These are the questions you should try to work through over the mid-term break. Attempt as much of it as you can after revising the related chapters.
If you get stuck leave the work and try a different question before you come back to it. Check here over the course of next week, as I will post some tips for each section.
In the two classes after the mid-term break, we will cover as much ground as possible and hopefully resolve any issues that you got stuck on.
I will add some tips for some of the questions over the next few days, in blue text like this.
I have now finished adding my notes - you can add a comment here if you need any more help.
Preparation for paper 1:
Question 1 and 2 - arithmetic, percentages, tax, number theory (primes etc), sets, indices, surds and that type of thing.
(Chapters 11, 13, 16, 17, 18)
Page 150 q5 (note part d on next page!) Remember to multiply across to get rid of the fractions in the ratio part a. In part b, change all the money amounts to either cent or euro and remember to sanity check your answer. For the last part you are trying to find what percentage one number is of another, and the rate of VAT should be reasonable. For part d, remember the layout of these questions that we used in class, and how you can navigate backwards from the net tax, to gross tax, to the 2 tax amounts (you can calculate the amount of tax paid at the lower rate).
Page 192 q2 (You'll need to revise all the symbols in set notation - book 1 ch 15).
Page 238 q4 (in ex 16.5) and q 12 (next page) (Look for factors that are square numbers. In q12, the denominator is the clue to the rest of the question, try expressing the other numbers as something × √2)
Page 247 q1, 2, 3, 7, 8, 9 and 18 (Refer to the rules on page 240/1 and practice using your calculator to check answers).
Page 253 q 17, 18 (Make sure you know how to do these on your calculator!)
Question 3 and 4 and 5- algebra
(Chapters 1 to 7 and 27)
Page 13 q12,15,18,21,24 (Last one has 3 factors. If you have trouble with these, practice a few more from the page.)
Page 19 q25 (Be careful with the "-" sign. The Lowest Common Denominator is the LCM of all 3 fractions. Note this isn't an equation, you are just rewriting a fraction, so you can multiply above and below by the same thing as in e.g. 1/3 = 2/6)
Page 22 q17 (I only set one Long Division question, but if you have trouble, practice some more. Remember that there is never a remainder.)
Page 26 q7, 13 (Another important skill to practice as it comes up on paper 1 and 2. In q13, you need to tidy it up by getting rid of the fractions. As it is an equation, you must multiply the left hand side and right hand side by the same thing.)
Page 38 q4
Page 40 q15, 17 ( Note that q15 doesn't look like a quadratic until you multiply across by everything under the line to get rid of the fractions (***). In q17 "correct to 2 decimal places" means you have to use the quadratic formula.)
Page 46 q12, 14 (First get rid of brackets, then treat it like an equation (adding to from both sides, dividing across by same thing etc) to arrive at e.g. x ≤ a number. Revise the rules for plotting x < or x ≤ and for plotting x ∈ N, Z or R).
Page 50 q9,10
Page 397 q2, 3 and next page q8, 10. (We got quite far in q10 in class.
i) number of rows needed first night = 600/x
ii) number of rows needed second night = 630/(x-2)
iii) the number of rows on first night + 5 = number of rows on second night.
or
the number of rows on first night = number of rows on second night - 5
Giving us
600/x + 5 = 630/(x-2) (this will be a quadratic when you tidy it up and get rid of fractions - see comment above (***))
These questions are tricky, but get easier with practice.
Question 5 and 6 - functions
(Chapters 14 and 15)
Page 226 q15 (If you rearrange f(x) to have the x² term first, then make sure you keep the minus signs in the right place. Your g(x) should be a straight line. For part (vii) you are looking for output values of f(x) that cannot occur. For example, if you look at the graph on page 223, f(x) = 6 or f(x) = 7 has no solution. You can never get those output values from the function f(x). In that case f(x) > 5.1 has no solution.)
Page 230 q 5 (The wording is a bit confusing here - they talk about the width of the rectangle being 10-x where they mean the height according to the diagram. If the perimeter is fixed at 20, then x(10-x) = 10x - x² is the area, regardless of what value x has. So if x = 5, you'd have a square with area = 25. If x = 2 you'd have a 2 × 8 rectangle with area 16. Follow the normal steps you go through for graphing functions. The output values of f(x) will be the various possible areas of the shape.
This is a simpler version of the "A4 page malteaser box" problem we looked at briefly in class.)
Preparation for Paper 2
Question 1 - Perimeter area and volume
(Chapter 12)
Page 178 q3 and page 180 q6 (Q3: Careful with diameter v radius in part a and b. In part b and c you need to set up an equation and solve it. E.g. Vol of cylinder × 6 = Vol of block. For Q6: you need to note what fraction of the pizza 135º represents i.e. simplify 135/360. Note also you are interested in how much pizza is left. For part c remember that the radius, height and slant-height of a cone form a right-angle triangle.)
Question 2 - Coordinate Geometry
(Chapter 10)
Page 123 q12, 13, 14 and page 125 q27 (Revise transformations especially translations. Remember that the corners of a parallelogram are always given in order either clockwise or anti-clockwise. Also remember how to change the slope of a line to find the slope of the perpendicular line - invert and change sign.)
