YEAR 10

REVISION BLASTER

GCSE MATHS REVISION

• UNIT 1 – all about DATA

• Probability

• Averages

• Cumulative frequency

• Histograms

• Percentages

Probability

KEY POINTS

•The probability scale goes from 0 to 1.

•Probability of a certain event = 1

•Probability of an impossible event= 0

•P(event)= No. of ways that event can happen

Total no. of ALL possible outcomes

22 Sweets in a bag, 7 are red, 6 are blue, 9 are green.

What is the probability of randomly picking a blue sweet?

•P(event)= No. of ways that event can happen

Total no. of ALL possible outcomes

P (BLUE) =

622

=

11

3

Calculate the Probability of NOT picking a blue sweet.

P(event Not happening)=1 – P(event happening)

P (NOT BLUE) = 1 – P(picking blue)

P (NOT BLUE) = 1 –

P (NOT BLUE) =

11

3

11

8

Calculating the Probability of combined events.STARTER:

P(garlic bread)=0.2; P(dough balls)= 0.8

MAIN:

P(pasta)=0.2; P(Salad)=0.2 P(Pizza) = 0.6

•Calculate the probability of having; garlic bread followed by pizza.

•Calculate the probability of having ANY starter followed by salad.

0.2

P(garlic bread)=0.2; P(dough balls)= 0.8P(pasta)=0.2; P(Salad)=0.2 P(Pizza) = 0.6

STARTER

Garlic Bread

Dough

Balls

0.2

0.8

Pizza

Pasta

Salad

0.2

0.2

0.6

Pasta

Salad

0.2

0.2

0.6Pizza

MAINCalculate the probability of having; garlic bread followed by pizza.

0.20.6

0.2 x 0.6 = 0.12

AND RULE (Going ALONG the branches)

We MULTIPLY the Probabilities.

Answer= 0.12

•Calculate the probability of having ANY starter followed by salad.

0.2

0.2 x 0.2 = 0.04

0.80.2 0.8 x 0.2 =

0.16

There are TWO possibilities here; garlic bread and then salad, OR dough balls and the salad.

OR RULE= ADD PROBABILITIES

SO we add together the possibilities:

0.04 + 0.16 = 0.20

Probability of having ANY starter followed by salad is 0.20

Relative Frequency: An estimate for theoretical probabilityRelative Frequency = Frequency of the event

Total Number of trials

Calculate the relative frequency of the toast landing jam side up.

No. of drops

No. of jam side up

Relative Frequency

12 6 12/6 =2

Relative Frequency = Frequency of the event

Total Number of trials

The relative frequency of seeing a red car go past your window is 0.6.

You watch 100 cars go past your window. How many would you expect to be red?

0.6

=Frequency of the event

100

0.6 x 100 = Frequency

60 = Frequency

Probability Answers• 1. (a) ½• (b) 300• 2. (a) (i) 11/12 or 0.92• (ii) 8/12 or 2/3• (b) (i) 15/22 (allow 14 or 15 on top 22 or 21 on

bottom)• (ii) 5/11 or any equivalent fraction

• 3. (a) 0.4 on first branch & all other branches correct

• (b) (i) 0.16 or 4/25 or 16 %• (ii) 0.84 or 21/25 or 84 %• (a)•• (b) 15• (c) More than expected with a suitable qualification

(allow expect 10)

20

3,

20

3,

20

3,

20

5,

20

3,

20

3

Averages

MediaMedia

nnM

od

Mod

ee

MEAMEA

NN

3, 5, 4, 3, 5, 4, 4, 6, 73, 3, 4, 4, 4, 5, 5, 6, 7

3, 3, 4, 4, 4, 5, 5, 6, 7

3+3+4+4+4+5+5+6+7=41 41

9= 4.55…

0 × 20 = 0

1 × 17 = 17

2 × 15 = 30

3 × 10 = 30

4 × 9 = 36

5 × 3 = 15

6 × 2 = 12

140

Mean = sum of ALL the values sum of frequencies

Calculate the mean.

26

3

9

10

15

17

20

Frequency

4

5

3

2

1

0

Numbers of sports played

TOTAL 76

Frequency x Numbers of sports played

= 140 = 1.84 76

2105263

1 × 7.5 = 7.5

8 × 12.5 = 100

12 × 17.5 = 210

10 × 22.5 = 225

3 × 27.5 = 82.5

1 × 32.5

1 × 37.5

= 32.5

= 37.5

695

7.5

12.5

17.5

22.5

27.5

32.5

37.5

Mean = sum of ALL frequency x midpoint sum of frequencies

= 695 = 19.3 36

055555…

135 ≤ d < 40

1

3

10

12

8

1

Frequency

30 ≤ d < 35

25 ≤ d < 30

20 ≤ d < 25

15 ≤ d < 20

10 ≤ d < 15

5 ≤ d < 10

Javelin distances in

metres

36TOTAL

Midpoint Frequency × midpoint

Calculate an estimate for the mean.

