(total for question is 3 marks) · 2019-06-06 · d is the point on ab, such that cd is...

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

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

(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.

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

(3)


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


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.


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


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.


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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)


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)


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.


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 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


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 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


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 = . . . . . . . . . . . . . . . . . . . . . . . . . . . .


-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 = . . . . . . . . . . . . . . . . . . . . . . . . . . .


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)


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)


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.

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(2)


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.

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(2)


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.


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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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.


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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.

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(2)


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)


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.

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(2)

(b) Find an estimate for the interquartile range of the lengths of the 80 calls.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . minutes

(2)


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


Length (x metres)Cumulative

frequency

0 ! x ! 1

0 ! x ! 2

0 ! x ! 3

0 ! x ! 4

0 ! x ! 5

0 ! x ! 6

(1)


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)


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)


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


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


(2)

(c) Find an estimate for the upper quartile of the weights of the 180 letters.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . grams

(2)


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)


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


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


(2)

(c) Use your graph to find an estimate for the lower quartile of the lengths of time the

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


Cumulative

frequency

200

150

100

50

O 40 80 160 200 240120

Time (t minutes)

160 students exercised.

(240+1)/4 = 40.5

(total for question is 3 marks) · 2019-06-06 · d is the point on ab, such that cd is...

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