1.The Diagram Shows a Cylinder. the Diameter of the Cylinder Is 10 Cm. the Height of The

1.The diagram shows a cylinder.
The diameter of the cylinder is 10 cm.
The height of the cylinder is 10 cm.

(a)Work out the volume of the cylinder.
Give your answer in terms of π.

......

Answer ...... cm3

(3)

(b)Twenty of the cylinders are packed in a box of height 10 cm.
The diagram shows how the cylinders are arranged inside the box.
The shaded area is the space between the cylinders.

Work out the volume inside the box that is not filled by the cylinders.
Give your answer in terms of π.

......

Answer ...... cm3

(4)

(Total 7 marks)

2.A large carton contains 4 litres of orange juice.
Cylindrical glasses of height 10 cm and radius 3 cm are to be filled from the carton.

How many glasses can be filled?
You must show all your working.

......

Answer ...... glasses

(Total 5 marks)

3.A cylinder contains 17 litres of water.
The radius of the cylinder is 15 cm.

Calculate the height of the water in the cylinder.

......

Answer ...... cm

(Total 4 marks)

4.A solid cube has a square hole cut through horizontally and a circular hole cut through vertically.

Both holes are cut centrally in the appropriate faces.

The dimensions of the cube and the holes are as shown in the diagram.

Calculate the volume remaining after the holes have been cut.

...... ……………………………………………………………………………..

Answer ......

(Total 5 marks)

5.A cuboid is made from centimetre cubes.
The area of the base of the cuboid is 5 cm2
The volume of the cuboid is 10 cm3.

Work out the surface area of the cuboid.
State the units of your answer.

......

Answer ......

(Total 4 marks)

6.The diagram shows a cylindrical tin of soup of diameter 7.5 cm and height 11.6 cm.

(a)Calculate the volume of the cylinder.

......

Answer ...... cm3

(3)

(b)A sheet of paper is wrapped around the curved surface of the tin with a 1 cm overlap along the dotted line shown in the diagram.

Calculate the area of the paper.

......

Answer ...... cm2

(4)

(Total 7 marks)

7.The diagram shows a cylindrical can of beans.
The height is 5.7 cm.
The radius of the base is 3.7 cm.

Calculate the total surface area of the can.

......

Answer ...... cm2

(Total 5 marks)

8.The diagram shows a cylinder.
The volume of the cylinder is 320 cm3.
The height of the cylinder is 3.2 cm.

Calculate the radius of the base of the cylinder.

...... ……………………………………………………………………………..

Answer ...... cm

(Total 3 marks)

9.A tin of diameter 7 cm and height 12 cm has a label around it.
The label is glued together using a 1 cm overlap.
There is a 1 cm gap between the label and the top and the bottom of the tin.

Find the length and the height of the label.

…......

Answer Length = ...... cm

Height = ...... cm

(Total 4 marks)

1.(a) (×) 52M1

Condone 3.1... × 52

 (×) 52 × 10
or their area × 10M1

Condone 3.1... × 52 × 10
Their area must contain (or 3.1...)

250 or 250 × or × 250A1

775 to 790 implies M2A0
Do not accept 250
Ignore fw
250 can be recovered in (b)

(b)40 × 50M1

10 × 10 × 10 40 × 50

their 2000 × 10

their 1000 – their 250 20 × their  (×) 52M1

20 × their 250M1

20 × their (1000 – 250) their 2000 – their 500

20 000 – 5000A1

oe
4290 – 4500 implies M3 A0
Ignore fw except 15000

[7]

2.Sight of 4000B1

B1 may be awarded later for dividing their cm3 answer by 1000

Vol of cup =  × 32 × 10M1

= 282.7 (433388)A1

Accept 280  vol  283

(their 4000) ÷ (their 282.7(....))DM1

= 14(.14710...)

= 14A1

[5]

3.24.05B1,M1,M1,A1

B1 for 17 litres = 17000 cm3
M1 for vol =  × 152 × h
H = 17000/( × 152)
A1 answer (allow 24 or 24.1 if supported)

[4]

4.Volume cube = 203 = 8000B1

Square hole = 102 × 20 = 2000B1

Circular holes = 2 × 42 × 5 = 502.7B1

Volume left = 5500B1 ft

ft 8000 – 2000 – their cylinders. Only ft if attempt made to find volume circular holes not using 20 cm as height (e.g. 5 or 4 or 10)

cm3B1

Units mark independent

[5]

5.1 by 5 by 2 identifiedB1

or height = 2 or base = 1 by 5

2× (1 × 5+ 1 × 2 + 2 × 5)M1

oearea of 6 faces attempted

34A1

cm2B1

[4]

6.(a)( 7.5)2 11.6M2

M1 for ( 7.5)2 or 44.1(…) seen
or 44.1786() or 44.1562(3.14)

512.2 to 512.5…A1

or 512

× (3. …)2 × 11.6 scores M1
(7.5)2 11.6  2048 to 2051… SC1

(b)(circumference =) 7.5M1

23.56… or 23.55 if used 3.14

(their 23.56) + 1M1 dep

(their 24.56) × 11.6M1 dep

284.78 to 285A1

or(their 23.56) × 11.6M1 dep

add 11.6M1 dep

284.78 to 285A1

[7]

7. × 3.72(= 43.0...)
or 2 × 3.72(= 86.0...)M1

or 42.9...
or 85.9...
If 43 multiplied by 5.7 at any stage... M0
unless also used as an add on

2 × 3.7 × 5.7 (= 132.5...)M1

or 132.4...

2 (their 43.0) + (their 132.5)M2 dep

M1 for top missing
Dep on both M1s

218 to 220A1

[5]

8.() × radius2 × 3.2 = (320 × ())M1

radius2 = 100M1

10A1

[3]

9.C =  × 7M1

C = 2 × 3.5 Must substitute numbers.
C = d or 2r is M0 until used.
NB  × 3.5 is M0 as wrong method (r)

= 21.98 – 22A1

3.14 × 7 = 21.98, × = 22

Length = 22.98 to 23A1 ft

ft their 21.99 + 1 if M1 awarded.

Height = 10 cmB1

Allow answers transposed.

[4]

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