Bridging Pack for pupils who have passed National 4 Mathematics and want to do National 5.

Unit 1: Preparation for National 5 Applications Before you begin National 5 you need to be sure that you can: 1. Use your knowledge of angles to calculate bearings. 2. Solve a basic trigonometric equation to find the size of an angle 3. Read and Plot Co-ordinates on a 2-Dimensional Grid. 4. Calculate percentage increase and decrease. 5. Convert between mixed numbers and improper fractions. 6. Find the mean, median, mode and range from a set of data. 7. Solve problems using Pythagoras Theorem

Before you begin National 5 Applications you must be sure that you can: 1. Use your knowledge of angles to calculate bearings. Examples 1. Calculate the bearing from HOME to every lettered point. Solutions 1. A = 070˚ B = 180 – 50 = 130˚ C = 360 – 60 = 300˚ * = 180 – 115 = 65, # = * = 65, Bearing of D = 360 – 65 = 295˚ $ = 80, Bearing of E = 180 – 80 = 100

A BEARING is measured clockwise from North. It would be useful to memorise the bearings of the 8 main compass points as shown on the diagram. Copy it into your jotter now. Before you calculate a bearing you should estimate what you think it is by comparing it to the compass. To calculate the bearing of a line going towards the right - write down the ‘top’ angle with a 0 before it. - subtract the ‘bottom’ angle from 180˚.

To calculate the bearing of a line going towards the left - subtract the ‘top’ angle from 360˚. - add the ‘bottom’ angle on to 180˚. Remember: Alternate angles are equal so a capital letter N (or backwards N) made by joining two north lines will contain two angles the same size. Starred angles are the same size.

Angles on the same side of a straight line will add up to 180˚. Angle marked with # = 180 – (*).

Practice 1. Estimate the bearing from A to B in each diagram below. 2. Now estimate the bearing of A from B in each diagram above. 3. Calculate the bearing from HOME to every lettered point.

Answers: 1. 020˚ - 040˚ 100˚ - 130˚ 320˚ - 355˚ 230˚ - 250˚ 2. 200˚ - 220˚ 280˚ - 310˚ 140˚ - 175˚ 050˚ - 070˚ 3. A = 073˚ B = 131˚ C = 203˚ D = 296˚ E = 329˚ F = 197˚ G = 148˚ H = 063˚ I = 247˚ J = 029˚ K = 251˚ L = 283˚ M = 033˚ P = 112˚

A

A B

B A

A B

B

Before you begin National 5 Applications you must be sure that you can: 2. Solve a basic trigonometric equation to find the size of an angle Examples 1. Solve these equations (a) sinx = 0.365 (b) tanx = 4/7 (c) cosx = -0.921 Solutions (a) x = sin-1(0.365) (b) x = tan-1(4/7) (c) x = cos-1(-0.921) x = 21.4˚ x = 29.7° x = 157° Practice 1. Solve these equations using the 2nd Function or SHIFT key on your scientific calculator. You must show your working (eg. write down what it says on your calculator screen before you press = )

(a) sinx = 0.3420 (b) tanx = 0.4663 (c) cosx = 0.921 (d) sinx = 0.5736

(e) tanx = 0.8391 (f) cosx = 0.7071 (g) sinx = 0.776 (h) tanx = 1.428

(i) sinx = (2/3) (j) tanx = (4/5) (k) cosx = (6/7) (l) sinx = (8/9)

(m) tanx = (9/7) (n) cosx = (4/9) (o) sinx = (2/7) (p) tanx = (3/8)

2. Some of the trig equations you have to solve in National 5 involve several steps. Do you still remember how to solve equations in steps? Solve these equations, showing your steps.

(a) 2x + 8 = 28 (b) 3x + 12 = 33 (c) 4x – 9 = 31 (d) 5x – 17 = 43

(e) 2x + 11 = 24 (f) 4x + 16 = 54 (g) 10x – 28 = 6 (h) 5x – 27 = 31

(i) x2 = 49 (j) x2 = 24 (k) x2 + 8 = 89 (l) x2 – 11 = 40 (m) 2x2 = 50

If you found these difficult you should find some more equations to solve before you go on to the next sheet in this pack. You will not cope with National 5 if you can’t solve equations.

Use your calculator to find the sine of 25˚. Provided your calculator is in degrees mode you should get the answer 0.422618…

Try it now to make sure you do. What buttons could press to get back to the number 25? If you can’t remember you should restudy SOH – CAH – TOA before you continue this sheet. To solve a trig equation you need to ask your calculator “What angle has a sine of 0.365?” or “What angle has a tangent of 4/7?” To ask this question you need to use the 2nd Function or SHIFT key.

Answers: 1(a) 20° (b) 25° (c) 30° (d) 35° (e) 40° (f) 45° (g) 50° (h) 55° (i) 41.8° (j) 38.7° (k) 31.0° (l) 62.7° (m) 52.1° (n) 63.6° (o) 16.6° (p) 20.6° 2(a) 10 (b) 7 (c) 10 (d) 12 (e) 6.5 (f) 9.5 (g) 3.4 (h) 11.6 (i) 7 (j) 4.9 (k) 9 (l) 6.4 (m) 5

Before you begin National 5 Applications you must be sure that you can: 3. Read and Plot Co-ordinates on a 2-Dimensional Grid. Examples 1. Plot the following points on a 2D coordinate grid. A = (4, 3) B = (-2, 1) C = (0, -5) 2. Plot a fourth point, D, so that ABCD is a square. 3. Draw the diagonals and write down the coordinates of the point where they cross. 4. What are the coordinates of the point (a) 6 boxes up from A (b) 8 boxes left of B? Solutions: 1. 2. 3. 4(a) A = (4, 3). ‘6 boxes up’ means ‘increase the y-coordinate by 6’ = (4, 9) x-coordinate doesn’t change (b) B = (-2, 1). ‘8 boxes left’ means ‘decrease the x-coordinate by 8’ = (-10, 3) y-coordinate doesn’t change

A 2D co-ordinate grid has a horizontal x-axis with positive numbers on the right and negative numbers on the left. The numbers go beneath the axis, one number straddles each vertical line as shown above. The grid also has a vertical y-axis with positive numbers at the top and negative numbers at the bottom. The numbers go on the left of the axis, one number straddles each horizontal line as shown on the right. The x- and y-axes cross at the ORIGIN, which is marked with an O. Co-ordinates are written as two numbers

between brackets, separated by a comma. eg. (3, 5) or (-6, 0)

The first number is the x-coordinate: it tells you how far along the x-axis you need to go before you go up or down. The second number is the y-coordinate: it tells you how far up or down you need to go to get to the point.

