Work and Rate Problems with Step-by-Step Solutions
This page presents a collection of work and rate problems in mathematics, covering scenarios such as lawn mowing, tank filling, swimming pools, and pumps. Each problem is accompanied by step-by-step solutions to help you understand the methods used to calculate time, rates, and combined work. These examples are ideal for students, teachers, and anyone looking to practice work and time exercises or improve problem-solving skills in work rate mathematics.
Table of Contents
Formula and Strategy Review
Core Idea: In work-rate problems, the fundamental relationship is:
\[ \text{Work} = \text{Rate} \times \text{Time} \]If a job takes \( t \) hours to complete alone, then the rate is \( \dfrac{1}{t} \) jobs per hour. Working together:
\[ r_1 + r_2 + \cdots = \frac{1}{t_1} + \frac{1}{t_2} + \cdots \]Strategy by Scenario:
- Two workers, together: \( t \left(\dfrac{1}{t_1} + \dfrac{1}{t_2}\right) = 1 \)
- Finding a third rate: \( T \left(\dfrac{1}{t_1} + \dfrac{1}{t_2} + r_3\right) = 1 \)
- With drainage: \( t \left(\dfrac{1}{t_{\text{fill}}} - \dfrac{1}{t_{\text{drain}}}\right) = 1 \)
- Delayed start: If the second starts \( h \) hours later, its time is \( t - h \).
Problem 1 — Lawn Mowing Time
It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it take Tim and Linda, working together, to mow the lawn?
View Step-by-Step Solution
We first calculate the rate of work of Tim and Linda (portion of the job completed per hour):
- Tim's rate: \( \dfrac{1}{1.5} = \dfrac{2}{3} \)
- Linda's rate: \( \dfrac{1}{2} \)
Let \( t \) be the time for Tim and Linda to mow the lawn together. The work done by Tim alone in \( t \) hours is:
\[ t \times \frac{1}{1.5} \]The work done by Linda alone in \( t \) hours is:
\[ t \times \frac{1}{2} \]When the two work together, their work is added to equal 1 complete job:
\[ t \times \frac{1}{1.5} + t \times \frac{1}{2} = 1 \]Multiply all terms by 6 to clear the decimals and fractions:
\[ 6 \left(t \times \frac{1}{1.5} + t \times \frac{1}{2} \right) = 6 \times 1 \] \[ 4t + 3t = 6 \implies 7t = 6 \]Verification: Tim completes \( \frac{6}{7} \times \frac{2}{3} = \frac{12}{21} = \frac{4}{7} \) of the lawn, and Linda completes \( \frac{6}{7} \times \frac{1}{2} = \frac{6}{14} = \frac{3}{7} \). Their sum is \( \frac{4}{7} + \frac{3}{7} = 1 \) ✓
Problem 2 — Pump Rates: Finding the Third Pump
It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. We want to use three pumps: A, B, and another pump C to fill the tank in 2 hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank?
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The individual rates of pumps A and B:
- Pump A's rate: \( \dfrac{1}{6} \)
- Pump B's rate: \( \dfrac{1}{8} \)
Let \( R \) be the rate of pump C. When working together for 2 hours, their combined work completes 1 tank:
\[ 2 \left( \frac{1}{6} + \frac{1}{8} + R \right) = 1 \]Divide by 2 and find a common denominator for the fractions:
\[ \frac{4}{24} + \frac{3}{24} + R = \frac{1}{2} \] \[ \frac{7}{24} + R = \frac{12}{24} \]Solve for \( R \):
\[ R = \frac{5}{24} \text{ tank per hour} \]The time it takes pump C, used alone, to fill the tank is the reciprocal of its rate:
\[ t = \frac{1}{R} = \frac{24}{5} = 4.8 \text{ hours} \]Verification: \( 2\left(\frac{1}{6} + \frac{1}{8} + \frac{5}{24}\right) = 2\left(\frac{4+3+5}{24}\right) = 2 \cdot \frac{12}{24} = 1 \) ✓
Problem 3 — Tank Filling with Drainage
A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours, each pipe working on its own. When the tank is full and a drainage hole is open, the water is drained in 20 hours. If initially the tank was empty and someone started the two pipes together but left the drainage hole open, how long does it take for the tank to be filled?
