Sign In | Starter Of The Day | Tablesmaster | Fun Maths | Maths Map | Topics | More
This is the Transum Newsletter for the month of December. I wonder if I will be the first to wish you a Merry Christmas and a Happy New Year? So with that festive thought in mind here is the puzzle of the month:
While Santa was busy delivering presents in Tran Towers, Prancer, one of his reindeer, went for a stroll to the base of Krimble Hill. Prancer then walked up the hill to see the view of Twinkle Town from the top. After enjoying the view, Prancer retraced his steps back to the Towers. His round trip took six hours. Prancer walked at an average of 4 km/h on level ground, 3 km/h uphill, and 6 km/h downhill. How far (in km) did Prancer walk?
# To the tune of God Rest Ye Merry, Gentlemen (an English traditional Christmas carol) #
That is the puzzle of the month for you to contemplate.
You'll need to wear your thinking cap and really concentrate.
And if you think you've solved it and your answer is correct,
Then send me an email of your thoughts and solution.
I'll be very pleased to hear from you!
gro.musnarT@rettelsweN
While you think about that here are some of the key resources added to the Transum website during the last month.
Christmas Tree Light Sum challenges pupils to choose the colours of the lights on a tree so that the pattern is balanced (that could mean symmetrical), while ensuring that the numbers on the lights add up to the randomly generated targets. Everyone in your class will undoubtedly come up with a different pattern after a suitable mental-maths workout.
I made a Christmas video! No comments about my singing please!
Christmas Puzzles is a festive collection of 12 engaging mathematical puzzles, most of which feature their own interactive web pages for an enhanced experience. Pause the video at any point to work through the solutions to the puzzles at your own pace. Perfect for festive puzzle enthusiasts aged 9 to 90, this video provides a fun and educational way to challenge your mind this holiday season!
Quartiles is a brand new, six-level exercise I wrote shortly after expressing my surprise in last month's Newsletter about the number of different methods there are for finding quartiles of discrete data according to Wikipedia. Well, I received a message from @themathsbazaar.bsky.social on social media, who informed me that there are actually many more methods. He very helpfully directed me to an interesting paper titled "Quartiles in Elementary Statistics" by Eric Langford, that I found a fascinating read.
Inspired by that, I created this new set of online exercises on finding quartiles, which is now ready to use. I’ve chosen what I consider to be the most common method, which also happens to be the one used by the textbook and the calculator used by my students. I refer to this as the "popular method."
For example, consider the discrete data set {1, 2, 3, 4, 5}. According to this popular method, the median is 3, the mean is 3, the range is 4, and the lower quartile is 1.5. The paper points out that if we duplicate each item in that data set to give {1, 1, 2, 2, 3, 3, 4, 4, 5, 5}, the median remains the same, and the mean and the range also remain unchanged. However, calculating the lower quartile using this popular method gives a different result of 2. Surely, it should remain the same?
So, I think this is an interesting fact that’s worth pointing out to the students you teach. It is important to know that most of the maths they will encounter is unquestionable but there are some concepts that can be described in different ways. Just ensure that they are learning the method that is consistent with their course.
If you are reading this on the 1st December, I can inform you that the first door of the Advent Calendar is now open. You won't get a chocolate but you will be drawn into the Christmaths Collection!
Don't forget you can listen to this month's podcast, which is the audio version of this newsletter. You can find it on all good podcast services.
Finally the answer to last month's puzzle which was:
There are four fictional planets orbiting the fictional star Vesperion:
How many moons does Nymeria have?
My method was to let n be the number of moons on Nymeria.
The total number of moons for the planetary system is:
7 + 10 + 12 + n
The mean number of moons for the four planets is:
Mean = (7 + 10 + 12 + n) / 4
According to the problem, Nymeria has twice as many moons as this mean:
n = 2 × (7 + 10 + 12 + n) / 4
Let's solve for n:
n = 2 × (29 + n) / 4
4n = 2 × (29 + n)
2n = 29 + n
2n - n = 29
n = 29
Therefore, Nymeria has 29 moons.
The first five people to provide correct answers were Mala, Leonard, Shreyak, Bardia and Rick. Thanks to everyone who took part. A couple of interesting methods can be found below.
That's all for now, jingle those bells!
John
P.S. Why can't Santa figure out the area of a rectangle using the formula L times W? Because in the Christmas alphabet there's Nöel.
P.P.S. I'm really scraping the barrel with these joke but maybe your Year 7 pupils will snigger at them?
Do you have any comments? It is always useful to receive feedback on this newsletter and the resources on this website so that they can be made even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
Leonard,
Saturday, November 2, 2024
"Looking for a non or minimal "calculation-based" solution, I came up with the following.
1. Consider all the moons as one large pie.
2. Due to the four planets, the average number of moons will be ¼ of the pie.
3. The number of moons of Nymeria will be twice this or ½ of the pie.
4. The remaining number of moons (29 = 7+10+12) represent the other ½ of the pie.
5. With the two halves equal, Nymeria must have 29 moons. "
Rick, United States
Tuesday, November 19, 2024
"Algebraic solution:
Let n be the number of moons for the planet Nymeria. Then we know the average (mean) is n/2, since Nymeria has twice as many moons as the mean number for the system. The mean is also the sum of the moons divided by four.
N / 2 = (n + 7 + 10 + 12) / 4 = (n + 29) / 4
Solving for n, we get 29 moons for Nymeria.
Intuition
We know the moon mean is half the number of moons for Nymeria. We know the mean is also the number of moons for Nymeria plus the sum of the moons from the other three planets and that this is divided by four. Hence the numerator needs to be double the number of moons for Nymeria, hence the number of moons for Nymeria must be equal to the number of moons for the other three planets, which is 29. "