Question 3 and 4 - Geometry
(Chapters 20 to 26 and 28) (NB We will cover ch 26 after the mocks!)
Page 311 q11, 16 (For q11: use the angles you are given and properties of opposite angles and quadrilaterals. For q16, look for which angles are equal and isosceles triangles).
Page 318 q1 (For part v, construct the missing lines and try to prove that the newly formed triangles are congruent and that the sum of alternate angles make the opposite angles equal.)
Page 332 q6, 9 (Note that ∠acb is not shown. Q9 i: ∠bap = 45º then find two angles that measure 90º For part ii, one way to solve it is to construct a line completing the triangle abp. Then you can use the fact that triangle abp is isosceles. Find another isosceles triangle ... hint: radius of a circle is a constant.)
Page 359 to 363 practice all constructions.
Theorems: pages 401 and 408. (Don't forget that the latter theorem boils down to proving that the lines are parallel - then you use the (not examinable) earlier theorem which I gave you a handout for).
Question 5 - Trigonometry
(Chapter 19)
Page 294 q13, 14 (Q13: For part a, try drawing a rough sketch, marking the sides you know, being careful to place them correctly with respect the the angle A which you need to mark. You then have to work out the missing side of a right angle (using pythagoras). sin²A is just the (sinA)². For part b i, note that you need two sides and the angle between them to use ½abSinC. Hint: think radius!!! For part b ii, you need to work out 80/360 times the area of the circle. For part c, draw a rough sketch and label both sides you know. You will need a compass. For part d, no right angles here, so you need to use the Sine rule twice. Remember that the angles in a triangle add to 180°
Q14: Part a, similar to part a in q13. Note that you can find the area of a triangle more easily if you know the base and the perpendicular height. For part c, use pythagoras, then the Sine rule. Don't assume any angles are 90° just because they look like it! Part d, first draw the CAST diagram to work out which quadrants have negative Cos.)
Question 6 - Statistics
(Chapter 9)
Page 92 q6 and page 95 q10 (don't forget part c on next page) (In part a of the first question remember that 45 seconds is ¾ or 0.75 of a minute. These questions will help you practice your statistics, especially histograms and ogives. In each type of graph, make sure you always put the frequency (no. of students or no. of people) on the vertical axis. If you have 100 people then the median person is the 50th, the lower quartile person is the 25th and the upper quartile person is the 75th. Find each of these on the vertical axis, draw a horizontal line across and drop a perpendicular from where it hits the curve.Remember that histograms are a little unintuitive, the area of each block reflects the frequency and there are no gaps between blocks.)
Friday, February 13, 2009
Thursday, February 12, 2009
Using Equations to solve problems
Finally, we get to the whole point of algebra.
You are given some text and you have to create an equation and then solve it. Sometimes you can work the easy answers out in your head, but you must write out the equation. That way you practice the technique and develop the skills to work out tougher problems also.
The work for the long weekend included these 2 questions:
p386 - q4 (make sure you only use a single unknown)
p389 - q1 (simultaneous equations)
and I also want you to try to solve the third type of problem you can enounter
p394 - q3 (this is solved using a quadratic equation)
We will look at each of these as well as a much trickier problem of the 3rd type (quadratics) tomorrow.
You are given some text and you have to create an equation and then solve it. Sometimes you can work the easy answers out in your head, but you must write out the equation. That way you practice the technique and develop the skills to work out tougher problems also.
The work for the long weekend included these 2 questions:
p386 - q4 (make sure you only use a single unknown)
p389 - q1 (simultaneous equations)
and I also want you to try to solve the third type of problem you can enounter
p394 - q3 (this is solved using a quadratic equation)
We will look at each of these as well as a much trickier problem of the 3rd type (quadratics) tomorrow.
Tuesday, February 10, 2009
Coordinate Geometry
P124 q 23
For parts (i) through (iv) first write down the relevant formula, then plug in your values.
Watch out for the negative coordinates - say for example, when you are finding slope x2-x1 is going to be 6 - (-2) = 6+2 = 8
For part (v) remember what "equation of a line" means. It is a rule for being on a line.
For example - is the point (2,1) on the line 2x + 3y = 7?
Sub in for x and y: 2(2) + 3(1) = 7 which is true, so the point (2,1) is on the line.
Is the point (5,-1) on the line. Sub in and make sure you can figure out that yes, it is.
For part (vi) remember the rule for perpendicular slopes. If two lines are perpendicular, their slopes multiply to give -1 (or put another way, the product of their slopes is -1).
So if you have the slope of one line, you find the slope of the perpendicular line by inverting and changing the sign. If the slope of the first line is -2, then the slope of the perpendicular line is ½.
Part (viii) can be done using simultaneous equations.
For part (ix) draw a rough diagram if you haven't already done so. Axial symmetry in the line ab means that the line ab acts as a mirror. Where will the image of d be? A clue is to use translations.