If the x-coordinate is 0 the point lies on the y-axis. eg. (0, 4) If the y-coordinate is 0 the point lies on the x-axis. eg. (7, 0) Vertex is the word we use for the corner of a shape. The plural is vertices (ie. corners).

Remember: coordinates are written inside brackets with a comma between.

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

8

D

D is the point (6, -3)

to get from corner to corner go along 6 and up 2

the point (-2, 1) is 2 boxes left of the origin and 1 box up

the point (4, 3) is 4 boxes right of the origin and 3 boxes up

the point (0, -5) is no boxes along and 5 boxes down.

E is the point (2, -1)

x O

-6

-5 -4

-3

-2 -1

1 2

3

4 5 6

y

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

8

x

A

B

C

O

-6

-5 -4

-3

-2 -1

1 2

3

4 5 6

y

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

8

x E O

-6

-5 -4

-3

-2 -1

1 2

3

4 5 6

y

the point (3, 5) is 3 boxes right of the origin and 5 boxes up

the point (3, -5) is 3 boxes right of the origin and 5 boxes down.

8 x

-6

the point (-6, 0) is 6 boxes left of the origin and no boxes up

O -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-5

-4

-3

-2

-1

1

2

3

4

5

6

y

x 0 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

-2

0

-6

-5

-4

-3

-1

1

2

3

4

5

6

y

Practice 1. Write down the coordinates of every point on this grid. 2. Answer these questions using the point on this grid (a) Name two pairs of points with the same x-coordinates. (b) Name two pairs of points with the same y-coordinates. (c) Which 4 points could be joined to form a square? (d) Which 4 points could be joined to form a kite? (e) What is the difference between the x-coordinate of point H and the x co-ordinate of point K? (f) What is the difference between the y-coordinate of point D and the y co-ordinate of point J? For each of the following questions you will need to draw a grid like the one above, with both axes numbered from -10 to 10. Remember: the axes should cover over lines on the page, the numbers straddle the lines (they don’t go in the spaces) and every point is placed where two lines cross. 3. Plot these points on a coordinate grid, joining them together as you go. (0, 6) (3, 6) (4, 5) (6, 2) (7, -1) (8, -9) (0, -8) (-3, -7) (-6, -5) (-7, -4) (-7, -1) (-5, 1) (-2, 1) (-2, 4) (0, 6) 4(a) Plot the following points on a new coordinate grid. A = (6, 9) B = (-4, 5) C = (0, -5) (b) Plot a fourth point, D, so that ABCD is a square. Write down the coordinates of D. (c) Draw the diagonals and write down the coordinates of the point where they cross. (d) What are the coordinates of the point (i) 6 boxes up from A (ii) 8 boxes left of B? 5(a) Plot the following points on a new coordinate grid. A = (-7, 2) B = (-10, -8) C = (6, 0) (b) Plot a fourth point, D, so that ABCD is a parallelogram (with two pairs of parallel sides). (c) Imagine the parallelogram was moved 10 boxes down. Write down the new coordinates of all 4 vertices. (d) Imagine the parallelogram was moved 10 boxes right. Write down the new coordinates of all 4 vertices. 6(a) Rhombus JKLM has vertices J = (3, 2) K = (-1, -4) L = (-5, 2). Find the co-ordinates of M. (b) Find the co-ordinates of N, the point where the diagonals cross. (c) Imagine the rhombus was moved 5 boxes up and 3 boxes right. Write down the new coordinates of all 4 vertices. 7. The line joining A (0, -7) and C (6, -3) is a diagonal of a square. Find the coordinates of the other two vertices, B and D.

Answers: 1. A = (-7, 4) B = (-4, 0) C = (0, 2) D = (3, 5) E = (4, 3) F = (5, 0) G = (7, 1) H = (-6, -3) I = (-2, -4) J = (0, -6) K = (2, -2) L = (4, -5) M = (6, -2) 2(a) C&J, E&L (b) B&F, K&M (c) BCKI (d) EKLM (e) 8 (f) 11 3. Your picture should resemble a symmetrical heart-shape tilted to the left. 4(b) D = (10, -1) (c) (3, 2) (d)(i) (6, 15) (ii) (-12, 5) 5(b) D = (9, 9) 5(c) A = (-7, -8) B = (-10, -18) C = (6, -10) D = (9, -1) (d) A = (3, 2) B = (0, -8) C = (16, 0) D = (19, 9) 6(a) M = (-1, 8) (b) N = (-1, 2) (c) J = (6, 7) K = (2, 1) L = (-2, 7) M = (2, 13) 7. (1, -2) (5, -8)

O -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 x

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

y

D

A

E

C

G

F B

H

M

L

I

J

K

Before you begin National 5 Applications you must be sure that you can: 4. Calculate percentage increase and decrease. Example 1 A town has 4 300 residents. The population increases by 7%. What is the new population? Solution

7% of 4 300 = 100

7 x 4 300 = 301 (7 ÷ 100 x 4 300)

New population : 4 300 + 301 = 4 601 residents. Example 2 A camera costs £240. In a sale there is a 15% discount. What is the sale price of the camera? Solution

15% of £240 = 100

15 x 240 = £36 (15 ÷ 100 x 240)

Sale Price : £240 - £36 = £204.

