have two different delivery routes. Driver A’s route is 80 km, and driver B’s route is 100 km. Driver B travels 10 km/h faster than driver A and finishes 10 minutes earlier. What are the speeds of each driver?
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Hint.
\(\frac{80}{x}-\frac{100}{x+10}=\frac{1}{6}\text{.}\)
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Answer.
Driver A’s speed is 30 km/h and driver B’s speed is 40 km/h.
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Checkpoint 7.7.24. Ted’s Trip.

Ted drove 275 km west at 10 km/h slower than when he drove 300 km east. His westward trip took 30 minutes longer. Find his speeds.
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Hint.
\(\frac{275}{x-10}=\frac{300}{x}+\frac{1}{2}\text{.}\)
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Answer.
Ted’s speed was 60 km/h east and 50 km/h west.
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Checkpoint 7.7.25. Ann and Ray.

Ann and Ray each ride 4 km. Ann starts 1 minute earlier, and Ray rides 1 km/h faster. They finish at the same time. Find Ann’s speed.
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Hint.
\(\frac{4}{x}-\frac{4}{x+1}=\frac{1}{60}\text{.}\)
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Answer.
Ann rides 15 km/h and Ray rides 16 km/h.
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Checkpoint 7.7.26. Bronwyn and Aaron.

Bronwyn rides her electric bicycle 10 km/h faster than Aaron. Bronwyn can travel 60 km in the same time that it takes Aaron to travel 40 km. Determine Bronwyn’s average speed and Aaron’s average speed.
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Hint.
\(\frac{60}{s+10}=\frac{40}{s}\text{.}\)
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Answer.
Aaron’s speed is 20 km/h and Bronwyn’s speed is 30 km/h.
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Checkpoint 7.7.27. Cows Travelling.

Cows travelled 70 km to Forde Lake, taking 4 days longer than travelling 60 km beyond the lake. They travelled 5 km/h slower before the lake. What was their speed beyond the lake?
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Hint.
\(\frac{70}{x-5}=\frac{60}{x}+96\text{.}\)
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Answer.
Their speed beyond the lake was about \(5.7\) km/h.
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Checkpoint 7.7.28. Airplane and Car.

The average speed of an airplane is ten times that of a car. It takes the airplane 18 hours less than the car to travel 1000 km. Determine the average speeds of the airplane and the car.
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Hint.
\(\frac{1000}{s}-\frac{1000}{10s}=18\text{.}\)
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Answer.
The car’s speed is 50 km/h and the airplane’s speed is 500 km/h.
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Checkpoint 7.7.29. Henry and Brandon.

Henry’s average running speed is 1 km/h greater than Brandon’s. In a 10-km practice race, Brandon finished 2 minutes behind Henry. Determine the average running speed of each person.
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Hint.
\(\frac{10}{s}-\frac{10}{s+1}=\frac{1}{30}\text{.}\)
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Answer.
Brandon’s speed is approximately \(16.8\) km/h and Henry’s speed is approximately \(17.8\) km/h.
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Subsection 7.7.3 Current/Wind Travel Problems

Example 7.7.30. Kayaking on a River.

You are kayaking on a river. You start paddling from a dock 8 km upstream to a campsite, then immediately turns around and paddles 8 km back downstream to the dock. The river’s current flows at 2 km/h. The total time for the trip is 3 hours. What is your paddling speed in still water?
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Let \(v\) be your speed in still water. Then, your speed upstream is \(v-2\text{,}\) because the 2 km/h current is pushing you backwards. Similarly, your speed downstream is \(v+2\text{,}\) because the 2 km/h current is helping you go faster. Then, the total time for the trip is 3 hours, so,
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\begin{gather*} \begin{pmatrix} \text{time} \\ \text{upsteam} \end{pmatrix} + \begin{pmatrix} \text{time} \\ \text{downstream} \end{pmatrix} = \begin{pmatrix} \text{total} \\ \text{time} \end{pmatrix} \end{gather*}
Then, to calculate time, use \(t = \frac{d}{v}\text{,}\)
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\begin{gather*} \frac{8}{v-2} + \frac{8}{v+2} = 3 \end{gather*}
To solve, multiply both sides by \((v-2)(v+2)\) (the LCD),
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\begin{align*} 8(v+2) + 8(v-2) \amp = 3(v-2)(v+2)\\ 8v + 16 + 8v - 16 \amp = 3(v^2 - 4)\\ 16v \amp = 3v^2 - 12\\ 0 \amp = 3v^2 - 16v - 12\\ 0 \amp = (2v+3)(v-6)\\ v \amp = -\frac{3}{2}, 6 \end{align*}
We keep only the positive solution, because paddling speed can’t be negative. Therefore, your paddling speed in still water is 6 km/h.
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If you like, to help organize the information, you can also make a table, with the distance, rate, and time going upstream and downstream,
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\begin{align*} \begin{array}{c|c|c|c} \amp d \amp r \amp t \\ \hline \text{upstream} \amp 8 \amp v-2 \amp \frac{8}{v-2} \\ \hline \text{downstream} \amp 8 \amp v+2 \amp \frac{8}{v+2} \end{array} \end{align*}
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Checkpoint 7.7.31. Two Kayakers.

