maths gcse paper 1 revision

Yr 11 Non-

Calculator Revision

Multiplication

• £1.82 x 15 =

• 3.28 x 13 =

• 4.56 x 1.8=Remove decimalsMultiple – box methodPut decimals back in

1

34 x 6430 4

60

4

x

1800 240

120 16

1800240120

162176

+

1

237 x 45200 30

40

5

x

8000 1200

1000

150

8000

1200

280

1000

150

35

10665

+

7

280

35

Given:

527 x 73 = 38471

1

What is: 52.7 x 73 = 3847.1

Given:

527 x 73 = 38471

2

What is: 52.7 x 7.3= 384.71

Given:

527 x 73 = 38471

3

What is: 5.27 x 7.3= 38.471

Given:

432 x 29 = 12528

1

What is:

12528 ÷ 2.9 =4320

Given:

432 x 29 = 12528

1

What is:

12528 ÷ 0.29=43200

Standard form• Write as an ordinary number

• 8.5 x 10 4

• 6.7 × 10–3

• Write in standard form

• 867000

• 0.00045

85000

0.0067

8.67 x 105

4.5 x 10 - 4

0.000512

Standard form

5.12 x 10-4

Normal Number

16

0.0000956

Standard form

9.56 x 10-5

Normal Number

17

0.73

Standard form

7.3 x 10-1

Normal Number

18

0.00103

Standard form

1.03 x 10-3

Normal Number

19

0.000423

Standard form

4.23 x 10-4

Normal Number

20

78000 Normal Number

7.8 x 104

Standard form

21

Standard form

• Calculate1. (8 x 104) x (4 x 105)

2. (8 x 104) ÷ (4 x 105)

8 x 4 104 x 105 x = 32 109x =

3.2 1010x =

(8 ÷ 4) (104 ÷ 105) x = 2 104-5x = 2 10-1x

3 x 102

1

2.5 x 104 X

3 x 2.5 102 x 104 x = 7.5 102+4x = 7.5 106x

3 x 107

1

5 x 109 X

3 x 5 107 x 109 x = 15 107+9x = 1.5x10 1016x = 1.5 1017x

4 x 106

1

6 x 1012 X

4 x 6 106 x 1012

x = 24 106+12x = 2.4x10 1018x = 2.4 1019x

6 x 108

1

2 x 105 ÷

(6 ÷ 2) (108 ÷ 105)

x = 3 108-5x = 3 103x

8 x 1012

1

4 x 103 ÷

(8 ÷ 4) (1012 ÷

103)x

= 2 108-5x = 2 103x

9 x 1015

1

2 x 104 ÷

(9 ÷ 2) (1015 ÷

104)x

= 4.5 1015-4x = 4.5 1011x

nth term

• 3 6 9 12.........

• 1 7 13 19........

• 5 12 19 26........

3n

6n - 5

7n - 2

3 5 7 9

nth term =

2n + 1

1

4 10 16 22

nth term =

6n - 2

2

2 5 8 11

nth term =

3n - 1

3

3 8 13 18

nth term =

5n - 2

4

Is 88 in the sequence

nth term =

2n + 1

1

3 5 7 9

2n + 1 = 882n = 88 - 1 2n = 87 n = 87 ÷ 2 n = 43.5

Because n is NOT a whole number 88 cannot be in the sequence

Is 75 in the sequence

nth term =

6n - 2

1

4 10 16 22

6n - 2 = 756n = 75 + 2 6n = 77 n = 77 ÷ 6 n = 12.....

Because n is NOT a whole number 75 cannot be in the sequence

Is 54 in the sequence

nth term =

7n - 4

1

3 10 17 24

7n - 4 = 547n = 54 + 4 7n = 58 n = 58 ÷ 7 n = 8. .....

Because n is NOT a whole number 54 cannot be in the sequence

Here is part of Gary's billWork out how much Gary has to pay for the units of electricity used

Units used

7155- 7095 60

Cost

60 x 15 = 900pence

= £9

Gas BillNew Reading 2060 units

Old Reading 1729 units

Price per unit 12p

Work out how much is paid for the gas used?

1

331x12 =3972p =£39.72

Electricity BillNew Reading 9488 units

Old Reading 9028 units

Price per unit 6p

Work out how much is paid for the electricity used?

