Rational Equation Word Problems: Guide, Practice, and Answer Key

Once a rational equation is written down, solving it follows a set routine. In a word problem, the harder part is deciding what each fraction stands for.

In tutoring sessions, I use a small table to keep the quantities and units straight: one row for each worker, each leg of a trip, or each group sharing a cost. Once the table is filled in, the equation comes from the sentence in the problem that compares the rows.

This guide shows that setup for work, travel, current, and cost-sharing problems. It includes four worked examples with short animations, an interactive simulator and setup trainer, 24 practice problems, and 3 challenge problems with answers.

Written by Berke Sahbazoglu · BA Biochemistry, Washington University in St. Louis · 10+ years of tutoring experience; teaches Algebra I and Algebra II · Burke Tutoring in Fremont, CA · Published September 23, 2026 · Free to print

The problems below are adapted from problem types I use in Algebra II and Precalculus tutoring sessions.

Part of a nine guide unit. See all the rational expressions and rational functions worksheets, listed in a recommended order.

What makes a word problem a rational equation

A word problem leads to a rational equation when the unknown ends up in a polynomial denominator, such as t, r + 2, or n − 5. The main types covered here are work, motion, and share or ratio problems.

Work

rate = 1 jobtime alone

A person who paints a room in t hours paints 1t of it each hour.

Motion

time = distancespeed

A 30 mile trip at r mph takes 30r hours.

Shares and ratios

share = totalcount

A $240 bill split n ways is 240n dollars each. Proportion = part ÷ whole, so 34 = 75%.

Assumptions used on this page. Rates and speeds are treated as constant, shared costs are divided equally, and quiz scores are equally weighted.

The setup table routine

Build the table before writing any equation.

  1. Name the unknown with its units. Write “let r = the kayak’s speed in still water, in miles per hour” rather than “let r = speed.”
  2. Build the table. Use one row per worker, leg, or group, and one column for each quantity in the formula: time alone, rate, and part done for work; distance, speed, and time for motion; total, count, and share for costs. Write the unknown into every cell that depends on it, such as r + 2 for the speed downstream.
  3. Fill the fraction column by dividing. Time is distance over speed, rate is 1 over time alone, and a share is total over count. This is the column where the unknown moves into a denominator.
  4. Find the sentence that compares the rows. Phrases such as “in the same time,” “working together,” “the whole trip takes,” or “each share drops by $10” tell you which cells to set equal, add, or subtract.
  5. Solve the rational equation. Write the excluded values first, multiply every term by the LCD, and solve the polynomial equation that remains.
  6. Check twice, then answer the question asked. Check each candidate in the equation and then in the table. Reread the question and give the quantity it asks for, with units, in a sentence.

Put units in every column heading. When the headings read miles, miles per hour, and hours, a time given in minutes stands out before it reaches the equation.

Phrase to equation chart for rational word problems

Each phrase below produces one piece of an equation. The letters match the worked examples and practice problems.

How common phrases in rational word problems translate into equations
Words in the problemWhat they tell youEquation piece
“Working together, they finish in 2 hours”The rates add to the rate together1a + 1b = 12
“The drain empties it in d hours”The drain’s rate is subtracted1a − 1d
“Hose B takes 4 hours longer than hose A”Compare the times, not the ratesB’s time = t + 4
“She works alone for 2 hours, then he joins”If T is the total elapsed time, she works T hours and he works T − 2 hours. The parts done add to 1.Ta + T − 2b = 1
“In the same time”The two time cells are equal9r + 2 = 5r − 2
“The round trip takes 2 hours 30 minutes”The two times add, in hours15r + 15r − 5 = 52
“Downstream” or “with a tailwind”The current adds to the speedr + c
“Upstream” or “into a headwind”The current subtracts from the speedr − c
“Arrives 30 minutes sooner”The times differ by 12 hour90r − 90r + 15 = 12
“Each person’s share drops by $10”Old share minus new share240n − 240n + 2 = 10
“Win the next x games in a row”Each win adds to the part and to the whole9 + x15 + x
“A number plus its reciprocal”The reciprocal of x is 1 over xx + 1x
“Average speed for the round trip”Total distance over total time2dd10 + dr

Four worked examples

Each example ends with a short animation of the situation. Examples 3 and 4 each produce a candidate that satisfies the equation and still has to be rejected.

