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1 x = a ×10n
where n is an integer and 10 ! a <10
Find, in standard form, an expression for x2.
Give your expression as simply as possible.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(Total for Question is 3 marks)
7 (a) x = 9 × 102m where m is an integer.
Find, in standard form, an expression for x
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(2)
(b) y = 9 × 102n where n is an integer.3
2Find, in standard form, an expression for y
Give your answer as simply as possible.
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(3)
(Total for Question is 5 marks)
Square root is equivalent to a power of a half
From part a we can see y =
Therefore y = (3 x 10 )
Coefficient must be less than 10 and greater than zero
3
! ?!5.82@!4A823@,/!1*5!*!/*@2B5!.-!%(C!4D!*3@!*!1,2910!.-!E(>!4D(
Work out the total surface area of the cylinder.
Give your answer correct to 3 significant figures.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm2
(Total for Question is 3 marks)
Diagram NOT
accurately drawn
3.7 cm
%(C!4D
10 A cylinder has diameter 12 cm and length 30 cm.
Work out the curved surface area of the cylinder.
Give your answer correct to 3 significant figures.
12 cm
30 cm
Diagram NOT
accurately drawn
Curved surface is equal to circumference x height
Circumference = x Diameter = 12
Height = 30
12 X 30 = 360 = 1130
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm2
(Total for Question is 3 marks)
1130
2 The diagram shows a solid cylinder.
30 cm
11 cm Diagram NOT
accurately drawn
The cylinder has a height of 30 cm and a radius 11 cm.
(a) Work out the total surface area of the cylinder.
Give your answer correct to 2 significant figures.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm2
(4)
(b) The height of the cylinder is 30 cm, correct to the nearest centimetre.
(i) Write down the lower bound of the height of the cylinder.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm
(ii) Write down the upper bound of the height of the cylinder.
. . . . . . . . . . . .29.5
2800(3sf)
Surface area of 1 + 3 :
2πr = 2 π (11)(11) = 242π
Area of 2:Circumference x height 2πr x h = 2π(11)(30) = 660π
2 + 1&3 = (660 + 242)π = 902π
30.5. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . cm
(2)
(Total for Question is 6 marks)
7
A cylinder has radius 5.4 cm and height 16 cm.
(a) Work out the volume of the cylinder.
Give your answer correct to the nearest whole number.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cm3
(2)
The radius 5.4 cm is correct to 2 significant figures.
(b) (i) Write down the upper bound of the radius.
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(ii) Write down the lower bound of the radius.
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(2)
(Total for Question is 4 marks)
Diagram NOT
accurately drawn
5.4 cm
16 cm
Volume of a cylinder is π r h
π x 5.4 x 5.4 x 16 = 1466
1466
5.45
5.35
3 Show that 6 − 8( )2 44= 24− 2
Show each stage of your working clearly.
(Total for Question is 3 marks)
4 3( 4+ () +a a ) = 17 + k a where a and k are positive integers.
Find the value of a and the value of k.
a = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
k = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(Total for Question is 3 marks)
5 +a a( )2
8 5= +4 b 2
a and b are positive integers.Find the value of a and the value of b.Show your working clearly.
a = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
b = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(Total for Question is 3 marks)
6 Given that 5 −( )2x y= 20− 2 where x and y are positive integers, find the value of x and the value of y.
x = . . . . . . . . . . . . . . . . . . . . . . . . . . . .
y = . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(Total for Question is 3 marks)
-x5x5 -5 5 - 5 + = y -2025 -10 +
Equating the surds leads us to the conclusion x = 8
7 Given that x and y are positive integers such that ( )1 3+ +(x x ) = y + 4 5 find the value of x and the value of y.
x = . . . . . . . . . . . . . . . . . . . . . . . . . . .
y = . . . . . . . . . . . . . . . . . . . . . . . . . . .
(Total for Question is 3 marks)
Expand using foil
Equating surds we find x to be 5
Therefore 3 + x = 8 =y
5
8
8 (a) Expand ( )+5 3 2 2
Give your answer in the form +a b 2( ) wh, ere a and b are integers.Show your working clearly.
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(2)
(b) +5 3( )28
2 p= +q , where p and q are integers.
Find the value of q .
(5 + 3 2 ) (5 + 3 2 ) Expand
5x5 + 5(3 2)+ 5(3 2) + (3 2)(3 2)
25 + 18 + 30 2
43 + 30 2
43 + 30 2
Equate irrational parts
q = 120. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(3)
(Total for Question is 5 marks)
9 (a) Write 132
as a power of 2
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(2)
(b) Show that 4 12+( ) 5( 3− ) 14= 6+ 3Show each stage of your working clearly.