2

460x6 = 2760p =£27.60

Electricity BillNew Reading 16155 units

Old Reading 16093 units

Price per unit 10p

Work out how much is paid for the electricity used?

3

62x10 = 620p = £6.20

Linear graphsCalculate the values form x = -2 to x = 2

• y = ½ x + 3

• y = 2 – 3x

x -2 -1 0 1 2y 2 2.5 3 3.5 3

x -2 -1 0 1 2y 8 5 2 -1 -4

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

1

2

3

45

6

7

-7

-6

-5

-4

-3

-2

-10

3y = ½x+ 1

x -3 -2 -1 0 1 2 3 4

y -0.5 0 0.5 1 1.5 2 2.5 3

xx

xx

xx

x

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

1

2

3

45

6

7

-7

-6

-5

-4

-3

-2

-10

4y = 5 - 2x

x -3 -2 -1 0 1 2 3 4

y 11 9 7 5 3 1 -1 -3

x

x

x

x

x

Ed has 4 cards.There is a number on each card.

  3, 7, x, 6 The mean of the 4 numbers on Ed’s cards is

5. Work out the number on the 4th card.

• The 4 Cards add to 5 x 4 = 20• We know 3 cards 3 + 7 + 6 = 16

• This means the 4th card = 20 -16 = 4

MeanA basket ball team plays 10 games and hasa mean score of 20 points. After theeleventh game the mean is 21. What wasthe score in the final game?

After 10 games total points = 10 x 20 = 200After 11 games total points = 11 x 21 = 231

So the score in the final game = 231 – 200 = 31 points

Inequalities

x is an integerState the possible values

1. -1 ≤ x < 3

2. 4 < x ≤ 7

3. -2 < x < 4

-1, 0 , 1, 2

5, 6, 7

-1, 0, 1, 2, 3

4n1 <<

2

n is an integer

List the possible values of n

2, 3

6n2 <≤

3

n is an integer

List the possible values of n

2, 3, 4, 5

9n5 ≤<

4

n is an integer

List the possible values of n

6, 7, 8, 9

4n1- ≤<

5

n is an integer

List the possible values of n

0, 1, 2, 3, 4

Solve the inequality

3x < 9 x < 3

4y > 12y > 3

2x + 2 < + 122x < 10X< 5

3y + 6 < 153y < 9Y < 3

1. 3x + 2 < 11

2. 4y – 2 > 10

3. 6x + 2 < 4x + 12

4. 3(y + 2) < 15

-3 < x ≤ 4

1

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

o

7n<

2

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

o

3n>

3

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

7n≤

4

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

3n≥

5

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

8n3 ≤<

o

6

Write down the inequality shown by:

0 1 2 3 4 5 6 7 8 9 10

8n3 <≤

o

Mr Mason asks 240 Year 11 students what theywant to do next year. 

15% of the students want to go to college. ¾ of the students want to stay at school.The rest of the students do not know. 

Work out the number of students who do not know.

10% of 240 = 24

5% of 240 = 15

So, 15% of 240 = 24 + 15 = 36

¾ of 240 = 240 ÷ 4 x 3 = 60 x 3

= 180

36 + 180 = 216Students who do not know= 240 – 216

= 24

6w – 10 > 20

w >

1

5

2x + 7 < 15

x <

2

4

3y + 14 > 20

y >

3

2

7a – 8 > 20

a >

4

4

5m – 9 > 31

m >

5

8

6w – 10 > 4w + 20

w >

1

15

7y – 8 < 3y + 20

y <

2

7

12x - 8 > 8x + 12

x >

3

5

ProbabilityThere are only red counters, blue counters, white counters and black counters in a bag. The table shows the probability that a counter taken at random from the bag will be red or blue.

The number of white counters in the bag is the same as the number of black

counters in the bag. Calculate the probability of getting a white or black counter?

