Why do we add rates instead of times?

A rate tells you the fraction of a job completed per unit time. If one person completes \frac{1}{6} of a job per hour and another completes \frac{1}{3} per hour, together they complete \frac{1}{6} + \frac{1}{3} = \frac{1}{2} per hour. Times themselves do not add: working 6 hours and 3 hours does not tell you how fast they work together.

What if the answer comes out to a fraction like \frac{35}{12}? Do I simplify it?

Yes, simplify if possible, but \frac{35}{12} is already in lowest terms because 35 and 12 share no common factors. You can convert it to a mixed number 2\tfrac{11}{12} or a decimal 2.92 hours, whichever the problem asks for, but the fraction itself is correct as is.

Can I use this formula if the times are given in different units, like one in hours and one in minutes?

No, t_1 and t_2 must be in the same units. If one person takes 2 hours and another takes 30 minutes, convert first: 2 hours = 120 minutes, so use t_1 = 120 and t_2 = 30, both in minutes. Then T will also come out in minutes.

What if I solve the problem and get a negative time?

A negative time is meaningless in this context and signals an error. Check that all your times t_1 and t_2 are positive, that you added the rates (not subtracted), and that you took the reciprocal correctly. Negative time should never occur in a Work Rate Problem with positive input times.

Work Rate Problems

Work Rate Problems help you find how long a job takes when people work together, using the sum of individual work rates.

\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T}

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Type the problem. The solver will use Work Rate Problems where Work Rate Problems is the right tool, and tell you when it is not.

What each symbol means

What Work Rate Problems takes

Symbol	Meaning
$t_1$	The time it takes the first person or machine to complete the entire job working alone, measured in hours, days, or whatever unit the problem uses; it must be positive and nonzero.
$t_2$	The time it takes the second person or machine to complete the entire job working alone, in the same units as $t_1$ ; it must be positive and nonzero.
$T$	The time it takes both people or machines working together to complete the entire job; it is always less than whichever of $t_1$ or $t_2$ is smaller, because working together is faster.

When to use it

Use this when you know how long each person takes to complete a task alone, and need to find how long it takes them working together.

Level

Usually taught in: Algebra I

Worked examples

1. Two painters working together

Problem

Sarah can paint a room in 6 hours. Tom can paint the same room in 3 hours. If they work together, how long does it take to paint the room?

$\frac{1}{6} + \frac{1}{3} = \frac{1}{T}$
Write the work rate for each person as a fraction of the job per hour, then set their sum equal to the combined rate.
$\frac{1}{6} + \frac{2}{6} = \frac{1}{T}$
Convert $\frac{1}{3}$ to have the same denominator as $\frac{1}{6}$ so you can add them: $\frac{1}{3} = \frac{2}{6}$ .
$\frac{3}{6} = \frac{1}{T}$
Add the fractions: $\frac{1}{6} + \frac{2}{6} = \frac{3}{6}$ .
$\frac{1}{2} = \frac{1}{T}$
Simplify $\frac{3}{6}$ by dividing numerator and denominator by 3.
$T = 2$
Since $\frac{1}{T} = \frac{1}{2}$ , take the reciprocal of both sides: $T = 2$ hours.

Answer: $T = 2$

This is the right method because work rates are additive: each person completes a fraction of the job per hour, and those fractions add together. Sarah does $\frac{1}{6}$ per hour, Tom does $\frac{1}{3}$ per hour, so together they do $\frac{1}{2}$ per hour, finishing in 2 hours.

2. Two pumps filling a tank

Problem

One pump can fill a tank in 5 hours. Another pump can fill the same tank in 7 hours. If both pumps run at the same time, how long does it take to fill the tank?

$\frac{1}{5} + \frac{1}{7} = \frac{1}{T}$
Each pump's rate is the reciprocal of the time it takes alone: the first pump's rate is $\frac{1}{5}$ tanks per hour.
$\frac{7}{35} + \frac{5}{35} = \frac{1}{T}$
Convert to a common denominator: $\frac{1}{5} = \frac{7}{35}$ and $\frac{1}{7} = \frac{5}{35}$ since $5 \times 7 = 35$ .
$\frac{12}{35} = \frac{1}{T}$
Add the numerators: $7 + 5 = 12$ .
$T = \frac{35}{12}$
Take the reciprocal of both sides. Since $\frac{1}{T} = \frac{12}{35}$ , then $T = \frac{35}{12}$ .
$T = 2\tfrac{11}{12} \text{ hours}$
Convert the improper fraction to a mixed number: $35 \div 12 = 2$ remainder $11$ , so $\frac{35}{12} = 2\tfrac{11}{12}$ , or about 2.92 hours.

Answer: $T = \frac{35}{12} \text{ or } 2\tfrac{11}{12} \text{ hours}$

Fractions with different denominators require a common denominator before adding. This problem is harder because 5 and 7 share no common factors, so the LCD is $5 \times 7 = 35$ . The reciprocal at the end is crucial: $\frac{12}{35}$ is the combined rate, not the time.

3. Garden plot fencing

Problem

Alex can fence a rectangular garden plot completely in 8 hours. Bella can fence the same plot in 12 hours. If they work together, how long will it take them to finish fencing the garden?

