Wrapping up geometry

We covered everything I planned in class today - we will repeat for the other half of the class on Tuesday.

Work for our next proper maths class is from the 2005 and 2006 papers.
In both papers, answer Q2 (c) and the complete question 3 and 4.

As it is a really short week maths-wise and there is a long weekend approaching, we will touch on functions and graphs and statistics so that you'll be able to do some useful work on these topics.

Planning notes

As alternative halves of the class will be out Monday/Tuesday - we will have a repeating class, covering the deductions of the circle theorem we did on Friday, the last circle theorem and an elegant proof of Pythagoras.
We will have only 1 proper maths class with everyone present this week, so we must finish the geometry.
According to the plan, we should be doing functions + graphs this week, so I will try to get this started during our Thursday class so that you can work through this topic over the long weekend.

We made our original plan at the end of February, let me know if you want to change the order of revision.

Geometry up as far as circle theorems

Today we went over the idea of a proof of a geometric problem as opposed to proof of a theorem.
Insofar as possible, keep the proof of problems as formal as the proofs of theorems:

• drawing a diagram
• stating what is to be proven in terms of your diagram
• doing a construction or a relabelling if needed
• justifying each step along the way.

You had 2 such problems for homework. Note tjhat in the rhombus question you should prove the shape is a rhombus using the principles of congruence.

You should know the first 6 theorems by now. It is important to remember the order so that you know which previous theorems you can use in the course of the theorem you are asked to prove.

Tomorrow we will look at the constructions and start proving the first of the circle theorems.

Geometry

Today we looked at the various transformations of the plane (translations, central and axial symmetry and the new one: rotations).
You need to know how to rotate a point or a line or a shape through a given angle and how to identify how many orders of rotational symmetry a shape has.
If you have a point a, its image under a given transformation will be called a'.

Homework is to look at some shapes and work out what their order of rotational symmetry is and to answer a question about rotations on the coordinated plane. It is important the the rotation is through 90degrees.
If you have are rotating about the point a(1,2) and you are rotating the point b(4,0) you need to find a point b' which is the same distance from a as the point b is, where ab is perpendicular (90 degrees) to ab'. See if you can work out a quick way of doing this, without using slope and distance formula.

We also started looking at an overview of the theorems which we will be covering.

Don't forget to bring you geometry set to each class from now on.

Arithmetic continued

Today we went over the topics of compound interest and income tax.
On compound interest we looked at how you could use the compound interest formula to work out the principal, given the amount of interest earned over a given number of years.
E.g. what principal will earn 315.35 interest at 5% over 3 years?
P +315.25 = P(1.05)³
P +315.25 = P(1.157625)
P(1.157625) - P = 315.25
Factor LHS:
P(1.157625 -1) = 315.25
P = 315.25/0.157625 = 2000

Income Tax questions are straightforward once you know the terms used and lay your work out clearly. Practice the questions where you have to work backwards to find the original Gross Income.
Also, watch out for older editions of the textbook. You do not need to know about Tax Free Allowance - just standard rate cut-off and tax credits.

Tomorrow we will get started on geometry.

Arithmetic

It is important not to overlook the arithmetic section of the course. The questions here can be subtly tricky.
These are the main pitfalls we went through in class:
Ratios:
If there are fractions, multiply the entire ratio by the Lowest Common Denominator before you start to work out the individual shares. So 1/2 : 1/4 : 1 becomes 2:1:4 when you multiply across by 4.
Distance Speed and Time:
Be careful subtracting 24hour clock questions - we talked about the different ways of doing this - stick with whatever works best for you, but double check your answer. Make sure that you don't treat 3h20mins as 3.2 hours. Work in fractions.
Finding average speed over a 2-part journey, don't add the 2 speeds and divide by 2, DO divide the total journey time by the total time taken.
Percentages + Tax
If you are going to use the % button on your calculator, make sure you know how to use it properly. If you want to find 19% of a number you can divide by 100 and multiply by 19 or just multiply by 0.19. To add 19% on in one step, just multiply by 1.19. You don't really need the percentage button at all.
For questions such as "a bill including VAT at 12% comes to €106.40, what was the bill before tax was added" you need to write down
106.4 = 112%
then divide accross by 112 to find 1% then multiply by 100 to find 100%.
The compound interest formula is useful for working out the original principle after n years of the same rate of interest being applied. One of the homework questions requires you to use it. Set up an equation using the formula and solve it.

