Question

Filling a Tub A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed?

Short answer

It will take 60 minutes to fill the tub with both faucets open and the stopper removed.

Step-by-step solution

- Determine individual rates

First, calculate the rate at which the tub fills when both faucets are open. It will fill in 15 minutes, so the rate is \(\frac{1}{15}\) of the tub per minute. Similarly, when the stopper is removed with both faucets closed, it empties in 20 minutes, so the emptying rate is \(\frac{1}{20}\) of the tub per minute.

- Combine the rates

When both faucets are open and the stopper is also removed simultaneously, the effective rate of filling the tub is given by the combined rate of filling and emptying. This can be expressed as \[ \text{Effective Rate} = \frac{1}{15} - \frac{1}{20} \]

- Simplify the combined rate

To simplify \[ \text{Effective Rate} = \frac{1}{15} - \frac{1}{20} \, \] find a common denominator. The least common multiple of 15 and 20 is 60. Using this, \(\frac{1}{15} = \frac{4}{60}\) and \(\frac{1}{20} = \frac{3}{60}\). Thus, \[\text{Effective Rate} = \frac{4}{60} - \frac{3}{60} = \frac{1}{60} \]

- Calculate the time to fill the tub

The effective rate of \(\frac{1}{60}\) means the tub fills at a rate of \(\frac{1}{60}\) of the tub per minute. Therefore, the time to fill the tub is the reciprocal of the effective rate, which is 60 minutes.

Key concepts

Rate of Change

When dealing with rate problems, it's essential to understand the concept of rate of change. The rate of change refers to how much one quantity changes with respect to another. In the context of filling or emptying a tub, we're looking at the rate at which the volume of water in the tub changes per unit of time.
For example, in this problem, the tub fills in 15 minutes, which gives a filling rate of \(\frac{1}{15}\) tubs per minute. Similarly, the tub empties in 20 minutes, resulting in an emptying rate of \(\frac{1}{20}\) tubs per minute. Combining these rates allows us to determine the net rate of change when both filling and emptying actions occur simultaneously.

Fraction Operations

Solving rate problems often involves operations with fractions. In this exercise, we combine the rates of filling and emptying the tub. Let's break down how to perform these operations step by step.
1. Identify the rates: Filling rate is \(\frac{1}{15}\) and emptying rate is \(\frac{1}{20}\).
2. Find a common denominator to combine the fractions: The least common multiple of 15 and 20 is 60. So we convert \(\frac{1}{15}\) to \(\frac{4}{60}\) and \(\frac{1}{20}\) to \(\frac{3}{60}\).
3. Subtract the fractions: Combining the rates gives us the effective rate \(\frac{4}{60} - \frac{3}{60} = \frac{1}{60}\).
Understanding fraction operations is crucial for dealing with these types of algebraic rates and simplifying them to find the effective rate of change.

Word Problems in Algebra

Word problems in algebra, like this tub-filling exercise, require translating a real-world scenario into mathematical equations. Here's a simple approach to tackle such problems:
1. Read the problem carefully to identify all given information and what you need to find.
2. Define variables for unknowns and set up equations based on the problem statement. In this case, we need to find the time to fill the tub when both filling and emptying occur.
3. Use relevant algebraic operations: We used the concept of rates and combined them using fraction addition and subtraction.
4. Solve the equations step by step: Calculate individual rates, combine them, simplify, and find the reciprocal to get the final answer.
Practice with different word problems helps in understanding how to break down and solve complex scenarios into manageable algebraic steps.