To calculate a percentage of a quantity Divide the percentage by 100 then multiply by the quantity Percentage increase To find the percentage increase

Calculate the percentage of the quantity

Add this to the original value Percentage decrease To find the percentage decrease

Calculate the percentage of the quantity

Subtract this from original value

Practice

1. Calculate (a) 4% of £540 (b) 8% of £6 700 (c) 72

1% of 600g

(d) 18% of 950m (e) 24

1% of 12.8cm (f) 6.5% of 70kg

2. Last year a painting cost £2800. This year the painting increased in value by 12%. How much is the painting worth this year? 3. John earned £13 000 last year. This year he got a 4% pay rise. How much did he earn this year? 4. A rabbit colony has 220 rabbits. As a result of disease 45% die. How many rabbits are left after the disease? 5. A pilot flying his plane at 36 000 feet decreased its height by 35% to avoid a storm. What is the new height of the plane? 6. A petri dish contains 3 700 bacteria. These increase overnight by 27%. How many bacteria are there the next morning? 7. A clothes shop has a sale and reduces the price of its items by 20%. How much is the sale price of a jacket originally costing £85? 8. 20% of the water in a full 5 lire jug evaporates. How much is still in the jug? 9. A car is worth £8 000 in 2007. By 2008 the car had lost 25% of its value. (a) How much was the car worth in 2008? In 2009 the car lost a further 12% of what its value was in 2008. (b) How much was the car worth in 2009?

Answers: 1 (a) £21.60 (b) £536 (c) 45g (d) 171m (e) 0.288cm (f) 4.55kg 2. £3 136 3. £13 520 4. 121 rabbits 5. 23 400 feet 6. 4 699 bacteria 7. £68 8. 4 litres 9 (a) £6 000 (b) £5 280

Before you begin National 5 Applications you must be sure that you can: 5. Convert between mixed numbers and improper fractions. Example 1

Change 35

2 to an improper fraction.

Solution There are 5 fifths in one whole, so there are 3 x 5 = 15 fifths in 3 whole ones, giving a total of

15 + 2 = 17 fifths. So 35

2 =

5

17

Example 2

Change 4

11 to a mixed number.

Solution

4

11 = 11 ÷ 4 = 2 r 3 = 2

4

3

A mixed number has a whole number part and a fraction part e.g. 23

2

An improper fraction has the numerator(top) bigger than the denominator (bottom) e.g. 3

8

Converting Mixed numbers to improper fractions

We can write 23

2as an improper fraction to show how many thirds we have altogether.

Each whole shape has three pieces (thirds), giving a total of 6 thirds, plus two thirds making eight thirds.

So 23

2can be written as

3

8 .

Improper fractions to mixed numbers

Opposite we have 9 quarters 4

9

We need 4 quarters to make a whole one. So with 9 quarters we can make 2 whole ones with 1 left over.

So 4

9 = 2

4

1

Practice 1. Change the mixed numbers to improper fractions

(a) 23

1 (b) 1

4

3 (c) 3

7

5 (d) 2

9

2 (e) 4

6

1

(f) 110

7 (g) 5

8

3 (h) 7

5

2 (i) 2

11

9 (k) 8

2

1

2. Change the improper fractions to mixed numbers

(a) 4

9 (b)

3

10 (c)

2

7 (d)

5

13 (e)

7

20

(f) 8

9 (g)

4

21 (h)

3

22 (i)

9

17 (k)

6

31

Reminder : to simplify a fraction divide the numerator and the denominator by the same number (the highest common factor). 3. Simplify the following fractions

(a) 6

3 (b)

12

3 (c)

12

8 (d)

25

20 (e)

27

18

(f) 16

12 (g)

80

20 (h)

77

33 (i)

56

49 (k)

35

14

Answers:

1 (a) 3

7 (b)

4

7 (c)

7

26 (d)

9

20 (e)

6

25 (f)

10

17 (g)

8

43 (h)

5

37 (i)

11

31 (j)

2

17

2 (a) 24

1 (b) 3

3

1 (c) 3

2

1 (d) 2

5

3 (e) 2

7

6 (f) 1

8

1 (g) 5

4

1 (h) 7

3

1 (i) 1

9

8 (j) 5

6

1

3 (a) 2

1 (b)

4

1 (c)

3

2 (d)

5

4 (e)

3

2 (f)

4

3 (g)

4

1 (h)

7

3 (i)

8

7 (j)

5

2

Before you begin National 5 Applications you must be sure that you can: 6. Find the mean, median, mode and range from a set of data. Example 1 Find the mean, median, mode and range of the following numbers 13 10 9 11 12 15 21 8 13 13 Solution Put the numbers in order of size 8 9 10 11 12 13 13 13 15 21 Mean = 125 ÷ 10 = 12.5 (total of numbers ÷ how many numbers) Median = 12.5 (middle number : half-way between 12 and 13 (12 + 13) ÷ 2 ) Mode = 13 (most common number) Range = 21 – 8 = 13 (biggest number subtract lowest number) Example 2 Steve and Jack are playing snooker. They play eight games. Shown below are the number of points Steve scored in one game. 22 38 23 52 46 18 44 45 (a) Find the median. (b) Find the range (c) The median number of points Jack scored is 23 and the range is 41. Make two comments comparing the number of points scored by Steve and jack. Solution Put the numbers in order of size 18 22 23 38 44 45 46 52 (a) Median = 41 (half-way between 38 and 44 (38 + 44) ÷ 2 ) (b) Range = 52 - 18 = 34 (c) On average Steve scored more than Jack. (compare median (average) – Steve’s is bigger than Jack’s). Steve’s scores varied less than Jack’s. (compare range (spread) – Steve’s is smaller than Jack’s).

Mean, median and mode are three different types of average. The range is a measure of spread. Mean The mean is the total of the numbers divided by how many numbers there are. Median The median is the middle value (when the numbers are in order of size). If there are two middle values, the median is half-way between them.

Mode

The mode is the value that appears the most.

Range

The range is the difference between the biggest and smallest number.

The mean average

is not always a

whole number.

If there are two middle

values find the mean of

them.

