Monthly Archives: July 2018

FiveThirtyEight Riddler Express Solution

I’d not peeked at FiveThirtyEight’s The Riddler in a while when I saw Neema Salimi ‘s post about the June 22, 2018 Riddler Express   Neema argued his solution was accessible to Algebra 1 students–not always possible for FiveThirtyEight’s great logic puzzles–so I knew it was past time to return.

After my exploration, I’ve concluded this is DEFINITELY worth posing to any middle school learners (or others) in search of an interesting problem variation.

Following is my solution, three retrospective insights about the problem, a comparison to Neema’s solution, and a proposed alternative numeric approach I think many Algebra 1 students might actually attempt.

THE PROBLEM:

Here is a screenshot of the original problem posted on FiveThirtyEight (2nd item on this page).

If you’re familiar with rate problems from Algebra 1, this should strike you immediately as a “complexification” of D=R*T type problems.  (“Complexification” is what a former student from a couple years ago said happened to otherwise simple problems when I made them “more interesting.”)

MY SOLUTION:

My first thought simplified my later algebra and made for a much more dramatic ending!!  Since Michelle caught up with her boarding pass with 10 meters left on the walkway, I recognized those extra 10 meters as irrelevant, so I changed the initial problem to an equivalent question–from an initial 100 m to 90 m–having Michelle catch up with her boarding pass just as the belt was about to roll back under the floor!

Let W = the speed of the walkway in m/s.  Because Michelle’s boarding pass then traveled a distance of 90 m in W m/s, her boarding pass traveled a total \displaystyle \frac{90}{W} seconds.

If M = Michelle’s walking speed, then her total distance traveled is the initial 90 meters PLUS the distance she traveled in the additional 90 seconds after dropping her boarding pass.  Her speed at this time was (M-W)  m/s (subtracting W because she was moving against the walkway), so the additional distance she traveled was D = (M-W) \cdot 90, making her her total distance D_{Michelle} = 90 + 90(M-W).

Then Michelle realized she had dropped her boarding pass and turned to run back at (2M+W) m/s (adding to show moving with the walkway this time), and she had \displaystyle \frac{90}{W} - 90 seconds to catch it before it disappeared beneath the belt.  The subtraction is the time difference between losing the pass and realizing she lost it.  Substituting into D = R*T gives

\displaystyle 90 + 90(M-W)=(2M+W)* \left( \frac{90}{W} - 90 \right)

A little expansion and algebra cleanup …

\displaystyle 90 + 90M - 90W = 180 \frac{M}{W} - 180M + 90 - 90W

\displaystyle 90M = 180 \frac{M}{W} - 180M

\displaystyle 270M = 180 \frac{M}{W}

And multiplying by \displaystyle \frac{W}{270M} solves the problem:

\displaystyle W = \frac{2}{3}

INSIGHTS:

Insight #1:  Solving a problem is always nice, but I was thinking all along that I pulled off my solution because I’m very comfortable with algebraic notation.  This is certainly NOT true of most Algebra 1 students.

Insight #2:  This made me wonder about the viability of a numeric solution to the problem–an approach many first-year algebra students attempt when frutstrated.

Insight #3:  In the very last solution step, Michelle’s rate, M, completely dropped out of the problem.  That means the solution to this problem is independent of Michelle’s walking speed.

Wondering if other terms might be superfluous, too, I generalized my initial algebraic solution further with A = the initial distance before Michelle dropped her boarding pass and B = the additional time Michelle walked before realizing she had dropped the pass.

\displaystyle A + B(M-W)=(2M+W)* \left( \frac{A}{W} - B \right)

And solving for gives \displaystyle W = \frac{2A}{3B}.

So, the solution does depend on the initial distance traveled and the time before Michelle turns around, and it was simplified in the initial statement with A=B=90.  That all made sense after a few quick thought experiments.  With one more variation you can show that the scale factor between her walking and jogging speed is relevant, but not her walking speed.  But now it was clear that in all cases, Michelle’s walking speed is irrelevant!

COMPARING TO NEEMA:

My initial conclusion matched Neema’s solution, but I really liked my separate discovery that the answer was completely independent of Michelle’s walking speed.  In my opinion, those cool insights are not at all intuitive.

AN ALTERNATIVE NUMERIC APPROACH:

While this approach is just a series of special cases of the generic approach, I suspect many Algebra 1 students would likely get frustrated quickly by the multiple variable and attempt

Ignoring everything above, but using the same variables for simplicity, perhaps the easiest start is to assume the walkway moves at W=1 m/s and Michelle’s walking speed is M=2 m/s.  That means her outward speed against the walkway is (2-1) = 1 m/s.  She drops the pass at 90 meters after 90 seconds.  So the pass will be back at the start after 90 seconds, the additional time that Michelle walks before realizing her loss.

I could imagine many students I’ve taught working from this through some sort of intelligent numeric guess-and-check, adjusting the values of M and W until landing at \displaystyle W=\frac{2}{3}.  The fractional value of W would slow them down, but many would get it this way.

CONCLUSION:

I’m definitely pitching this question to my students in just a few weeks.  (Where did summer go?)  I’m deeply curious about how they’ll approach their solutions.  I”m convinced many will attempt–at least initially–playing with more comprehensible numbers.  Such approaches often give young learners the insights they need to handle algebra’s generalizations.

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