Expert: Josh - 3/10/2006

Question
Hello:

I appreciate your efforts in attempting to answer my question. I, however, am somewhat confused by your calculations. My mathematical level is not as advanced as yours. Intermediate algebra is the highest level of mathematics that I have had. I personally do not find it "pointless" to assign units to numbers as I have done in my first question.  I'll need to study your reply so that I may find some value from it. I think that you need to know that it is quite common to attach units to numbers especially with fractions so that one can see what gets canceled out from the multiplication. I believe that in mathematics and physics it has a specific name "dimensional analysis".  When I try to solve a mathematical calculation, I try to use a simple calculation, if possible, to determine the answer. There is no need to use a shotgun to kill a fly on the wall, so to speak. Once again, I thank you for your reply.
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Followup To
Question -
Hello:

If 3 people can rake a yard in 7 hours, how long will it take 5
people to rake the same yard?

Solution: (3 people X 7 hours)/5 people = 3 people X (7 hours)/(5
people) = 21/5 hours or 4 1/5 hours.

My question is how does someone know what units belong with the other
units so that the answer is correct?

For example, (3 people X 7 hours)/5 people will give the desired
answer, but (5 people X 7 hours)/3 people will give a different
answer for the above question because the number and units do not
belong together to determine the desired answer.

I thank you for your reply.

Answer -
Matching units somewhat arbitrarily is rather pointless in my opinion. That's not the way I would go about doing this type of problem. I would prefer to look at it this way. The following gives you much more intuition. By the end, I hope you will see that the questions you asked me previously are essentially the same sort of problem.

Remember, the amount of work done by person n (which I write w[n], this stands for some unknown quantity) is always the rate at which the work is done (r[n], again, this stands for some unknown quantity) multiplied by the time this person has spent working on the task (t[n]). Mathematically, we can express this as
w[n]=r[n]*t[n] ...[#1]

More generally, the following principle always holds.
w=r*t ...[#2]
where w=total amount of work, r=effective work rate, and t=time interval (whether for an individual or group depends on the context of the problem).

If N people contribute to this work, then,
w=w[1]+w[2]+...+w[N] --- a linear sum comprising N terms
=r[1]*t[1]+r[2]*t[2]+...+r[N]*t[N] ...[#3]
Without loss of generality, each variable may be different.

In your question, the first scenario involves three people doing W amount of work. So, N=3. From [#3], we have

W=r[1]*t[1]+r[2]*t[2]+r[3]*t[3] ...[#4a]

Furthermore, since they start and finish the work at the same time, t[1]=t[2]=t[3]=t, for some time interval t. Thus,
W=(r[1]+r[2]+r[3])*t, ...[#4b]

Assuming that each person gets the same amount of work done in a given period of time, that is to say, they work at the same rate, we can further simplify [#4b] as

W=3*r*t ...[#4c]

using the assumption that r[1]=r[2]=r[3]=r. In [#4c], we interpret 3r=r[1]+r[2]+r[3] as the effective work rate.

From [#4c], we deduce that the individual work rate must be r=W/(3t) ...[#5]

In the second scenario, we implicitly assume that N'=5 people work at the same rate, for equal length of time.
We may let r[1]=r[2]=r[3]=r[4]=r[5]=r as before, and let t[1]=t[2]=t[3]=t[4]=t[5]=t', where t'<t (new time interval shorter than the previous time interval, as more people work on the same problem).

Holding W constant, and r constant, from [#3], we must have

W=r[1]*t[1]+r[2]*t[2]+r[3]*t[3]+r[4]*t[4]+r[5]*t[5]
=5rt'    Observe: t' is different and smaller than t.
t'=W/(5r)...[#6]

Substituting for r (using [#5]) in expression [#6], we have
t'=W/(5*W/(3t))
 =(W/5)*(3t/W)
 =(3/5)*t.

Known fact from your question: t=7 (hours).

So, the reduced work hours, t'=(3/5)*7 = 4.2 hours

I only substitute for the known value of t at the last minute, to preserve the meaning of variables in the algebraic manipulation. This is standard practice.

P.S. I am not sure that putting units in an equation is really a good idea. It just adds to the confusion sometimes. Does that help you in any way?

Answer
Don't be upset. I am not saying that attaching units to numbers is always pointless, but I just don't see any advantage in doing so for the problem you were looking at. I know that in certain branches of science, "dimensional analysis" plays an important role. For instance, this is the case in chemistry, when performing entropy calculations in thermodynamics (far more complex problems which involve several quantities and multiple units). I certainly recognize that dimensional analysis is a powerful tool, but your situation does not call for it.

If your goal is to get better at manipulating fractions, then, that's fine. But recall that it was the idea of cancelling the units that got you confused. This is because both "(3 people X 7 hours)/5 people" and "(5 people X 7 hours)/3 people" are viable options in dimensional analysis. Dimensional analysis is like a check sum, it alone does not solve a problem.

It would be much simpler to look at the question in terms of a rate change problem. This notion may be further developed in calculus.

The only idea that I tried to describe is

e.g.1, distance = speed * time
e.g.2, work = work_rate * time

Cancelling units is purely a mechanical process. To me, a physical principle would offer far greater insight.

If you look at the units in the two examples, there are only two units involved.
for e.g. 1: we may have [m] ~ [m/sec] * [sec]
for e.g. 2: we may have [Joules] ~ [Joules/sec] * [time]

As you've said, why make life more complicated than necessary. Isn't it easier to remember a principle and recognize similar problems as being fundamentally the same?

If there are 3 or more units involved, then, for sure, the labels would help avoid confusion. But here, you are only multiplying two numbers on the right hand side of an equation. The meaning of these two quantities are well defined.

In summary, dimensional analysis is not called for in your situation. It is an "over-kill" of a simple physics problem. Understanding "rate change over time" is the key to clearing your confusion. "Work=Rate*Time" is in fact a simple and beautiful concept. I do not recommend killing a fly on a wall with a shot-gun.

Regards.