Practice 1. Find the (i) mean (ii) median (iii) mode and (iv) range of the following (a) 3 1 2 16 7 5 1 (b) 11 14 12 26 18 12 12 (c) 6.6 6.8 7.0 6.6 6.6 6.9 7.0 6.9 7.0 6.6 (d) 12 000 15 000 17 000 18 000 15 000 (e) 24 36 18 34 44 18 34 24 24 8 2. The weights of 6 boys are shown 48kg 42kg 49kg 47kg 42kg 60kg (a) Find the range of their weights. (b) Calculate the mode and the median weights. 3. The price of a pint of milk in 6 local shops is 56p 45p 55p 50p 62p 42p (a) Calculate the median and the range of prices. In 6 supermarkets the median price of a pint of milk is 48p and the range of prices is 10p. (b) Make two comments comparing the price of milk in the local shops and the supermarkets. 4. Before training the number of sit-ups eight athletes could do in a minute were recorded 60 48 45 39 52 57 62 37 (a) Calculate the mean and range. After a training programme the mean number of sit-ups recorded was 58 and the range was 30. (b) Make two comments comparing the athletes before and after training.

Answers: 1 (a) (i) 5 (ii) 3 (iii) 1 (iv) 15 (b) (i) 15 (ii) 12 (iii) 12 (iv) 15 (c) (i) 6.8 (ii) 6.85 (iii) 6.6 (iv) 0.4 (d) (i) 15 400 (ii) 15 000 (iii) 15 000 (iv) 6 000 (e) (i) 26.4 (ii) 24 (iii) 24 (iv) 36 2 (a) 18kg (b) mode = 42kg median = 47.5kg 3 (a) Median = 52.5p and range = 20p. (b) On average milk is cheaper in the supermarket. The price of milk is more consistent in the supermarket. 4 (a) Mean = 50 and range = 25. (b) On average the athletes did more sit-ups after training. The athletes were less consistent after training.

Before you begin National 5 Applications you must be sure that you can: 7. Solve problems using Pythagoras Theorem Examples 1. Find the length of the line marked a in each shape below. Solutions 1. 2. 3. 4. Make the line the longest

side of a right-angled triangle. Count the boxes to discover the length of the other two sides. Use Pythagoras to find a.

Pythagoras Theorem is used to find the length of one side of a right-angled triangle when you already know the lengths of the other two.

If you want to find the length of the longest side you use this formula a2 = b2 + c2

If you want to find the length of either shorter side you use this formula a2 = b2 – c2

Provided you follow these steps you will get the right answer. - write down the correct Pythagoras formula. - write it again with the correct lengths from the question in place of the b and c. - evaluate the right-hand side by typing it into your calculator and pressing = . - press the √ button to get the answer. In more difficult questions you may have to calculate the length of the triangle before you can use Pythagoras. Even if the question gives you more than 2 numbers you should only use 2 lengths in your Pythagoras calculation.

a2 = b2 + c2

a2 = 92 + 72

a2 = 130

a = √130

a = 11.4cm

a2 = b2 - c2

a2 = 92 - 72

a2 = 32

a = √32

a = 5.66cm

b = 15 – 9 = 6

c = 12 – 7 = 5

a2 = b2 + c2

a2 = 62 + 52

a2 = 61

a = √61

a = 7.81cm

9 cm

7 cm a

9 cm

7 cm a

c°

b = 4, c = 7

a2 = b2 + c2

a2 = 42 + 72

a2 = 65

a = √65

a = 8.06 units

a

c°

c°

c°

Practice 1.Calculate the length of the missing side of each triangle. 2. Calculate the length of each lines. (a) (b) (c) 3. Calculate the perimeter of this trapezium.

4(a) A ladder stands 1m from the base of a wall. The ladder is 4m long. How far up the wall does the ladder reach? (b) if the wall is 5m tall how far is it from the top of the ladder to the top of the wall? Answer in centimeters. 5. Find the length of the shorter sides in this right-angled isosceles triangle whose longest side is 8.5m.

Answers: 1(a) 15m (b) 10.2m (c) 17.2m (d) 8.5m (e) 10.6m 2(a) 9.2units (b) 7.2units (c) 9.2units 3. height = 10, right slope = 12.5, perimeter = 130cm 4(a) 3.87m (b) 113cm 5. 6.01m

12m

9m

11m

15m 10m

14m

12m

16m

17m

19m (a)

(b)

(c)

(d)

(e)

26cm

24cm 30cm 7.5cm

8.5m

Bridging Pack for pupils who have passed National 4 Mathematics and want to do National 5.

Unit 2: Preparation for National 5 Expressions and Formulae Before you begin National 5 you need to be sure that you can: 1. Multiply out single brackets 2. Multiply out single brackets and simplify 3. Factorise by taking out the Highest Common Factor 4. Understand Indices and roots 5. Convert between and simplify different numerical fractions 6. Add, subtract, multiply and divide numerical fractions

Before you begin National 5 Exp&Formulae you must be sure that you can: 1. Multiply out single brackets Examples (a) )13(4 x (b) )52(6 x (c) (3 2)x x (d) 4 (3 2 6)x x y

Solutions

(a) xx 1234 (b) xx 1226 (c) 233 xxx (d) 24 3 12x x x 4 1 4 3056 2 2x x 4 2 8x y xy

(Be careful of double negative) 4 6 24x x

= 412 x = 3012 x = 23 2x x 212 8 24x xy x

Practice 1. Multiply out these brackets (a) 6( 2)a (b) 5( 3)x (c) 5( 8)s (d) 3(6 )v (e) 9(2 3)b

(f) 7( 1)x (g) 3( 2)m (h) 4(4 3)p (i) 5(2 3)a (j) 5(3 4 )w

(k) )23(6 ba (l) )4(5 x (m) )48(5 p (n) )6( vv (o) )3( ba

(p) )864(7 yx (q) )2(3 m (r) )34( 22 ppp (s) )3(5 a (t) )63( ww

To multiply out of a set of brackets you need to make sure EVERYTHING in the bracket is multiplied by the term in front of the brackets.

