[2] - Amazon Web Servicesverulam.s3.amazonaws.com/resources/ks3/maths/level 8 ans.pdf ·...

Hertfordshire La Schools 1

KS3 Revision work answers

Level 8

1. 100 1

6 1 [2]

2. (a) Gives a correct explanation 1

The most common correct explanations:

Show that the values 6, 8 and 10 work using

Pythagoras’ theorem

eg

• 62 + 8

2 = 36 + 64

= 100

= 102

• 102

– 82 = 100 – 64

= 36

= 62

Do not accept: explanation uses only accurate or scale

drawing

Accept : minimally acceptable explanation

eg

• 62 + 8

2 = 10

2

• 36 + 64 = 100

• The square of the longest side is equal

to the sum of the squares of the other

two sides

Do not accept: incomplete explanation

eg

• 62

+ 82

• 36 + 64


State or imply that the triangle is an enlargement of a 3, 4, 5

right-angled triangle

eg

• A 3, 4, 5 triangle is right-angled and 3 × 2 = 6, 4 × 2 = 8

and 5 × 2 = 10

• It’s just a 3, 4, 5 triangle with the lengths of the sides doubled

• Because 6, 8 and 10 make a Pythagorean triple

Accept: minimally acceptable explanation

eg

• It’s an enlarged 3, 4, 5 triangle

• 3 × 2 = 6, 4 × 2 = 8 and 5 × 2 = 10


eg

It’s like a 3, 4, 5 triangle

(b) Gives a correct justification 1

eg

• 6

9.6 × 8 = 9.2

• 8 × 1.15 = 9.2

• 9.2 ÷ 1.15 = 8

• 6.9 ÷ 9.2 = 4

3

6 ÷ 8 = 4

3

• 6 6.9 is a 15% increase

8 × 0.15 = 1.2

8 + 1.2 = 9.2

• tan–1

6

8 = 53.1...

6.9 × tan 53.1… = 9.2



eg

• 6

6.9 × 8

• 8 × 1.15

• 9.2

6.9 =

8

6


eg

• 9.2 ÷ 1.15

Do not accept: explanation attempts to use

Pythagoras’ theorem

eg

• 6.92 + 9.2

2 = 11.5

2

(c) Shows the digits 115 1

eg

• 1.15 × 108

• 115 000 000

• 11.5

Shows the correct value in standard form, 1

ie 1.15 × 108

! Zero(s) given after the last decimal place within standard form

notation

Condone

eg, for both marks in part (c) accept

• 1.150 × 108

[4]

3. Gives an integer value between 16 500 and 17 000 inclusive 2

eg

• 17 000

• 16 700

• 16 667

! Gives a non-integer value within the

correct range

eg

• 16 666.(…)

Condone


or Shows the digits 166(…) or 167 1

or

Shows a complete correct method with not more than one computational

or rounding error

eg

• 5000 ÷ 0.3

• 5000 ÷ 3 × 10

• 30

100 × 5000

• 5000 ÷ 30 = 200 (premature rounding),

200 × 100 = 20 000 [2]

question

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

5y

4

.

(

a

)

(

+

)

2

0

a

n

d

–

2

0

,

i

n

e

i

t

h

e


r

o

r

d

e

r

1

A

c

c

e

p

t

a

n

s

w

e

r

o

f

±

2

0

Accept answer of ± 20

(b) Gives a correct explanation 1

eg

The denominator is zero, and fractions with

denominators of zero are not defined

0

60 isn’t defined

Accept minimally acceptable explanation

eg

The denominator would be zero

You can’t divide by 0

There’s nothing to divide 60 by

0

60

! Use of ‘infinity’

Condone

eg, accept

The closer the denominator gets to 0, the

more the fraction tends towards infinity


Anything divided by 0 = infinity

0

60 =

Do not accept incomplete or incorrect

explanation

eg

It’s 0

60 and that’s impossible

Because 10 – 10 = 0

You cannot divide by zero and you

cannot find the square root of zero

The denominator would be zero but

0

60 = 60

0

60 = 0

(c) Gives a value less than 10 1

Accept correct set of values described

eg

x < 10

Less than 10 [3]

