How can this be? Simplify (600x)/(300x - 300); find time for 4/5 delivery

[allegansveritatem]

Here is the problem:

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72. It is estimated that in a certain community the number of hours required to distribute new telephone books to x percent of the households can be represented by

. . . . .\(\displaystyle \dfrac{600x}{300x\, -\, 300}\)

Simplify this rational expression. Use the simplified form to find the number of hours necessary to distribute telephone books to four-fifths of the population.

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Here is what I get when I work this out:

. . . . .\(\displaystyle \dfrac{600x}{300x\, -\, 300}\, =\, \dfrac{100\, \times\, 6x}{100\, \times\, (3x\, -\, 3)}\, =\, \dfrac{6x}{3x\, -\, 3}\)

. . . . .\(\displaystyle \dfrac{4}{5}\, =\, 0.8\)

. . . . .\(\displaystyle \dfrac{6\, (0.8)}{3\, (0.8)\, -\, 3}\, =\, \dfrac{4.8}{0.6}\, =\, -8\)

How can it take -8 hours to distribute the phone books?

 

[JeffM]

Here is the problem:

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72. It is estimated that in a certain community the number of hours required to distribute new telephone books to x percent of the households can be represented by

. . . . .\(\displaystyle \dfrac{600x}{300x\, -\, 300}\)

Simplify this rational expression. Use the simplified form to find the number of hours necessary to distribute telephone books to four-fifths of the population.

---

Here is what I get when I work this out:

. . . . .\(\displaystyle \dfrac{600x}{300x\, -\, 300}\, =\, \dfrac{100\, \times\, 6x}{100\, \times\, (3x\, -\, 3)}\, =\, \dfrac{6x}{3x\, -\, 3}\)

. . . . .\(\displaystyle \dfrac{4}{5}\, =\, 0.8\)

. . . . .\(\displaystyle \dfrac{6\, (0.8)}{3\, (0.8)\, -\, 3}\, =\, \dfrac{4.8}{0.6}\, =\, -8\)

How can it take -8 hours to distribute the phone books?

I suspect there is a mistake in the statement of the problem. Notice that it would take infinite time to distribute 1 percent of the phone books, a time that seems implausible to me.

Perhaps the problem should read

\(\displaystyle \dfrac{600x}{300x + 300} = \dfrac{300(2x)}{300(x + 1)} = \dfrac{2x}{x + 1}.\)

\(\displaystyle x = \dfrac{4}{5} \implies \dfrac{2 * \dfrac{4}{5}}{\dfrac{4}{5} + 1} = \dfrac{\dfrac{10}{5}}{\dfrac{9}{5}} = \dfrac{10}{9}.\)

 

[pka]

Here is the problem: View attachment 10229

Here is what I get when I work this out: View attachment 10228
How can it take -8 hours to distribute the phone books?

I would say that this is a faulty question. Percent here seems to imply a positive quantity.
If \(\displaystyle x<1\) then the denominator is negative. That just does not fit the description.

 

[ksdhart2]

All other (perfectly valid) concerns about this problem aside, you've also made a small error in reading the problem. It reads "...distribute new telephone books to x percent of the households..." The problem then goes on to ask how long it would take to deliver to 4/5 of the households. Since 4/5 is 80 percent, you'd actually want x = 80, not x = 4/5. Making the simplification:

\(\displaystyle \dfrac{600x}{300x-300} = \dfrac{2x}{x - 1} \text{ for } x \neq 1\)

and plugging in x = 80 yields:

\(\displaystyle \dfrac{2(80)}{80 - 1} = \dfrac{160}{79} \approx 2.0253 \approx 2.0333...\)

In other words, it would take a smidge under 2 hours and 2 minutes to deliver phone books to 4/5 of the population.

 

[allegansveritatem]

All other (perfectly valid) concerns about this problem aside, you've also made a small error in reading the problem. It reads "...distribute new telephone books to x percent of the households..." The problem then goes on to ask how long it would take to deliver to 4/5 of the households. Since 4/5 is 80 percent, you'd actually want x = 80, not x = 4/5. Making the simplification:

\(\displaystyle \dfrac{600x}{300x-300} = \dfrac{2x}{x - 1} \text{ for } x \neq 1\)

and plugging in x = 80 yields:

\(\displaystyle \dfrac{2(80)}{80 - 1} = \dfrac{160}{79} \approx 2.0253 \approx 2.0333...\)

In other words, it would take a smidge under 2 hours and 2 minutes to deliver phone books to 4/5 of the population.

