Transition to university

Christian Lawson-Perfect on Find and use the formula for a geometric sequence 8 years, 5 months ago

Gave some feedback: Has some problems

Christian Lawson-Perfect on Find and use the formula for a geometric sequence 8 years, 5 months ago

Saved a checkpoint:

The gaps are all set as "mathematical expression", but there's no reason they shouldn't be "number entry" (apart from the formula part, obv.)

The parts of this question aren't coherent; you could keep the same sequence throughout, and ask the student to:

• Find the common ratio
• Give the formula for the $n$th term
• Calculate a given term.

Hannah Aldous on Arithmetic sequences in an ice cream shop 8 years, 5 months ago

Gave some feedback: Needs to be tested

Chris Graham commented on Weight of a scoop in two ice cream parlours 8 years, 5 months ago

Good question. Just a couple of things:

In the statement you say that the shops will be referred to as $s$ and $t$, but this notation is not used in the question itself, just the advice, so why not introduce this shorthand at the start of the advice instead.

In your advice, you have several calculations involving fractions inside the align environment. This is fine, however you will notice that they appear quite squashed with the default line spacing. You could add [1pt] (play around with the number) immediately after the line-break (the \\) to add additional line spacing, e.g.

\overline{s} &= \frac{\sum s}{n} \\[2pt]
...

Chris Graham on Weight of a scoop in two ice cream parlours 8 years, 5 months ago

Gave some feedback: Has some problems

Christian Lawson-Perfect on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Gave some feedback: Has some problems

Christian Lawson-Perfect on Finding the formula for the $n^{\text{th}}$ term of linear sequences 8 years, 5 months ago

Saved a checkpoint:

I've split this into two parts - why make up your own part headers when Numbas can do it for you? It's important to let the student submit their answer to one part before moving on to another.

This question could very easily have steps - identify the first term, identify the common difference, and then you can write down the formula.

I've added a sentence to the top of the advice describing what needs to be done. A "plan of action" like this makes it easier to follow the more detailed parts of the solution.

I'd work this out by drawing up a table:

$n$	1	2	3
$a_n$	18	27	36
First differences		9	9

The formula is $a_n = a_1 + (n-1)d$. I can read off from the table that $a_1 = 18$ and $d = 9$.

There's a reason you're taught to do this stuff using tables at school - visually, you can see how $n$, $a_n$, and the common difference relate to each other as the sequence progresses. Without this, it can all seem like aimlessly shuffling numbers about.

Elliott Fletcher on Identify independent events 8 years, 5 months ago

Gave some feedback: Needs to be tested

Christian Lawson-Perfect commented on Finding the $n^{\text{th}}$ Term of a Quadratic Sequence 8 years, 5 months ago

Also, note that I used "first differences" and "second differences" - I thnk Lauren also spotted that it's tricky to keep track of all the differences floating about.

Christian Lawson-Perfect on Finding the $n^{\text{th}}$ Term of a Quadratic Sequence 8 years, 5 months ago

Gave some feedback: Has some problems

Chris Graham commented on Find the equation of a line through two points - negative gradient 8 years, 5 months ago

Brad, why not split this into two parts, the first asking for the gradient and intercept and the second the equation. The student can then see if they are on the right lines. You could also then use adaptive marking (ro replace m and c) in the second part.

Christian Lawson-Perfect on Finding the $n^{\text{th}}$ Term of a Quadratic Sequence 8 years, 5 months ago

Saved a checkpoint:

The layout of the advice is very hard to follow.

I would work this out by drawing up a table with rows for $n$, $a_n$, and first and second differences. 

For example,

$n$	1	2	3	4
$a_n$	15	22	33	48
First differences		7	11	15
Second differences			4	4

Because the second difference is $4$, I know that the coefficient of $x^2$ is $4/2 = 2$.

Then I can add a row for $a_n - 2n^2 = bn+c$:

$n$	1	2	3	4
$a_n$	15	22	33	48
$2n^2$	2	8	18	32
$a_n-2n^2$	13	14	15	16

So $bn+c = n+12$, and the formula for the whole sequence is $a_n = 2n^2+n+12$.

Sometimes, a good layout can save a lot of verbiage!

Bradley Bush commented on Find the equation of a line through two points - negative gradient 8 years, 5 months ago

Thank you for the feedback. I have acted on all the points you made there and I will apply these to the other 3 versions of this question too.

Stanislav Duris on Calculate the original price before a decrease 8 years, 5 months ago

Gave some feedback: Needs to be tested

Stanislav Duris commented on Calculate the original price before a decrease 8 years, 5 months ago

I have fixed everything. There is a different price range for every item, I made sure these are all realistic compared to the actual market right now.

All the original prices now end with .99. It would not be random enough if I still wanted the discounted price to end with .99 as well and still have integer percentages, so now, the new discounted price ends with .x9. I think this is good and realistic enough.

Hannah Aldous on Using Laws for Addition and Subtraction of Logarithms 8 years, 5 months ago

Gave some feedback: Needs to be tested

Aiden McCall on Use formulae for the area and volume of geometric shapes 8 years, 5 months ago

Gave some feedback: Has some problems

Stanislav Duris on Calculate a student discount 8 years, 5 months ago

Gave some feedback: Needs to be tested

Stanislav Duris commented on Calculate a student discount 8 years, 5 months ago

I've changed the percentages to be multiples of 5 and adjusted the advice.

I believe the line in the advice about not rouding can be correct. This is because in case the third decimal place of the calculated discount is 5 and all the following decimal places are zeros (so it would be xxx.xx5), we would round up and therefore subtract this rounded up value. However, the shop would technically round this down as they calculate the price by simply multiplying the original price by the percentage and they would round up there, cause they would also get 5 as the third decimal place and the rest would be zeros. So rounding the discount results in an error of 1 penny in some cases

This happened on occasions with previous percentages but seems not to happen anymore with multiples of 5 so I removed that line.