There are 240 counters in the bag how many are white

Colour Red Blue White Black

Probability

0.2 0.5

0.2 + 0.5 = 0.7 1- 0.7 = 0.3 0.3 ÷ 2 = 0.15

Number of White counters = 0.15 x 240= 36

3

The Probability the spinner lands on 4

= 0.35

Number 1 2 3 4

Probability 0.1 0.2

The table shows the probabilities that the spinner will land on 1 or on 3

The probability that the spinner will land on 2 is the SAME as the probability that the spinner will land on 4

4

The Probability of picking a Blue = 0.25

Colour Red Green Blue Black

Probability 0.1 0.4

The table shows the probabilities that a counter picked will Red or Green

The probability of picking a Blue or Black are the SAME

5

The Probability the spinner lands on 4

= 0.1

Number 1 2 3 4

Probability 0.6 0.2

The table shows the probabilities that the spinner will land on 1 or on 3

The probability that the spinner will land on 2 is the SAME as the probability that the spinner will land on 4

1

200 x 0.3 = 60

Number 1 2 3 4

Probability 0.2 0.4 0.3 0.1

John is going to spin the spinner 200 times.Work out an estimate for the number of times the spinner will land on 3

2

Colour Red Green Blue Black

Probability 0.2 0.4 0.3 0.1

There are 500 counters in the bag.

Work out an estimate for the number of Green counters.

500 x 0.4 = 200

3

300 x 0.2 = 60

Number 1 2 3 4

Probability 0.2 0.4 0.3 0.1

John is going to spin the spinner 300 times.Work out an estimate for the number of times the spinner will land on 1

4

Colour Red Green Blue Black

Probability 0.2 0.4 0.3 0.1

There are 400 counters in the bag.

Work out an estimate for the number of Red counters.

400 x 0.2 = 80

NOTES : SCATTER GRAPHS

• POSITIVE CORRELATION• NEGATIVE CORRELATION• NO CORRELATION

•When drawing a line of best fit try to have an equal number of points above and below the line

(a) On the scatter graph, plot the information from the table.

An apartment is 2.8 km from the city centre.(c) Find an estimate for the monthly rent for this apartment.   £ .......................

Distance from the city centre (km)

2 3.1

Monthly rent (£) 250 190

(b) Describe the relationship between the distance from the city centre and the monthly rent. 

................................................................Negative Correlation

xx

220

(a) On the scatter graph, plot the information from the table.

An apartment is 2.8 km from the city centre.(c) Find an estimate for the monthly rent for this apartment.  

£ .......................

Distance from the city centre (km)

2 3.1

Monthly rent (£) 250 190

(b) Describe the relationship between the distance from the city centre and the monthly rent. 

................................................................

6 0

5 0

4 0

3 0

2 0

1 0

00 1 0 2 0 3 0 4 0 5 0 6 0

M ark in m a th em atic s te s t

M arkinsc ien cete s t

Mark in mathematics test

14 25 50 58

Mark in science test 21 23 38 51(a) On the scatter graph, plot the information from the table.

Josef was absent for the mathematics test but his mark in the science test was 45(c) Estimate Josef’s mark in the mathematics test..  

.......................

(b) Describe the correlation between the marks in the mathematics test and the marks in the science test. 

................................................................

Positive Correlation Negative Correlation

x x

x

xx

x

50 marks 220

A box of grass seed covers 20m2

How many boxes are needed to cover the lawn

Area = 9 x 12 = 108 m2

Area = 9 x 6 = 54 ÷ 2 = 27 m2

Total area = 108 + 27= 135m2

Seven boxes required

20 cm

14 cm

AREA = 140cm2

2

25 cm

12 cm

AREA = 150cm2

4

32 cm

16 cm

AREA = 256cm2

6

10 cm

7 cm

AREA = 35cm2

8

2m

6m

5m

2m

Mrs Philips is going to cover the floor with floor boardsOne pack covers 2.5 m2

How many packs does she need?

