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AR · lesson 10 of 12 · 35 min

Work and Combined-Rate Problems

This advanced 35-minute lesson takes on the problems many test takers skip: jobs done by more than one worker, machine or pipe. It begins by treating work rate as a fraction of a job and applying work equals rate times time, then shows how two workers together combine their rates and how common denominators make the arithmetic cleaner.

Video lesson: Work and Combined-Rate Problems

Video transcript

0:00 All right, let's jump right into this explainer. If you're gearing up for the ASVAB and the math sections got you feeling a little on edge, hey, take a deep breath, you are in exactly the right place. Today, we're going to totally conquer ASVAB math, and we're going to do it without touching a single calculator. Seriously, how? Well, by mastering a really simple, incredibly friendly concept that I like to call the power of one. Here is our game plan for today. We'll start with work rates and combined teams, then hit shifting teams and opposing rates. After that, sizing, scaling, and estimation, step-by-step worked examples, common traps to avoid, and we'll wrap it all up with a final review and some practice questions. Okay, section one, work rates and combined teams. Let's get our foundational rules down for any combined work problem you're going to face. Use the whole job as the number one. It doesn't matter if the task is building a giant fence, painting a massive house, or filling a gigantic tank. We just treat that entire full task as exactly one.

1:01 We use the formula w equals r times t. That just means work done equals the worker's rate multiplied by their time. By making this whole intimidating job equal to just one, we never ever have to worry about the actual physical size of the wall or the tank. It keeps the math so incredibly simple. Let's actually look at this concept in action. Say you've got a mechanic who can finish a job in five hours and another mechanic who takes 20 hours. Their hourly rates are simply fractions of that whole job, one fifth and one 20th. When they work together as a team, we just add those fractional rates, one fifth plus one 20th. Using a common denominator of 20, that becomes four 20th plus one 20th, which gives us five 20th. Simplify that down, and you get one fourth. Now to find the total combined time, you just invert that one fourth fraction. Literally flip it upside down, and boom, you get exactly four hours total. Shifting teams and opposing rates. We're going to add a little twist here and see what happens when a team changes mid job or when things are actually working against each other.

2:08 Let's patiently break down a partial work scenario. Imagine a worker who needs 12 hours alone, but they start the job completely by themselves and work for four hours before any backup arrives. Step one, find the work they did alone. Four hours out of 12 is four 12ths, which simplifies to one third. Step two, find the remaining work. Well, if one third is done, exactly two thirds of our whole one job is left. Step three, a helper who needs six hours alone finally shows up. Their combined team rate is one 12th plus one sixth, which gives us a shiny new team rate of one fourth. Finally, step four. Divide that remaining work by the new team rate. Two thirds divided by one fourth is exactly the same as two thirds multiplied by four over one, giving us eight thirds hours. You just split the job into stages, and track what gets done in each stage. Now, picture an inlet pipe filling up a water tank while a drain at the very bottom is letting the water out. If a fill pipe adds a positive one third per hour, but a drain removes a negative one sixth per hour, you simply subtract that drain rate from the total.

3:10 One third minus one sixth leaves us with a net rate of positive one sixth. The tank is absolutely still filling up. It just takes six hours with both of those pipes open. Sizing, scaling, and estimation. Let's look at a pretty cool advanced tactic for handling tricky numbers when you don't have that calculator to lean on. When working together, a team is always faster than the fastest individual worker, but they're going to be slower than half of that fastest worker's time. For example, if you have a four-hour worker and a five-hour worker, the fastest is four hours, right? Working together, they obviously have to finish in under four hours. But even if they were both equally that fast, like two four-hour workers, it would still take two hours. Therefore, the combined time absolutely must fall somewhere between two and four hours. This rule is basically a super power. It lets you instantly eyeball and eliminate wrong multiple choice answers without doing a single bit of fractional math. All right, section four. Step-by-step worked examples. It is time to execute our strategy and actually put these formulas to work exactly as you'll see them on the test.