Worked example 1 · Work with a drain

An inlet pipe fills a tank in 4 hours. When the drain is left open, the same pipe takes 6 hours to fill the tank. How long would the open drain take to empty a full tank?

  1. Name the unknown. Let d = the number of hours the open drain takes to empty a full tank.
  2. Build the table.
    Time (hours)Rate (tank per hour)
    Inlet pipe414
    Draind−1d
    Both running616
  3. Write the equation. With both running, the pipe’s rate minus the drain’s rate is the net rate: 14 − 1d = 16.
  4. Solve. Excluded value: d ≠ 0. Multiply every term by 12d: 3d − 12 = 2d, so d = 12.
  5. Check. In the equation: 14 − 112 = 212 = 16. In the story: the drain has to be slower than the pipe, or the tank would not fill, and 12 hours is longer than 4 hours. Over 6 hours the pipe adds 64 = 1.5 tanks and the drain removes 612 = 0.5 tank, which leaves exactly one full tank.

Answer: the open drain would empty a full tank in 12 hours.

Animation · pipe and drain rates

The streams are drawn in proportion to the rates, so the pipe’s stream is three times as wide as the drain’s. The bucket holds half a tank, and it fills at the same moment the tank does.

Worked example 2 · Current, with equal times

A kayak travels 9 miles downstream in the same time it travels 5 miles upstream. The river’s current is 2 mph. What is the kayak’s speed in still water?

  1. Name the unknown. Let r = the kayak’s speed in still water, in mph.
  2. Build the table.
    Distance (miles)Speed (mph)Time (hours)
    Downstream9r + 29r + 2
    Upstream5r − 25r − 2
  3. Write the equation. “In the same time” sets the two time cells equal: 9r + 2 = 5r − 2.
  4. Solve. Excluded values: 2 and −2. With one fraction on each side, cross multiply: 9(r − 2) = 5(r + 2), so 9r − 18 = 5r + 10, 4r = 28, and r = 7.
  5. Check. The kayak is faster than the current (7 > 2), so it can make progress upstream. Downstream: 9 miles at 9 mph is 1 hour. Upstream: 5 miles at 5 mph is 1 hour. The times match.

Answer: the kayak’s speed in still water is 7 mph.

Animation · downstream and upstream in the same hour

Both kayaks leave the dock together and reach their flags at the same moment, one hour later. The buoy drifts with the current, and each kayak ends 7 miles from the buoy, which is the still-water speed.

Worked example 3 · Round trip with minutes, and a positive root that fails

A cyclist rides 15 miles out and 15 miles back on the same route. Riding back into a headwind, her average speed is 5 mph slower than on the way out. The riding time for the whole trip is 2 hours 30 minutes. How fast did she ride on the way out?

  1. Name the unknown and convert units. Let r = her speed on the way out, in mph, so the return speed is r − 5. The total time is 2 hours 30 minutes = 52 hours.
  2. Build the table.
    Distance (miles)Speed (mph)Time (hours)
    Out15r15r
    Back15r − 515r − 5
  3. Write the equation. “The riding time for the whole trip” means the two times add: 15r + 15r − 5 = 52.
  4. Solve. Excluded values: 0 and 5. Multiply every term by 2r(r − 5): 30(r − 5) + 30r = 5r(r − 5), so 60r − 150 = 5r2 − 25r. That gives 5r2 − 85r + 150 = 0. Divide by 5: r2 − 17r + 30 = 0, which factors as (r − 15)(r − 2) = 0.
  5. Check. Neither 15 nor 2 is excluded, and both satisfy the equation. For r = 2: 152 + 15−3 = 7.5 − 5 = 2.5. The table rules it out, though: the return speed would be 2 − 5 = −3 mph, so r = 2 is nonviable. For r = 15: 1 hour out and 1510 = 1.5 hours back, 2.5 hours in all.