View Step-by-Step Solution
The rates of the two pipes and the drainage hole:
- Pipe A's rate: \( \dfrac{1}{5} \)
- Pipe B's rate: \( \dfrac{1}{8} \)
- Drainage rate: \( \dfrac{1}{20} \)
Let \( t \) be the time for the tank to be filled. The pipes add water into the tank while the drainage hole removes water, so:
\[ t \left( \frac{1}{5} + \frac{1}{8} - \frac{1}{20} \right) = 1 \]Find a common denominator (40):
\[ t \left( \frac{8}{40} + \frac{5}{40} - \frac{2}{40} \right) = 1 \] \[ t \left( \frac{11}{40} \right) = 1 \]Verification: In \( \frac{40}{11} \) hours, pipe A fills \( \frac{40}{55} = \frac{8}{11} \), pipe B fills \( \frac{40}{88} = \frac{5}{11} \), drainage removes \( \frac{40}{220} = \frac{2}{11} \). Net: \( \frac{8+5-2}{11} = \frac{11}{11} = 1 \) ✓
Problem 4 — Swimming Pool with Delayed Start
A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pipe working on its own. At 9:00 am pipe A is started. At what time will the swimming pool be filled if pipe B is started at 10:00 am?
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The rates of the two pipes:
- Pipe A's rate: \( \dfrac{1}{3} \)
- Pipe B's rate: \( \dfrac{1}{6} \)
If pipe A works for \( t \) hours, then pipe B works \( t - 1 \) hours (it was started 1 hour later):
\[ t \times \frac{1}{3} + (t - 1) \times \frac{1}{6} = 1 \]Multiply the entire equation by 6 to clear the fractions:
\[ 2t + (t - 1) = 6 \] \[ 3t - 1 = 6 \implies 3t = 7 \] \[ t = \frac{7}{3} \text{ hours} = 2 \text{ hours } 20 \text{ minutes} \]The swimming pool will be filled 2 hours and 20 minutes after pipe A started at 9:00 am:
Verification: Pipe A runs for \( \frac{7}{3} \) hours filling \( \frac{7}{9} \) of the pool; pipe B runs for \( \frac{4}{3} \) hours filling \( \frac{4}{18} = \frac{2}{9} \). Total: \( \frac{7}{9} + \frac{2}{9} = 1 \) ✓
Summary Table of Answers
| # | Scenario | Answer |
|---|---|---|
| 1 | Tim and Linda mow the lawn together | \( \frac{6}{7} \) hours ≈ 51.4 minutes |
| 2 | Pump C filling a tank with pumps A and B | Rate \( \frac{5}{24} \) tank/hr; 4.8 hours alone |
| 3 | Tank filling with drainage hole open | \( \frac{40}{11} \) ≈ 3.63 hours |
| 4 | Pool filled by A (9:00) and B (10:00) | 11:20 am |
Frequently Asked Questions
How do you calculate combined work rate?
Find the individual rate of work for each person or machine (\( 1 \) divided by the time each takes alone), add these rates together to get the combined rate, then set combined rate × time = 1 to solve for the combined time.
How do you solve a work problem with a drainage hole?
Add the filling rates together and subtract the drainage rate: \( \left(\dfrac{1}{t_{\text{fill1}}} + \dfrac{1}{t_{\text{fill2}}} - \dfrac{1}{t_{\text{drain}}}\right) \times t = 1 \). Solve for \( t \). This works because the drainage removes water at its own rate.
How do you solve a work problem where one worker starts later?
Let \( t \) be the time the first worker has worked. The second worker has worked for \( (t - \text{delay}) \) hours. Write the work equation using these times and set the total equal to 1, then solve for \( t \).
What is the formula for work-rate problems?
The fundamental formula is \( \text{Work} = \text{Rate} \times \text{Time} \). If a job takes \( t \) hours alone, the rate is \( 1/t \) (jobs per hour). When multiple workers combine, their rates add: \( \dfrac{1}{t_1} + \dfrac{1}{t_2} + \cdots = \dfrac{1}{T} \).
How do you find the rate of a third pump given the combined time?
Set up the equation \( T \times (r_1 + r_2 + r_3) = 1 \), where \( T \) is the combined time and \( r_1, r_2, r_3 \) are the individual rates. Solve for \( r_3 \). The time for the third pump alone is \( 1/r_3 \).