5. (a) Draws the correct triangle in any orientation 1

eg

▪

(b) Draws a correct shape in any orientation, ie 1


or

or

or

! Lines not ruled or accurate

Accept provided the pupil’s intention is clear

! Side lengths labelled

Ignore, even if incorrect [2]


6. (a) Gives a correct explanation that shows the correct application 1

of Pythagoras’ theorem

eg

(52 + 5

2) = (25 + 25)

5 × 5 + 5 × 5 = 50, so y = 50

y2 = 5

2 + 5

2

y2 = 50

y = 50

50

25

5

y 255

so = 50and = 50

yy

2

It’s an enlargement of a 1, 1, 2 triangle, so

it’s 52 and 52 = 50


eg

(52 + 5

2)

52+ 5

2

(25 + 25)

(2 × 25)

2 × 25 = 50 [with no evidence of a

misconception, eg about area]

5 2 = (52 × 2)

! Throughout the question, incorrect notation

or incorrect further working alongside a

correct explanation

Condone

eg, for part (a) accept

y2 = 5

2 + 5

2

y

2 = 50

y2 (error) = 50

5 × 5 + 5 × 5 = 50,

so length is 50 = 7.5 (error)


Do not accept incomplete or incorrect explanation

eg

y2 = 50

Use Pythagoras

25 + 25

It’s 2 × 25

5 × 10 = 50 and y = 50

Area = 5 × 5 × 2

= 50

5 × 5 = 25 which is half the square,

so 25 × 2 = 50

(b) Indicates 200 and gives a correct explanation 1

eg

250 = 4 × 50 = 200

The sides would be 10cm

(102 + 10

2) = 200

102 = 100 × 2 = 200

100 is 10, 10 ÷ 2 = 5 but the length of the diagonal of the small

square is >5

100 = 10, but 50 ≠ 5


eg

2 50 = 4 × 50

(102 + 10

2)

10 2 + 10

2

10 × 10 = 100, 100 × 2


eg

200 = 2 50

10 2

100 = 10

50 × 2 then × 2 again

Area = 10 × 10 × 2

= 200 [2]


7. Expressions

(a) For 2m indicates a correct simplified expression, eg: 2

18x3

18 (x3)

For 2m do not accept x3 not expressed as a power eg:

‘x × x × x × 18

For only 1m shows a correct simplified expression for the cross-sectional area, eg:

6x2

For 1m, the response need not state that 6x2 relates to the area, but do not accept 6x2 derived from incorrect methods.

or Shows a correct partially simplified expression for the volume, eg:

x × x × x × 18

6x2 × 3x

(8x2 – 2x2) × 3x

3x × (4 x2 + 2 x2)

(4 x2 + 2x × x) 3x

3x (2x × 4x – 2x × x)

4x2 × 3x + 6x2 × x

8x2 × 3x – 2x × x × 3x

Do not accept omission of brackets where required eg:

‘8x2 – 2x2 × 3x’

Do not accept an expression that has no simplification eg:

‘2x × 4x × 3x – 2x × x × 3x’

or Simplifies fully but makes one computational error, eg:

(2x × 2x + 2x × x) 3x = (2x2 + 2x2) 3x = 4x2 × 3x = 12 x3

Do not accept incorrect multiplication by x2 or x as a computational error.

(b) Indicates 90 1

Accept values of 90 multiples of 360, provided 90 is also shown.


