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The song Frosty the Snowman goes like this:
"Frosty the Snowman
Was a jolly happy soul
With a corncob pipe and a button nose
And two eyes made out of coal
Frosty the Snowman"
An extract from Clement Clarke Moore's "A Visit from St. Nicholas":
"Twas the night before Christmas, when all through the houseNot a creature was stirring, not even a mouse;The stockings were hung by the chimney with care,In hopes that St. Nicholas soon would be there;
...I knew in a moment he must be St. Nick.More rapid than eagles his coursers they came,And he whistled, and shouted, and called them by name:"Now, Dasher! now, Dancer! now Prancer and Vixen!On, Comet! on, Cupid! on, Donner and Blitzen!To the top of the porch! to the top of the wall!"
Holly has the red berries, mistletoe the white. Don't forget or you could be hanging around, puckered up under the Holly for a while...
Unscramble each of the following anagrams into a common festive word or phrase.
Most elite Expected answer:
Romantic Christmas greenery.
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Thick massacre Expected answer:
Hopefully, if you make one for pudding, it will not be described as such.
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Honesty warms font Expected answer:
He could laugh and play, just the same as you and me.
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The rich Mr Fat Ass Expected answer:
Creepy man who sneaks around at night.
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You can't do it? That's just out of order!
Numbas has many options for numeric input, including restrictions on precision and options to allow, for example, input as a fraction. You might need a calculator for this one!
The chimney at 52 Festive Road has internal dimensions of 55 x 40 cm. Santa has arrived with presents on Christmas morning.
Assuming that Santa has a big, perfectly circular tummy, what is its maximum circumference, in order to fit down the chimney?
Expected answer: cm Round your answer to 1 decimal places.
The circumference of a circle is given by $\pi d$, where $d$ is the diameter. Now think about the maximum diameter for a circle that would fit inside the rectangle at the top of the chimney.
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The circumference of a circle is given by $\pi d$, where $d$ is the diameter.
The trick here is to realise that the maxmum diameter corresponds to the shortest length of the rectangular chimney, 40cm.
Therefore the maximum circumference of his tummy is $\pi\times\var{a} = \var{precround(pi*a,1)}$cm.
According the National Obesity Forum, he is at substantially increased risk of conditions such as coronary heart disease.
Note that this question is randomised, including the graph (try clicking "Try another question like this one"), and that you get a preview of the answer when you type in an expression. One more thing: the correct answer is written in terms of the variable n, go on try a different letter, like x, it separates the validation from the marking to help you out.
Santa is reviewing the personnel requirements for the North Pole's UK division. He requires 40 elves to do the basic administrative jobs, plus 25 elves for every 1 million children, to manage toy making and the like.
There are roughly 11 million children in the UK. How many elves does Santa need?
Expected answer: elves
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Santa makes a plot of the number of elves he requires, $E$, against the number of children, $n$ (in units of million children):
What is the equation of the line?
$E(n)=$ Expected answer:
The equation of a straight line is traditionally written
\[ y = mx+c\text{,}\]
where $m$ is the gradient of the line, and $c$ the intercept with the y axis.
In our case, the variables are labelled $E$ and $n$ (rather than $y$ and $x$ respectively). The intercept $c$ is the number of elves required for $0$ children, i.e. the base number, $\var{c}$. And the gradient $m$ is the number of elves required for each unit $n$, which is given as $\var{m}$.
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The number of elves Santa requires is
\[ ({m}\times 11) + {c} \text{.}\]
The equation of a straight line is traditionally written
\[ y = mx+c\text{,}\]
where $m$ is the gradient of the line, and $c$ the intercept with the y axis.
In our case, the variables are labelled $E$ and $n$ (rather than $y$ and $x$ respectively). The intercept $c$ is the number of elves required for $0$ children, i.e. the base number, $\var{c}$. And the gradient $m$ is the number of elves required for each unit $n$, which is given as $\var{m}$.