Area = 5 x 2 = 10 m2 Area = 4 x 2 = 8 m2

Total area = 10 + 8= 18 m2

Eight boxes required

3

10cm

22cm

12cm

6cm

6 x 10= 60cm212cm

12 x 12= 144cm2

Total Area = 144 +

60= 204cm2

4

14cm

20cm

10cm15cm

14 x 10

= 140cm2

6cm

6 x 15= 90cm2

Total Area = 140 +

90= 230cm2

5

30cm

20cm

4cm

8cm8 x 20

4 x 30

= 160cm2

= 120cm2

Total Area = 160 +

120= 280cm2

Calculate the Volume of the Prism

Calculate Area of Front

Area = 4 x7 = 28 cm2

Area = 5 x 2 = 10 c m2

CSA = 10 + 28 = 38cm2

Volume = 38 x 10 = 380cm3

1

Find the Volume of Compound Shapes

Area of the Front =

21 + 10= 31

Volume of Prism =

31 x 4

= 124cm3

8cm

2cm7cm

3cm

4cm 7 x 3 = 21

5 x 2 = 10

5

2

12cm

4cm8cm

5cm

5cm

8 x 5 = 40 7 x 4 =

28

7

Area of the Front =

40 + 28= 68

Volume of Prism =

68 x 5

= 340cm3

Find the Volume of Compound Shapes

3

Find the Volume of Compound Shapes

11cm

4cm7cm

5cm

20cm

7 x 5 = 35 6 x 4 =

24

6

Area of the Front =

35 + 24= 59

Volume of Prism =

59 x 20

= 1180cm3

3 x 10

= 30cm2

5cm

3cm 10cm

5 x 10= 50cm2

3 x 5= 15cm2

= 2 x( 15+50+30)

Surface Area

= 190cm2= 2 x 95

Find the Surface Area of the Cuboid

3

2 x 15

= 30cm2

12cm

2cm 15cm

12 x 15= 180cm2

2 x 12= 24cm2

= 2 x( 24+180+30)

Surface Area

= 468cm2= 2 x 234

Find the Surface Area of the Cuboid

4

7 x 12

= 84cm2

10cm

7cm 12cm

10 x 12= 120cm2

7 x 10= 70cm2

= 2 x( 70+120+84)

Surface Area

= 548cm2= 2 x 274

Find the Surface Area of the Cuboid

Find the Surface Area of the Triangular Prism

15cm

10cm

½ x 8 x 15

= 60cm2

15 x 10= 150cm2

8 x 10= 80cm2

8cm

17 x 10= 170cm2

17cm

Surface Area= 60+60+80+150+170= 520cm2

4

Find the Surface Area of the Triangular Prism

24cm

20cm

½ x 7 x 24

= 84cm2

24 x 20= 480cm2

7 x 20= 140cm2

7cm

25 x 20= 500cm2

25cm

Surface Area= 84+84+140+480+500= 1288cm2

5

Find the Surface Area of the Triangular Prism

12cm

20cm

½ x 9 x 12

= 54cm2

12 x 20= 240cm2

9 x 20= 180cm2

9cm

15 x 20= 300cm2

15cm

Surface Area= 54+54+180+240+300= 828cm2

Calculate the Volume of the Prism

4 cm5 cm

3 cm

7 cm

D iag ram accu ra te ly d raw n

N O T

Calculate Area of Front

4 x 3 = 12 ÷ 2 = 6 cm2

Volume = 6 x 7 = 42 cm3

3

8cm

10cm6cm 10cm

Find the Volume of the Triangular Prism

½ x 6 x 8 = 24cm2

Volume= 24 x 10 = 240cm3

x

4

12cm

20cm5cm 13cm

Find the Volume of the Triangular Prism

½ x 5 x 12

= 30cm2

Volume= 30 x 20 = 600cm3

x

Translate shape P by the vector 52

.

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

6

TRANSLATION-6

-7

A

B

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

7

TRANSLATION-8

-3

A

B

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

8

TRANSLATION5

4

A

B

Enlargement

Scale Factor 2.5Centre of enlargement (0,0)

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

1

ENLARGEMENT

A

B

Scale Factor=

Centre

3 (0,0)

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

2

ENLARGEMENT

A B

Scale Factor=

Centre

2 (0,1)

8 9 10 11 12 13 1471 2 3 4 5 6

8

10

11

12

13

14

9

7

1

2

3

4

5

6

0

3

ENLARGEMENT

AB

Scale Factor=

Centre

2 (0,0)

x = -1

0 1 2 3 4-1-2-3-4

y

x

1

2

3

4

-1

-2

-3

-4

Reflect triangle a in the line x = -1

A

0 1 2 3 4-4

y

x

y = x

1

2

3

4

-1

-2

-3

-4

Reflect triangle a in the line x = -1

A

NOTES : QUESTIONNAIRES / DATA COLLECTION

• Questionnaire or Data Collection Table• Look out for

Time frame: day, week, month

Overlapping intervals

Is there an option for 0 or

none

Is there an option for More

than..