To set up the calculation we multiply everything inside the bracket by the term outside:

2 1

4 ( a + 8 )

To answer we follow the rule shown by the arrow:

4 x a = 4a 4 x 8 = 32

Read down the calculation to find the end result of:

4a + 32 Even if there are 3 or more terms inside the brackets, everything inside the brackets is multiplied by the term in front.

Before you begin National 5 Exp&Formulae you must be sure that you can: 2. Multiply out single brackets and simplify Examples (a) 15)3(6 t (b) )2(23 xww

Solutions (a) 15186 t (collect like terms) (b) xww 423 = 36 t = xw 45 Practice 1. Multiply out the brackets and simplify (a) 1)4(3 z (b) 4)2(7 q (c) 12)3(8 b (d) aa 3)7(4

(e) 5( 4) 8x x (f) 6(3 2) 3y y (g) 8(5 2 ) 20a a (h) 4(5 6) 13b

(i) )2(77 pp (j) )2(4 xx (k) 20 3( 5)s (l) )8(310 r

2. For each of the following multiply out both sets of brackets and collect like terms. (a) 2( 2) 3( 1)a a (b) 4(2 3) 7( 6)m m (c) 5(3 1) 4(2 3)p p

(d) )1(4)2(3 aa (e) )6(7)2(5 mm (f) )32(4)13(2 pp

(g) ( 2) (2 3)x x x x (h) (3 4) (5 6)x x x x (i) )22()2( xxxx

The same format applies as before, make sure EVERYTHING in the bracket is multiplied by the term in front of the brackets. Anything outside the brackets DOES NOT get multiplied. To simplify the expression collect the like terms.

3 (z + 4 ) + z

Term before bracket Bracket

Carry out the same process as before. 2

1 3 ( z + 4 ) + z

Complete the process by multiplying out the bracket and simplifying.

3z + 12 + z = (3z + z) + 12

= 4z + 12

Before you begin National 5 Exp&Formulae you must be sure that you can: 3. Factorise by taking out the Highest Common Factor Examples

1(a) 497 x (b) 4515 c (c) 40248 cc (d) mnm12 (e) 28 10x x Solutions 1(a) 7( ) (b) 15( ) (c) 8( ) (d) m( ) (e) 2x( ) )7(7 x )3(15 c )53(8 cc )12( nm 2x( 4x - 5 )

Practice 1. Factorise each of the following: (a) 124 x (b) 366 a (c) 328 p (d) 4411 w

(e) 77 f (f) 3010 b (g) 549 m (h) 4515 a

2(a) ba 153 (b) 48 x (c) 248 g (d) dc 2550

(e) 1218 w (f) xzxy (g) mnm9 (h) qpq 12

(i) 24 12x x (j) 210 15y y (k) 212 21x x (l) 218 24p p

3. This time try with 3 terms. Factorise each of the following: (a) rqp 333 (b) tsr 1263 (c) cba 8124 (d) bbcab

(e) yxw 203540 (f) dcb 493514 (g) ptpspr (h) bebdbc 1282

Previously, you learned how to multiply out brackets, this time we are going to do the reverse. We are going to FACTORISE the expression. Factorise 12 36x Steps to follow:

1) Find the highest number which can go into 12x and 36 (it is 12) 2) Write this number down followed by brackets 12( ) 3) Now divide each term inside the bracket by 12 12 12x x

36 12 3 4) Read down the answers to step 3 to fill in the brackets 12( 3)x Final Answer

Answers to Relationships Sheets 1 – 4

Answers: 3. Factorise by taking out the Highest Common Factor

1 (a) )3(4 x (b) )6(6 a (c) )4(8 p (d) )4(11 w

(e) )1(7 f (f) )3(10 b (g) )6(9 m (h) )3(15 a

2 (a) )5(3 ba (b) )12(4 x (c) )3(8 g (d) )2(25 dc

(e) )23(6 w (f) )( zyx (g) )9( nm (h) )12( pq (i) 4 ( 3)x x (j) 5 (2 3)y y (k) 3 (4 7)x x (l) 6 (3 4)p p

3 (a) )(3 rqp (b) )42(3 tsr (c) )23(4 cba (d) )1( bab

(e) )478(5 yxw (f) )752(7 dcb (g) )( tsrp (h) )34(2 edcb

Answers: 1. Multiply out single brackets

(a) 6 12a (b) 5 15x (c) 5 40s (d) 18 3v (e) 18 27b

(f) 7 7x (g) 3 6m (h) 16 12p (i) 10 15a (j) 15 20w

(k) 18 12a b (l) 5 20x (m) 40 20p (n) 26v v (o) 3ab a

(p) 28 42 56x y (q) 3 6m (r) 4 24 3p p (s) 5 15a (t) 23 6w w

Answers: 2. Multiply out single brackets and simplify 1(a) 3 12z (b) 7 18q (c) 8 12b (d) 28a

(e) 13 20x (f) 15 12y (g) 40 + 4a (h) 20 37b

(i) 14 14p (j) 5 2x (k) 5 – 3s (l) 34 3r

2(a) 5 7a (b) 15 54m (c) 23 17p (d) 7 2a

(e) 12 32m (f) 2 14p (g) 23 5x x (h) 28 10x x (i) 2x

Answers: 4. Understand Indices and roots 1 a) 49 (b) 64 (c) 121 (d) 81 (e) 100 (f) 144 (g) 169 2 a) 343 (b) 512 (c) 1331 (d) 729 (e) 1000 (f) 1728 (g) 2197 3 a) 2401 (b) 4096 (c) 14641 (d) 6561 (e) 10000 (f) 20736 (g) 28561 4 a) 16807 (b) 32768 (c)161051 (d) 59049 (e) 1000000 (f) 2985984 (g) 4826809 5 a) 1 (b) 8 (c)11 (d) 9 (e) 0 (f) 0 (g) 0 6(a) 7 (b) 12 (c) 20 (d) 4.90 (e) 4.12 (f) 10.58 (g) 0.5 (h) 3.1 (i) 1 7(a) 3 (b) 5 (c) 9 (d) 1 (e) 2.57 (f) 4.82 (g) 0.5 (h) 1.5 8(a) 3 (b) 10 (c) 5 (d) 2 (e) 7 (f) 16 (g) 2 (h) 1

