Tuesday, November 16, 2010

CONTEST! Just Another "Rate-Time-Distance" Problem?

CONTEST IS OFFICIALLY OVER AND THE WINNER IS ----- NO ONE! Guess I should have offered a 64GB 3G IPad! to be awarded on Black Friday...

The floor is now open for David, Curmudgeon, and my other faithful readers to offer their own solutions.  

And the next contest is...

This is a contest so students must work alone and this needs to be verified by a teacher or parent. No answer will be posted at this time. Deadline is Wed 11-17-10 at 4 PM EST.

Here's a variation on the classic motion-type problems we don't see as often in Algebra I/II but still appear on the SATs. I found this in some long-forgotten source of excellent word problems to challenge NINTH graders! 

Barry walks barefoot in the snow to school in the AM and back over the same route in the PM.  The trip to school first goes uphill for a distance, then on level ground for a distance and finally a distance downhill.  Barry's rate on any uphill slope is 2 mi/hr, any downhill slope is 6 mi/hr and 3 mi/hr on level ground.  If the round trip took 6 hours (hey, these are the old days in the 'outback'), what was the total number of miles walked?

First five correct answers  with complete detailed solutions emailed to me at dmarain@gmail.com will receive a downloaded copy of my new book of Challenge Problems for the SATs and Beyond when it becomes available. Both the student and teacher(s) will receive this.  (Illegal to reproduce or send electronically!). Read further...

Submission by email must include (Number these in your email and copy the validation as well).

1.  Answer and complete detailed solution. If answer is correct but method is sketchy or flawed,      the submission will be rejected.
2.  Full name of student
3.  Grade of student
4.  Math course(s) currently taking
5.  Math teacher's name(s) and parent's name(s)
6.  Name, Complete Address of School; Principal's Name & Email address (if known) 
7.  Email addresses of teacher(s),  parents, student 
8.  Phone number (in case I need to call you) - Optional
9.  How your or your teacher or parent became aware of MathNotations.


I certify that my student (child) did the work independently.


Name of Teacher or Parent (if work done at home)

"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)

 "You've got to be taught To hate and fear, You've got to be taught From year to year, It's got to be drummed In your dear little ear. You've got to be carefully taught." --from South Pacific


David said...

Very clever!

Curmudgeon said...

A nice problem. It bears little resemblance to an SAT problem but it is a good algebra question.

Dave Marain said...

Thanks, David and Curm...
I'm disappointed that students haven't submitted solutions yet. Surely this problem is not beyond many of our students. I'm leaving the contest open at this point to allow students to try the problem. Hopefully some teachers will encourage this.

Curmudgeon, this was not intended to be a sample SAT problem! Rate problems, however, are still tested but are typically more routine. Since many of our students have less exposure to these, even easier questions stop most. Of course the values of the rates and total time were chosen purposefully to allow the algebra to work out in a surprising manner. These are the kinds of problems I included in my new book. They are intended to provide teachers with a resource of nonroutine and challenging questions to promote deeper thinking in the classroom. Students who learn some of the strategies needed to solve these should be able to transfer them to standardized tests.

Eric Jablow said...

Actually, this is a very clever problem with a very clever simplification, which I doubt a high school student would ever think of.

Suppose the path from home to school consists of l miles of level travel, u miles of uphill travel, and d miles of downhill travel. The problem asks to solve for 2l + 2u + 2d, and the only constraint is a linear constraint on the values of l, u, and d. I am deliberately not finding the constraint here.

Now, this is an underdetermined system of equations, and yet it's been made clear that there's a unique solution.This can occur only if the constraint is a multiple of 2l+2u+2d. In that case, it doesn't matter what proportion of the road is uphill or is downhill; we can choose the values of u and d arbitrarily. Let them be 0, and so we have 2l/3 = 6 and so l = 9 and 2l = 18.

Now, what is the relationship between the three speeds that make this conclusion possible? It's that 3 is the harmonic mean of 2 and 6. Change any one of the speeds, and you can't conclude anything.

Here's a related idea, taken from the unimaginatively-named "How to Solve It: Modern Heuristics", by Michalewicz and Fogel, published by Springer-Verlag in 2002.

Two men meet by chance for the first time in 20 years. One says to the other, a mathematician, that "All three of my sons celebrate their birthday this very day. So, can you tell me how old each of them is?"

The mathematician needs more info.

:The product of their ages is 36."

He needs more info.

"The sum of the ages equals the number of facing windows in that building over there."

He needs still more info.

"My oldest son has blue eyes."

The mathematician then explains the solution. Why?

There are many different ways to solve the first step: 36, 1, 1; 18, 2, 1; etc.

The second step can't provide a unique solution, and so almost all solutions fail ; only 9, 2, 2 and 6, 6, 1 are eligible.
That there is an oldest son in this case foRces the solutions to be 9, 2, 2.


Dave Marain said...

Thank you for the reference - great problem!

However, you may be underestimating our stronger students. Algebraically, I think some would set up the equation with 3 variables even though there is only one condition and see where it goes. This comes from success which breeds confidence. Some just go with the flow:

(u/2 + d/6 + L/3) + (L/3 + d/2 + u/6) = 6 -->
3u + d + 2L + 2L + 3d + u = 36 -->
4(d+L+u) = 36 --> d +L+u = 9, etc...

I've had these kinds of students! Now I sincerely believe that the other students in the class can gain much from seeing this. But is it part of our "prescribed" curriculum? Of course not! A standardized curriculum doesn't dictate what should NOT be taught. It merely tells us what SHOULD be covered!