State two things wrong with this question

1)

2)

Design a better question

No timescale

Response boxes overlap

1. What time did Jasmine leave home?

2. For how long did Jasmine stop to talk to her friend on the way to the park

3. Jasmine stayed at the park for ½ hour, then walked home at a steady rate of 7.5 km/hr

Jasmine's journey to the park

9.10am

10 mins

The travel graph shows Amy’s distance from home during a cycle journey to the shops and back.

a) How far away from home was Amy at 1:45pm? ___________

b) What happened between 2pm and 2:45pm?

__________________________________

c) Between which times was Amy cycling fastest?

BETWEEN ________ and _________

d) Calculate Amy’s average speed during this time?

___________km/h

The travel graph shows Jim’s distance from home during a walk.a) How far had Jim walked in 1½ hours?

___________

b) What does the part of the graph BC represent?

__________________________________c) After walking 9km, Jim turned round and

walked straight back home. It took him 2 hours.Draw a line on the grid to show this?

d) Work out the average speed of the return journey?

___________km/h

15kmShe stopped

1:30pm

2pm

S = D T

2030min

200.5 hr 40

4.5km

He stopped

S = D T

92 hrs 4.

5

1 2 0 °

A

B

C

D

G

1 2 0 °

E

F

H

7 .5 cm

8 cm

6 cm 9 cm

Find:a) Length of BC

b) Area of Shape ABCD

12 cm

Area of EFGH = 15 cm2

Find:a) Length of QR

b) Length of AB

Area of ABC = 10 cm2

Find:a) Length of BC

b) Area of Shape DEF

2 cm 8 cm

20 cm

Volume of shape A = 30 cm3

Find the Volume of shape B.

s.f. = 2

= 2 x 8= 16cm

= 22 x 15

= 60cm2

= 4 x 15

s.f. = 1.5

= 12 x 1.5

= 18cm= 15 ÷

1.5= 10cm

s.f. = 4

= 20 ÷ 4= 5cm

= 42 x 10

= 160cm2

= 16 x 10

s.f. = 2

= 23 x 30

= 240cm3

= 8 x 30

OP

S

T

Tangents and radius = 90°

32°

Find angle TOS

Angles in quadrilaterals add up to 360 °

TOS = 180 – 32 = 148°

41°

ab

O

Find angles a & bAngles in semi circle = 90° at the circumference

a = 180 – 90 – 41 = 39° b = 90 – 39 = 41°

140

x

A

C B

The angle subtended at the centre is twice the angle at the circumference

Find angle x

Star Trek Symbol

x = 140 ÷ 2 = 70°

B

C

D

A

Opposite angles in a cyclic quadrilateral add up to 180 degrees

Find angle x

80°

x

x = 180 – 80 = 100°

Find the value of x 3x - 15 2x 2x+ 2x + 24 9x +9

Angles in a quadrilateral = 360o

So 9x + 9 = 360 9x = 351 x = 39

1

Write an EXPRESSION for the PERIMETER of the triangle?

x + 52x + 1

3x - 2

2x + 1 + 3x - 2 + x + 5

= 6x + 4

=

1

Write an EXPRESSION for the PERIMETER of the quadrilateral?

2x + 8

5x + 1

3x - 2

5x + 1 + 3x - 2 +2x + 8

= 12x + 10

=

2x + 3

+2x + 3

1

Write an EXPRESSION for the PERIMETER of the rectangle?

2x - 3

2x - 3 + 2x - 3 + 3x - 1

= 10x - 8

=

3x - 1

+ 3x - 1

3x - 1

2x - 3

1

Write an EXPRESSION for the PERIMETER of the square?