Before you begin National 5 Exp&Formulae you must be sure that you can: 4. Understand Indices and roots Examples Evaluate the following, giving your answer as a whole number (a) 92 (b) 33 (c) 54 (d) 45 (e) 76 (b) 121 (c) 500 Solutions (a) 9×9 = 81 (b) 3×3×3 = 27 (c) 5×5×5×5 = 625 (d) 4×4×4×4×4 = 1024 (e) 12 (f) 1

because anything to the power of 0 equals 1

Definition of Index: What does it mean? The index of a number says how many times to use the number in a multiplication. It is written as a small number to the right and above the base number. The plural of index is indices. Other names for index are exponent or power. Example : The 2 is the Index or power or exponent

72

7 is the base number This can be read “seven squared” or “seven to the power of two”. In 72 the "2 - index" says to use 7 twice in a multiplication, i.e. 72 = 7 7 = 49 Other examples:

The “3 – index” says to use 7 three times, i.e. 73 = 7 7 7 = 343 This can be read “seven cubed” or “seven to the power of three”.

The “4 – index says to use 7 four times, i.e. 74 = 7 7 7 7 = 2401 This is read “seven to the power of four”.

What if the Index is 1, or 0?

1 If the index is 1, then you just have the number itself (example 91 = 9)

0 If the index is 0, then you get 1 (example 90 = 1)

But what about 00 ? It could be either 1 or 0, and so we say it is "indeterminate".

73

74

Practice

1 (a) 72 (b) 82 (c) 112 (d) 92 (e) 102 (f) 122 (g) 132

2(a) 73 (b) 83 (c) 113 (d) 93 (e) 103 (f) 123 (g) 133

3(a) 74 (b) 84 (c) 114 (d) 94 (e) 104 (f) 124 (g) 134

4(a) 75 (b) 85 (c) 115 (d) 95 (e) 106 (f) 126 (g) 136

5(a) 11 (b) 81 (c) 111 (d) 91 (e) 100 (f) 120 (g) 130

Roots (AKA Radicals)

The opposite of squaring a number is finding the square-root. You should have memorised the first 10 square numbers. 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 62 = 36 72 = 49 82 = 64 92 = 81 102 = 100 Working in reverse from above gives: √1 = 1 √4 = 2 √9 = 3 √16 = 4 √25 = 5 √36 = 6 √47 = 7 √64 = 8 √81 = 9 √100 = 10 Only square numbers have a whole number square-root. The square-root of other numbers is called a SURD which can be approximated as a decimal using this button on your calculator. √ e.g. √10 = 3.1622776 √19 = 4.3588989 √95 = 9.7467943

------------------------------------------------------------------------------------------- The opposite of cubing a number is finding the cube-root. Can you find the cubed-root button on your calculator? You should memorise the first 10 cubed numbers. 13 = 1 23 = 8 33 = 27 43 = 64 53 = 125 63 = 1296 73 = 343 83 = 512 93 = 729 103 = 1000 Working in reverse from above gives: 3√1 = 1 3√8 = 2 3√27 = 3 3√64 = 4 3√125 = 5 3√1296 = 6 3√343 = 7 3√512 = 8 3√729 = 9

Only cube numbers have a whole number cube-root. The cube-root of other numbers is called a SURD which can be approximated as a decimal using this button on your calculator. 3√ e.g. 3√10 = 2.15443469 3√39 = 3.391211443 3√95 = 4.562902635

Practice

6. Use the square-root button on your calculator to evaluate the following: (round surds to 2 decimal places)

(a) √49 (b) √144 (c) √400 (d) √24 (e) √17 (f) √112 (g) √0.25 (h) √9.61 (i) √1

7. Use the cube-root button on your calculator to evaluate the following:

(a) 3√27 (b) 3√125 (c) 3√729 (d) 3√1 (e) 3√17 (f) 3√112 (g) 3√0.125 (h) 3√3.375

8. Use the other root button on your calculator to evaluate the following: (If you don’t have that button

you will just need to work out “What number, if I multiply it by itself that many times, will make the number in the question?”)

(a) 4√81 (b) 5√100000 (c) 6√15625 (d) 7√128 (e) 3√343 (f) √256 (g) 4√16 (h) 30√1

Bridging Pack for pupils who have passed National 4 Mathematics and want to do National 5. Unit 3: Preparation for National 5 Relationships Before you begin National 5 you need to be sure that you can: Draw straight lines on a 2-Dimensional Grid using the gradient (m) Solve linear equations Substitute numbers into expressions Change the subject of the formula Use Trigonometric formulae to find missing sides and angles in Right-angled Triangles. Identify Similar Figures and use Scale Factors to calculate missing lengths. Use Pythagoras’ Theorem to find a missing side of a triangle Use Pythagoras’ Theorem to find the distance between two coordinates

Before you begin National 5 Relationships you must be sure that you can: Draw straight lines on a 2-Dimensional Grid using the gradient (m) Examples B 1. Find the gradient of the line AB A 24 2. Find the gradient of AB, where A = (3, 2) and B = (11, 6) 3. Draw the line y = 3x 4. Draw the line y = 2x + 1 Solutions

1. Gradient = = = 2. m = = = =

3. y = 3x x -2 -1 0 1 2 y = 3x 3(-2) 3(-1) 3(0) 3(1) 3(2) = -6 -3 0 3 6 (-2, -6) (-1, -3) (0, 0) (1, 3) (2, 6) y = 3x

8

The gradient of a line is a number that measures the slope of the line. Gradient = distance up or down distance along m = 1 Positive gradients are ‘uphill’, from left to right. Negative gradients are ‘downhill’, from left to right

m = -

Gradient can also be described as vertical horizontal B (x2, y2)

Using points, the gradient can be written as m =

A (x1, y1) To draw a straight line on the grid, make a table of values to give you the coordinates to plot. x -2 -1 0 1 2 y Multiply the x co-ordinate by the gradient in the equation and add on or subtract the second number