(3)
(Total for Question is 5 marks)
32 = 2
10 (a) Show that 3 2+( )2 4( − 2) 8= 5+ 2
Show your working clearly.
(2)
(b) Rationalise the denominator and simplify fully10 + 3 2
2
Show your working clearly.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(Total for Question is 4 marks)
As shown
LHS
LHS
5 2 +3
10 + 3 2 Multiply top and bottom by 2
2
2 10 + 3 2 = 10 2 + 3(2) = 5 2 +3
22
(3)
Given that c is a prime number,
(b) rationalise the denominator of 3 −c c
c
Simplify your answer.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(Total for Question is 5 marks)
11 (a) Show that 5 −8( ) 7 +2( ) = 31 − 9 2Show each stage of your working.
Expand and simplify
Multiply by which is equivalent to 1
10
Diagram NOT
accurately drawn
A
B C
D
59°
16.5 m
10°
The diagram shows a vertical flagpole in Chennai, India.
The point A is at the top of the flagpole.
The point B is at the foot of the flagpole.
There is a platform at the point D on the flagpole.
B and C are points on horizontal ground.
AD = 16.5 m
The angle of elevation of A from C is 69°
The angle of elevation of D from C is 59°
Calculate the height, AB, of the flagpole.
Give your answer correct to 3 significant figures.
6
The diagram shows triangle ABC.
D is the point on AB, such that CD is perpendicular to AB.
AC = 8.3 cm.
AD = 4.7 cm.
BD = 7.5 cm.
Calculate the size of angle ABC.
Give your answer correct to 1 decimal place.
C
DBA
7.5 cm4.7 cm
8.3 cm
Diagram NOT
accurately drawn
AC = CD + AD , 8.3 - 4.7 = c = 6.841... ( using Pythagoras)
Using tan with CD as the opposite angle and DB and the adjacent
Tan(x) = 6.84/ 7.5
X = arctan( 6.84/7.5) X = 42.4
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(Total for Question is 4 marks)
42.4
3
PS iR s a straight line.
Angle PSQ = 90°
PS = 8.4cm
Angle QPS = 38°
Angle SQR = 44°
Work out the length of QR.
Give your answer correct to 3 significant figures.
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(Total for Question is 4 marks)
Q
P
Diagram NOT
accurately drawn
S
38°
R8.4 cm
44°
Using :
Substituting equation 1 into equation 2
1 The table shows information about the weights of 80 parcels.
Weight (w kg) Frequency
0 < w 2 8
2 < w 4 14
4 < w 6 26
6 < w 8 17
8 < w 10 10
10 < w 12 5
(a) Work out an estimate for the total weight of the 80 parcels.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . kg
(3)
(b) Complete the cumulative frequency table.
Weight (w kg)Cumulative
frequency
0 < w 2
0 < w 4
0 < w 6
0 < w 8
0 < w 10
0 < w 12
(1)
(c) On the grid, draw a cumulative frequency graph for your table.
(2)
(d) Use the graph to find an estimate for the number of parcels which weighed less than
5.2 kg.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(Total for Question is 8 marks)
2
Weight (w kg)
Cumulative
frequency
O
20
40
60
80
4 6 8 10 12
2 The grouped frequency table gives information about the ages of 200 elephants.
Age (t years) Frequency
t 10 55
t 20
t 30
t 22
t 50 13
t 10
(a) Complete the cumulative frequency table.
Age (t years)Cumulative
frequency
t 10
t 20
t 30
t
t 50
t
(1)
115
155
177
190
200
55
(b) On the grid, draw a cumulative frequency graph for your table.
(2)
(c) Use the graph to find an estimate for the number of elephants with ages of more than
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(Total for Question is 5 marks)
xO
50
10 20 50 60
100
Cumulative 150
frequency
30 40
Age (t years)
200
62
Cf = 138
Frequency greater:20-138 = 62
3 The cumulative frequency graph gives information about the lengths, in minutes, of
80 telephone calls.
5O
20
40
60
80
10 15 20
Length of call (minutes)
Cumulative
frequency
25 30
(a) Find an estimate for the number of calls which were longer than 15 minutes.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(b) Find an estimate for the interquartile range of the lengths of the 80 calls.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . minutes
(2)
(Total for Question is 4 marks)
4 The grouped frequency table gives information about the lengths of 160 pythons.
Length (x metres) Frequency
0 ! x ! 1 4
1 ! x ! 2 8
2 ! x ! 3 16
3 ! x ! 4 32
4 ! x ! 5 72
5 ! x ! 6 28
(a) Complete the cumulative frequency table.