4:17 The good news is the rules do not change, even if we start adding more workers. In example two here, we have three machines that empty a pond in 10, 15, and 30 hours alone. Step one, write their rates as one 10th, one 15th, and one 30th. Step two, grab a common denominator of 30, which gives us three 30ths plus two 30ths plus one 30th. Step three, add them all up for a sum of six 30ths, which beautifully reduces to one fifth. Step four, just invert that fraction to get five hours. Look at that. We just took three super intimidating fractions and turned them into one incredibly simple answer. Now, here's another classic ASVAB setup in example three, finding an unknown worker. We know the team rate is four hours, which means one fourth of the job per hour. We know one solo worker takes five hours, meaning one fifth. So how do we find the unknown partners time? We just subtract the solo rate from the total team rate. Step three shows one fourth minus one fifth. With a common denominator of 20, that gives us a result of one 20th.

5:18 Invert that result and we instantly see the unknown worker takes exactly 20 hours alone. Common traps to avoid. We've built an awesome mathematical foundation so far, but now we really need to flag the specific pitfalls the ASVAB uses to try and trip you up. Okay, let's run through these traps so you don't fall for them. First, adding times instead of rates. Two workers taking three and six hours do not need nine hours together. Guys, a team is always faster. Second trap, averaging the times. No, also completely wrong. Third, forgetting to invert at the end. If your combined rate is one fourth, the time is four hours, not a quarter of an hour. Fourth, adding a drain rate instead of subtracting it. Always remember drains work against you. And fifth, giving the wrong elapsed time. On those shifting team problems we talked about, carefully read whether they're asking for the total elapsed time from the very beginning or just the time after the extra help showed up. Final review and practice questions. We are going to lock in all this knowledge right now with two exact practice questions from our material.

6:26 Let's run the gauntlet. Okay, question one. A landscaper can mow a large park alone in eight hours. After working for three hours, what fraction of the park is still unmowed? A is five eighths, B is three eighths, C is one eighth, D is three fifths. I invite you to pause the explainer right here. Really try solving this partial work question without your calculator. Then hit play when you're ready. All right, let's see how you did. The correct answer is A, five eighths. Why? Well, each hour, the landscaper mows one eighth of the park. So three hours of work covers three eighths of the park. But because the question specifically asks what is still unmowed, we take our whole one park and subtract the three eighths that's already finished. That leaves exactly five eighths remaining. Here is our final challenge. Question two. One clerk files a batch in four hours, a second clerk files it in 12 hours. Working together, how long will it take them? A is eight hours, B is 16 hours, C is three hours, D is one third of an hour.

7:27 This is the ultimate classic combined rate question. Go ahead, pause the explainer one last time to test your skills. And we're back. The correct answer is C, three hours. Let's break it down. Step one, add the hourly rates one fourth plus one 12th. Step two, convert to a common denominator of 12, making it three 12ths plus one 12th. Step three, the sum is four 12ths, which simplifies perfectly down to one third. Step four, we invert that one third fraction to get a simple solid three hours. Do you see how smoothly one fourth plus one 12th becomes three hours when we just follow those steps? And that brings us to the end. The power of one means that every huge intimidating task is just one single job, waiting to be conquered one simple fraction at a time. No matter how complex the ASVAB math looks, treating the whole job as one makes you completely unstoppable. And you can do it entirely without a calculator. So I'll leave you with this. What other complex overwhelming challenges in your life could you conquer just by breaking them down into simpler fractions?

8:30 Think about it. Thanks for joining me on this explainer and keep practicing.

Visual lesson

Lesson slides

14 slides
Work and Combined-Rate Problems – slide 1 of 14

Lesson Goal

This lesson shows how to treat a job as one whole unit and each worker or machine as finishing a fraction of it per hour. You will combine rates for teams, find a missing worker's rate, handle work done before help arrives, deal with pipes that fill while drains empty, and work with output counted in parts or pages.

