Problem 22 - Intermediate Algebra

Question

Fire Drill. If the east and west exit doors of a banquet hall are open, the occupants can clear out in 2 minutes. It takes 3 minutes longer to clear the hall if just the east door is open as it does if just the west door is open. How long does it take to clear the hall if just the west door is open?

Step-by-Step Solution

Answer

It takes 3 minutes to clear the hall if just the west door is open.

Step 1: Understand the given information

We are told that both doors open allow clearing the hall in 2 minutes. Let \( W \) be the time taken with only the west door open, and \( E \) the time with only the east door open. Additionally, \( E = W + 3 \) because it takes 3 more minutes using just the east door.

Step 2: Set up the equations

According to the problem, with both doors open, the clearing time is 2 minutes. The effective rate of clearing the hall with both doors open is: \( \frac{1}{W} + \frac{1}{E} = \frac{1}{2} \). Since \( E = W + 3 \), substitute \( E \) in the equation: \( \frac{1}{W} + \frac{1}{W+3} = \frac{1}{2} \).

Step 3: Simplify and Solve the Equation

Combine the fractions and solve the equation: \( \frac{(W+3) + W}{W(W+3)} = \frac{1}{2} \). This simplifies to \( \frac{2W + 3}{W^2 + 3W} = \frac{1}{2} \). Cross-multiply to get: \( 4W + 6 = W^2 + 3W \). Rearrange this to form a quadratic: \( W^2 - W - 6 = 0 \).

Step 4: Solve the Quadratic Equation

To solve \( W^2 - W - 6 = 0 \), we factor it: \( (W - 3)(W + 2) = 0 \). This gives solutions \( W = 3 \) and \( W = -2 \). Since time cannot be negative, discard \( W = -2 \). Thus, \( W = 3 \).

Step 5: Verify the Solution

Substitute \( W = 3 \) back into the relationship \( E = W + 3 \) to find \( E = 6 \). Check: \( \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \), which matches the open-both-exit criteria. Therefore, the solution is verified.

Key Concepts

Time and motion problems

Time and motion problems often involve understanding how various rates of working together or separately lead to different outcomes. In the fire drill situation, knowing how long it takes to clear the hall depends on the rate at which each door facilitates movement. This is similar to solving problems about cars traveling at different speeds or workers completing tasks.

• When both doors open, they work together to clear the hall faster. Their combined rate gives us an overall faster time.
• If only one door opens, the process takes longer due to limited capacity.
• The east door alone introduces an additional challenge, taking three more minutes than one with just the west door. This time difference is crucial for setting up an equation that reflects their efficiency.

Understanding these dynamics is key to setting up equations where you can add rates, reflect additional time, and solve for unknowns effectively.

Algebraic expressions

Algebraic expressions represent mathematical statements with varying parts, like numbers, variables, and operation symbols. In tackling our problem with the doors, we use algebraic expressions to model situations and relationships.

• Letting variables like \( W \) and \( E \) denote time for the west and east doors allows us to set up easy-to-handle statements.
• The relationship \( E = W + 3 \) captures an essential understanding that the east door takes longer.
• Combining these into expressions like \( \frac{1}{W} + \frac{1}{E} \) presents these times as rates, making it simpler to see their combined and individual contributions to the total clearing time.

These expressions transform word problems into solvable mathematical equations, revealing insights through strategic substitution and evaluation.

Factoring polynomials

Factoring polynomials is a crucial technique in algebra, simplifying complex equations to reveal their roots. Here, we navigated algorithmically from the equation \( W^2 - W - 6 = 0 \).

• Polynomials like this one contain terms with different powers and are commonly rewritten as products of simpler expressions.
• Recognizing this form helped us identify values for \( W \) by rewiring the quadratic into \( (W - 3)(W + 2) = 0 \).
• This factoring uncovers the possible values for \( W \), offering solutions \( W = 3 \) and \( W = -2 \), with "3" being logical as it fits real-world contexts, because time can't be negative.

Mastery of this skill not only aids in finding solutions but also offers appreciation for the interconnected nature of algebraic forms and their applications in everyday scenarios.