For part (x) you are finding the points where the line crosses the X and Y axes. Remember that every point on the X axis has the coordinate (something,0).
You can work out the "something" by subbing y=0 in the equation of the line ab (see ans to part iv).
Finally, make sure that you use this question as an opportunity to revise all the formulae you need to know.
Really finally, don't forget your geometry sets tomorrow!
For parts (i) through (iv) first write down the relevant formula, then plug in your values.
Watch out for the negative coordinates - say for example, when you are finding slope x2-x1 is going to be 6 - (-2) = 6+2 = 8
For part (v) remember what "equation of a line" means. It is a rule for being on a line.
For example - is the point (2,1) on the line 2x + 3y = 7?
Sub in for x and y: 2(2) + 3(1) = 7 which is true, so the point (2,1) is on the line.
Is the point (5,-1) on the line. Sub in and make sure you can figure out that yes, it is.
For part (vi) remember the rule for perpendicular slopes. If two lines are perpendicular, their slopes multiply to give -1 (or put another way, the product of their slopes is -1).
So if you have the slope of one line, you find the slope of the perpendicular line by inverting and changing the sign. If the slope of the first line is -2, then the slope of the perpendicular line is ½.
Part (viii) can be done using simultaneous equations.
For part (ix) draw a rough diagram if you haven't already done so. Axial symmetry in the line ab means that the line ab acts as a mirror. Where will the image of d be? A clue is to use translations.
For part (x) you are finding the points where the line crosses the X and Y axes. Remember that every point on the X axis has the coordinate (something,0).
You can work out the "something" by subbing y=0 in the equation of the line ab (see ans to part iv).
Finally, make sure that you use this question as an opportunity to revise all the formulae you need to know.
Really finally, don't forget your geometry sets tomorrow!
Monday, February 9, 2009
Theorems
Don't forget to use the pro-forma (template) layout when proving theorems).
A diagram, labelled followed by
Given:
To prove:
Construction:
Proof:
... and the last line of your Proof: should be the same as your To prove: line.
Look over the first 2 theorems on page 401 and practice writing these out.
Look over the theorem we did today - I will give you a simplified version of it which is from the marking scheme (2004) in class tomorrow.
Then practice your constructions from the Ordinary level course (you'll find more detail on them in book 1).
You should be able to draw parallel lines, bisect an angle, construct the perpendicular bisector of a line and divide a line into 3 or more equal parts.
These constructions take some practice, so it is worthwhile spending time.
Remember that the construction lines aren't supposed to be hidden - so make sure that any arcs or other marks you make on the page are visible.
Don't forget your geometry sets for Tuesday!!
You need at least a compass and ruler, preferably a set square also.
A diagram, labelled followed by
Given:
To prove:
Construction:
Proof:
... and the last line of your Proof: should be the same as your To prove: line.
Look over the first 2 theorems on page 401 and practice writing these out.
Look over the theorem we did today - I will give you a simplified version of it which is from the marking scheme (2004) in class tomorrow.
Then practice your constructions from the Ordinary level course (you'll find more detail on them in book 1).
You should be able to draw parallel lines, bisect an angle, construct the perpendicular bisector of a line and divide a line into 3 or more equal parts.
These constructions take some practice, so it is worthwhile spending time.
Remember that the construction lines aren't supposed to be hidden - so make sure that any arcs or other marks you make on the page are visible.
Don't forget your geometry sets for Tuesday!!
You need at least a compass and ruler, preferably a set square also.
Sunday, February 8, 2009
Indices Problems
Taking a closer look at page 248 question 22:
(i) You are asked to express √2 as 2 to the power of something.
Remember that 9 to the power of ½ is the same as √9 which is = 3.
Make sure you know how to check this on your calculator.
(ii) Know your square and cube numbers.
Square numbers are 1,4,9,16,25,36,49,64,81,100,121,144,169,196 (up as far as 14²)
Cube numbers are 1,8,27,64,125,216 (up as far as 6³)
It is also useful to know your powers of 2.
2²=4, 2³=8 2^4 = 16 etc.
(iii)Looking at 8√2 you need to use your answers for the previous two parts to work this out. 8 = 2³ and √2 = 2 to the power of ½
So you need to apply the first rule of indices on the list on page 240, adding the two powers (even if one is a fraction).
Finally solving the equation.
Leave the LHS as it is.
Write down the RHS in the form that you arrived at in part (iii).
Now you have:
2 to the power of something (3x-1) = 2 to the power of something else (ans from part iii)
So you ignore the 2s and equate the powers to solve for x.
Note that you can check your answer by putting
2 ^ (3×x +1) into your calculator (subbing for x) and then check if you get the same result for 8√2.
(i) You are asked to express √2 as 2 to the power of something.
Remember that 9 to the power of ½ is the same as √9 which is = 3.
Make sure you know how to check this on your calculator.
(ii) Know your square and cube numbers.
Square numbers are 1,4,9,16,25,36,49,64,81,100,121,144,169,196 (up as far as 14²)
Cube numbers are 1,8,27,64,125,216 (up as far as 6³)
It is also useful to know your powers of 2.
