Fine Dining. It takes a waiter 5 minutes less time than a...

Name: Problem 21, Chapter 6: Intermediate Algebra
Uploaded: 2020-08-08T13:36:13-08:00
Duration: 3 min 38 s
Channel: Kelsey Dondelinger
Description: Fine Dining. It takes a waiter 5 minutes less time than a busboy to fold the napkins used for the dinner seating in an upscale restaurant. Working together, they can fold the napkins in 6 minutes. How long would it take each person working alone to fold the napkins?

Fine Dining. It takes a waiter 5 minutes less time than a busboy to fold the napkins used for the dinner seating in an upscale restaurant. Working together, they can fold the napkins in 6 minutes. How long would it take each person working alone to fold the napkins?

Key Concepts

Work Rate

This concept involves understanding how much of a task an individual can complete per unit of time. Work rate is typically expressed as the reciprocal of the time taken to complete a job, providing a measure of efficiency that can be used in various related problems.

Reciprocal Relationship between Time and Rate

A key idea in work problems is that the rate at which a job is completed is the reciprocal of the time taken. This relationship allows one to convert time to work rate and vice versa, which is essential when combining different workers’ contributions.

Combined Work Rate

When multiple individuals work together, their individual work rates add up to produce a combined work rate. This aggregate rate can be used to determine how long the group takes to complete the task, making it a crucial concept for solving tandem work problems.

Setting Up Systems of Equations

To tackle problems that involve differences in work times or rates, forming and solving a system of equations is often necessary. This approach allows for the representation of the relationships between the various quantities and the efficient computation of unknowns.

Transcript

00:01 So in this example, we have a bus boy and a waiter who are folding napkins.

00:04 So the bus boy can do it in, let's just say x minutes because we don't know how long it takes.

00:10 The waiter can do it five minutes faster than that.

00:13 So x minus five minutes.

00:16 And then together, they can do it in six minutes.

00:20 So we want to know how long it's going to take for each person if they didn't do it together.

00:27 So if we look at this, when we think about our work problems, we usually convert to per minute.

00:32 So we're looking at the busboy doing one x of the job per minute, the waiter doing 1 over x minus 5 per minute, in total they can do 1 6 of the job per minute.

00:44 So we can write this as an equation.

00:46 So 1 over x for the bus boy, plus 1 over x minus 5 for the waiter is 1 over 6 together.

00:54 So i'm trying to solve for x.

00:56 Our common denominator here is 6.

01:02 6, let's see, 6 times x times x minus 5, we're going to multiply all of our denominators together.

01:16 So in the first one, when i multiply that by 1 over x, the x is will cancel, so i'm left with the 6 and the x minus 5.

01:24 In the second fraction, the x minus 5s will cancel, so i'm left with the 6x.

01:30 Then the last fraction, the 6 is will cancel, so i'm left with x times x minus 5.

01:36 And now we're going to multiply this all out using distributive property...

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