Two kayakers paddle 18 km downstream with the current in the same time it takes them to go 8 km upstream against the current. The rate of the current is 3 km/h. Find their speed in still water.
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Hint.
\(\frac{18}{v+3}=\frac{8}{v-3}\text{.}\)
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Answer.
\(\frac{39}{5}=7.8\) km/h.
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Checkpoint 7.7.32. Boat Travel Times.

The current of a river is \(c\) miles per hour. A boat’s speed in still water is 12 miles per hour. Write expressions for the time to travel 150 miles downriver, the time to travel 150 miles upriver, and the difference between these times.
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Hint.
Use \(t=\frac{d}{r}\) with speeds \(12+c\) and \(12-c\text{.}\)
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Answer.
(a) downriver: \(\frac{150}{12+c}\) hours, (b) upriver: \(\frac{150}{12-c}\) hours, (c) difference: \(\frac{150}{12-c}-\frac{150}{12+c}\) hours.
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Checkpoint 7.7.33. Plane with Wind.

A plane flies 1200 miles with the wind and then 1000 miles against the wind. The wind speed is 50 miles per hour. If the total travel time for the round trip is 4 hours, what is the speed of the plane in still air?
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Hint.
\(\frac{1200}{v+50}+\frac{1000}{v-50}=4\text{.}\)
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Answer.
550 mi/h.
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Checkpoint 7.7.34. Rose Paddling.

Rose paddles at a speed of 5 km/h in still water. She travels 12 km upstream, and it takes her 1 hour longer than it does to paddle the same distance downstream. Find the speed of the current in the river.
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Hint.
\(\frac{12}{5-c}=\frac{12}{5+c}+1\text{.}\)
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Answer.
1 km/h.
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Checkpoint 7.7.35. Boat Upstream and Downstream.

A boat can travel at 25 km/h in still water. The boat travels 30 km upstream (against the current) then turns around and travels the same distance back (with the current). The total trip took 5 hours. Find the speed of the current.
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Hint.
\(\frac{30}{25-c}+\frac{30}{25+c}=5\text{.}\)
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Answer.
\(5\sqrt{13}\approx 18.0\) km/h.
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Checkpoint 7.7.36. Round Trip to Town.

The speed of the current in a river is 2 km/h. A boat made a round trip to a town 24 km away in a total of 5 hours. What was the speed of the boat in still water?
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Hint.
\(\frac{24}{v+2}+\frac{24}{v-2}=5\text{.}\)
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Answer.
10 km/h.
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Checkpoint 7.7.37. Canoe on a Lake.

The time it takes for a canoe to go 3 miles upstream and back 3 miles downstream is 4 hours. The current in the lake is 1 mile per hour. Find the average speed of the canoe in still water.
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Hint.
\(\frac{3}{v-1}+\frac{3}{v+1}=4\text{.}\)
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Answer.
The canoe’s speed in still water is 2 miles per hour.
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Checkpoint 7.7.38. Plane Against Wind.

A plane travels 500 km against the wind and then 600 km with the wind in the same time. The wind speed is 20 km/h. Find the speed of the plane in still air.
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Hint.
\(\frac{500}{v-20}=\frac{600}{v+20}\text{.}\)
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Answer.
220 km/h.
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Checkpoint 7.7.39. Plane Flying 500 km.