2²=4, 2³=8 2^4 = 16 etc.
(iii)Looking at 8√2 you need to use your answers for the previous two parts to work this out. 8 = 2³ and √2 = 2 to the power of ½
So you need to apply the first rule of indices on the list on page 240, adding the two powers (even if one is a fraction).
Finally solving the equation.
Leave the LHS as it is.
Write down the RHS in the form that you arrived at in part (iii).
Now you have:
2 to the power of something (3x-1) = 2 to the power of something else (ans from part iii)
So you ignore the 2s and equate the powers to solve for x.
Note that you can check your answer by putting
2 ^ (3×x +1) into your calculator (subbing for x) and then check if you get the same result for 8√2.
Friday, February 6, 2009
Work for the weekend
As promised ...!
First, do the correct questions from the surds chapter.
P238 Q 1,2,3,4,5 of the chapter test
In question 1 and 2 you get a clue to the answer in the question - you know that you can rewrite √108 as √something × √3
In question 3 rewrite "Three and one sixteenth" as a top-heavy fraction, then the answer will become clearer.
Remember in all these questions that you need to look out for square numbers: 1,4,9,16 etc.
Next, the new stuff:
Page 247 q 13,14,17,22
In each of these, refer back to the rules and examples from pag 240 and remember how to solve equations with exponents.
In particular q22 is one that you should try hard to work through. Parts (i) (ii) and (iii) are needed to solve the main part of the question.
We will spend most of next week revising geometry.
As preparation for this, try the following questions. Remember how to write out a formal proof, in particular naming angles properly.
Page 310 q 9
Page 319 q7
For coordinate geometry revision, read over chapter 10 and revise all formulae.
Then answer
Page 124 q 23
The last topic to look at is the final chapter in the book - using equations to solve problems.
Have a go at these 2 types of questions:
p386 - q4 (make sure you only use a single unknown)
p389 - q1 (simultaneous equations)
That is the lot!
I will post some help for some of these questions over the weekend.
First, do the correct questions from the surds chapter.
P238 Q 1,2,3,4,5 of the chapter test
In question 1 and 2 you get a clue to the answer in the question - you know that you can rewrite √108 as √something × √3
In question 3 rewrite "Three and one sixteenth" as a top-heavy fraction, then the answer will become clearer.
Remember in all these questions that you need to look out for square numbers: 1,4,9,16 etc.
Next, the new stuff:
Page 247 q 13,14,17,22
In each of these, refer back to the rules and examples from pag 240 and remember how to solve equations with exponents.
In particular q22 is one that you should try hard to work through. Parts (i) (ii) and (iii) are needed to solve the main part of the question.
We will spend most of next week revising geometry.
As preparation for this, try the following questions. Remember how to write out a formal proof, in particular naming angles properly.
Page 310 q 9
Page 319 q7
For coordinate geometry revision, read over chapter 10 and revise all formulae.
Then answer
Page 124 q 23
The last topic to look at is the final chapter in the book - using equations to solve problems.
Have a go at these 2 types of questions:
p386 - q4 (make sure you only use a single unknown)
p389 - q1 (simultaneous equations)
That is the lot!
I will post some help for some of these questions over the weekend.
Monday, February 2, 2009
Revised Plan Leading to Mock Exams
If you have completed the work you were set last week (see the list) that is great. If you haven't, then try to complete it over the next few days, using the tips on here or asking me in class if you are really stuck. We will look at the solutions later in the week.
This week we will wrap up the last two topics in statistics (Histograms and Ogives) and deal with two small topics which look a lot more difficult than they really are.
Next week will be devoted to Geometry and co-ordinate Geometry.
We will use the 2 days after the mid-term break for general revision.
--------------------------------------------
Mon 2 Tue 3 Wed 4 Thu 5 Fri 6
Stats Stats Surds Indices
--------------------------------------------
Mon 9 Tue 10 Wed 11 Thu 12 Fri 13
Theorems Geometry Coordinate Geom
--------------------------------------------
Mon 16 Tue 17 Wed 18 Thu 19 Fri 20
M I D T E R M B R E A K
--------------------------------------------
Mon 23 Tue 24 Wed 25 Thu 26 Fri 27
General revision
--------------------------------------------
This week we will wrap up the last two topics in statistics (Histograms and Ogives) and deal with two small topics which look a lot more difficult than they really are.
Next week will be devoted to Geometry and co-ordinate Geometry.
We will use the 2 days after the mid-term break for general revision.
--------------------------------------------
Mon 2 Tue 3 Wed 4 Thu 5 Fri 6
Stats Stats Surds Indices
--------------------------------------------
Mon 9 Tue 10 Wed 11 Thu 12 Fri 13
Theorems Geometry Coordinate Geom
--------------------------------------------
Mon 16 Tue 17 Wed 18 Thu 19 Fri 20
M I D T E R M B R E A K
--------------------------------------------
Mon 23 Tue 24 Wed 25 Thu 26 Fri 27
General revision
--------------------------------------------
Scientific Notation
I added these questions in to give you a refresher course in using this function on your calculator.
