THis problem sounded like the problem other students had been working on, which was a standard issue “rate together” question in which, to my discerning eye, the middle part wasn’t even necessary. However, that question had “t” and “t+2,” and then stated that in 5 hours, the fractional part of the job done was 5/t + 5/(t + 3). I am not sure that the last part was the same… I think it had a different number of hours, or may just have asked how many hours it would take them…

So, first, I told the student “it’s a cut and paste error — don’t bother with it.” Then… I figured it might be solvable anyway.

I turned this into a rather nasty problem that turned into a cubic equation, and told the student “ask your teacher, and get back to me.” (Oh, he said it had stumped all the peer tutors, too.)

“One person can clean a house in t hours, while another person needs t + 3 hours. In 5 hours, the fractional part of the job done is 6/t + 6/(t+3). Determine how many houses can be cleaned in 8 hours.”

Ignoring the fact taht the problem doesn’t say whether either and/or both folks will be working on it, let’s just assume they both are. HOw would *you* tackle that problem?

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@JustinAion

October 21, 2013

AAAAAAAAAAAAAAAlright! Here’s how I would tackle the problem:

I wouldn’t!

This is a terrible problem for several reasons. These are the three big ones that I see.

First, it takes something pretty basic (two people working at different speeds) and makes it’s absurdly complicated! There are enough complex problems out there that making a simple one into a hard one is insane.

Second, just looking at this problem makes my head hurt. I’m not sure how to set it up or even that there is enough information to find out the answer. After a few minutes of thinking about it, I can only imagine that it would be set up as a proportion, but again, with the “fractional part of the job” being as complex a fraction as it is, that proportion would be ugly. Making something purposely ugly may have connections to reality, but it’s the quickest way to make students second guess their work, erode their confidence and make them think that math is difficult. Too often, we let our students confuse difficult with ugly.

Third, combining the first two, this problem takes a simply problem and makes it super ugly, transforming a legitimate question into the epitome of why students hate math.

Without knowing the concept that this problem is suppose to focus on, I would think that a better alternative might be something like:

“Mike takes 2 hours longer than Jim to clean a house. After 5 hours of working together, they have cleaned 3 houses. If Speedy Jim decides to ditch Slow Mike and clean on his own, how many houses could he clean in 8 hours?”

Cleargrace

October 22, 2013

How about drawing out t and t+3? The fraction for the first problem

5 hrs = 5/t + 5/t+3: let it be

5/[][][][][] + 5/[][][][][][][][], or 1+5/8

So 1 house plus 5/8 of a house gets painted in 5 hours. One house would take

Second problem

5hrs is 6/t + 6/t+3 , draw it out:

5hrs= 6/[][][][][][] + 6/[][][][][][][][][], or 1+6/9, so

8 hrs =

xiousgeonz

October 22, 2013

THey were the *same* problem.

Cleargrace

October 22, 2013

Oh. Did I misunderstand?

xiousgeonz

October 22, 2013

(see my reply below — they weren’t two differentproblems at all. It was all written as one problem.)

Cleargrace

October 22, 2013

What do you think about this approach?

5hrs= 5/t + 5/t+3

Draw

5/[][][][][] + 5/ [][][][][][][][] = 1+ 5/8 or 13/8

5hrs = 13/8, so 1 hour = 1/5 of 13/8 or 13/40, which is .325

Working together, it would take them about 3 hr and 4 1/2 min (1/.325=3.076923) to paint one house.

To apply to second problem

5 hrs = 6/t + 6/t+3, we draw again:

6/[][][][][][] + 6/[][][][][][][][][] = 1+ 6/9

5hr = 1 2/3

1hr = 1/3

8 hr = 8/3, or 2 full houses and 2/3 of another house.

Only two houses could be completed in 8 hours if both painters worked together.

Cleargrace

October 22, 2013

What do you think about this approach?

5hrs= 5/t + 5/t+3

Draw

5/[][][][][] + 5/ [][][][][][][][] = 1+ 5/8 or 13/8

5hrs = 13/8, so 1 hour = 1/5 of 13/8 or 13/40, which is .325

Working together, it would take them about 3 hr and 4 1/2 min (1/.325=3.076923) to paint one house.

To apply to second problem

5 hrs = 6/t + 6/t+3, we draw again:

6/[][][][][][] + 6/[][][][][][][][][] = 1+ 6/9

5hr = 1 2/3

1hr = 1/3

8 hr = 8/3, or 2 full houses and 2/3 of another house.

Only two houses could be completed in 8 hours if both worked together.

xiousgeonz

October 22, 2013

Thing is, these were the same problem. “One person can clean a house in t hours while another person need st + 3 hours. In 5 hours, the fractional part of the job done is 6/t and 6/(t + 3).”

Here’s my thinking: The first situation doesn’t tell us what t is, so we honestly don’t *know* how many they can do in 8 hours. After all, it just depends. If, working together, they could clean a house in 8 hours, then we could figure t out.

However, we don’t know.

We do know, however, that 1/t + 1/(t + 3) should give us 1/(time for both together).

And we are also told that the rate for five hours is 6/t and 6/(t + 3). SO! If I add those two together and then divide that by five (6/5t) + 6/(5t + 15)… and reciprocate the answer, then I could plop that on the other side of the equals sign of the original expression.

THEN I could figure out what T was — except that when I did (granted, not all that carefully; could have made a mistake) I got a cubic equation and I *know* they haven’t gone there… and I just wasn’t willing to go there, even if, in fact, this was one of those “well, we give them harder problems on the homework to challenge them so the quizzes and tests seem easy” which is actually like saying “well, we put piranhas in the water so that it would seem nicer just dealing with the leeches later” to the students. They’re mangled, now, thank you.

I had kinda hoped it was solvable… but I refuse to dignify it with going back over it, especially since about 80% of the time I’m dead-on right about stuff like this.

goldenoj

October 22, 2013

It should be 5/t+ 5/(t+3) in story problem logic, they say it’s 6/t+6/(t+3), so if (5(t+3)+5t)/(t(t+3))=(6(t+3)+6t)/(t(t+3)), then 10t+15=12t+18, so t= -1.5 hours. That cleaner has a Tardis. We should also check what happens at t=0 and t=-3 hours, in story problem land.

@JustinAion

October 22, 2013

…You are awesome.

Rachel Baron

October 22, 2013

I also thought this looked like a proportion, with a combined 2t+3 hours needed to clean 2 houses and 5 hours needed to clean 6/t+6/(t+3) houses. Using cross products, that means that 10 = (2t+3)(6/t+6/(t=3)), which turned into one of those rabbit holes I remember as being a signal that I did something wrong in high school math class. Someone with more quadratic confidence might be able to make it work, but I’m not making any promises that it will make sense.

xiousgeonz

October 22, 2013

As written, assuming that the question is asking how many houses can be cleaned by both people, then yes, it turns into a nice, linear equation (when I set it up before at the end of the busy day, I threw the 5 at the bottom of the second situation and then tried to solve it w/ cross multiplying and stopped before things simplified back to linear. Using the more popular “Death to denominators! Clear the fractions!” method, then — they can clean -8/1.5 houses in 8 hours, or negative 5 1/3 houses. My only “wish” is that I had had the cognitive juices to calculate that out so that they could drolly stroll in and give him that answer.

(I’m also wondering if taking Java is affecting me and inspiring me to persevere to Get An Answer with the full knowledge that it is useless except to point out that by following formulae you can get answers and so it’s up to the human to recognize whether they make sense or not… now back to Lab 7 and *hoping* I can find my Java book, because I started planning the class “Math Problem” last night…)