Answer: she rode 15 mph on the way out and 10 mph on the way back.

Why reject 2 if it works in the equation? It makes the return speed 2 − 5 = −3 mph. Problem 12 produces a second positive candidate the same way.

Animation · a faster leg out and a slower leg back

Both legs cover the same 15 miles, but the slower return leg takes more of the timeline: 1.5 hours against 1 hour.

Worked example 4 · Cost sharing

A robotics team splits the $240 cost of a rental van evenly. Two more students sign up for the trip, and each person’s share drops by $10. How many students were on the original list?

  1. Name the unknown. Let n = the number of students on the original list.
  2. Build the table.
    Total ($)StudentsShare each ($)
    Original list240n240n
    After sign-ups240n + 2240n + 2
  3. Write the equation. “Drops by $10” means the original share minus the new share is 10: 240n − 240n + 2 = 10.
  4. Solve. Excluded values: 0 and −2. Multiply every term by n(n + 2): 240(n + 2) − 240n = 10n(n + 2), so 480 = 10n2 + 20n. Divide by 10: n2 + 2n − 48 = 0, which factors as (n + 8)(n − 6) = 0.
  5. Check. The candidate −8 satisfies the equation but is not a possible number of students, so it is nonviable. For n = 6: $240 ÷ 6 = $40 each, then $240 ÷ 8 = $30 each, a drop of $10.

Answer: 6 students were on the original list.

Animation · the same cost split more ways

As two students join, each slice narrows from $40 to $30, and the amber strip above the bar grows to the $10 drop.

Two checks before you write the answer

An answer has to pass two checks: one against the equation and one against the story.

Check 1 · The equation

Compare each candidate with the excluded values, then substitute it into the equation you wrote, with the denominators still in it. Reject any candidate that makes an original denominator zero; it lies outside the equation’s domain.

Check 2 · The story

Substitute each remaining candidate into every cell of the table, not only the variable. A negative time, speed, or count, a count that is not a whole number, or a current as fast as or faster than the boat makes the candidate nonviable, even though it satisfies the equation.

What the equation excludes and what the story requires in the four worked examples
Worked exampleExcluded by the equationWhat the story requiresCandidate removed by the story
1. Pipe and draind = 0d > 4, a drain slower than the pipenone
2. Currentr = 2, r = −2r > 2, faster than the currentnone
3. Round tripr = 0, r = 5r > 5, so the return speed r − 5 is positiver = 2
4. Cost sharingn = 0, n = −2n is a positive whole numbern = −8

A quick estimate for work problems

Two workers who take a and b hours alone finish together in T = aba + b hours. That time is shorter than the faster worker’s time alone and at least half of it. For a 3 hour worker and a 6 hour worker, T has to land between 1.5 and 3 hours, and it does: T = 2.

It works in reverse too: a helper’s time alone has to be longer than the time together.

Work Rate Simulator

Set how long A and B take alone, whether B helps or drains, and when B starts. The green line shows total progress. Before B starts, its slope is A’s rate; after B starts, its slope is the combined rate, or the net rate when B drains. A dashed line shows A working alone; in help mode, a second dashed line shows B working alone.

Work Rate SimulatorTwo workers, one job, and the equation that connects them
Load a setup:

Set B to drain with a drain time below A’s time. Once B starts, the green line falls toward zero, or stays at zero if the drain starts immediately. With B helping instead, set a late start longer than A’s time alone, and A finishes before B begins.

Equation Setup Trainer

Each problem comes with four equations. One matches the story, and each of the other three contains one setup error, such as adding times, swapping the sign of a current, or mixing minutes with hours. Pick the equation that fits, then solve it.

Equation Setup TrainerSeven problem types, new numbers every time

Step 1. Which equation models this situation?