Length (x metres)Cumulative
frequency
0 ! x ! 1
0 ! x ! 2
0 ! x ! 3
0 ! x ! 4
0 ! x ! 5
0 ! x ! 6
(1)
(b) On the grid, draw a cumulative frequency graph for your table.
O
40
80
120
160
1 2 5 63 4
Length (x metres)
Cumulative
frequency
(2)
(c) Use your graph to find an estimate for the median length of the pythons.
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(2)
(Total for Question is 5 marks)
5 The cumulative frequency table shows information about the diameters of 60 oranges.
Diameter
(d mm)
Cumulative
frequency
%J!N!d ! 60 12
%J!N!d ! 70 42
%J!N!d ! 80 %=
%J!N!d ! 90 %>
%J!N!d !"100 %M
%J!N!d ! 110 60
(a) On the grid, draw a cumulative frequency graph for the table.
(2)
(b) Use your graph to find an estimate for the median diameter of the 60 oranges.
Cumulative
frequency
80
Diameter (d mm)
60
40
20
0%J 60 70 90 100 110
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . mm
(2)
(Total for Question is 4 marks)
6 The grouped frequency table gives information about the weights of 180 airmail letters.
Weight (w grams) Frequency
0 < w ! 20 15
20 < w ! 40 25
40 < w ! 60 47
60 < w ! 80 70
80 < w ! 100 18
100 < w ! 120 5
(a) Complete the cumulative frequency table.
Weight (w grams)Cumulative
frequency
0 < w ! 20
0 < w ! 40
0 < w ! 60
0 < w ! 80
0 < w ! 100
0 < w ! 120
(1)
15
40
87
157
175
180
Cumulative frequency is
current + all previous
frequencies
(b) On the grid, draw a cumulative frequency graph for your table.
(2)
(c) Find an estimate for the upper quartile of the weights of the 180 letters.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . grams
(2)
(Total for Question is 5 marks)
200
Cumulative
frequency
Weight (w grams)
150
100
50
O20 40 60 80 100 120
Upper quartile is 3/4 x total cumulative frequency3/4 x 180 = 135
74
74
7 A box contains 80 tea bags.
The table shows information about the weight of each tea bag.
Weight (w grams)Number of
tea bags
2.8 < w 2.9 2
2.9 < w 3.0 4
3.0 < w 3.1 22
3.1 < w 3.2 32
3.2 < w 3.3 14
3.3 < w 3.4 6
(a) Work out the percentage of the 80 tea bags that weigh more than 3.1 grams.
65. . . . . . . . . . . . . . . . . . . . . . %
(2)
(b) Work out an estimate for the total weight of the 80 tea bags.
Weights highlighted in orange are the ones greater than 3.1
32 + 14 + 6 = 52
Percentage = 52/80 x 100
Use halfway values of 2.85 grams, 2.95 grams, ...
Sum of Midpoint * frequency gives an aproximate value for total weight
2.85 x 2 + 2.95 x 4 + 3.05 x 22 + 3.15 x 32 + 3.25 x 14 + 3.35 x 6
= 251
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . grams
(3)
. . . . . . . .
251
Here is a cumulative frequency graph for the weights of the 80 tea bags.
(c) Use the graph to find an estimate for the number of tea bags which weighed more
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
(2)
(d) Use the graph to find an estimate for the interquartile range of the weights of
the tea bags.
Cumulative
frequency
3.1
Weight (w grams)
80
60
40
20
02.8 2.9 3 3.2 3.3 3.4
than 3.25 grams.
From the graph we read the value of cumulative frequency 79
Frequency of values above this are 80 ( the total) - 79 = 11
11
Lower quartile is (80+1)/4 = 20.25
Upper quartile is 3(80+1)/4 = 80.75
0.13. . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . grams
(2)
(Total for Question is 9 marks)
Reading off graph the interquartile range is 3.2 -3.07 = 0.13
8 The grouped frequency table gives information about the lengths of time 160 students
exercised one day.
Time (t minutes) Frequency
0 < t 40 20
40 < t 80 35
80 < t 120 60
120 < t 160 33
160 < t 200 7
200 < t 240 5
(a) Complete the cumulative frequency table.
Time (t minutes)Cumulative
frequency
0 < t 40
0 < t 80
0 < t 120
0 < t 160
0 < t 200
0 < t 240
(1)
20
55
115
148
155
160
(b) On the grid, draw a cumulative frequency graph for your table.
(2)
(c) Use your graph to find an estimate for the lower quartile of the lengths of time the
. . . . . . . . . . . . . . . . . . . . . . . . . . . minutes(2)
(Total for Question is 5 marks)
Cumulative
frequency
200
150
100
50
O 40 80 160 200 240120
Time (t minutes)
160 students exercised.
(240+1)/4 = 40.5