There are differences between calculators so make sure that you know how to input numbers in scientific notation and how to convert to/from scientific notation on yours!
It is really important that you always practice with the calculator you will have with you during your exams.
There are differences between calculators so make sure that you know how to input numbers in scientific notation and how to convert to/from scientific notation on yours!
It is really important that you always practice with the calculator you will have with you during your exams.
Sets
Read over the chapter on sets in Book 1. It gives a good overview and covers all of the notation well. There is a little bit of higher level work on maximising and minimising sets which is very simple, but we'll do it after the mocks.
For now, make sure that you know all the notation and that you know how to deal with unknowns using a Venn diagram.
Don't forget that you fill a Venn diagram from the centre outwards.
For question 4:
Draw two circles marked A and B with a decent overlap.
The overlap area represents A ∩ B and the number of elements in the area is 6.
That means that the shape representing A not B or A\B is going to contain the number 13-6=7 and the 3rd region represented by B\A will contain the number 14-6= 8.
So reading left to right (assuming you've laid your work out that way with A on the left and B of the right) you will have the numbers 7 then 6 then 8.
That means that the total number of elements in A∪B is 7+6+8=21.
The total number of elements in A∪B is written as #(A∪B).
General point:
If you have numbers with dots beside them in a region of a Venn diagram then those numbers are the elements of a set.
If you have a single number and no dot beside it, then that number is the number of elements in that region.
For now, make sure that you know all the notation and that you know how to deal with unknowns using a Venn diagram.
Don't forget that you fill a Venn diagram from the centre outwards.
For question 4:
Draw two circles marked A and B with a decent overlap.
The overlap area represents A ∩ B and the number of elements in the area is 6.
That means that the shape representing A not B or A\B is going to contain the number 13-6=7 and the 3rd region represented by B\A will contain the number 14-6= 8.
So reading left to right (assuming you've laid your work out that way with A on the left and B of the right) you will have the numbers 7 then 6 then 8.
That means that the total number of elements in A∪B is 7+6+8=21.
The total number of elements in A∪B is written as #(A∪B).
General point:
If you have numbers with dots beside them in a region of a Venn diagram then those numbers are the elements of a set.
If you have a single number and no dot beside it, then that number is the number of elements in that region.
Perimeter Area and Volume
The 1st question in the chapter test is a perfect example of an exam type question, so you should take about 25 minutes to answer it.
For part a)
Use the area of a circle formula (area = πr²) twice for the larger circle and for the smaller one, making sure to do as you're told with π = 22/7.
Do not fall into the trap of thinking you can subtract the radii and find the area of a circle with radius = 7cm. If in doubt, sketch out the circles.
For part b)
You need to use the formulae for volume of a sphere and volume of a cylinder.
Part iii of this question is a little confusing. They are looking for the volume of air that is left in the cylinder when it contains the 3 balls.
Fort part c)
This is a tricky question. You are told the radius of the hemisphere so first step is to work out its volume. Note that the question doesn't mention whether π should be 3.14 or 22/7. That is because you can actually work out the answer conveniently without setting π equal to anything, just leave your work in terms of π.
When you find the volume of the hemisphere, you know that the volume of a cone is half of this.
So work out half the volume of the hemisphere (in terms of π) and then set up an equation with the formula for volume of a cone on the left (subbing r = 6) and the volume you just calculated on the right.
An example (not accurate for this question). Lets say the volume of the hemisphere turns out to be 20π. Then write down
1/3π r²h = 10π and sub in the value of the radius on the left.
To simplify your equation, first divide across by π (gets rid of both πs) and multiply across by 3 (gets rid of the fraction). Then solve for h, the only unknown.
Don't lose sight of what you were originally asked. The height of the entire shape is equal to this height plus the radius of the hemisphere. Make sure you can understand why.
General point:
If you are working out e.g. πr² make sure that you understand fully that it is only r which gets squared and that the result of r² gets multiplied by π.
You never square or cube π.
For part a)
Use the area of a circle formula (area = πr²) twice for the larger circle and for the smaller one, making sure to do as you're told with π = 22/7.
Do not fall into the trap of thinking you can subtract the radii and find the area of a circle with radius = 7cm. If in doubt, sketch out the circles.
For part b)
You need to use the formulae for volume of a sphere and volume of a cylinder.
Part iii of this question is a little confusing. They are looking for the volume of air that is left in the cylinder when it contains the 3 balls.
Fort part c)
This is a tricky question. You are told the radius of the hemisphere so first step is to work out its volume. Note that the question doesn't mention whether π should be 3.14 or 22/7. That is because you can actually work out the answer conveniently without setting π equal to anything, just leave your work in terms of π.
When you find the volume of the hemisphere, you know that the volume of a cone is half of this.
So work out half the volume of the hemisphere (in terms of π) and then set up an equation with the formula for volume of a cone on the left (subbing r = 6) and the volume you just calculated on the right.