Whole numbers, decimals such as 7.5, and fractions such as 15/2 are accepted, and a count of people or games has to be a whole number. An entry that cannot be read is not scored.

Six setup errors and their fixes

Each one shows an incorrect setup, the corrected line, and what to do differently.

1. Adding the times

Incorrect45 + 36 = 81 minutes for both printers together
Correct145 + 136 = 1T, so T = 20 minutes

A second printer should not make the job take longer. Times do not add when two workers share a job. Rates do, and each rate is 1 over the time alone.

2. Minutes and hours in the same equation

Incorrect15r + 15r − 5 = 150
Correct15r + 15r − 5 = 52

Speeds in miles per hour give times in hours, so convert 2 hours 30 minutes to 52 hours before writing the equation.

3. The current on the wrong leg

Incorrect9r − 2 = 5r + 2
Correct9r + 2 = 5r − 2

Downstream, the current carries the boat, so the speed is r + c. In equal times, the longer distance needs the faster speed, so the 9 miles belongs over the sum.

4. A drain written as a helper

Incorrect14 + 1d = 16
Correct14 − 1d = 16

The incorrect version solves to d = −12. A negative time in a pipe problem is a signal to check the sign on the drain’s rate.

5. “Longer” translated in the wrong direction

IncorrectHose B takes 4 hours longer, so 1t + 1t − 4 = 23
Correct1t + 1t + 4 = 23, with t as hose A’s time

Test the relationship with a made-up number before writing it. If hose A took 5 hours, hose B would take 9, so B’s time is t + 4.

6. Keeping a root because it is positive

Incorrectr = 2 or r = 15, so she rode 2 mph or 15 mph
Correctr = 2 gives a return speed of −3 mph, so only r = 15 is viable

Substitute each candidate into every cell of the table. In worked example 3, r = 2 fails in the return speed cell, r − 5.

One-page setup card

This is page one of the printable worksheet, without the phrase chart shown above.

Rational Equation Word Problems: Setup Card
Burke Tutoring in Fremont, CA · burketutoringinfremont.com

Three formulas

  • Work: rate = 1 ÷ time alone; part done = rate × time worked.
  • Motion: time = distance ÷ speed. With a current c: r + c one way, r − c the other.
  • Shares and ratios: share = total ÷ count; proportion = part ÷ whole, so 34 = 75%.

The routine

  1. Name the unknown with its units.
  2. Build a table: one row per worker, leg, or group.
  3. Fill the fraction column by dividing.
  4. Turn the sentence that compares the rows into the equation.
  5. List the excluded values, multiply by the LCD, solve.
  6. Check in the equation, then in the table. Answer with units.

Two checks

  • Equation: reject any candidate that makes an original denominator zero; it lies outside the equation’s domain. Substitute the rest into the original equation.
  • Story: times, speeds, and counts must be positive, counts must be whole, and a current must be slower than the boat. While a pipe and a drain both run, the filling rate must exceed the draining rate for the tank to keep filling. A candidate that fails here is nonviable.

Work estimate

Two workers who take a and b hours alone finish together in T = aba + b. That is less than the faster time alone and at least half of it.

Worked in one line

9 miles downstream and 5 upstream in the same time, current 2 mph: 9r + 2 = 5r − 2 → 9r − 18 = 5r + 10 → r = 7 mph.

Printable worksheet and setup card

Download the PDFs

Both files print in black and white on letter paper. The worksheet has a blank setup table and workspace beside each problem. The answer key gives an equation and answer for every problem, along with relevant restrictions and story checks.

Worksheet with answer keyPDF · 27 problems · 14 pages Setup cardPDF · 1 page
Free to copy for your classroom or your own kitchen table. If you share it, please link to this page rather than rehosting the files, so students land on the current version.

24 practice problems

For each problem, name the unknown with units, fill a table, write the equation, and answer in a sentence. Answers are below each part.