A plane is flying 500 km each way. Its air speed (speed in still air) on both trips is 165 km/h. On the flight there, there is a tail wind, and on the way back there is a head wind. Calculate the total time for the round trip for a wind speed of 30 km/h and for a wind speed of 40 km/h.
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Hint.
For wind speed \(w\text{,}\) use \(T(w)=\frac{500}{165+w}+\frac{500}{165-w}\text{.}\)
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Answer.
For \(w=30\) km/h, \(T=\frac{2200}{351}\approx 6.27\) hours; for \(w=40\) km/h, \(T=\frac{264}{41} \approx 6.44\) hours.
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Checkpoint 7.7.40. Boat Travel Comparison.

A boat travels 108 km downstream in the same time it travels 78 km upstream. The current is 10 km/h. Find the boat’s speed in still water.
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Hint.
\(\frac{108}{v+10}=\frac{78}{v-10}\text{.}\)
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Answer.
The boat’s still-water speed is 62 km/h.
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Checkpoint 7.7.41. Boat Speed.

A boat travels 40 km downstream in the same time it takes to travel 30 km upstream. If the current flows at 6 km/h, what is the speed of the boat in still water?
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Hint.
\(\frac{40}{v+6}=\frac{30}{v-6}\text{.}\)
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Answer.
The speed of the boat in still water is 42 km/h.
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Checkpoint 7.7.42. Speed of Current.

The speed of a boat in still water is 10 mph. The boat travels 24 miles upstream and back downstream in a total of 5 hours. What is the speed of the current?
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Hint.
\(\frac{24}{10-c}+\frac{24}{10+c}=5\text{.}\)
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Answer.
The speed of the current is 2 mph.
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Checkpoint 7.7.43. Two Friends in a Canoe.

Two friends paddle a canoe at 6 km/h in still water. They travel 2 km upstream and then 2 km back downstream in a total of 1 hour. Find the speed of the current.
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Hint.
\(\frac{2}{6-c}+\frac{2}{6+c}=1\text{.}\)
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Answer.
\(2\sqrt{3}\approx 3.5\) km/h.
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Checkpoint 7.7.44. Boat Average Speed.

A boat travels 4 km upstream in the same time that it takes the boat to travel 10 km downstream. The average speed of the current is 3 km/h. What is the average speed of the boat in still water?
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Hint.
\(\frac{10}{s+3}=\frac{4}{s-3}\text{.}\)
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Answer.
7 km/h.
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Subsection 7.7.4 Average Speed and Mixed-Rate Problems

Checkpoint 7.7.45. Driving to Work.

A woman drives to work at an average speed of 50 mph. The average speed of the return trip home is 30 mph. What is the average speed for the round trip?
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Hint.
\(\frac{d}{50}+\frac{d}{30}=\frac{2d}{v}\text{,}\) and \(d\) cancels out.
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Answer.
\(37.5\) mph.
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Checkpoint 7.7.46. Round Trip Speed.

On a 100 km round trip, Jessica averages 40 km/h to her destination and 60 km/h returning. What is the average speed for the entire trip?
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Hint.
\(\frac{d}{40}+\frac{d}{60}=\frac{d}{x}\text{.}\)
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Answer.
48 km/h.
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Checkpoint 7.7.47. Priya’s Trip.

Priya decides to go to the store which is 8 km from her house. She first ran at a rate of 7 km/h, then walked the rest of the trip at a speed of 3 km/h. If the total trip took 2 hours, how many kilometers did she walk?
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Hint.
\(\frac{8-x}{7}+\frac{x}{3}=2\text{.}\)
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Answer.
\(4.5\) km.
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Checkpoint 7.7.48. Reading a Book.

You have 21 days to read a 518-page book. After reading half, you need to increase your reading rate by 12 pages/day to finish on time. What was your initial rate?
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Hint.
if the initial rate is \(x\) then \(\frac{259}{x}+\frac{259}{x+12} = 21\text{.}\)
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Answer.
\(\frac{19+\sqrt{1693}}{3} \approx 20.0\) pages/day.
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Checkpoint 7.7.49. Car Average Speed.

A car travels to work at 30 km/h and returns at 60 km/h. Find the average speed for the round trip.
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Hint.
\(\frac{d}{30}+\frac{d}{60}=\frac{2d}{x}\text{.}\)
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Answer.
The average speed is 40 km/h.
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Checkpoint 7.7.50. Ann Cycling and Walking.