An example (not accurate for this question). Lets say the volume of the hemisphere turns out to be 20π. Then write down
1/3π r²h = 10π and sub in the value of the radius on the left.
To simplify your equation, first divide across by π (gets rid of both πs) and multiply across by 3 (gets rid of the fraction). Then solve for h, the only unknown.
Don't lose sight of what you were originally asked. The height of the entire shape is equal to this height plus the radius of the hemisphere. Make sure you can understand why.
General point:
If you are working out e.g. πr² make sure that you understand fully that it is only r which gets squared and that the result of r² gets multiplied by π.
You never square or cube π.
Income Tax Questions
You should not have any problems with these questions.
Find a layout you like and stick to it.
E.g.
Amount above cut-off @ higher rate % = part 1 of tax
Total income <
Amount up to cut-off @ lower rate % = part 2 of tax
Add to give Gross Tax
Subtract Tax credits (minus)
Results in Net Tax
Find a layout you like and stick to it.
E.g.
Amount above cut-off @ higher rate % = part 1 of tax
Total income <
Amount up to cut-off @ lower rate % = part 2 of tax
Add to give Gross Tax
Subtract Tax credits (minus)
Results in Net Tax
Linear Inequalities
You are used to solving equations e.g. finding the value of x for which this is true
3x - 5 = 7
Treat it like a weighing scales, and do the same to both sides to solve the equation.
So
3x = 12 (+5 to both sides)
x = 4 (÷ 3 on both sides)
The same weighing scales analogy works for inequalities. If you have
3x - 5 ≤ 7
3x ≤ 12
x ≤ 4
In other words, the inequality is true for any number x as long as it is less than or equal to 4.
Try 4:
3(4) - 5 = 7 and 7 is ≤ 7
Try 2:
3(2) - 5 = 1 and 1 is ≤ 7
Try a number bigger than 4 e.g. 5 :
3(5) - 5 = 10 and 10 is not ≤ 7
When you solve an inequality you have to check which set of numbers the solution is in.
For example if you were told x ∈ N that means it is a natural number (whole number ≥ 0)
Then the solution set would be x = {0,1,2,3,4}
However, if it was an integer i.e. x ∈ Z, then the negative whole numbers could also be in the solution set.
Finally if you were told x ∈R, that means x can be any number (fraction, decimal or whole number).
In the last two cases you can't write out the solution set, so you will normally be asked to graph it on the number line.
To do this draw a numberline and indicate with dots and arrows which set of numbers are included.
NB Note that there is one important difference between solving equations and finding solution sets to inequalities.
If you have
-x = 3
then multiplying across by -1 gives you
x = -3
However, if you have
-x < 3
when you multiply accross by -1, you have to reverse the inequality
x > -3
The reason for this is simple. -1 < 2 but multiplying across by -1 gives 1 on the left and -2 on the right so the sign must turn around 1 > -2
3x - 5 = 7
Treat it like a weighing scales, and do the same to both sides to solve the equation.
So
3x = 12 (+5 to both sides)
x = 4 (÷ 3 on both sides)
The same weighing scales analogy works for inequalities. If you have
3x - 5 ≤ 7
3x ≤ 12
x ≤ 4
In other words, the inequality is true for any number x as long as it is less than or equal to 4.
Try 4:
3(4) - 5 = 7 and 7 is ≤ 7
Try 2:
3(2) - 5 = 1 and 1 is ≤ 7
Try a number bigger than 4 e.g. 5 :
3(5) - 5 = 10 and 10 is not ≤ 7
When you solve an inequality you have to check which set of numbers the solution is in.
For example if you were told x ∈ N that means it is a natural number (whole number ≥ 0)
Then the solution set would be x = {0,1,2,3,4}
However, if it was an integer i.e. x ∈ Z, then the negative whole numbers could also be in the solution set.
Finally if you were told x ∈R, that means x can be any number (fraction, decimal or whole number).
In the last two cases you can't write out the solution set, so you will normally be asked to graph it on the number line.
To do this draw a numberline and indicate with dots and arrows which set of numbers are included.
NB Note that there is one important difference between solving equations and finding solution sets to inequalities.
If you have
-x = 3
then multiplying across by -1 gives you
x = -3
However, if you have
-x < 3
when you multiply accross by -1, you have to reverse the inequality
x > -3
The reason for this is simple. -1 < 2 but multiplying across by -1 gives 1 on the left and -2 on the right so the sign must turn around 1 > -2
Solving Quadratics
There are 2 ways to Solve quadratics
1 - By factoring
2 - Using the quadratic formula.
Solving Quadratics by Factoring:
The main difference between solving quadratics on the Higher Level course is that you will often have a number in front of the x² term, and the quadratic won't always be presented as a nice tidy x² + 4x + 4 = 0 format.
Dealing with the number in front of the x² term:
Looking at q 13
7x² + 29x + 4 = 0
As 7 is prime, the two brackets will be of the form
(7x .... )(x .....)=0
The numbers that multiply to make 4 are 4 × 1 and 2 × 2.
Trying e.g.