Work Rate as Fraction of a Job

If a task takes a worker h hours alone, the worker completes of the task each hour. A mechanic who rebuilds an engine in 5 hours does of the rebuild per hour. Thinking in fractions of a job lets you add, subtract and compare workers easily.

Work = Rate × Time

The amount of a job completed equals the rate times the time worked.

A crew that does of a fence per hour finishes of it in 4 hours, leaving of the fence still to build.

Fence in 9 parts: 4/9 built in 4 hours, 5/9 still to build.

Two Workers Together

When two people work at the same time, add their rates and then invert the sum to get the time. Workers needing 5 hours and 20 hours alone have a combined rate of

so the job takes 4 hours together.

Common Denominators for Work Rates

To add rates, rewrite the fractions over a common denominator. For 6 hours and 10 hours:

The time is hours, or 3 hours 45 minutes.

Three Workers or Machines

The same idea works for any number of machines. Copiers that take 4, 6 and 12 hours alone combine to

of the job per hour, so all three together need 2 hours.

Partial Work Before Help Arrives

Worker A needs 12 hours alone and works 4 hours, finishing

so remains. Worker B, who needs 6 hours alone, then joins. Together they do

per hour. Time for the rest:

hours

or 2 hours 40 minutes.

Changing Teams Mid-Job

Split the job into stages and track how much is done in each. Two cleaners who take 9 and 18 hours alone work together for 3 hours at one-sixth of the job per hour, finishing half. Then the slower one leaves. The faster cleaner needs 4.5 more hours for the remaining half, so the whole job takes 7.5 hours.

18-part job bar: both cleaners 3 h (1/2), faster alone 4.5 h (1/2), 7.5 h total.

Fill and Drain Problems

A fill pipe adds a positive rate and a drain adds a negative rate. A pipe that fills a tank in 3 hours and a drain that empties it in 6 hours give

tank per hour, so the tank fills in 6 hours with both open.

Opposing Production Rates

Quantities work the same way as fractions of a job. A conveyor delivers 40 boxes per minute while packers remove 25 boxes per minute. The pile grows by 15 boxes per minute, so a backlog of 300 boxes builds up in 20 minutes.

Pile gains 40 and loses 25 boxes per minute: net 15, 300 boxes in 20 minutes.

Finding an Unknown Worker Rate

Subtract the known rate from the team rate. A team needs 6 hours; A alone needs 10:

So B alone needs 15 hours.

Work Rate From Quantity Output

When output is counted in items, find each machine's items per unit time and add. Printer A makes 540 pages in 9 minutes (60 per minute); printer B makes 280 in 7 minutes (40 per minute). Together they make 100 pages per minute, so 1,500 pages take 15 minutes.

Printer A 60 + printer B 40 = 100 pages per minute; 1,500 ÷ 100 = 15 minutes.

Different Job Sizes

A rate measured on one job size can be scaled to another. A crew that paints 3 rooms in 4 hours works at 0.75 room per hour, so 9 rooms take 12 hours. If one job is twice as large as another at the same rate, it takes twice as long. Always state what one job means before combining rates.

Use the Whole Job as 1

Call the full task 1, whatever it is: a wall, a tank, an order. Each worker's rate is 1 divided by solo time, and the remaining work is 1 minus what is finished. This keeps every problem in one form, so you never need the actual size of the wall or tank.

Choose a Convenient Common Job Size

If fractions feel slow, invent a job size that both times divide evenly, usually their least common multiple. For workers needing 6 and 9 hours, call the job 18 units. They do 3 and 2 units per hour, so 5 together, and the job takes

18-unit job at 3 + 2 = 5 units per hour: 18 ÷ 5 = 3.6 hours.

Estimate Combined Time

Two workers together are always faster than the faster worker alone. If both were as fast as the faster one, the time would be half of that worker's time. So the answer lies between those two values. For 4 and 5 hours, it must be between 2 and 4 hours; the exact value is about 2.2 hours.

