Problem 16
Question
Each exercise is a problem involving work. A pool can be filled by one pipe in 3 hours and by a second pipe in 6 hours. How long will it take using both pipes to fill the pool?
Step-by-Step Solution
Verified Answer
It will take 2 hours to fill the pool using both pipes.
1Step 1: Understand the rates of work
Let's understand the rates at which each pipe fills the pool. The first can fill it in 3 hours, so its rate is 1/3 of the pool per hour. The second fills it in 6 hours, thus its rate is 1/6 per hour.
2Step 2: Add the individual rates
The combined rate at which the two pipes can fill the pool is the sum of their individual rates. So, add 1/3 and 1/6 together which is \( \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \) of the pool per hour.
3Step 3: Calculate the time
As we now know the combined rate, we can find out the time it will take both pipes to fill the pool. If they can fill half the pool in an hour, they will take 2 hours to fill the full pool.
Key Concepts
Work Word ProblemsAlgebraic RatesAdding RatesRational Expressions
Work Word Problems
Work word problems in algebra involve figuring out how long it takes for one or more workers (or machines, pipes, etc.) to complete a particular job. A typical example, like the one involving the two pipes filling a pool, requires us to understand the concept of work rates. These rates indicate the fraction of work done by an individual or a combined effort per unit of time. In this specific case, we measure the rate at which each pipe fills the pool and then determine how long it will take for both pipes to fill the pool together. It's crucial to consider that the work being done is cumulative, meaning the work completed by one pipe adds to the work completed by the other. Understanding the underlying principle of work problems can help in solving similar problems in different contexts, such as multiple workers contributing to a single project.
Here's an improvement tip: Always clearly define what one unit of 'work' represents (e.g., one filled pool) before solving the problem.
Here's an improvement tip: Always clearly define what one unit of 'work' represents (e.g., one filled pool) before solving the problem.
Algebraic Rates
Algebraic rates are expressions that show the relationship between the amount of work completed and the time taken to do the work. These are usually expressed in the form of fractions or ratios. In the context of our exercise, the rate at which the first pipe fills the pool is represented as \( \frac{1}{3} \) pools per hour, and the second pipe as \( \frac{1}{6} \) pools per hour. The algebraic rate is essential as it allows us to use mathematical operations to combine and manipulate the rates to solve work-related problems. Algebraic rates can be complex, but they're the cornerstone for understanding how different factors affect the time it takes to complete a job when varying work rates are involved.
To enhance student understanding, it is advised to provide multiple examples with varying rates and to explain the concept of 'unit rate'—the amount of work per one unit of time.
To enhance student understanding, it is advised to provide multiple examples with varying rates and to explain the concept of 'unit rate'—the amount of work per one unit of time.
Adding Rates
When faced with a scenario where multiple entities contribute to a single task, we often need to combine their work rates to find the total rate of work being done. Adding rates algebraically is necessary when solving work problems like the pool and pipe scenario. By adding the rates of the two pipes, we determine the rate at which they fill the pool together. In this example, we have \( \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \) pools per hour as the combined rate. When adding rates, it's important to find a common denominator to combine the fractions properly. This application of basic algebraic principles helps to create a single equation that can be solved to find the desired result, such as the total time taken for a job.
To build a solid foundation, you should practice adding rates with different denominators by finding the least common denominator (LCD) and then solving addition.
To build a solid foundation, you should practice adding rates with different denominators by finding the least common denominator (LCD) and then solving addition.
Rational Expressions
Rational expressions are fractions that involve polynomials in the numerator and the denominator. They appear often in work word problems when rates are expressed as fractions of work per unit time. In our exercise, the individual work rates (\( \frac{1}{3} \) and \( \frac{1}{6} \) pools per hour) are rational expressions. To solve these problems, we carry out operations with these expressions such as finding a common denominator and simplifying. It is crucial to understand how to manipulate these rational expressions algebraically because it enables us to combine rates accurately and come up with a solution. A good grasp of how to add, subtract, multiply, and divide rational expressions is invaluable in solving a wide array of algebra problems.
As an improvement tip, ensure that you're comfortable simplifying complex fractions and converting mixed numbers to improper fractions, which can often be part of work word problems.
As an improvement tip, ensure that you're comfortable simplifying complex fractions and converting mixed numbers to improper fractions, which can often be part of work word problems.
Other exercises in this chapter
Problem 15
Multiply as indicated. $$\frac{y^{2}-7 y-30}{y^{2}-6 y-40} \cdot \frac{2 y^{2}+5 y+2}{2 y^{2}+7 y+3}$$
View solution Problem 15
add or subtract as indicated. Simplify the result, if possible. $$\frac{4 x+1}{6 x+5}+\frac{8 x+9}{6 x+5}$$
View solution Problem 16
Use the four-step procedure for solving variation problems given on page 551 An object's weight on the moon, \(M,\) varies directly as its weight on Earth, \(E
View solution Problem 16
Find all numbers for which each rational expression is undefined. If the rational expression is defined for all real numbers, so state. $$\frac{y+8}{6 y^{2}-y-2
View solution