Application

Upon discovering that I had been paying much more for my coffee than I realized, I was faced with a dilemma only algebra could solve.   Or, at least help analyze.

You see, my local coffee shop offers a simple reward program for regular customers:  purchase 10 bags of coffee, and your eleventh bag is free.  At the time of my discovery, I had credit for four bags of coffee.  Thus, the dilemma:  do I continue to buy the over-priced coffee six more times in order to get the free bag?  Or do I just start buying cheaper coffee somewhere else?

The coffee at my local shop was costing me around $12 for 12 ounces.  I knew I could get comparable coffee at another shop for around $10 a pound.  So should I stay or switch?

If I bought 6 more bags from the local shop, I’d spend $72 and get seven 12-ounce bags of coffee (the six I bought, and the free one).  Thus, I would get 84 ounces of coffee for $72.

Now $72 dollars at the new place would get me 7.2 pounds of coffee, or around 115 ounces.  So I’d be getting an additional 31 ounces, or almost two pounds of coffee, by switching!

Needless to say, I swtiched.  But it got me thinking:  at what point would it have been better to continue buying the over-priced coffee?  I turned to algebra for the answer.

Let x = the number of bags of coffee that have already been purchased.  In order to get a free bag of coffee from my local shop, I’d need to buy ten total bags; thus, I’d need to purchase (10 – x) more bags of coffee.  After that, I’d receive one free bag, so in the end I’d get (10 – x + 1) = (11 – x) total bags of coffee.

Each 12-ounce bag of coffee costs $12, so I’d spend a total of $12 * (10 – x) to get (11  – x) * 12 ounces of coffee.

This works out to a price of

dollars per ounce

for the coffee I’d get from the old place.

I know I can get coffee at the other place for $10 per pound, or dollars per ounce.  Thus, I want to find x so that

which would mean that the coffee from the old place would be cheaper per ounce than the coffee from the new place.

Now we simplify our inequality:

,

,

,

Thus, I should continue buying coffee at the old place if I’d already bought more than eight bags of coffee there.  Otherwise, I’d be better off switching.

I probably didn’t need algebra to tell me to stick with the old place if I’d already bought 9 bags, but algebra did show me just how hopeless my situation was!

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• More Clever Accounting

While riding the rails around Portugal, I frequently saw passengers buying tickets directly from the conductor on the train.  It got me thinking about how high the penalty should be for not buying your ticket ahead of time.  That is, how much more should you be charged for a ticket purchased on the train than in the station?

You see, if a rider could evade the conductor at a consistent rate, it might make mathematical (if not ethical) sense to gamble on paying the higher fare every so often.  For example, let’s say you can successfully sneak a free ride once every three attempts.  If the ticket in the station costs $5, then the price of the on-board ticket should be at least $7.50 to discourage you from attempting this cheat.

I never found out the price difference in Portugal.  But I do know how it works on the Long Island Rail Road.

Returning from vacation, we were rushing from the airport to the train station.  We didn’t have time to purchase tickets from the machine beforehand as the train was literally pulling into the station as we arrived.  After a long day’s travel, we were happy just to make our connection and get home as quickly as possible.  We figured whatever increase we’d have to pay was worth it.

And it turned out to be nearly a 100% increase.  Instead of the usual $6.25, the on-board charge was $12.  I guess that means they think fare-evaders can get away with it a little less than half the time?

We were happy to get home in a timely manner.  And I was happy to have one more open mathematical question resolved!