Averages Answers

1. 99.7 (allow 100)

2. (a) 131.6 (allow 131 to 132)(b) 110 £ t < 130

3. (a) 26.4 (allow 26)(b) Frequency polygon from (10, 42) to (70, 3) joined with approximately straight lines(c) Comment - eg. Higher mean on Saturday, or

Larger range on Saturday, orMore money spent on Saturday

AD BREAKYear 10 Unit 1 Exam. Tuesday 9th November-

TOMORROW!!

Don’t Forget: CALCULATORS

PEN

PENCIL

RULER

REVISE

CUMULATIVE

FREQUENCY

Cumulative Frequency

The cumulative frequency is the

RUNNING TOTAL OF FREQUENCIES

Used for finding the MEDIAN and UPPER AND LOWER QUARTILES

Marks in a Yr 10 Maths TestMarks Frequency Cumulative

Frequency

11-20

21-30 11

31-40 19

41-50 36

42

61-70 31

71-80 13

81-90 6

110 people scored 60 marks OR less

2 21332

68110141154160

51-60

Plot the top value in each group against cumulative frequency

Median shows that ½ of people score less

than 53 marks and ½ score more

Lower Quartile shows 25% of people scored 44 marks or

lessUpper Quartile shows 75% of people scored 65 marks or

less

Interquartile range =65 -44

=21

Cumulative Frequency Answers

1. (a) Any value from 55 to 57 inclusive(b) 6 ± 0.4

2. (a) 12, 27, 56, 72, 83, 90(b) Correct graph ‘increasing’ (S shape)(c) (i) Median line from 45(.5)

e.g. 26-28(ii) IQR lines from Q1 (22.5) and Q3 (67.5)

and subtract their answers (eg 16 – 20) (d) Approx 48

HISTOGRAMS

Finding lengths given areas

• If the width is 10– What must the height be

to make an area of 25?

10

252.5As 10 x 2.5 = 25

Bar Graphs

• Bar graphs are great to use when you have equal class intervals:

Histograms

• Histograms are used when you have different sized class intervals.

• The area of the rectangle represents the frequency.

• The width is the size of the class interval.

• The height is what we call the frequency density.

Number of kick-ups

Frequency Frequency

Density

0-5 10

5-15 25

15-30 45

30-40 15

The table below shows the number of kick-ups completed in a competition at a local primary school.

We have different sized class intervals

Our class interval width is 5, and our frequency (area) is 10.

So, Frequency Density (height) = 10/5

2

2.53

1.5

Number of

kick-ups

Frequency Frequency

Density

0-5 10 25-15 25 2.515-30 45 330-40 15 1.5

Estimate how many students managed ten kick-ups or less.

Frequency =5 x 2.5 = 12.5

Frequency=5 x 2 = 10

So, 10 + 12.5 = 22.5We can estimate that 22.5 students made ten kick-upsor less.

Comparing Histograms and Bar Graphs

Histograms Answers

1. (a) 0.4, 0.4, 0.5, 0.1

(b) 21

2. (a) Correct histogram:

Widths: 15, 5, 5, 10, 15

Heights: 0.6, 4.2, 4.8, 3.1, 1

(b) 42

Percentage Increase

and Decrease

Increase £20 by 15%

Method 1Find 15%10% = £2 5% = £1 so 15% = £3

Now add it on to £20£20 + £3 = £23

Increase by 15% so

1.15 is the multiplier

Method 2Method 2

I need to find 115% I need to find 115% OF original amountOF original amount115% of £20 is 115% of £20 is 1.15 x £20 = £231.15 x £20 = £23

Decrease £60 by 5%

Method 1

Find 5% of £60

10% = £6 5% = £3

Subtract from £60

£60 - £3 = £57

Method 2

A decrease of 5% is same as 95% of original amount

So 0.95 x £60 = £57 95% of original amount means 0.95 is the multiplier

Compound Interest

• £1,000 in bank earns 5% interest per year. How much will you have in 3 years?

After 1 year 1000 x 1.05 = 1050

After 2 years 1050 x 1.05= 1102.50

After 3 years 1102.50 x 1.05 = 1157.63

OR 1000 x 1.05 x 1.05 x1.05=

1000 x 1.05³=£1157.63

1.05 is the multiplier!

Percentages Answers

1.£2140 (allow £140)

2.(a) 1.029

(b) 1 087 401 937 (allow 1 087 000 000)

3.1st year is £20; 2nd year is £20.80

Interest is £40.80 so Amir is wrong

4.(a) 2.04

(b) 6 (windmills)