Part A · Work and rate problems
Problems 1 to 6. A worker’s rate is 1 job divided by the time alone. Rates add when workers share a job, and a drain’s rate is subtracted.
  1. Maya can weed the school garden in 3 hours. Jordan can weed it in 6 hours. How long does the job take if they work together?
  2. One printer finishes a batch of flyers in 45 minutes. With a second printer running at the same time, the batch takes 20 minutes. How long would the second printer take alone?
  3. Hose B takes 4 hours longer than hose A to fill a pool. With both hoses running, the pool fills in 1 hour 30 minutes. How long does each hose take alone?
  4. An inlet pipe fills a water tank in 6 hours. With the drain open as well, the tank takes 10 hours to fill. How long would the open drain take to empty a full tank?
  5. Ana can paint a fence alone in 8 hours. She paints alone for 2 hours, then Ben joins her, and together they finish 2 hours later. How long would Ben take to paint the fence alone?
  6. Three pumps running together empty a flooded basement in 2 hours. Pump A alone takes 6 hours and pump B alone takes 5 hours. How long would pump C take alone?
Show answers for Part A
  1. Let t = hours together. 13 + 16 = 1t. Excluded: t ≠ 0. LCD 6t: 2t + t = 6, so t = 2. Estimate check: 2 hours is less than 3 hours and at least half of it.Answer: 2 hours
  2. Let p = minutes for the second printer. 145 + 1p = 120. Excluded: p ≠ 0. LCD 180p: 4p + 180 = 9p, so 5p = 180 and p = 36. Story check: 36 minutes is longer than the 20 minutes together.Answer: 36 minutes
  3. Let t = hours for hose A, so hose B takes t + 4. 1 hour 30 minutes is 32 hours, so the rate together is 23: 1t + 1t + 4 = 23. Excluded: 0, −4. LCD 3t(t + 4): 3(t + 4) + 3t = 2t(t + 4), so 6t + 12 = 2t2 + 8t and 2t2 + 2t − 12 = 0. Divide by 2: t2 + t − 6 = 0, which factors as (t + 3)(t − 2) = 0. The candidate −3 satisfies the equation but is a negative time, so it is nonviable.Answer: Hose A 2 hours, hose B 6 hours
  4. Let d = hours for the drain. 16 − 1d = 110. Excluded: d ≠ 0. LCD 30d: 5d − 30 = 3d, so d = 15. Story check: the drain has to be slower than the pipe for the tank to fill, and 15 > 6.Answer: 15 hours
  5. Let b = Ben’s hours alone. Ana works 4 hours in all and does 48 of the fence. Ben works 2 hours and does 2b. The parts add to one fence: 48 + 2b = 1. Excluded: b ≠ 0. Then 2b = 12, so b = 4.Answer: 4 hours
  6. Let c = hours for pump C. 16 + 15 + 1c = 12. Excluded: c ≠ 0. LCD 30c: 5c + 6c + 30 = 15c, so 4c = 30 and c = 152.Answer: 7.5 hours (7 hours 30 minutes)
Part B · Distance, speed, and current
Problems 7 to 12. Time is distance divided by speed. A current or wind adds to the speed in one direction and subtracts in the other.
  1. A cyclist rides 30 miles in the same time a runner covers 12 miles. The cyclist’s average speed is 9 mph faster than the runner’s. Find both speeds.
  2. A boat that goes 12 mph in still water travels 45 miles downstream in the same time it travels 27 miles upstream. How fast is the current?
  3. A small plane flies 600 miles with a tailwind in the same time it flies 480 miles into a headwind of the same speed. The wind is 20 mph. What is the plane’s speed in still air?
  4. A kayaker paddles 6 miles upstream and 6 miles back down. The current is 2 mph, and the round trip takes 4 hours of paddling. What is her speed in still water?
  5. A car and a bus each travel 90 miles. The car averages 15 mph faster than the bus and arrives 30 minutes sooner. Find both speeds.