2x + 1

2x + 1 + 2x + 1 +2x + 1

= 8x + 4

=

2x + 1

+2x + 1

2x + 1

2x + 1

The perimeter of the shape is 38 cm

Calculate the area

2x – 1 2x 3x -2+ 3x + 1 10x -2

10x -2 = 3810x = 40

x = 4 calculate lengths

8

7

13

Area = 8 x7 =56m2

Area = 8 x 6 = 48 ÷ 2 = 24cm2

Area = 56 + 24 = 80c m2

Calculate the value of x – give reasons

53 – Corresponding angles are EQUAL

180 – 53 = 127Angles on straight lineAdd to 180

28Vertically opposite angles are EQUAL

180 – 127 – 28 = 25Angles in a triangle add to 180

NOTES : ANGLES

• Angles in a triangle Add to 180

• Angles in a Quadrilateral Add to

360

• Angles around a point Add to 360

• Angles on a Straight line Add to

180

• The base angles in an ISOSCELES Triangle are Equal

r = 121°

r

59°

c = 53°

c

53°

w = 81°

w

81°

p = 52°

p

128°

g = 99°

g

81°

d = 139°

d

139°

f = 37°

f

37°

Exterior angle of Hexagon = 360 ÷ 6 = 60

Exterior angle of Octagon = 360 ÷ 8 = 45

x = 60 + 45 = 105

NOTES : EXTERIOR ANGLES• Exterior Angle: Outside

• Interior Angle: Inside

• Exterior Angles = 360 ÷ Number of Sides

• Exterior Angle + Interior Angle = 180

Compare the distributionsMedian girls = 29 boys = 25.5 / Girls have highest median IQR girls = 8 boys = 9 / Boys have highest IQR

Median

Range

Interquartile Range

40

50

30

1

Median

Range

Interquartile Range

45

45

15

2

Median

Range

Interquartile Range

65

55

25

3

+ 12

1227

45

+ 27 + 45+ 57

57 60

CHECK

x

x

x

x x

Lower Quartile = 32

Upper Quartile = 50

MEDIAN= 41

1

Height cm Frequency

Cumulative Freq.

150 < x < 160 12 12

160 < x < 170 7 19

170 < x < 180 4 23

180 < x < 190 5 28

190 < x < 200 2 30

2

Marks Frequency

Cumulative Freq.

10 < x < 20 6 6

20 < x < 30 9 15

30 < x < 40 12 27

40 < x < 50 9 36

50 < x < 60 14 50

3

Weight kg Frequency

Cumulative Freq.

50 < x < 60 11 11

60 < x < 70 29 40

70 < x < 80 18 58

80 < x < 90 15 73

90 < x < 100 17 90

USE THE MID-POINTS

1DRAW A FREQUENCY POLYGON

FREQUENCY

150 160 170 180 190

MP155165

175185

195x

2

200

4

6

8

10

12

14

16

18

HEIGHT

x

xx

x

ESTIMATE

• 31 x 9.87 0.509

• 84 x 10 23

• 26 x 33 0.235

• 30 x 10 = 300 = 600 0.5 0.5

• 80 x 10 = 800 = 40 20 20

• 30 x 30 = 900 = 4500 0.2 0.2

0.463.8 + 54.1

0.5 4+ 50

=

0.554

= 108=

ESTIMATE9

0.4539.5 + 176.4

0.510 + 200

=

0.5210

= 420=

ESTIMATE10

0.233.8 + 16.4

0.2 4+ 20

=

0.224

= 120=

ESTIMATE11

16 gingerbread 8 gingerbread 24 gingerbread

180 flour 90 flour 270 flour

40 ginger 20 ginger 60 ginger

110 butter 55 butter 165 butter

30 sugar 15 sugar 45 sugar

1 + 3 + 5 = 9

180 = 20 9

Need

Cement 1 x 20 = 20 kgSand = 3 x 20 = 60 kgGravel = 5 x 20 = 100 kg

He does not have enough cement, as he needs 20 kg and he has 15 kg

2 9

3 1 6 5 3 9

4 3 9 3 4 6 2 8

5 2 4 5

2 9

3 1 3 5 6 9

4 2 3 3 4 6 8 9

5 2 4 5

KEY

2 9 = 29

Rotating a Shape

1. Use tracing paper and trace the shape

2. Put pencil at the coordinate

3. Rotate (turn) the shape the number of degrees anticlockwise

T

Rotate the shaded shape, 900, clockwise, centre (0,0).

Label it T

Q

Rotate the shaded shape, 1800, centre (0,1). Label it Q

Translations

Simplify Fully • m4 x m2

• 5m2t3 x 2m4t

• h9÷ h3

• 35e7x2

5e5x

• (m2)3

• x0

• 16 ½

• (4x6)1/2

m6

10m6t4

h6

7e2x1

m6

14 (power of ½ means square root)2x3

Factorise Fully

• 8x2 + 4x

• 6x2 + 12x

• 3xy2 + 9xy

• 2x + x2

4x(2x + 1)

6x(x + 2)

3xy(y + 3)

x(2 + x)

NOTES : NEGATIVE NUMBERS

• Draw a number line to help you with these questions.