Number line: time together for 4-hour and 5-hour workers lies between 2 and 4 hours, about 2.2.

Worked Example 1 — Two Workers

Problem: A plumber needs 8 hours to do a job and an apprentice needs 24 hours. How long does it take them working together?

Step 1: Write each rate as the part of the job done in one hour: the plumber does and the apprentice does .

Step 2: Use the common denominator 24. Since

8 × 3 = 24

multiply the top and bottom of 1/8 by 3.

Step 3: Add the rates.

Step 4: Simplify by dividing top and bottom by 4.

Together they do one-sixth of the job per hour, so the whole job takes 6 hours.

Answer: They finish in 6 hours.

24-part job bar: plumber 3/24 plus apprentice 1/24 each hour fills the job in 6 hours.

Check: In 6 hours the plumber does

6/8 = 3/4

of the job and the apprentice does

6/24 = 1/4
3/4 + 1/4 = 1

whole job. And 6 hours is less than the plumber's 8 hours alone, as it should be.

Worked Example 2 — Three Machines

Problem: Three pumps empty a pond in 10, 15 and 30 hours when working alone. How long do all three take together?

Step 1: Each pump's hourly rate is 1 divided by its time: , and .

Step 2: Rewrite each with the common denominator 30.

Step 3: Add the three rates.

Step 4: Simplify by dividing top and bottom by 6.

All three together empty one-fifth of the pond per hour, so the whole pond takes 5 hours.

Answer: All three pumps take 5 hours.

30-part pond: pumps do 3/30, 2/30 and 1/30 each hour, empty in 5 hours.

Check: In 5 hours the pumps do

5/10 + 5/15 + 5/30 = 15/30 + 10/30 + 5/30 = 30/30

one whole pond. And 5 hours beats the fastest pump alone (10 hours).

Worked Example 3 — Unknown Worker

Problem: Together two typists finish a job in 4 hours. One alone takes 5 hours. How long does the other take alone?

Step 1: The team does of the job per hour, and the first typist does . The other typist's rate is what is left when you subtract.

20-part bars: team 5/20 per hour, first typist 4/20, other typist ? per hour.

Step 2: Use the common denominator 20.

Step 3: Subtract.

Step 4: The other typist does one-twentieth of the job per hour, so alone the job takes 20 hours.

Answer: The other typist alone takes 20 hours.

Check: Add the rates back:

1/5 + 1/20 = 4/20 + 1/20 = 5/20 = 1/4

That is the team rate, so together they finish in 4 hours.

Worked Example 4 — Partial Work

Problem: A welder can finish a frame alone in 15 hours and works by himself for 5 hours. A partner who could do the whole frame in 10 hours then joins. How long is the job in total?

Step 1: Find how much the welder finished alone. He does per hour for 5 hours.

Step 2: Find what remains.

Step 3: Find the team rate. Use the common denominator 30.

Step 4: Time = work ÷ rate. To divide by a fraction, multiply by its flip.

Step 5: Add the time alone and the time together.

Answer: The job takes 9 hours in total.

30-part frame: welder alone 5 hours (1/3), then both 4 hours at 1/6 per hour (2/3), 9 hours total.

Check: The welder works all 9 hours:

9/15 = 3/5

The partner works 4 hours:

4/10 = 2/5
3/5 + 2/5 = 1

whole frame.

Worked Example 5 — Fill and Drain

Problem: An inlet fills a tank in 6 hours and an open drain empties it in 15 hours. With both open, how long does the tank take to fill?

Step 1: The inlet adds of a tank per hour and the drain removes . The drain works against the inlet, so subtract.

Step 2: Use the common denominator 30.

Step 3: Simplify by dividing top and bottom by 3.

The tank gains one-tenth of its volume each hour, so it fills in 10 hours.

Answer: With both open, the tank fills in 10 hours.