$\frac{1}{8} + \frac{1}{12} = \frac{1}{T}$
Alex's rate is $\frac{1}{8}$ of the fence per hour; Bella's rate is $\frac{1}{12}$ of the fence per hour. Add them to get the combined rate.
$\frac{3}{24} + \frac{2}{24} = \frac{1}{T}$
Find the LCD of 8 and 12. Since $8 = 2^3$ and $12 = 2^2 \times 3$ , the LCD is $2^3 \times 3 = 24$ . Convert: $\frac{1}{8} = \frac{3}{24}$ and $\frac{1}{12} = \frac{2}{24}$ .
$\frac{5}{24} = \frac{1}{T}$
Add: $3 + 2 = 5$ , so the combined rate is $\frac{5}{24}$ of the fence per hour.
$T = \frac{24}{5}$
Reciprocal both sides: if $\frac{1}{T} = \frac{5}{24}$ , then $T = \frac{24}{5}$ .
$T = 4\tfrac{4}{5} \text{ hours}$
Convert to a mixed number: $24 \div 5 = 4$ remainder $4$ , so $T = 4\tfrac{4}{5}$ hours, or 4.8 hours.

Answer: $T = \frac{24}{5} \text{ or } 4\tfrac{4}{5} \text{ hours}$

Word problems require you to first identify what $t_1$ and $t_2$ represent. Here, Alex's solo time is 8 hours and Bella's is 12 hours. The combined time is always less than the faster person's solo time, which checks out: $4\tfrac{4}{5}$ hours is less than 8 hours.

Common mistakes

Where Work Rate Problems usually goes wrong

The mistake	Why it is wrong	The fix
Writing $T = t_1 + t_2$ or adding the times directly, such as $T = 6 + 3 = 9$ hours.	This ignores that work rates add, not times; if two people work together, they finish faster than either person alone, not slower.	Always use the formula $\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T}$ , which adds rates (reciprocals of times), not the times themselves.
Forgetting to take the reciprocal at the end and writing $T = \frac{12}{35}$ instead of $T = \frac{35}{12}$ .	The equation $\frac{12}{35} = \frac{1}{T}$ means the combined rate is $\frac{12}{35}$ per hour; to find time, you must flip it.	After solving for $\frac{1}{T}$ , always take the reciprocal to get $T$ : if $\frac{1}{T} = \frac{12}{35}$ , then $T = \frac{35}{12}$ .
Adding fractions without finding a common denominator, such as writing $\frac{1}{5} + \frac{1}{7} = \frac{2}{12}$ .	Fractions can only be added if they have the same denominator; adding numerators and denominators separately gives a meaningless result.	Always find the LCD first: for $\frac{1}{5} + \frac{1}{7}$ , the LCD is 35, so $\frac{7}{35} + \frac{5}{35} = \frac{12}{35}$ .

Tips and when to use something else

The combined rate is always faster than either person working alone, so $T$ must be less than $\min(t_1, t_2)$ . If your answer is not, check that you took the reciprocal.
When one person finishes much faster than the other (e.g., 2 hours vs. 20 hours), the combined time will be close to the faster time, because the faster person does most of the work.
If a problem has three or more workers, extend the formula: $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{T}$ . The principle is the same.
For problems where one person works at the job and another undoes it (e.g., one fills a pool while another drains it), use subtraction instead: $\frac{1}{t_1} - \frac{1}{t_2} = \frac{1}{T}$ ; see Rational Equations for more detail.

Related formulas

Rational EquationsA rational equation has a polynomial fraction equal to zero; solve by setting the numerator to zero, then verify the denominator is nonzero.Literal EquationsLiteral equations rearrange formulas to solve for any variable, letting you find missing measurements when you know the others.ProportionUse Proportion when two ratios are equal to find a missing value by cross-multiplying to solve equations with fractions.Distance Rate TimeCalculate distance, rate, or time in any uniform motion problem using the formula d = rt, where distance equals rate multiplied by time.

Frequently asked questions

Why do we add rates instead of times?: A rate tells you the fraction of a job completed per unit time. If one person completes $\frac{1}{6}$ of a job per hour and another completes $\frac{1}{3}$ per hour, together they complete $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ per hour. Times themselves do not add: working 6 hours and 3 hours does not tell you how fast they work together.
What if the answer comes out to a fraction like $\frac{35}{12}$ ? Do I simplify it?: Yes, simplify if possible, but $\frac{35}{12}$ is already in lowest terms because 35 and 12 share no common factors. You can convert it to a mixed number $2\tfrac{11}{12}$ or a decimal 2.92 hours, whichever the problem asks for, but the fraction itself is correct as is.
Can I use this formula if the times are given in different units, like one in hours and one in minutes?: No, $t_1$ and $t_2$ must be in the same units. If one person takes 2 hours and another takes 30 minutes, convert first: 2 hours = 120 minutes, so use $t_1 = 120$ and $t_2 = 30$ , both in minutes. Then $T$ will also come out in minutes.
What if I solve the problem and get a negative time?: A negative time is meaningless in this context and signals an error. Check that all your times $t_1$ and $t_2$ are positive, that you added the rates (not subtracted), and that you took the reciprocal correctly. Negative time should never occur in a Work Rate Problem with positive input times.

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Reviewed 2026-09-18