I am going to work this out in the morning....no more brain left right now. But I guess this would make sense. Just wish the problem had been more explicit. 4/5, in my mathematical lexicon. is synonymous with .8....but he did say per cent, didn't he?

 

[allegansveritatem]

All other (perfectly valid) concerns about this problem aside, you've also made a small error in reading the problem. It reads "...distribute new telephone books to x percent of the households..." The problem then goes on to ask how long it would take to deliver to 4/5 of the households. Since 4/5 is 80 percent, you'd actually want x = 80, not x = 4/5. Making the simplification:

\(\displaystyle \dfrac{600x}{300x-300} = \dfrac{2x}{x - 1} \text{ for } x \neq 1\)

and plugging in x = 80 yields:

\(\displaystyle \dfrac{2(80)}{80 - 1} = \dfrac{160}{79} \approx 2.0253 \approx 2.0333...\)

In other words, it would take a smidge under 2 hours and 2 minutes to deliver phone books to 4/5 of the population.

I am going to work this out in the morning. It surprises me that any sort of simplification would produce an ultimately different result.

 

[allegansveritatem]

I would say that this is a faulty question. Percent here seems to imply a positive quantity.
If \(\displaystyle x<1\) then the denominator is negative. That just does not fit the description.

That's how it seemed to me too. But it has been suggested that, although he used a fraction in the body of the question, he meant for it to be translated into a non decimal percent. Shoulda said so, seems to me. But I guess the idea is, once burned, you will know better next time.

 

[Dr.Peterson]

I suspect there is a mistake in the statement of the problem. Notice that it would take infinite time to distribute 1 percent of the phone books, a time that seems implausible to me.

Perhaps the problem should read

\(\displaystyle \dfrac{600x}{300x + 300} = \dfrac{300(2x)}{300(x + 1)} = \dfrac{2x}{x + 1}.\)

\(\displaystyle x = \dfrac{4}{5} \implies \dfrac{2 * \dfrac{4}{5}}{\dfrac{4}{5} + 1} = \dfrac{\dfrac{10}{5}}{\dfrac{9}{5}} = \dfrac{10}{9}.\)

I think we need to take x not as a percentage in the sense that x=100 means 100% (which is the assumption in saying there is a vertical asymptote at 1%), but as a percentage in the sense that x=1 means 100% (as you did in applying the modified function), so that the infinite time occurs at 100% rather than 1%. It makes sense that it would get harder and harder to reach everybody, so that it becomes impossible to reach 100% of the population.

So my suggested correction is that they just got the sign wrong; it should be

\(\displaystyle \dfrac{600x}{300 - 300x} = \dfrac{2x}{1 - x}\) .

This way, the function is increasing, as it surely has to be, with f(0) = 1 and f(1) = infinity.

Then the answer would be 8 days. For 90%, it would be 18 days, and so on. Probably not very realistic, but more so than any alternative I see, most of which involve a decreasing function.

 

[pka]

All other (perfectly valid) concerns about this problem aside, you've also made a small error in reading the problem. It reads "...distribute new telephone books to x percent of the households..." The problem then goes on to ask how long it would take to deliver to 4/5 of the households. Since 4/5 is 80 percent, you'd actually want x = 80, not x = 4/5. Making the simplification:n.

Although I spent fully half my working life dealing with testing companies, I have no experience with this sort of question.
Now with that said, I have real concerns. This question was posted at mathematics help site.
Here is a simple question: saying \(\displaystyle x\) is 80 percent means what?
In mathematics percent means per-hundred. So 80 percent is \(\displaystyle \dfrac{80}{100}\) which converts to \(\displaystyle 0.8\)
So it would seem reasonable that \(\displaystyle x=0.8\) But that is not, not what we are asked to use here.
\(\displaystyle x\) is 80 percent means \(\displaystyle x=80\). REALLY? Good grief, I am glad I never had to deal with business school testing.

 

[allegansveritatem]

I think we need to take x not as a percentage in the sense that x=100 means 100% (which is the assumption in saying there is a vertical asymptote at 1%), but as a percentage in the sense that x=1 means 100% (as you did in applying the modified function), so that the infinite time occurs at 100% rather than 1%. It makes sense that it would get harder and harder to reach everybody, so that it becomes impossible to reach 100% of the population.