Tank: inlet +1/6 per hour, drain −1/15 per hour, net 1/10 per hour, full in 10 hours.

Check: In 10 hours the inlet adds 10/6 tanks and the drain removes 10/15 tanks.

10/6 − 10/15 = 25/15 − 10/15 = 15/15 = 1

full tank. And 10 hours is slower than the inlet alone (6 hours), as it should be.

Worked Example 6 — Quantity Production

Problem: Machine A makes 210 bolts in 3 hours; Machine B makes 150 bolts in 5 hours. Working together, how long do they take to make an order of 850 bolts?

Step 1: Find each machine's hourly rate. Machine A:

Machine B:

Step 2: Add the rates to get the combined rate.

Step 3: Divide the order size by the combined rate.

Answer: The order takes 8.5 hours (8 hours 30 minutes).

Check: In 8.5 hours, A makes

70 × 8.5 = 595

bolts and B makes

30 × 8.5 = 255

bolts.

595 + 255 = 850

Common Combined-Work Traps

Adding times instead of rates: two workers taking 3 and 6 hours do not need 9 hours together. Add and invert.

A alone 3 h, B alone 6 h, together 2 h, not 9 h: 1/3 + 1/6 = 1/2 job per hour.

Averaging times: the team time is shorter than either worker's time, never the average.

Forgetting to invert: a combined rate of job per hour means 4 hours, not one-quarter hour.

Adding a drain: a drain's rate is subtracted.

Giving the wrong time: when help arrives partway, check whether the question asks for time after help joins or total elapsed time.

Quick Review

Treat a job as 1. A worker who needs h hours has a rate of 1/h per hour, and work done equals rate times time. Add rates for people or machines working together, then invert the total to get the time. Subtract the rate of a drain or opposing process. Find an unknown worker by subtracting the known rate from the team rate. For staged jobs, find what is left after each stage. With item counts, add items per hour directly. A convenient job size such as the LCM of the times avoids fractions, and the team must always beat the fastest single worker.

Quick review

A worker finishes a job alone in h hours. What is the work rate?
of the job per hour, with the whole job counted as 1. A mechanic who rebuilds an engine in 5 hours does of it each hour.
What is the work formula?
: work done (fraction of the job) equals rate (jobs per hour) times time (hours). At per hour, 4 hours completes , leaving .
How do you find the time for two workers working together?
Add their rates, then invert the sum. For 6 and 10 hours: , so the time is hours, or 3 hours 45 minutes.
A team needs 6 hours and A alone needs 10. How long does B need alone?
Subtract the known rate from the team rate: , so B alone needs 15 hours.
A pipe fills a tank in 6 hours; a drain empties it in 15. Both open?
The drain's rate is subtracted: tank per hour, so the tank fills in 10 hours.
Welder (15 h alone) works 5 hours, then a partner (10 h alone) joins. Total time?
Done: , so remains. Team rate: . Rest: hours. Total: 9 hours.
How can a convenient job size help you avoid fractions?
Let the job be the LCM of the solo times. For 6 and 9 hours, call it 18 units: the rates are 3 and 2 units per hour, 5 together, so hours.
Workers need 3 and 6 hours alone. Why is the team time not 9 or 4.5 hours?
Never add or average times. Add rates: , so 2 hours. The team time must fall between half the faster worker's time and that worker's time.

What next?

← Previous: Distance, Rate and Time

About this lesson

Later sections extend the method to three workers or machines, partial work before help arrives, changing teams mid-job, fill and drain problems and opposing production rates. You also learn to find an unknown worker rate, compute work rate from quantity output and handle different job sizes. Strategy sections on using the whole job as 1, choosing a convenient common job size and estimating combined time help you answer faster on Arithmetic Reasoning, an AFQT subtest.

Six worked examples cover two workers, three machines, an unknown worker, partial work, fill and drain, and quantity production. Review the common combined-work traps and Quick Review, then take the lesson test.

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