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Then I hit this one:
19. A commercial jet can fly from San Francisco to Dallas in 3 h. A private jet can make the same trip in 3 1/2 h. If the two planes leave San Francisco at noon, after how many hours is the private jet twice as far from Dallas as the commercial jet? (page 246)I couldn't do it, and I didn't enjoy not being able to do it.
I'm not even going to look at #20.
Until tomorrow.
11 comments:
I won't spoil the fun and explain how I did it, but I got 2.625 hours.
I got the same number; wasn't sure how to set it up, so I plugged in some numbers. I chose 2100 miles because I could easily divide it by 3 and by 3.5. Once I plugged in numbers, I was able to do it.
I got the same, using d for distance and t for time.
I kept mis-reading the question: I interchanged the words "private" and "commercial" in the last sentence, and kept coming up with 4.25. Since Kate and James both got a different answer I went back and re-read the problem, and found my mistake.
time remaining
commercial 3 - t
private 3.5 - t
Is that enough of a hint (without spoiling it, I hope)
Jonathan
Catherine, the aspect of this problem that makes it seem harder than other rate problems is that you have to consider the distance remaining rather than the distance traveled. More over, we don't know the actual distance.
There are probably several ways to this, but a natural way is to start by writing down everything you know about the problem, including quantities that you don't have specified. For example, we don't know the distance between the cities. I will call it L and use it in full confidence that this will help me but that I will never need to know the value of L. Similar things work well in many proportional problems. The advantage of doing things this way is that you don't need to know the ins and outs of a particular class of problem, you just need to start translating the problem into equations, and following your nose.
(1) so let's say the distance is L.
(2) speeds. The commercial jet travels L/3 mph. The private jet travels L/3.5 = L/(7/2) = (2/7)xL mph
(3)We want the time T such that the remaining distance is such and such ... we want to consider distances such as L - T*speed
(4) in particular we want T hours such that
L - T*(2/7)xL = 2x(L - T*(1/3)*L)
distribute and collect and end up with
T x (8/21) x L = L
Cancel the Ls and get
T = 2 + 5/8 hours
= 2.625 hours
keywords: translate from English to math
It took me forever to do this problem too (word problems are still my nemesis, with only the tank inflow/outflow problems frustrating me more than the planes and trains!).
I always start with d = rt.
I define my variables:
D is the distance between SF and Dallas
Rf = rate of the fast plane = D/3.0
Rs = rate of the slow plane = D/3.5
Thus D = 3.0*Rf = 3.5*Rs
So Rf/Rs = 3.5/3.0 = 7/6
which allows me to restate Rf in terms of Rs, or vice versa. I figured this could be useful.
Now we need to define t, which we'll be solving for:
t = the time into the flight(s) where the distance between the fast plane and Dallas is twice that of the distance between the slow plane and Dallas
That is,
X = distance of the fast plane from Dallas at time t.
2X = distance of the slow plane from Dallas at time t.
Then using d=rt,
For the fast plane:
X = Rf * (3.0 - t)
For the slow plane:
2X = Rs * (3.5 - t)
For the fast plane we can substitute in Rf = (7/6)* Rs:
X = (7/6)*Rs * (3.0 - t)
Then let's multiply that equation by 2 and subtract it from the equation for the slow plane.
2*(7/6)*Rs*(3.0-t) = Rs * (3.5-t)
Simplify
7 - (7/3)t = 3.5-t
3.5 = (4/3) * t
t = 10.5/4 = 2.625 hr
I sure know how to make these problems hard!
Problem: A fast plane can fly from Here to There in 3 hours, but a slow plane takes 3 1/2 hours. If the two planes leave Here at the same time, after how many hours will the slow plane be twice as far from There as the fast plane?
I don't know whether anybody will appreciate my attempt to reduce this problem to its essence, but here goes...
Step 1.
Choose the fundamental unit of time to be one "hour", and the fundamental unit of distance to be one "trip" (from Here to There). I call these fundamental units because I don't anticipate needing to know, for example, how many minutes are in one hour, or how many miles in one trip from Here to There. Here are some of the givens in the problem, restated:
speed of fast plane = 1/3 trips per hour;
speed of slow plane = 2/7 trips per hour.
Step 2.
Use "distance covered equals speed times elapsed time":
distance covered by fast plane in first T hours = (1/3)T trips;
distance covered by slow plane in first T hours = (2/7)T trips.