(c) For 2m indicates 5 2

For only 1m shows a correct value for x3, eg:

x3 = 125

For 1m, accept implied values eg:

‘500/0.5 = 1000, ‚ 8 = 125’

or Makes one computational error then correctly follows through to solve for x, eg:

8x3 × ½ = 500, 8x3 = 250, x3 = 31.25, so x = 3.15

Values of x may be rounded or truncated to 1 or more d.p.

or Uses an incorrect value for sin a then correctly follows through to solve for x, eg:

8x3 × 0.45 = 500, so 8x3 = 1111 x3 = 139 so x = 5.2

Accept an incorrect value for sin a provided 0 < sin a < 1 [5]

8. (a) States a value between 30.50 and 32.00 inclusive, eg: 1

31

31.11

32

Accept a response given to one decimal place provided it is between 30.5 and 32.0 inclusive eg:

31.5

(b) States a value between 82 and 87 inclusive, eg: 1

85

(c) Shows on the grid vertical lines drawn down from the cumulative 1 frequency graph at the points corresponding to cumulative frequency values of 10 (or 10.5 or 10.25) and 30 (or 30.5 or 30.75) to meet the horizontal axis.

or

Shows on horizontal axis points corresponding to cumulative frequency values of 10 (or 10.5 or 10.25) and 30 (or 30.5 or 30.75).


Indicates a value between 9.50 and 11.50 inclusive. 1

Horizontal lines need not be drawn across from the vertical axis at points 10 (or 10.5 or 10.25) and 30 (or 30.5 or 30.75) to meet the cumulative frequency graph.

Ignore any lines drawn to find the median, and other lines drawn, provided it is clear that they do not relate to the interquartile range.

Do not accept a range given eg:

26 to 37

(d) For 2m draws a graph passing through the points (25,0), (30,1), (35,3), (40,6), 2 (45,10), (50,20), (55,27) and (60,30)

For only 1m draws a graph passing through seven of the eight points, eg:

Graph passes through all eight points apart from (25,0)

Graph passes correctly through seven of the eight points but goes through (55,28) instead of (55,27)

The graph may be a curve or a series of straight lines.

Do not accept two or more graphs drawn.

For 1m accept all eight points marked correctly but not all joined.

For 1m accept graph drawn through all eight points consistently 1 square to the left of the correct positions [apart from( 25,0)] ie

Graph passes through (24,0), (29,1), (34,3), etc.

Graph passes through (25,0), (29,1), (34,3), etc.

Ignore additional points marked on the grid through which no graph is drawn.

(e) Indicates statement A is true and statements B and C are false. 1

Do not allow follow through from an incorrect graph drawn as the correct information is given in the tables.

[7]

9. Solving x

For 2m indicates a correct value, eg: 2

22.5

For 2m accept 22 or 23 provided there is evidence of

a correct method.


For only 1m finds, in terms of x, at least 2 of the missing angles as shown below:

A

D

B C

5x

2x

2x

3x3x

or Forms a correct equation, eg:

8x = 180

2x + 3x + 3x = 180

The angles may be shown on the diagram or written elsewhere. Accept any unambiguous indication eg:

‘Angle A is 2x’

‘The other angle at B is 2 × x’

‘y = 2x’ (with angle DBC shown as y)

Accept an angle written as its complement from 180, eg:

‘180 – 6x’ for 2x

‘180 – 5x’ for 3x

Ignore incorrect angles.

Accept the correct computation as evidence of a correct equation eg:

‘180 ‚ 8’ [2]

10. (a) Gives a value between 0.65 and 0.68 inclusive 1

or equivalent probability

eg

1000

660 [0.66]


(b) Gives a value between 0.5 and 0.61 inclusive or 1

equivalent probability

eg

290

160 [0.5517…]

290

150 [0.5172…]

300

160 [0.5333…]

[2]

11. (a) 9.43 × 1012

1

! Zero(s) given after the last decimal place within standard form

notation

eg for part (a)

