maths gcse paper 1 revision
TRANSCRIPT
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Yr 11 Non-
Calculator Revision
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Multiplication
• £1.82 x 15 =
• 3.28 x 13 =
• 4.56 x 1.8=Remove decimalsMultiple – box methodPut decimals back in
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1
34 x 6430 4
60
4
x
1800 240
120 16
1800240120
162176
+
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1
237 x 45200 30
40
5
x
8000 1200
1000
150
8000
1200
280
1000
150
35
10665
+
7
280
35
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Given:
527 x 73 = 38471
1
What is: 52.7 x 73 = 3847.1
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Given:
527 x 73 = 38471
2
What is: 52.7 x 7.3= 384.71
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Given:
527 x 73 = 38471
3
What is: 5.27 x 7.3= 38.471
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Given:
432 x 29 = 12528
1
What is:
12528 ÷ 2.9 =4320
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Given:
432 x 29 = 12528
1
What is:
12528 ÷ 0.29=43200
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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
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0.000512
Standard form
5.12 x 10-4
Normal Number
16
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0.0000956
Standard form
9.56 x 10-5
Normal Number
17
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0.73
Standard form
7.3 x 10-1
Normal Number
18
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0.00103
Standard form
1.03 x 10-3
Normal Number
19
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0.000423
Standard form
4.23 x 10-4
Normal Number
20
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78000 Normal Number
7.8 x 104
Standard form
21
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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
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3 x 102
1
2.5 x 104 X
3 x 2.5 102 x 104 x = 7.5 102+4x = 7.5 106x
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3 x 107
1
5 x 109 X
3 x 5 107 x 109 x = 15 107+9x = 1.5x10 1016x = 1.5 1017x
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4 x 106
1
6 x 1012 X
4 x 6 106 x 1012
x = 24 106+12x = 2.4x10 1018x = 2.4 1019x
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6 x 108
1
2 x 105 ÷
(6 ÷ 2) (108 ÷ 105)
x = 3 108-5x = 3 103x
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8 x 1012
1
4 x 103 ÷
(8 ÷ 4) (1012 ÷
103)x
= 2 108-5x = 2 103x
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9 x 1015
1
2 x 104 ÷
(9 ÷ 2) (1015 ÷
104)x
= 4.5 1015-4x = 4.5 1011x
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nth term
• 3 6 9 12.........
• 1 7 13 19........
• 5 12 19 26........
3n
6n - 5
7n - 2
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3 5 7 9
nth term =
2n + 1
1
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4 10 16 22
nth term =
6n - 2
2
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2 5 8 11
nth term =
3n - 1
3
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3 8 13 18
nth term =
5n - 2
4
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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4n1 <<
2
n is an integer
List the possible values of n
2, 3
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6n2 <≤
3
n is an integer
List the possible values of n
2, 3, 4, 5
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9n5 ≤<
4
n is an integer
List the possible values of n
6, 7, 8, 9
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4n1- ≤<
5
n is an integer
List the possible values of n
0, 1, 2, 3, 4
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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
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-3 < x ≤ 4
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1
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
o
7n<
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2
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
o
3n>
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3
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
7n≤
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4
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
3n≥
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5
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
8n3 ≤<
o
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6
Write down the inequality shown by:
0 1 2 3 4 5 6 7 8 9 10
8n3 <≤
o
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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
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6w – 10 > 20
w >
1
5
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2x + 7 < 15
x <
2
4
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3y + 14 > 20
y >
3
2
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7a – 8 > 20
a >
4
4
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5m – 9 > 31
m >
5
8
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6w – 10 > 4w + 20
w >
1
15
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7y – 8 < 3y + 20
y <
2
7
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12x - 8 > 8x + 12
x >
3
5
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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
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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
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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
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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
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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
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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
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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
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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
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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
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(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
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(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
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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
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20 cm
14 cm
AREA = 140cm2
2
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25 cm
12 cm
AREA = 150cm2
4
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32 cm
16 cm
AREA = 256cm2
6
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10 cm
7 cm
AREA = 35cm2
8
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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
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3
10cm
22cm
12cm
6cm
6 x 10= 60cm212cm
12 x 12= 144cm2
Total Area = 144 +
60= 204cm2
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4
14cm
20cm
10cm15cm
14 x 10
= 140cm2
6cm
6 x 15= 90cm2
Total Area = 140 +
90= 230cm2
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5
30cm
20cm
4cm
8cm8 x 20
4 x 30
= 160cm2
= 120cm2
Total Area = 160 +
120= 280cm2
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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3
8cm
10cm6cm 10cm
Find the Volume of the Triangular Prism
½ x 6 x 8 = 24cm2
Volume= 24 x 10 = 240cm3
x
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4
12cm
20cm5cm 13cm
Find the Volume of the Triangular Prism
½ x 5 x 12
= 30cm2
Volume= 30 x 20 = 600cm3
x
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Translate shape P by the vector 52
.