4. y = 2x + 1 x -2 -1 0 1 2 y 2(-2)+1 2(-1)+1 2(0)+1 2(1)+1 2(2)+1 = -3 -1 1 3 5 (-2, -3) (-1, -1) (0, 1) (1, 3) (2, 5) y = 2x + 1

Practice 1. Calculate the gradient of the line AB for: a) A(5, 8) and B(2, 2) b) A(2, 6) and B(1, 9) c) A(5 , –1) and B(1, –3) d) A(–4, 0) and B(5, –6) e) A(–2, –8) and B(–7, –3) f) A(0, –4) and B(–8, 2) 2. For the following diagrams, work out the gradient of each line a) b) c) 3. Draw lines for the following: a) y = 3x + 2 b) y = 3x – 1 c) y = –x + 2

d) y = –x – 3 e) y = x + 1 f) y = x + 2

Answers:

1. a) 2 b) – 3 c) d) e) –1 f)

2. a) m = b) m = =

c) m = =

3. a)Line goes through (0,2) and (3, 11) b) Line goes through (0, -1) and (3 , 8) c) Line goes through (0, 2) and (3 , -1) d) Line goes through (0, -3) and (3 , -6) e) Line goes through (0, 1) and (4 , 3)

f) Line goes through (0, 2) and (6 , 4)

Before you begin National 5 Relationships you must be sure that you can: Solve linear equations Examples

1.Solve 2x = 37 2. Solve = 15 3. Find the value of x in 4. Solve 4(x – 3) = 60

3x + 4 = 52 Solutions

1. 2x = 37 2. = 15 3. 3x + 4 = 52

divide both sides by 2 multiply both sides by 3 subtract 4 from both sides x = 18.5 x = 45 3x = 48 divide both sides by 3 4. Solve 4(x – 3) = 60 x = 16 multiply out brackets 4x – 12 = 60 add 12 to both sides 4x = 72 divide both sides by 4 x = 18 Practice 1. Solve the following equations a) 5x = 40 b) 4y = 112 c) 7z = 161 d) 8w = 4 2. Find the value of x in each of the following equations a) 2x + 1 = 47 b) 3x + 6 = 108 c) 8x – 4 = 132 d) 10x – 5 = 135 3. Solve the following equations a) 4(x – 1) = 36 b) 3(2 + y) = 105 c) 7(z + 9) = 119 d) 2(2w – 1) = 34 4. Find the value of x in each of these equations

a) = 24 b) = 72 c) = 18 d) = 18

e) + 1 = 17 f) + 2 = 22 g) 3(x + 7) – 8 = 34 h) (x – 8) = 23

When solving an equation, the aim is to get the letter on its own e.g. x = ….. or y = …… Do this by always doing the same operation to both sides of the equation – it is the reverse operation of the operation given In general, follow these steps:-

1. remove any fractions by multiplying both sides by a suitable number 2. multiply out any brackets 3. simplify each side by collecting like terms 4. get the letter on its own by adding/subtracting extra terms 5. get one of the letter by dividing

Answers: 1. a) x = 8 b) y = 28 c) z = 23 d) w = 0.5 2. a) x = 23 b) x = 34 c) x = 17 d) x = 14 3.a) x = 10 b) y = 33 c) z = 8 d) w = 9 4.a) x = 48 b) x = 360 c) x = 162 d) x = 27 e) x = 32 f) x = 50 g) v = 7 h) x = 54

Before you begin National 5 Relationships you must be sure that you can: Substitute numbers into expressions Examples 1.Evaluate x + 2y, when x = 4 and y = 2 2. Evaluate x2 – 5y, when x = 7 and y = –2 3. Find 2y3 – 3x2 + 8z when x = 5, y = 4 and z = –3 4. Find the value of (x – 4)(x + 2) when x = 3 Solutions 1. x + 2y 2. x2 – 5y 3. 2y3 – 3x2 + xz 4. (x – 4)(x + 2) =4 + 2(2) = (7)2 – 5(–2) = 2(4)3 – 3(5)2 + (5)(-3) = (3 – 4)(3 + 2) =4 + 4 = 49 + 10 =2(64) – 3(25) – 15 = (– 1)(5) = 8 = 59 = 128 – 75 – 15 = – 5 = 38 Practice 1. Find the value of the following expressions, when a = 2 and b = 5 a) 2a + b b) 3a + 2b c) a + b2 d) (b – a)2 2. Evaluate these expressions, when x = 5, y = – 3 and z = – 1 a) y2 – xz b) xyz c) x2y + yz3 d) 5z – 4x + 3y 3. Calculate the values when s = 12, t = 13, v = – 5 a) 3(s – 1) b) 4(2 + t) c) (v + 9)(v – 7) d) 2(s + t)(s – t) 4. Find the value of the following, when x = 25

a) b) c) d)

When substituting values into an expression, the aim is to replace the letters with numbers and then work out the final value. In general, follow these steps:-

1. Write out the expression 2. Replace each letter with the appropriate value 3. Calculate the value of the expression, remembering to use BIDMAS where necessary

Answers: 1. a) 9 b) 16 c) 27 d) 9 2. a) 14 b) 15 c) – 72 d) – 34 3. a) 33 b) 60 c) – 48 d) – 50 4.a) 12.5 b) 7.5 c) 9 d) 312.5

Before you begin National 5 Relationships you must be sure that you can: Change the subject of the formula Examples 1. Change the subject to x 2. Change the subject to x 3. Change the subject to x

x + a = b = b 5x = b

4. Change the subject to x 2x + 3 = c Solutions

1. x + a = b 2. = b 3. 5x = b 4. 2x + 3 = c

– a – a x a x a ÷ 5 ÷ 5 – 3 – 3

x = b – a x = ab x = 2x = c – 3

÷ 2 ÷ 2 x = c – 3 2 Practice 1. Change the subject to x in each of the following equations a) x + y = z b) x – y = t c) x + 8 = k d) x – t = 4 e) 2 – x = y 2. Change the subject to p in each of the following equations

a) 12p = Q b) = Q c) = 7 d) = t

3. Change the subject to x a) 4x – 1 = y b) 3(2 + x) = yz c) 7(x – 1) = 3y + 2 d) 2(2x – 1) = x + y + 4z 4. Change the subject to x

a) = 3y + 1 b) + 3 = y c) – 4y = 10 d) + y = 5z

When changing the subject of a formula, the aim is to get a different letter on its own

e.g. y = mx + c can be rewritten as c = y – mx (c is the subject) OR x = (x is the subject)