  6. A delivery driver travels 40 miles to a stop and 40 miles back along the same route. Traffic on the way back cuts the average speed by 10 mph. The driving time for the round trip is 1 hour 48 minutes. What was the speed on the way out?
Show answers for Part B
  1. Let r = the runner’s speed. 12r = 30r + 9. Excluded: 0, −9. Cross multiply: 12(r + 9) = 30r, so 108 = 18r and r = 6. Each travels for 2 hours.Answer: Runner 6 mph, cyclist 15 mph
  2. Let c = the current. 4512 + c = 2712 − c. Excluded: 12, −12. 45(12 − c) = 27(12 + c) gives 540 − 45c = 324 + 27c, so 72c = 216 and c = 3. Story check: 0 < 3 < 12. Each leg takes 3 hours.Answer: 3 mph
  3. 600r + 20 = 480r − 20. Excluded: 20, −20. 600(r − 20) = 480(r + 20) gives 600r − 12000 = 480r + 9600, so 120r = 21600 and r = 180. Each leg takes 3 hours.Answer: 180 mph
  4. 6r − 2 + 6r + 2 = 4. Excluded: 2, −2. LCD (r − 2)(r + 2): 6(r + 2) + 6(r − 2) = 4(r2 − 4), so 12r = 4r2 − 16 and 4r2 − 12r − 16 = 0. Divide by 4: r2 − 3r − 4 = 0, which factors as (r − 4)(r + 1) = 0. A speed of −1 is nonviable. Check: 3 hours up plus 1 hour back is 4 hours.Answer: 4 mph
  5. Let r = the bus speed. The bus takes 12 hour longer: 90r − 90r + 15 = 12. Excluded: 0, −15. LCD 2r(r + 15): 180(r + 15) − 180r = r(r + 15), so 2700 = r2 + 15r, or r2 + 15r − 2700 = 0, which factors as (r − 45)(r + 60) = 0. A speed of −60 is nonviable. Times: 2 hours and 1.5 hours.Answer: Bus 45 mph, car 60 mph
  6. 1 hour 48 minutes is 95 hours. 40r + 40r − 10 = 95. Excluded: 0, 10. LCD 5r(r − 10): 200(r − 10) + 200r = 9r(r − 10), so 400r − 2000 = 9r2 − 90r and 9r2 − 490r + 2000 = 0, which factors as (r − 50)(9r − 40) = 0. The candidate 409 satisfies the equation, but the return speed would be 409 − 10, which is negative, so it is nonviable. Check: 0.8 hour + 1 hour = 1.8 hours.Answer: 50 mph out, 40 mph back
Part C · Ratios, percentages, and shares
Problems 13 to 18. A share is a total divided by a count. A proportion is a part divided by a whole (3/4 = 75%), and each new success adds to both.
  1. A volleyball team has won 9 of its first 15 matches. How many matches in a row must it win to raise its winning percentage to 75%?
  2. A softball player has 14 hits in 50 at-bats, a batting average of .280. Batting average is hits divided by at-bats, as defined in the MLB glossary. How many hits in consecutive at-bats would raise her average to .400?
  3. A group splits the $600 cost of a party bus evenly. Five people drop out, and each remaining person pays $10 more. How many people were in the original group?
  4. Last month a café spent $180 on bags of coffee beans. This month the price per bag is $3 higher, so the same $180 buys 3 fewer bags. How many bags did $180 buy last month?
  5. A lab has 8 liters of a solution that is 25% acid. How many liters of pure acid must be added to make the solution 40% acid?
  6. Kai’s quizzes are equally weighted, and his average so far is 80. If he scores 100 on each of the next 3 quizzes, his average will rise to 85. How many quizzes has he taken so far?
Show answers for Part C
  1. 9 + x15 + x = 34. Excluded: −15. 4(9 + x) = 3(15 + x), so 36 + 4x = 45 + 3x and x = 9. Check: 1824 = 0.75.Answer: 9 matches
  2. Each hit is also an at-bat. 14 + h50 + h = 25. Excluded: −50. 5(14 + h) = 2(50 + h), so 70 + 5h = 100 + 2h and h = 10. Check: 2460 = .400.Answer: 10 hits