Step 3.
Use "distance remaining = total distance - distance covered in first T hours":
distance remaining for fast plane = 1 - (1/3)T trips;
distance remaining for slow plane = 1 - (2/7)T trips.
(Total distance is 1 trip for both planes!)
Step 4.
distance remaining for slow plane = twice the distance remaining for fast plane:
1 - (2/7)T = 2(1 - (1/3)T).
Step 5.
This equation is a linear equation with single unknown T. It will have one solution, which will be the solution to the problem. The answer will be 2 5/8 hours.
Interesting point to ponder: We have solved this problem without knowing how many miles constitutes one "trip" from Here to There. So even if Here is Chicago Midway airport and There is Chicago O'Hare, the total trip for either plane will still take 2 5/8 hours and then some. But it is common sense that the flight from Here to There should take just a few minutes (or seconds?). Explain!
I tried to resist.
I liked Vicky's approach because it showed her thinking process. One thing I never liked about math and CS texts is that you often see a neat, cleaned-up solution that makes the process look obvious. (Makes you feel dumb that you didn't figure it out yourself.)
I normally like to have the math do the problem for me. I don't like to think too much. I'm also not afraid of equations and variables. I think many students get stuck trying to create fewer equations. Some even think that they have to come up with one equation in one unknown in their head.
I also think math classes should more strongly emphasize the role of governing equations. Using the DRT equation should be obvious in this case, but selecting a proper governing equation in other cases might be more difficult.
The approach I take is to just draw pictures, define variables, and then start writing equations that I know are true.
I know that:
D = Rf * 3
and
D = Rs * 3.5
I can also figure out (with a drawing and a little thinking) that
D - d = Rf * T
and
D - 2*d = Rs * T
Where 'd' is the distance that the fast plane has yet to go.
I know I'm creating a lot of variables that may not be necessary, but I don't care. I feel very comfortable that these equations are true. (I should get some partial credit!)
Now I'm stuck, because I have 4 equations and 5 unknowns. I need one more equation, but none jump out at me. Darn, I'll have to start to think.
One of the problems is that you might come up with another equation and not realize that it's just a variation of one or two that you've already defined. Actually, this would be a good problem to give a student; to show linear dependence and what happens when you try to plug the equations together. (Assuming that they don't know about matrices or rank yet.)
OK, I'm stalling. I need one more equation. I don't see any number I haven't used yet, so it's not going to be easy.
I think I need something that relates to the full time for one or both of the planes. It's 3 hours for one and 3.5 hours for the other. I know that the rates of the planes are constant, and one of my variables is 'd', the distance the faster plane has yet to go, and I know that the time it takes to go the 'd' distance is (3-T). So,
d = Rf(3 - T)
I now have one more equation without having to define any new variables.
Now, just turn the crank.
I push this because I've seen so many real world problems solved the same way; select a governing equation(s), define variables and start writing equations. Perhaps later, you can consolidate variables or find better equations. If you don't have enough equations, you need to find some way to come up with the rest. These extra equations are often defined using boundary conditions. That would make a good problem; requiring the student to come up with their own boundary condition equation.
Another approach is to leave some missing eqations, but search for a solution that is optimal using some sort of cost or merit function. That's really for a later math class.
VickyS' solution was the closest to mine. Somewhere I also made use of the constant ratio of the rates and the constant ratio of the distance remaining (3.5/3 and 2/1, respectively). I like how everyone who wrote out solutions did them differently - there's more than one way to skin a math problem. And also, I agree with SteveH, multiple approaches and false starts should be expected, it's part of the process - I filled up 2 full sheets of blank scrap paper.
Like Steve, my preferred approach is to let the math do the work for me. I can hardly really think until I start scribbling something on paper. When my older son is stumped on a precalc problem and asks me to look at it, it does me no good when he just hands me the book--I need to have pencil and paper before it's worth my time to even read the problem. This frustrates him to no end! He doesn't understand that my pencil leads my thinking; it usually doesn't follow it.
This is a fascinating exercise because we show, like the fuzzies mantain, that there are lots of different approaches to a problem like this.
However what we all have in common is the process of defining variables, articulating relationships between the data, and moving through the problem in a logical step by step fashion.
I suppose that means we haven't evidenced understanding.
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