Question

Dear Mitch,

With some free time during Christmas break, I revisited your site and enjoyed seeing a new balance problem for the holidays. However, I do not arrive at the same answer as you. I worked from two different directions and came up with one star equaling two trees, which would mean that the answer should be four trees. Would you show me how you solve the problem to get your answer of 6 trees?

Happy New Year.

Thank you.

Colette Henderson

P.S. I am enjoying using the booklet I purchased from you last year with the 100 ways to help students remember different math concepts, "110% Recall", which I believe has since been renamed, "The Extremely Sloppy Booklet of Very Neat Mnemonics".

Answer

Dear Colette Henderson,

You are ABSOLUTELY CORRECT!

I made an error. (I am more than willing to take the blame for any problem sent in to the site which I check too quickly for my own good, as surely that is my job, and my responsibility alone.) So, I now am going to post * your* correction along with an explanation of how others can go about solving it.

First, a quick recap of the problem, which was called "EARLY CHRISTMAS CHALLENGE":

There were three balance scales presented vertically, scale A, scale B, and scale C. Scale A was balanced, with a depiction of one star and one Christmas tree on its left-hand side, and a depiction of two Christmas trees and four Christmas ornaments on the right-hand side of that scale. Scale B depicted another balanced scale, with twelve ornaments on its left-hand side and a depiction of one Christmas tree and one star on its right-hand side.

And scale C, which was not yet balanced, had two stars depicted on its left-hand side and nothing on its right-hand side, with the left-hand side obviously weighing down the entire scale. And the question asked was this: "Based on what you can learn from the perfectly balanced scales A and B, how many Christmas trees would you have to place on the right-hand side of scale C in order to balance the two stars on scale C's left-hand side?"

So, to change things into fairly easy math language, I let the letter 's' represent each star, the letter 't' represent each tree, and -- since I did not want to run the risk of making things unnecessarily confusing by using an 'o' to represent each ornament (as one could easily lose track of things and start seeing the o as a zero) -- I chose the letter 'r' to represent each ornament.

So, scale A is represented by the equation:

s + t = 2t + 4r

scale B is represented by the equation:

12r = s + t

And since the left-hand side of A = the right-hand side of B,

12 r = 2t + 4r

subtracting 4r from both sides we get:

8r = 2t

and dividing both sides by 2 we get:

4r = t

So: (from scale A), substituting one t for 4r, we get s + t = 3t

and subtracting one t from each side we get:

s = 2t

So 2s = 4t,

and...

so, Mrs. Henderson,

you are absolutely right!

Two stars are balanced with four Christmas trees.

Thank you for taking the time to help me find such an unnecessary error on my website, Adler-n-subtract.com. As a small token of my appreciation for your time and effort to not only solve the problem on your own, but send me even a hint of the fruits of your labor, I will soon send you a little Adler-n-subtract.com gift. I hope you like it.

And I hope this helps others,

Mitch