• 9.430 × 1012

Condone

(b) 7.35(54) × 1013

or 7.36 × 1013

or 7.4 × 1013

1

! For part (b), follow through

Accept 7.8 × their (a) provided this is

written correctly in standard form to at

least 2 s.f. [2]

heading 3;annotation text;annotation subject;Balloon

Text;mark;indent2;indent1;right;table;indent3;graph;graph Char Char Char

Char;question(a);heading 1;heading 2;heading

6;Default;CM30;CM31;CM32;CM33;CM1;CM34;CM37;CM4;CM5;CM6;CM7;CM36;

CM39;CM8;CM9;CM40;CM41;CM10;CM42;CM11;CM12;CM43;CM13;CM44;CM35;C

M45;CM46;CM47;CM48;CM14;CM49;CM50;CM15;CM51;CM16;CM17;CM52;CM53;

CM18;CM54;CM19;CM55;CM20;CM21;CM22;CM23;CM56;CM24;CM38;CM25;CM26

;CM27;CM28;CM57;CM29;CM58;CM3;CM2;macro;question(a)(i);indent1(a);indent1(a)(i)

;annotation reference;Mark Char Char; 12. 1.6 3

For 3m, do not accept equivalent fractions

or decimals


or Shows the value 98.4, 98.3(...) or 98 2

or

Shows or implies a correct method even if there are rounding

or truncation errors

eg

100 – 87.49

10034.297.20

20.97 × 2.34 = 49.07

49.87 – 49.07 = 0.8

87.49

8.0

(97.20

87.49 – 2.34) ×

87.49

97.20 × 100

34.2

87.49 = 21.(…),

.(...)21

97.20–.(...)21

Gives an answer that rounds or truncates to 1.6,

or is equivalent to 1.6

Shows the digits 16(...)

or Shows the number of people who did live in households 1

eg

49.0698 million

49.1 million

49.0(...) million

or

Shows the number of people who did not live in households

eg

0.8(...) million

800 200

800 000


or

Shows the number of households there would have been

if every person had lived in one

eg

21.3(...) million

For 1m, accept ‘million’ omitted

! Value of 49 (million) given as the number of

people who did live in households

For 1m, do not accept unless a correct

method or a more accurate value is seen [3]

13. Triangle

Forms a correct equation for the equal sides, and shows a correct first step of 1

algebraic manipulation, eg

a = 4b

b = a/4

8b = 2a

Forms a correct equation for the perimeter of the triangle, and simplifies, eg 1

3a + 14b = 91

5a + 6b = 91

22b + a = 91

26b = 91

6½ × a = 91

Gives both correct values, ie a = 14 and b = or equivalent, even if these 1

do not follow from a correct algebraic method

! Correct equation for the equal sides implied by equation for

the perimeter but not stated explicitly, eg

26b = 91

6½ × a = 91

Award both the first and second marks []

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

1 mark


(b) A fild vole is 40 weeks old.

Estimate the probability that it will live to be at least 50 weeks old.

14. (a) Gives a correct explanation 1


Show or imply that the median for group A is 26, and for group B is 29

eg

Median A – median B = 29 – 26

= 3

26 + 3 = 29 and A is 26, B is 29

! Median line referred to as the ‘middle’

or ‘centre’

Condone

eg, accept

The lines in the middle are at 26 and 29

The centre points of the boxes are

3mm apart


eg

26, 29

A is 29 – 3

B is 26 + 3

Do not accept incomplete explanation

eg

29 – 3

26 + 3

Indicate, in words or on the diagram, the locations of the medians

for A and B

eg

The vertical lines on the shaded part of the box plots represent the

medians and they are 3mm apart on the graph


eg

The lines in the shaded bit are 3 apart

The lines in the boxes are the medians

Arrows indicating both medians on the

diagram


Do not accept incomplete explanation

eg

The vertical lines are 3mm apart on the

graph

The lines for the medians are 3mm apart

on the graph

! Throughout the question, incorrect units

Condone

eg, for part (a) accept

The lines in the boxes are 3 cm apart

! Throughout the question, ambiguous notation

eg, for part (a)