• + x + = +• - x - = +• + x - = -• - x + = -

Factorise

• x2 + 7x + 10

• x2 - 6x + 8

• x2 + 3x -4

(x + 2)(x + 5)

(x - 4)(x -2)

(x + 4)(x -1)

Factorise

• x2 – 36

• y2 - 64

• x2 - 81

x2 – 62 = (x – 6) (x + 6)

y2 - 82 = (x – 8) (x + 8)

x2 - 92 = (x – 9) (x + 9)

Expand and simplify

• 3(4x + y)

• 3y(y – 3)

• (y + 8)(y – 3)

• (2x – 3)2

12x + 3y

3y2 – 9y

y2 -3y +8y -24 = y2 + 5y -24

(2x -3)(2x-3) = 4x2 -6x -6x + 9 = 4x2 -12x+ 9

STEP 6

Substitute your answer from STEP 5 into equation (1) or (2) in order to find the other solution.

STEP 1

Write both equations underneath each other and label (1) and (2)

STEP 2: Decision 1

Do you want to get rid of x or y?

STEP 3

Make the coefficients of the letter you want rid of the same by multiplying equation (1) or (2) or both.

STEP 4: Decision 2

Do I Add or subtract equations (1) and (2)?

S T O P: Same take away / Opposite plus(This refers to the signs)

STEP 5

Form and solve one equation in order to find one of the values

SIMULTANEOUS EQUATIONS

Express 120 as a product of its prime factors.

Find the highest common factor (HCF) of 90 and 120

9.(a) Express the following numbers as products of their prime factors.

(i) 60,

.............................

(ii) 96.

.............................(4)

(b) Find the Highest Common Factor of 60 and 96.

.............................(1)

(c) Work out the Lowest Common Multiple of 60 and 96.

............................(2)

302152

3 5

96482

2422 12

2 62 3

60 =2 x 2 x 3 x 596 =2 x 2 x 2 x 2 x 2 x 3

= 2 x 2 x 3 =12

= 12 x 2 x 2 x 2 x 5 =480

8

132

Write 132 as the Product of Primes

662

332

3 11

132 = = 22 x 3 x 11

2 x 2 x 3 x 11

9

486

Write 486 as the Product of Primes

2432

813

9 9

486 = = 2 x 35

2 x 3 x 3 x 3 x 3 x 3

3 33 3

10

525

Write 525 as the Product of Primes

1055

215

525 = = 3 x 52 x 7

3 x 5 x 5 x 7

3 7

• Janice asks 100 students if they like biology or chemistry or physics best.

• 38 of the students are girls.• 21 of these girls like biology best.• 18 boys like physics best.• 7 out of the 23 students who like chemistry

best are girls.

• Work out the number of students who like biology best.

Biology Physics Chemistry

Total

Girls 21 10 7 38

Boys 28 18 16 62

Total 49 28 23 100

The area of the triangle is equal to the area of the square

What is the perimeter of the square

Area triangle = 9 x 8 = 72÷ 2 = 36 cm2

Area square = 36 cm2

Length of each side of square = 6 cm

6 + 6 + 6+ 6 = 24 cm

V = 3b + 2b2

Find the value of V when b = –4

s = 3d + 8Write d in terms of s

V = 3 x (-4) + 2 x (-4)2

V = - 12 + 2 x 16

V = - 12 + 32

V = 20

d → x 3 → +8 → s

d ÷3 -8 s

S – 8 = d 3

Express the area in terms of

7cm

7cm

10cm

Area = r2

= x 7 x 7 cm 2

= 49 cm 2

Volume = 4 9 x 10

= 490 cm3

1

Shoe Size Frequency

3 12 36

4 5 20

5 4 20

6 5 30

7 2 14

30 120

Mean = = 4 120 ÷ 30

1

Hair Colour Frequency Angle

Black 8 800

Brown 6 600

Red 12 1200

Blond 7 700

None 3 300

36 3600

2

Bird Frequency Angle

Sparrow 6 300

Magpie 12 600

Dove 14 700

Pigeon 24 1200

Robin 16 800

72 3600

3

Car Frequency Angle

Ford 24 960

Astra 20 800

Honda 12 480

VW 14 560

Volvo 20 800

90 3600