We will try to finish off Arithemtic in the next class.

Algebra test results

I just finished correcting today's test.

A couple of points before we move on to the next topic.

On the question about the buffet meal, everyone managed to get the two expressions representing the old price per person (160/x) and the new price per person (160/(x+4)).
The tricky part is putting the equation together.
To do this you need to say that the old price minus the new price is 2 euro (as they tell you that the new price is cheaper).
(160/x) - (160/(x+4)) = 2
And solve.
[You could also work it out by saying the new price + 2 euro = old price]

A few people made worrying mistakes on BEMDAS.
In ....

3(-1)²

it is the -1 that gets squared, not the 3, so the answer is 3(1) = 3

With the double inequality, you can solve it all in one go but it is easier to follow if you split it into two inequalities and solve them separately.

Most of you managed the long division nicely.

In the equation with algebraic fractions many of you made it very complicated, looking for a large common multiple ... instead of just multiplying both sides of the equation by the smallest number that would make all the fractions disappear.

Finally, many of you managed to express 3/(2x-1) - 2/(3x+1) as a single fraction, but completely missed the point of verifying your answer by setting x = 1.
If you just sub x=1 into your answer you are not verifying (i.e. proving) anything.
The point is that if you replace x with 1 in the above expression you get 3/(2-1) - 2/(3+1) = 3 - 1/2 = 2.5
and if you replace x with 1 in the answer you got for the single fraction
(should be (5x + 5)/(2x-1)(3x+1), you'll get (5+5)/(2-1)(3+1) = 10/4 = 2.5

You verify that you haven't changed the value of the expression by subbing x=1 into the original version as you were given and your new improved version of the expression and getting the same answer in both.

Note that you can do this verification to check that you haven't made a mistake anytime you are asked to simplify an expression, write as a single fraction etc.

Tomorrow we will start on arithmetic.

Using algebra to solve problems

We focussed today on the type of questions where an unknown is parcelled up in two different ways and a relationship between them is given.

There is no doubt about it - these problems can be tricky. Partly because you have to parse through a lot of text, and partly because the equation that you end up with can be awkward.

Some good examples are in the 2006, 2005 and 2004 papers.

In the 2006 q, you make 2 expressions
540/x gives us the number of days the first lot of hay will last.
300/x+1 gives us the number of days the second lot will last
The sum of these two expressions is 90 days.

When you build your equation and simplify it you get

3x² -25x - 18 = 0

After you have factored, don't forget to discard the negative root and write your answer as x = 9 bales of hay and to work out the last part of the problem ... how many days it took to complete the first set of 540 bales. In other words, evaluate the first expression with x = 9 and write down the answer as 60 days.

Long Division in Algebra

The main things to remember are :

DMSB - the divide, multiply, subtract, bring-down repeated sequence is the same as long division.
If you have something like x³ + 2x²- 12, you have to add in the missing term to make it x³ + 2x² +0x - 12 before you start dividing.
The expressions will always divide in equally. You should not get a remainder.

Practice as many of these as you can.

Quadratic Equations

Quadratic equations are equations that involve a squared term.
If

x² = 4

then x could be +4 or -4.

So quadratic equations usually have 2 possible solutions.

We solve quadratics by factoring or by using the quadratic formula.

The more you practice factoring of quadratic trinomials, the more easily the answer will jump off the page for you when you do them.

The quadratic formula also takes practice, plus you need to commit it to memory. Practice writing it out regularly and always check in the book to ensure that you have got it exactly right before you use it.

For homework you are doing a selection of quadratics.

• a straightforward equation that has a coefficient in front of the x²
• one that will only take the ax² + bx + c = 0 format after you have done some manipulation
• one that requires the use of the formula
• one that needs a second round of equation solving after you have worked out possible values for x.

In tomorrow's class, we will continue with quadratics and look at the worded problems which involve forming and solving a quadratic equation.

Algebra

We have just started revising Algebra.
We started with simultaneous equations. These are really important as they test a lot of mathematical skills and they can come up in co-ordinate geometry (intersection of 2 lines) also.

We will look at some worded problems which lead to simultaneous equations in the next class.
The next topic we'll cover is called changing the focus of en equation.
E.g. a = 2(b+c) Express b in terms of a and c.