To do this always reverse the operation of the one given in the question In general, follow these steps:-

4. remove any fractions by multiplying both sides by a suitable number 5. multiply out any brackets 6. simplify each side by collecting like terms 7. get the letter on its own by adding/subtracting extra terms 8. get one of the letter by dividing

Answers: 1. a) x = z – y b) x = t + y c) x = k – 8 d) x = 4 + t e) x = 2 – y 2. a) p =

b) p = 5Q c) p = d) p = 3.a) x = b) x = c) x = d) x =

4.a) x = 2(3y + 1) b) x = 5(y – 3) c) x = 3(y + 10) d) x =

Before you begin National 5 Relationships you must be sure that you can: Use Trigonometric formulae to find missing sides and angles in Right-angled Triangles. Examples Find the missing Step 1 Sketch and label sides Opp, hyp then adj Dimensions below Step 2 SOH CAH TOA Step 3 Tick what you want to find and what you know Step 4 Select ratio and solve Solutions: 1. 2.

Trigonometry is a huge branch of Mathematics and one of the most important we will study at National 5 level. We already know from Pythagoras’ Theorem how the sides on a right-angled triangle are connected. Trigonometry looks at the connection between angles and the three sides. We know how to find the hypotenuse on a right-angled triangle but the other sides also have special names related to a given angle.

x

Opposite

Hypotenuse

Adjacent

Imagine this was a ramp for jumping over on a bike. If we were to keep angle xº the same but make the length (adjacent) longer, then the height of the ramp (opposite) would increase. Trigonometry will help us calculate how the angle and sides are linked. The three ratio’s

Adj

Oppx

Hyp

Adjx

Hyp

Oppx tancossin

7cm

11cm

y

Opp

Adj

Hyp

5.50

11

7cos

11

7cos

cos

TOA CAH SOH

1

y

y

y

hyp

adjx

8cm

x

65

Opp

Adj

Hyp

cm3.7

65sin8

865sin

sin

TOA CAH SOH

x

x

x

hyp

oppx

3. 4. Practice

1. Find the missing dimensions. a) b) c) d)

2. At 57” from the base of a building you need to look up at 55° to see the top of a building. What is the height of the building?

3. If the angle of elevation to the top of a tree is 20°. How far are you away from the tree, if the tree's height is 31 feet?

Answers: 1. a) 66.4⁰ b) 8.1 c) 13.8 d) 50⁰ 2. 81.4” 3. 85.2 feet

57

w

7m Adj

Opp

Hyp

mw

w

w

adj

oppx

8.10

57tan7

757tan

tan

TOA CAH SOH

m

17m 17m

8m

4.76

17

4cos

17

4cos

cos

TOA CAH SOH

1

y

y

m

hyp

adjx

Before you begin National 5 Relationships you must be sure that you can: Identify Similar Figures and use Scale Factors to calculate missing lengths. Examples

1. Find the value of x in the following pair of triangles.

Similar figures have the same shape (but not necessarily the same size) and the following properties:

Corresponding sides are proportional. That is, the ratios of the corresponding sides are equal.

Corresponding angles are equal. For example, consider the following squares.

Thus the squares are similar figures as their corresponding sides are proportional and their corresponding angles are equal.

2. Find the unknown lengths for these similar triangles. n m 7.5cm 15cm 17cm 8cm

Solutions

1.

2.

Practice

1. Calculate the value of a and b in triangle S.

2. Calculate the value of x in each pair of triangles. a) b)

c) d)

e) f)

Answers: 1. a = 4.8, b = 5.6 2. a) 4.5cm b) 12cm c) 8cm d) 8cm e) 4cm f) 12cm

Before you begin National 5 Relationships you must be sure that you can: Use Pythagoras’ Theorem to find a missing side of a triangle Examples 1. Find x a) b)

3 x x 15

4 8 Solutions

1.a) x2 2 23 4 b) x2 2 215 8

x2 25 x2 161

x 25 x 161

x 5 x 12 7 Practice 1. Calculate the length of the missing side of each triangle. a) b) c) d) 2. A ladder is 4 m long. It rests against a wall so that the foot of the ladder is 1.2 m away from the base of the wall. How high up the wall does the ladder reach? 3. A boat sails north for 450m and then east for 300m. How far will it have to sail to go straight back to its starting point?

Pythagoras came up with a simple rule that connects the three sides of a right angled triangle. The longest side of the right angled triangle is called the HYPOTENUSE. In this triangle Pythagoras’ Theorem says

2 2 2c a b

Hint: To find the hypotenuse square and add the other two sides. To find a shorter side square and subtract the other two sides

Answers: 1. a) 12m b) 26.9m c) 7.5m d) 23.3m 2. 3.8m 3. 540.8m

26m 9m

7m

15m

10m 12.5m

12m 20m

a cm c cm

b cm

Before you begin National 5 Relationships you must be sure that you can: Use Pythagoras’ Theorem to find the distance between two coordinates Example Calculate the distance between the points A(2, 6) and B( 6, 3).

Practice Calculate the length of the line joining the points (i) A(-1, 4) and B( 3, 2)

(ii) C(-2, 1) and D(4, 5) (iii) E(-3, 8) and F(4, 3)

To calculate the distance between two coordinates we need to

plot the points on a coordinate diagram and join them up.

draw in a horizontal line and a vertical line to form a right angled triangle.

count the squares to find the lengths of the sides you have just drawn.

use Pythagoras’ Theorem to calculate the line joining the points.

AB2 = 32 + 42

AB2 = 25

AB = 25

AB = 5 units

Answers: ( i) 4.5 units (ii) 7.2 units (iii) 8.6 units