  3. 600n − 5 − 600n = 10. Excluded: 0, 5. LCD n(n − 5): 600n − 600(n − 5) = 10n(n − 5), so 3000 = 10n2 − 50n. Divide by 10: n2 − 5n − 300 = 0, which factors as (n − 20)(n + 15) = 0. A group of −15 is nonviable. Check: $30 each, then $40 each.Answer: 20 people
  4. Let n = bags last month. 180n − 3 − 180n = 3. Excluded: 0, 3. LCD n(n − 3): 180n − 180(n − 3) = 3n(n − 3), so 540 = 3n2 − 9n. Divide by 3: n2 − 3n − 180 = 0, which factors as (n − 15)(n + 12) = 0. A count of −12 bags is nonviable. Check: $12 a bag, then $15 a bag.Answer: 15 bags
  5. The solution holds 0.25 × 8 = 2 liters of acid. 2 + x8 + x = 25. Excluded: −8. 5(2 + x) = 2(8 + x), so 3x = 6 and x = 2. Check: 410 = 40%.Answer: 2 liters
  6. Let n = quizzes so far, so his points total 80n. 80n + 300n + 3 = 85. Excluded: −3. 80n + 300 = 85n + 255, so 5n = 45 and n = 9. Check: 102012 = 85.Answer: 9 quizzes
Part D · Mixed problems
Problems 19 to 24. Numbers and reciprocals, two science formulas, an average speed problem, and one problem (23) where a positive candidate still fails the story.
  1. The sum of a number and three times its reciprocal is 4. Find every such number.
  2. The denominator of a fraction is 5 more than its numerator. When 3 is added to both the numerator and the denominator, the result is 23. Find the original fraction.
  3. Two resistors wired in parallel have a combined resistance R given by 1R = 1R1 + 1R2. One resistor is 12 ohms and the combined resistance is 4 ohms. Find the other resistor.
  4. The thin lens equation 1f = 1do + 1di relates a lens’s focal length to the object distance and the image distance (see HyperPhysics for the physics). A converging lens with focal length 12 cm forms a real image whose distance from the lens is three times the object distance. How far is the object from the lens?
  5. Two copiers running together finish a print job in 4 minutes. The faster copier alone takes 6 minutes less than the slower one. How long does each copier take alone?
  6. A cyclist rides out at 10 mph and returns on the same route. Her average speed for the whole round trip is 12 mph. How fast did she ride back?
Show answers for Part D
  1. x + 3x = 4. Excluded: 0. Multiply by x: x2 − 4x + 3 = 0, so (x − 1)(x − 3) = 0. Both candidates check.Answer: x = 1 or x = 3
  2. Let n = the numerator, so the original fraction is nn + 5 and n ≠ −5. n + 3n + 8 = 23. Excluded from this equation: −8. 3(n + 3) = 2(n + 8), so n = 7, which is neither −5 nor −8. Check: 1015 = 23.Answer: 712
  3. 14 = 112 + 1R2. Excluded: R2 ≠ 0. LCD 12R2: 3R2 = R2 + 12, so R2 = 6.Answer: 6 ohms
  4. Let d = the object distance, so the real image is 3d from the lens. 112 = 1d + 13d. Excluded: d ≠ 0. LCD 12d: d = 12 + 4, so d = 16. Check: 116 + 148 = 448 = 112.Answer: 16 cm, with the image at 48 cm
  5. Let t = the slower copier’s minutes, so the faster one takes t − 6. 1t + 1t − 6 = 14. Excluded: 0, 6. LCD 4t(t − 6): 4(t − 6) + 4t = t(t − 6), so 8t − 24 = t2 − 6t and t2 − 14t + 24 = 0, which factors as (t − 12)(t − 2) = 0. The candidate 2 satisfies the equation, but it gives the faster copier 2 − 6 = −4 minutes, so it is nonviable.Answer: Slower 12 minutes, faster 6 minutes
  6. Let d = the one-way distance and r = the return speed. Average speed is total distance over total time: 2dd10 + dr = 12. The d cancels, leaving 2110 + 1r = 12, so 110 + 1r = 16. Excluded: 0, −10. LCD 30r: 3r + 30 = 5r, so r = 15. The answer is not 14, because the slower leg takes more of the time. Challenge problem 27 on the complex rational expressions worksheet derives the general formula.Answer: 15 mph