Therefore the equation of Santa's line is
\[ E(n) = \var{m}x + \var{c}\text{.}\]
To learn more about the equation of a straight line, here is Professor Robin Johnson solving a slightly trickier problem:
At Christmas dinner, your pal's cracker contains an Amazing Magic Trick.
Your pal asks you to pick a whole number between 1 and 30, then shows you the following five cards and asks which cards contain your number.
7 | 25 | 3 | 29 |
13 | 31 | 21 | 23 |
19 | 11 | 15 | 5 |
9 | 17 | 27 | 1 |
7 | 14 | 30 | 23 |
31 | 22 | 19 | 18 |
15 | 11 | 26 | 10 |
6 | 3 | 27 | 2 |
6 | 15 | 28 | 5 |
22 | 29 | 20 | 14 |
31 | 30 | 13 | 12 |
21 | 23 | 7 | 4 |
31 | 14 | 27 | 24 |
9 | 25 | 13 | 29 |
11 | 28 | 12 | 10 |
26 | 30 | 15 | 8 |
17 | 23 | 25 | 27 |
26 | 28 | 24 | 31 |
18 | 21 | 22 | 20 |
19 | 29 | 30 | 16 |
Amazingly, your pal is able to magically divine the number you're thinking of every time!
To get an idea of how the trick works, try picking a number between 1 and 30 and clicking all the cards containing that number. When you submit this part, we'll guess which number you were thinking of. It's important that you make sure you check every card containing your number.
You can try this as many times as you like.
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When you think you've worked out how the trick works, your pal offers to let you try it on them.
Your pal has thought of a number, and says it's on the cards highlighted below:
7 | 25 | 3 | 29 |
13 | 31 | 21 | 23 |
19 | 11 | 15 | 5 |
9 | 17 | 27 | 1 |
7 | 14 | 30 | 23 |
31 | 22 | 19 | 18 |
15 | 11 | 26 | 10 |
6 | 3 | 27 | 2 |
6 | 15 | 28 | 5 |
22 | 29 | 20 | 14 |
31 | 30 | 13 | 12 |
21 | 23 | 7 | 4 |
31 | 14 | 27 | 24 |
9 | 25 | 13 | 29 |
11 | 28 | 12 | 10 |
26 | 30 | 15 | 8 |
17 | 23 | 25 | 27 |
26 | 28 | 24 | 31 |
18 | 21 | 22 | 20 |
19 | 29 | 30 | 16 |
You can work it out by looking very carefully at all the cards, but there's a quicker way!
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This is one for SPSS users (or in Excel, R, or similar; the data is small and just for demonstration purposes). The question provides a CSV file containing data. The clever bit is that the data file is randomised by Numbas (in this case the number of years of data, but could easily have been the values themselves).
The so called 'Santa Rally' is the pattern of positive rises in the financial markets in the month of December. There are many suggestions for the cause, including the mood of traders and the tax benefits, which cause a surge of investments before the end of the calendar year.
Download this {download_csv(ftse_given,rows)}, which contains data from the last {rows} Decembers, up to 2016, and open it in your favourite program.
Calculate the percentage change in the FTSE 100 index for each of the given months to find the average change:
The average change was Expected answer: % Round your answer to 2 decimal places.
The percentage change for a given month is:
\[ \frac{\text{(closing value) - (opening value)}}{\text{(opening value)}}\times \text{100%.}\]
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Which year was most volatile (had the maximum difference between the high and low value)?
The most volatile year was Expected answer: with a difference of Expected answer: .
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Firstly, create a new column of data containing the percentage change, given by:
\[ \frac{\text{(closing value) - (opening value)}}{\text{(opening value)}}\times \text{100%.}\]
The following video explains how to compute the mean of a column in SPSS. (Video from BrunelASK)
To find the most volatile month, create a new column of data:
\[ \text{(high value) - (low value),} \]
to find the maximum value and corresponding year.