Ann cycles 6 km to return a friend’s bicycle. She then walks home. Her total time for the trip is 90 minutes. Ann cycles four times as fast as she walks. Determine Ann’s average speeds for walking and for cycling.
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Hint.
\(\frac{6}{4v}+\frac{6}{v}=1.5\text{.}\)
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Answer.
Ann’s walking speed is 5 km/h and her cycling speed is 20 km/h.
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Subsection 7.7.5 Number Problems Involving Reciprocals

Recall that the reciprocal of a number \(a\) is \(\frac{1}{a}\text{.}\) For example, the reciprocal of 3 is \(\frac{1}{3}\text{,}\) and the reciprocal of \(\frac{2}{5}\) is \(\frac{5}{2}\text{.}\)
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Checkpoint 7.7.51. Consecutive Numbers.

The sum of the reciprocals of two consecutive numbers is equal to \(\frac{11}{30}\text{.}\) Find the two numbers.
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Answer.
5 and 6.
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Checkpoint 7.7.52. Sum of Reciprocals.

The sum of two numbers is 25, and the sum of their reciprocals is \(\frac{1}{4}\text{.}\) Find the two numbers.
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Hint.
\(\frac{1}{x} + \frac{1}{25-x} = \frac{1}{4}\text{.}\)
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Answer.
5 and 20.
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Checkpoint 7.7.53. Sum of Two Numbers.

The sum of two numbers is 12, and the sum of their reciprocals is \(\frac{3}{8}\text{.}\) Find the two numbers.
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Answer.
4 and 8.
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Checkpoint 7.7.54. Reciprocal Equality.

Find the value(s) of \(x\) in the numbers \(x\text{,}\) \(x+1\text{,}\) and \(x+2\) such that the reciprocal of the smallest number equals the sum of the reciprocals of the other two numbers.
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Hint.
\(\frac{1}{x} = \frac{1}{x+1} + \frac{1}{x+2}\text{.}\)
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Answer.
\(x=-\sqrt{2}\) or \(x=\sqrt{2}\text{.}\)
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Checkpoint 7.7.55. Difference in Reciprocals.

The sum of two integers is 12, and the difference in their reciprocals is \(\frac{2}{9}\text{.}\) Find the two integers.
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Answer.
3 and 9.
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Checkpoint 7.7.56. Natural Numbers.

A natural number is 4 more than another natural number. When the reciprocal of the greater number is subtracted from the reciprocal of the lesser number, the difference is \(\frac{1}{15}\text{.}\) What are the two numbers?
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Hint.
\(\frac{1}{x}-\frac{1}{x+4}=\frac{1}{15}\text{.}\)
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Answer.
6 and 10.
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Checkpoint 7.7.57. Sum with Reciprocal.

The sum of a number and its reciprocal is \(\frac{13}{6}\text{.}\) Find the number.
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Hint.
\(x+\frac{1}{x}=\frac{13}{6}\text{.}\)
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Answer.
\(\frac{2}{3}\) or \(\frac{3}{2}\text{.}\)
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Checkpoint 7.7.58. Integer and Reciprocal.

The sum of an integer and its reciprocal is \(\frac{65}{8}\text{.}\) Find the integer.
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Hint.
\(x+\frac{1}{x}=\frac{65}{8}\text{.}\)
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Answer.
8.
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Checkpoint 7.7.59. Consecutive Integers.

Find two consecutive integers whose reciprocals add to \(\frac{7}{12}\text{.}\)
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Hint.
\(\frac{1}{x}+\frac{1}{x+1}=\frac{7}{12}\text{.}\)
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Answer.
3 and 4.
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Checkpoint 7.7.60. Consecutive Odd Integers.

Find two consecutive odd integers whose reciprocals add to \(\frac{8}{15}\text{.}\)
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Hint.
\(\frac{1}{x}+\frac{1}{x+2}=\frac{8}{15}\text{.}\)
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Answer.
3 and 5.
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Checkpoint 7.7.61. Consecutive Even Integers.

Find two consecutive even integers whose reciprocals add to \(\frac{7}{24}\text{.}\)
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Hint.
\(\frac{1}{x}+\frac{1}{x+2}=\frac{7}{24}\text{.}\)
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Answer.
6 and 8.
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Checkpoint 7.7.62. Quotient Problem.

The quotient of a number and 5 minus \(\frac{3}{2}\) is the quotient of the number and 2. Find the number.
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Hint.
\(\frac{x}{5}-\frac{3}{2}=\frac{x}{2}\text{.}\)
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Answer.
\(-5\text{.}\)
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Checkpoint 7.7.63. Number Product Problem.