??? (7x + 2)(x + 2)= 0
If we multiply the inside pair and outside pair to check the x term we get
inside pair = 2x
outside pair = 14x
Total = 16x ≠ 29x
Trying again
(7x + 1)(x +4) = 0
Gives x + 28x = 29x so these factors are correct.
Factors are (7x + 1)(x +4) = 0
Roots are
7x + 1 = 0
x = -1/7
x+4=0
x = -4
Ans: Roots of equation are x = -1/7 or -4
With question 16
x(x-4) = 21
you have to tidy it up first
x² - 4x = 21
x² - 4x -21 = 0 (Note the =0 part is critical, because we will be exploiting the zero law to solve the factored equation).
Then factor as usual.
With questions 41 and 44 there is a lot more tidying up to do. Get rid of brackets and then put each equation in the form ....x² ....x ... number = 0
While you are looking at this topic, note the questions in the next section. There is no sign of a quadratic in e.g. question 5 but when you multiply accross by x to get rid of the fraction, you end up with x(x) = x²
Solving Quadratics Using the Quadratic Formula
If the question asks for the solution correct to 1 or 2 etc places of decimals, you need to use the quadratic formula.
Note that all the tidying up stuff mentioned above still applies, so you need to make sure the equation is in the form ax² + bx + c = 0
Looking at p35 q 13 you have
4x² + x -1 = 0
Which means that
a = 4, b = 1 (because x is the same as 1x) and c = -1 (note the sign).
Plugging numbers into the formula you get x =
-1 ± √( 1² - 4(4)(-1))
----------------------
2(4)
= -1 ± √17
--------
8
The results are (-1 + 4.123)/8 = 0.3903 ≈ 0.39
or (-1 - 4.123)/8 = -0.6403 ≈-0.64
Checking your answers in quadratics
Checking your answer is always a good idea, especially if you have time in an exam. Sometimes you may be asked to verify your answer. This means taking the result you got and plugging it back in to the original equation for x.
That means that in the examples above:
1 - By factoring
2 - Using the quadratic formula.
Solving Quadratics by Factoring:
The main difference between solving quadratics on the Higher Level course is that you will often have a number in front of the x² term, and the quadratic won't always be presented as a nice tidy x² + 4x + 4 = 0 format.
Dealing with the number in front of the x² term:
Looking at q 13
7x² + 29x + 4 = 0
As 7 is prime, the two brackets will be of the form
(7x .... )(x .....)=0
The numbers that multiply to make 4 are 4 × 1 and 2 × 2.
Trying e.g.
??? (7x + 2)(x + 2)= 0
If we multiply the inside pair and outside pair to check the x term we get
inside pair = 2x
outside pair = 14x
Total = 16x ≠ 29x
Trying again
(7x + 1)(x +4) = 0
Gives x + 28x = 29x so these factors are correct.
Factors are (7x + 1)(x +4) = 0
Roots are
7x + 1 = 0
x = -1/7
x+4=0
x = -4
Ans: Roots of equation are x = -1/7 or -4
With question 16
x(x-4) = 21
you have to tidy it up first
x² - 4x = 21
x² - 4x -21 = 0 (Note the =0 part is critical, because we will be exploiting the zero law to solve the factored equation).
Then factor as usual.
With questions 41 and 44 there is a lot more tidying up to do. Get rid of brackets and then put each equation in the form ....x² ....x ... number = 0
While you are looking at this topic, note the questions in the next section. There is no sign of a quadratic in e.g. question 5 but when you multiply accross by x to get rid of the fraction, you end up with x(x) = x²
Solving Quadratics Using the Quadratic Formula
If the question asks for the solution correct to 1 or 2 etc places of decimals, you need to use the quadratic formula.
Note that all the tidying up stuff mentioned above still applies, so you need to make sure the equation is in the form ax² + bx + c = 0
Looking at p35 q 13 you have
4x² + x -1 = 0
Which means that
a = 4, b = 1 (because x is the same as 1x) and c = -1 (note the sign).
Plugging numbers into the formula you get x =
-1 ± √( 1² - 4(4)(-1))
----------------------
2(4)
= -1 ± √17
--------
8
The results are (-1 + 4.123)/8 = 0.3903 ≈ 0.39
or (-1 - 4.123)/8 = -0.6403 ≈-0.64
Checking your answers in quadratics
Checking your answer is always a good idea, especially if you have time in an exam. Sometimes you may be asked to verify your answer. This means taking the result you got and plugging it back in to the original equation for x.
That means that in the examples above:
- if you substitute x = -4 into 7x² + 29x + 4 you should get zero.
- if you substitute x = 0.39 into 4x² + x -1 you should get very close to zero.
Writing as a single fraction
Example 1:
If you are asked to add
2/3 + 1/4
you have to first convert them into the same type of fraction i.e. find a new denominator which is the LCM of 3 and 4 => 12
2/3 = 8/12 (you have to multiply the bottom by 4 to go from 3 to 12, so you must do the same with the top)
1/4 = 3/12
So the sum becomes
8/12 + 3/12 = 11/12
Example 2:
The exact same principle applies when you have to add (or write as a single fraction)
5x-1 - 2x + 3
----- -------
4 5
The new denominator will be 20.