26 – 29

eg, for part (b)

24 – 29 > 27 – 31

Condone

(b) Indicates A and gives a correct explanation 1


Show or imply that the inter-quartile range for A is 5 and for B is 4

eg

For A the IQ range is 29 – 24 = 5,

for B the IQ range is 31 – 27 = 4

The distance between 24 and 29 is greater than that between 27 and 31

The IQR is 1mm bigger for group A

! Inter-quartile range referred to as ‘range’

Condone

eg, accept

Range for A = 5, range for B = 4

The boxes show the range and A’s

is longer

Accept minimally acceptable explanation eg

5, 4

29 – 24 > 31 – 27

1 more



eg

5 is the larger inter-quartile range

31 – 27 is less

The inter-quartile range for A is 4 cm and for B is

3.2 cm [scale ignored]

Indicates, in words or on the diagram, the sizes of the inter-quartile ranges for A and B

eg

The shaded box in A is longer than in B, so A has a bigger inter-quartile range

The box for group A covers 6 whole numbers, but for B only 5


eg

The box is bigger

Distances between lower and upper

quartiles for both A and B indicated

It covers 6 numbers, the other covers 5

(c) Gives a correct reason 1

The most common correct reasons:

Refer to possible differences in the conditions of the two samples

eg

The two groups could have collected the samples at different times of year

Group A could have picked from one side of the tree and group B from the other side

One group could have picked from the tree, the other from the ground

Group B may have collected first and taken most of the larger ones

Accept minimally acceptable reason

eg

Different times

Different areas of the tree

B’s acorns may have had more

sunlight

Do not accept incomplete or incorrect reason

eg

Different areas

They used different trees


Refer to possible differences in the sizes of the two samples

eg

One group could have collected a much larger number of acorns than the other

One sample may be less representative as they didn’t collect enough

Accept minimally acceptable reason

eg

Different numbers of acorns

You don’t know how many acorns

Do not accept incomplete reason

eg

You don’t know how many

One group could have spent longer

There could have been more people to

collect acorns in one of the groups [3]

15. 36

1 or equivalent probability 2

! For 2m or 1m, values rounded or truncated

For 2m, accept 0.03, 0.028 or 0.027(...), or

the percentage equivalents

For 2m, do not accept 0.02 unless a correct

method or a more accurate value is seen

or Shows or implies a complete correct method, even if values 1

are rounded or truncated

eg

6

1

6

1

6

6

1 × 6

1

6

1

6

1

6

1

3

6

1

× 6

0.17 × 0.17

0.02


or

Shows or implies a correct method to find the

total number of possible outcomes

eg

216

6 × 6 × 6

3

6

1

or

Shows a correct method that uses explicitly the

fact that, in this case, the outcome of one dice is irrelevant

eg

It doesn’t matter what you throw on the first dice,

but the other two dice must

match it, so it’s 6

1 then

6

1

! For 2m or 1m, values rounded or truncated

For 2m, accept 0.03, 0.028 or 0.027(…),

or the percentage equivalents

For 2m, do not accept 0.02 unless a correct

method or a more accurate value is seen

For 1m, accept 0.17 or 0.16(…) for 6

1, or

the percentage equivalents

For 1m, do not accept 0.2 for 6

1 unless a

more accurate value is seen [2]

16. Angle proof

(a) Gives a correct explanation, eg 1

U1

BO and OA are radii, so triangle OBA is isosceles, so ABO = BAO. The same

is true of CO and OB, so BCO = OBC

Both triangles are isosceles because two of their sides are radii of the circle


Triangle AOB is isosceles because O is the centre and A and B are on the

circumference, and so is triangle BOC

Accept minimally acceptable explanation, eg

BO and OA are radii, so ABO = BAO. The same is true

of CO and OB

! Explanation correct but only refers to one of ABO or CBO

As the explanations are essentially the same for both angles, condone

Do not accept incomplete explanation that does not explain why

the triangles are isosceles, eg

OC = OB, so OCB = OBC, and the same for ABO

Both triangles are isosceles

(b) Gives a correct proof, eg 1

U1

x + x + y + y = 180

2x + 2y = 180

so x + y = 90 = CBA

Accept minimally acceptable justification, eg

x + x + y + y = 180, so x + y = 90 [4]


17.