A number added to the product of 6 and the reciprocal of the number is \(-5\text{.}\) Find the number.
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Hint.
\(x+\frac{6}{x}=-5\text{.}\)
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Answer.
\(-3\) or \(-2\text{.}\)
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Checkpoint 7.7.64. Reciprocal Ratio Problem.

If 12 times the reciprocal of a number is added to the ratio of 7 times the number minus 5 and the number plus 1, the result is 8. What is the number?
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Hint.
\(\frac{12}{x}+\frac{7x-5}{x+1}=8\text{.}\)
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Answer.
\(-4\) or 3.
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Checkpoint 7.7.65. Twice Reciprocal.

The sum of a number and twice its reciprocal is \(\frac{9}{2}\text{.}\) Find the number.
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Hint.
\(x+\frac{2}{x}=\frac{9}{2}\text{.}\)
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Answer.
\(x=4\) or \(x=\frac{1}{2}\text{.}\)
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Subsection 7.7.6 Mixture Problems

Checkpoint 7.7.66. Acid Solution.

Suppose that \(x\) mL of pure acid are added to 100 mL of a \(35\%\) acid solution. How many mL of pure acid must be added to produce a new solution that is \(75\%\) acid?
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Hint.
\(\frac{x+0.35\cdot100}{x+100}=0.75\text{.}\)
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Answer.
\(x=160\) mL.
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Checkpoint 7.7.67. Pure Acid Addition.

How much pure acid must be added to 50 mL of a \(35\%\) acid solution to produce a mixture that is \(75\%\) acid?
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Hint.
\(\frac{x+0.35\cdot50}{x+50}=0.75\text{.}\)
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Answer.
\(x=80\) mL.
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Checkpoint 7.7.68. Bleach Solution.

How much bleach should be added to 47 L of water to make a solution that is \(6\%\) bleach?
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Hint.
\(\frac{x}{x+47}=\frac{6}{100}\text{.}\)
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Answer.
3 L.
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Checkpoint 7.7.69. Acid Volume.

What volume of pure acid must be added to 125 mL of a \(60\%\) acid solution to obtain a solution that is \(83\%\) acid?
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Hint.
\(\frac{x+0.6\cdot125}{x+125}=0.83\text{.}\)
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Answer.
\(x=\frac{2875}{17}\approx169.1\) mL.
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Subsection 7.7.7 Misc Word Problems

Checkpoint 7.7.70. Box Dimensions.

(a)
If the length of the base decreases by \(x\) cm, write an expression for the new height of the box.
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Answer.
\(\frac{1280}{(16-x)^2}\text{.}\)
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(b)
Write an expression for the increase in height of the box.
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Answer.
\(\frac{1280}{(16-x)^2} - 5\text{.}\)
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(c)
If the base becomes a 12 cm square, find the increase in height.
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Hint.
if the base length is 12 cm, what is \(x\text{?}\)
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Answer.
\(\frac{35}{9} \approx 3.9\) cm.
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Checkpoint 7.7.71. Golf Driving Range.

(a)
If Jacob hits \(x\) buckets of balls in one month, write an expression that represents how much he would pay per month.
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Answer.
\(15+x\text{.}\)
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(b)
Write an expression that represents the actual cost per bucket for a month (including the monthly fee).
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Answer.
\(\frac{15 + x}{x}\text{.}\)
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(c)
Write and solve an equation to determine the number of buckets of balls Jacob would have to hit in a month for the actual cost per bucket to be $1.25.
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Hint.
\(\frac{15+x}{x}=1.25\text{.}\)
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Answer.
60 buckets.
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Checkpoint 7.7.72. T-shirt Profit.

Rima bought a case of concert T-shirts for $450. She kept two T-shirts for herself and sold the rest for $560, making a profit of $10 on each T-shirt. How many T-shirts were in the case?
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Answer.
18 T-shirts.
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Checkpoint 7.7.73. French Club Trip.

A French club collected the same amount of money from each student for a trip. Six students cancelled, so each remaining student paid $3 extra. The total cost was $540. How many students went on the trip?
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Answer.
30 students.
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Checkpoint 7.7.74. Polygon Angles.

The measure \(d\) (in degrees) of each angle in a regular polygon with \(n\) sides is given by the equation \(d = 180 - \frac{360}{n}\text{.}\) When each angle in a regular polygon is \(162^\circ\text{,}\) how many sides does the polygon have?
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Hint.
\(162 = 180 - \frac{360}{n}\text{.}\)
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Answer.
20 sides.
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