You multiply 4 by 5 to get 20, so you must multiply 5x-1 by 5 also giving 5(5x-1) = 25x - 5
Repeat the corresponding operation on the other fraction and then add/subtract like terms.
Example 3:
In question 17 you have to find the LCM of 2x+1 and 2x-1.
This will be (2x+1)(2x-1) and exactly the same principle as in examples 1 and 2 applies.
Remember you don't have to multiply out the denominator (expression on the bottom).
If you are asked to add
2/3 + 1/4
you have to first convert them into the same type of fraction i.e. find a new denominator which is the LCM of 3 and 4 => 12
2/3 = 8/12 (you have to multiply the bottom by 4 to go from 3 to 12, so you must do the same with the top)
1/4 = 3/12
So the sum becomes
8/12 + 3/12 = 11/12
Example 2:
The exact same principle applies when you have to add (or write as a single fraction)
5x-1 - 2x + 3
----- -------
4 5
The new denominator will be 20.
You multiply 4 by 5 to get 20, so you must multiply 5x-1 by 5 also giving 5(5x-1) = 25x - 5
Repeat the corresponding operation on the other fraction and then add/subtract like terms.
Example 3:
In question 17 you have to find the LCM of 2x+1 and 2x-1.
This will be (2x+1)(2x-1) and exactly the same principle as in examples 1 and 2 applies.
Remember you don't have to multiply out the denominator (expression on the bottom).
Sunday, February 1, 2009
Factoring
When a question is written as "Factor the following expression" or simply "Factor ..." all you have to do is re-write the expression given as 2 numbers or expressions multiplied by each other.
Factor 12:
12 = 2 × 6 or 2(6)
or
12 = 3 × 4 or (3)(4)
... where the brackets imply multiplication.
You can always check your answer by multiplying it out - but make sure you know the difference between your answer and your check.
Factor 2x + xy
Do this by finding the HCF of both parts of the expression (in this case x) and then rewriting the expression as
x(2+y) (<= This is the answer)
x(2+y) = 2x + xy (<= This is the check)
Look back in the orange and purple books for a revision of factoring.
For question 7:
Remember that you have to put (3x ....) into the first bracket and (x ....) into the second one. When you multiply these out that'll give you 3x².
Then remember that when you are looking for 2 numbers that multiply to give -8, you will be multiplying one by 3 and one by 1 to get them to add to +2.
For questions 8 and 9:
Both must be difference of 2 squares. Remember that 1 = 1².
For question 10:
In this one, you might not realise that you are finished when you take out the HCF - but that is all there is to it.
I'll start question 31 from the set work here.
2x(2x-y)+y(2x-9y)
= 4x² - 2xy + 2xy - 9y²
-2xy + 2xy cancel each other out so you are left with a familiar looking
4x² - 9y² to factor.
Note also the questions from 22 to 30:
The tip is at the top of the page - we don't have a single rule for factoring something like this is one go, so see if you can simplify it by factoring out something as a first step. This means we'll end up with 3 factors. Like saying that 24 = 2 × 3 × 4.
So for 5x² - 20y² we first take the HCF which is 5 and write down
5x² - 20y²
= 5(x² - 4y²)
= 5((x)² - (2y)²)
= 5(x-2y)(x+2y)
Factor 12:
12 = 2 × 6 or 2(6)
or
12 = 3 × 4 or (3)(4)
... where the brackets imply multiplication.
You can always check your answer by multiplying it out - but make sure you know the difference between your answer and your check.
Factor 2x + xy
Do this by finding the HCF of both parts of the expression (in this case x) and then rewriting the expression as
x(2+y) (<= This is the answer)
x(2+y) = 2x + xy (<= This is the check)
Look back in the orange and purple books for a revision of factoring.
For question 7:
Remember that you have to put (3x ....) into the first bracket and (x ....) into the second one. When you multiply these out that'll give you 3x².
Then remember that when you are looking for 2 numbers that multiply to give -8, you will be multiplying one by 3 and one by 1 to get them to add to +2.
For questions 8 and 9:
Both must be difference of 2 squares. Remember that 1 = 1².
For question 10:
In this one, you might not realise that you are finished when you take out the HCF - but that is all there is to it.
I'll start question 31 from the set work here.
2x(2x-y)+y(2x-9y)
= 4x² - 2xy + 2xy - 9y²
-2xy + 2xy cancel each other out so you are left with a familiar looking
4x² - 9y² to factor.
Note also the questions from 22 to 30:
The tip is at the top of the page - we don't have a single rule for factoring something like this is one go, so see if you can simplify it by factoring out something as a first step. This means we'll end up with 3 factors. Like saying that 24 = 2 × 3 × 4.
So for 5x² - 20y² we first take the HCF which is 5 and write down
5x² - 20y²
= 5(x² - 4y²)
= 5((x)² - (2y)²)
= 5(x-2y)(x+2y)
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