100

0

0 10

Draws a complete correct curve within the tolerance 3

as shown above

For 3m, do not accept points joined with

straight lines for a curve

or Draws a curve within the tolerance as shown above 2

between (2, 50) and (5, 20), even if the curve is incorrect or

omitted elsewhere

or

Indicates at least 5 correct points on the graph, even if the points

are not joined or joined with straight lines


or Indicates at least 3 correct points on the graph 1

or

Gives the coordinates of at least 5 correct points with x values

greater than 0 but less than or equal to 10

! For 2m or 1m, points inaccurately plotted

Accept provided the pupil’s intention

is clear

! For 2m or 1m, points not explicitly plotted

Accept unambiguous indications of the

locations of points on the graph, for example

the tops of vertical lines

Note to markers:

The five points with integer coordinates are

(1, 100), (2, 50), (4, 25), (5, 20) and (10, 10) [3]

18. (a) Indicates False and gives a correct explanation 1

eg

▪ The median was about 44.5

▪ The median is at the 2500th value and when you read the graph down from that

value you can see it is greater than 40

▪ Only 1750 pupils got up to 38 marks and you need 2500 for the median

▪ About 1750 pupils scored 38 or less which is the 35th percentile

▪ Up to 38 is only 1750 pupils and that’s less than half

! Range of values

For the median on paper 1, accept 44 to 45 inclusive

For the position of the median, accept

2500 or 2500.5

For a value corresponding to a mark of 38,

accept 1700 to 1800 inclusive,

or 34% to 36% inclusive



eg

• 44 to 45 inclusive seen

• Correct value for the median on paper 1

marked on x-axis

• The 2500th mark is bigger than 38

• 1750 and 2500 seen

• 1750 and 35% seen


eg

• The 2500th value is not 38

• 38 is not in the middle of the cumulative

frequency

• 38 is too small to be the median

• Most pupils scored more than 38

(b) Indicates True and gives a correct explanation 1

eg

▪ The LQ is about 33.5

The UQ is about 56.5

56.5 – 33.5 = 23

or

Indicates either True or False and gives evidence that the inter-quartile

range is between 22 and 24 inclusive, excluding 23

eg

▪ The LQ is about 33

The UQ is about 57

57 – 33 = 24


! Range of values

For the lower quartile on paper 1, accept 33 to 34 inclusive

For the upper quartile on paper 1, accept 56 to 57 inclusive

For the position of the lower and upper quartiles,

accept 1250 or 1250.25 and 3750 or 3750.75 respectively


eg

• Correct values for the lower and upper quartiles on paper

1 marked on x-axis

• 33 to 34 inclusive and 56 to 57 inclusive seen

• From the 1250th to the 3750th marks is about 23


eg

• The lower quartile taken away from the upper quartile

gives 23 [no indication of quartiles on graph]

(c) Indicates False and gives a correct explanation 1

The most common correct explanations: U1

Use values from the graph

eg

▪ The median on paper 1 is 44.5, the median on paper 2 is 51.5, so paper 1 is harder

▪ About 850 pupils got less than 30 marks on paper 1 but only about 250 did on paper

2

▪ About 400 pupils got more than 65 marks on paper 1, but about 600 did on paper 2

Use or interpret the relative positions of the lines

eg

▪ The graph for paper 2 is always lower

▪ The dotted line is always on the right of the other line

▪ The marks on paper 2 were higher

! Range of values



For any other values on the x-axis, accept the correct

values ± 0.5

For corresponding values on the y-axis, accept the correct

values ± 50



eg

• The median on paper 1 is lower than the median

on paper 2

• More people got lower marks

[paper 1 implied]