So my suggested correction is that they just got the sign wrong; it should be

\(\displaystyle \dfrac{600x}{300 - 300x} = \dfrac{2x}{1 - x}\) .

This way, the function is increasing, as it surely has to be, with f(0) = 1 and f(1) = infinity.

Then the answer would be 8 days. For 90%, it would be 18 days, and so on. Probably not very realistic, but more so than any alternative I see, most of which involve a decreasing function.

I have no learning experience with vertical asymtotes. However, I know what an asymtotic (sic) state is and understand the concept describe here and yes, that is what is happening. But...I found after having done the calculations again using 80 instead of .8 that the result seems to me to be reasonable, i.e., something a little over 2 hours. But, however high the percentage I plug in, the difference it makes in the result is in the order of thousandths of a hour. Thus for 80% it would take 2.022 hours. For 90, it % would take 2.025. Huh? ( I am using the reduced fraction I already posted in the first entry here.)

 

[JeffM]

I think we need to take x not as a percentage in the sense that x=100 means 100% (which is the assumption in saying there is a vertical asymptote at 1%), but as a percentage in the sense that x=1 means 100% (as you did in applying the modified function), so that the infinite time occurs at 100% rather than 1%. It makes sense that it would get harder and harder to reach everybody, so that it becomes impossible to reach 100% of the population.

So my suggested correction is that they just got the sign wrong; it should be

\(\displaystyle \dfrac{600x}{300 - 300x} = \dfrac{2x}{1 - x}\) .

This way, the function is increasing, as it surely has to be, with f(0) = 1 and f(1) = infinity.

Then the answer would be 8 days. For 90%, it would be 18 days, and so on. Probably not very realistic, but more so than any alternative I see, most of which involve a decreasing function.

Yes, that looks reasonable.

 

[JeffM]

I have no learning experience with vertical asymtotes. However, I know what an asymtotic (sic) state is and understand the concept describe here and yes, that is what is happening. But...I found after having done the calculations again using 80 instead of .8 that the result seems to me to be reasonable, i.e., something a little over 2 hours. But, however high the percentage I plug in, the difference it makes in the result is in the order of thousandths of a hour. Thus for 80% it would take 2.022 hours. For 90, it % would take 2.025. Huh? ( I am using the reduced fraction I already posted in the first entry here.)

I am now 99% convinced that you were correct that 0.8 was the proper input and that Dr. Peterson is correct that the problem should have specified a denominator of 300 - 300x. If you work that out in the same manner as you did, you get

\(\displaystyle h = \dfrac{600x}{300 - 300x} = \dfrac{300(2x)}{300(1 - x)} = \dfrac{2x}{1 - x}.\)

\(\displaystyle \therefore x = \dfrac{4}{5} = 0.8 \implies h = \dfrac{2 * 0.8}{1 - 0.8} = \dfrac{1.6}{0.2} = 8.\)

The point about asymptotes is that as x gets very close to 1, the graph of the function above increases without limit while drawing closer and closer to the vertical line x = 1.

You did very well to apply a reasonableness check to your work. I am merely sorry that I was not able to guide you to what I am now morally certain to be the intended problem.

 

[allegansveritatem]

I am now 99% convinced that you were correct that 0.8 was the proper input and that Dr. Peterson is correct that the problem should have specified a denominator of 300 - 300x. If you work that out in the same manner as you did, you get

\(\displaystyle h = \dfrac{600x}{300 - 300x} = \dfrac{300(2x)}{300(1 - x)} = \dfrac{2x}{1 - x}.\)

\(\displaystyle \therefore x = \dfrac{4}{5} = 0.8 \implies h = \dfrac{2 * 0.8}{1 - 0.8} = \dfrac{1.6}{0.2} = 8.\)

The point about asymptotes is that as x gets very close to 1, the graph of the function above increases without limit while drawing closer and closer to the vertical line x = 1.

You did very well to apply a reasonableness check to your work. I am merely sorry that I was not able to guide you to what I am now morally certain to be the intended problem.

So I guess we will have to leave it at that...somehow the variable got stuck on the wrong side of the negative sign. A little thing that makes a huge difference.
Thanks to all who replied.