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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
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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
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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
![Page 100: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/100.jpg)
Enlargement
Scale Factor 2.5Centre of enlargement (0,0)
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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)
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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)
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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)
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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
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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
![Page 106: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/106.jpg)
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..
![Page 107: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/107.jpg)
State two things wrong with this question
1)
2)
Design a better question
No timescale
Response boxes overlap
![Page 108: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/108.jpg)
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
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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
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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
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OP
S
T
Tangents and radius = 90°
32°
Find angle TOS
Angles in quadrilaterals add up to 360 °
TOS = 180 – 32 = 148°
![Page 112: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/112.jpg)
41°
ab
O
Find angles a & bAngles in semi circle = 90° at the circumference
a = 180 – 90 – 41 = 39° b = 90 – 39 = 41°
![Page 113: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/113.jpg)
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°
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B
C
D
A
Opposite angles in a cyclic quadrilateral add up to 180 degrees
Find angle x
80°
x
x = 180 – 80 = 100°
![Page 115: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/115.jpg)
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
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1
Write an EXPRESSION for the PERIMETER of the triangle?
x + 52x + 1
3x - 2
2x + 1 + 3x - 2 + x + 5
= 6x + 4
=
![Page 117: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/117.jpg)
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
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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
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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
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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
![Page 121: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/121.jpg)
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
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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
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r = 121°
r
59°
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c = 53°
c
53°
![Page 125: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/125.jpg)
w = 81°
w
81°
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p = 52°
p
128°
![Page 127: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/127.jpg)
g = 99°
g
81°
![Page 128: MATHS GCSE PAPER 1 REVISION](https://reader033.vdocument.in/reader033/viewer/2022061609/558c0aa4d8b42a5b568b4704/html5/thumbnails/128.jpg)
d = 139°
d
139°
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f = 37°
f
37°
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Exterior angle of Hexagon = 360 ÷ 6 = 60
Exterior angle of Octagon = 360 ÷ 8 = 45
x = 60 + 45 = 105
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NOTES : EXTERIOR ANGLES• Exterior Angle: Outside
• Interior Angle: Inside
• Exterior Angles = 360 ÷ Number of Sides
• Exterior Angle + Interior Angle = 180
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Compare the distributionsMedian girls = 29 boys = 25.5 / Girls have highest median IQR girls = 8 boys = 9 / Boys have highest IQR
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Median
Range
Interquartile Range
40
50
30
1
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Median
Range
Interquartile Range
45
45
15
2
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Median
Range
Interquartile Range
65
55
25
3
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+ 12
1227
45
+ 27 + 45+ 57
57 60
CHECK
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x
x
x
x x
Lower Quartile = 32
Upper Quartile = 50
MEDIAN= 41
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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
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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
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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
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USE THE MID-POINTS
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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
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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
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0.463.8 + 54.1
0.5 4+ 50
=
0.554
= 108=
ESTIMATE9
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0.4539.5 + 176.4
0.510 + 200
=
0.5210
= 420=
ESTIMATE10
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0.233.8 + 16.4
0.2 4+ 20
=
0.224
= 120=
ESTIMATE11
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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
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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
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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
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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
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T
Rotate the shaded shape, 900, clockwise, centre (0,0).
Label it T
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Q
Rotate the shaded shape, 1800, centre (0,1). Label it Q
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Translations
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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
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Factorise Fully
• 8x2 + 4x
• 6x2 + 12x
• 3xy2 + 9xy
• 2x + x2
4x(2x + 1)
6x(x + 2)
3xy(y + 3)
x(2 + x)
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NOTES : NEGATIVE NUMBERS
• Draw a number line to help you with these questions.
• + x + = +• - x - = +• + x - = -• - x + = -
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Factorise
• x2 + 7x + 10
• x2 - 6x + 8
• x2 + 3x -4
(x + 2)(x + 5)
(x - 4)(x -2)
(x + 4)(x -1)
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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)
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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
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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
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Express 120 as a product of its prime factors.
Find the highest common factor (HCF) of 90 and 120
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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
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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
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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
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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
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• 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
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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
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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
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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
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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
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1
Hair Colour Frequency Angle
Black 8 800
Brown 6 600
Red 12 1200
Blond 7 700
None 3 300
36 3600
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2
Bird Frequency Angle
Sparrow 6 300
Magpie 12 600
Dove 14 700
Pigeon 24 1200
Robin 16 800
72 3600
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3
Car Frequency Angle
Ford 24 960
Astra 20 800
Honda 12 480
VW 14 560
Volvo 20 800
90 3600
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