Challenge problems

Challenge set · Problems 25 to 27
  1. Return to problem 13. Show that no winning streak raises the volleyball team’s percentage to exactly 100%. Then find the percentage after 30 straight wins, and describe what happens to the percentage as the streak gets longer.
  2. Pipes A and B together fill a tank in 4 hours, pipes B and C together fill it in 6 hours, and pipes A and C together fill it in 5 hours. How long do all three pipes take together? How long does pipe A take alone?
  3. A boat that goes 8 mph in still water travels 15 miles upstream and 15 miles back in 4 hours. (a) Find the speed of the current. (b) For a current c and a still-water speed r, show that when 0 < c < r, a round trip of d miles each way takes longer than it would in still water.
Show answers for the challenge set
  1. 9 + x15 + x = 1 gives 9 + x = 15 + x, which reduces to 9 = 15. No value of x works, so there is no solution. After 30 wins the record is 3945 ≈ 86.7%. As x grows the fraction gets closer to 1 without reaching it, which is the horizontal asymptote y = 1 of y = 9 + x15 + x. The page on graphing rational functions covers horizontal asymptotes.Answer: No streak reaches 100%
  2. Let a, b, and c be the rates in tanks per hour. a + b = 14, b + c = 16, and a + c = 15. Adding the three equations gives 2(a + b + c) = 3760, so a + b + c = 37120. Pipe A: a = 37120 − 16 = 17120.Answer: All three: 12037 hours, about 3 hours 15 minutes. Pipe A: 12017 hours, about 7 hours 4 minutes
  3. (a) 158 − c + 158 + c = 4. Excluded: 8, −8. LCD (8 − c)(8 + c): 15(8 + c) + 15(8 − c) = 4(64 − c2), so 240 = 256 − 4c2 and c2 = 4. The candidate −2 is nonviable. (b) The round trip takes dr − c + dr + c = 2drr2 − c2. Since 0 < c < r, the denominator satisfies 0 < r2 − c2 < r2, so this is greater than 2drr2 = 2dr, the still water time.Answer: (a) 2 mph

Where this fits in the unit

If a step above felt unfamiliar, one of these pages covers it.

Berke Sahbazoglu, founder and tutor at Burke Tutoring in Fremont, California

About the author

Berke Sahbazoglu · Founder and tutor, Burke Tutoring in Fremont, CA

I have more than ten years of tutoring experience and teach Algebra I, Algebra II, and Precalculus in homes across Fremont, Newark, and Union City. I hold a BA in Biochemistry from Washington University in St. Louis and an MS in Bioinformatics from the University of Maryland Global Campus. The worksheets I publish here are built from the notes and homework sets I use in sessions.

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Working through word problems with a tutor

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Standards and further reading

  • Common Core standard HSA.CED.A.1 asks students to create equations in one variable, including equations from simple rational functions, and use them to solve problems. HSA.CED.A.3 adds interpreting solutions as viable or nonviable in a modeling context, which is the story check above.
  • EngageNY Algebra II, Module 1, Lesson 27, Word Problems Leading to Rational Equations, a student lesson with winning percentage, work, speed, and mixture exercises.
  • CSU East Bay Student Center for Academic Achievement, Word Problems: Mixture, Uniform, and Work Problems, a handout that sets up each type with a table.
  • Purdue University MA 15910, Job or Work Problems, a one-page summary of the rate and part-of-the-job formulas.
  • HyperPhysics, Thin Lens Equation, the physics behind problem 22.