• Fewer people got lower marks on paper 2

• More people got better marks on paper 2

• The line for paper 1 is higher

Do not accept: incomplete or incorrect explanation

eg

• Paper 2 was easier

• Everybody’s score is higher in paper 2 than in paper 1 [3]

19. Tiles

Gives a complete correct justification that encompasses all four 3

conditions below:

1. For the octagon, shows or implies that the interior angle is 135°,

or the exterior angle is 45°

2. For the square, shows or implies that the interior or exterior angle is 90°

3. For the hexagon, shows or implies that the interior angle is 120°, or

the exterior angle is 60°


4. Justifies why the hexagon will not fit

eg

135

120

135 + 120 + 90 360

135

135

135 120

4590

90 + 45 = 135°

which is 15° too big

135

240

135 + 90 = 225

but it should be 240

! Explanation does not identify, on the diagram or otherwise,

whether interior or exterior angles are being considered, or

to which shape the angles belong

For 3m, accept only if there is no redundant information and

the justification is unambiguous

eg, accept

90 + 135 = 225, 360 225 = 135 but the angle in a

hexagon is 120

360 (90 + 135) > 120


or

Shows at least one correct value from each of the following three sets 2

of angles, even if it is not clear to which shape the angle belongs

135 or 45

90

120 or 60

or

Shows or implies the ‘gap’ is 135°

eg

90 + 45 = 135

45

Shows at least one correct value from two of the following three sets of 1

angles, even if it is not clear to which shape the angle belongs

135 or 45

90

120 or 60

Accept 90 implied by a right angle symbol

! Explanation confuses the terminology of interior

and exterior angles

For 2m or 1m, Condone

Do not accept for 2m, incorrect angles marked or further

working indicates confusion between interior and exterior

angles

eg

Angle of 135 marked as 45

or

Shows at least one correct value from each of the following three sets of

angles, even if the angles are ascribed to incorrect shapes

135 or 45

90

120 or 60 U1 [6]


20. Gives all three correct values, ie 2

3

6

9

or Gives two correct values 1

! Incomplete processing

Withhold only 1m for the first occurrence

eg, for 1m accept

• 3

2 × 3

3 × 3

! For 1m, follow through

For the second value, accept their first value × 2, provided this

does not give a value of 0 or 2

For the third value, accept

their first value × 3 or

their second value × 2

3,

provided this does not give a value of 0 or 3 [2]

21. Births

(a) 1920 1

Accept unambiguous indication

eg

1.13 106

(b) 4.5 104 2

or

Shows or implies the value 45 000 1

eg

45 000

45 103

0.45 105

Do not accept incorrect value

eg

45 104

4.54

[3]


22. (a) Indicates y = –10 1

Accept equivalent equations.

Do not accept responses not given as an equation.

(b) Indicates, in either order, A and B. 1

Accept any unambiguous indication of the correct line eg:

through ‘(0, 15)’ and ‘(15, 0)’

(c) Indicates, on the diagram, the line y = x 1

Accept an unlabelled line only if no other lines have been drawn on the diagram.

The line must meet or cross BA and EF.

Accept a line which is not completely accurate as long as the pupil’s intention is clear.

(d) Indicates a correct equation, eg: 1

x = 0

y = 0

x = –y

y = –x

x + y = 0

Accept equivalent equations eg:

y = 0

y = x ×

x = y – 2y

x = 10 – 10

Ignore the original line given alongside correct equations. Otherwise, do not accept a restatement of the original line, y = x

Do not accept responses not given as a equation.

(e) For 2m indicates x = 35 and y = 20 2

For only 1m indicates either x = 35 or y = 20

or

Shows a correct method to find both variables, making only error.


(f) Indicates (35, 20) 1

Allow follow through from part (e) for numerical values only.

Do not accept unconventional notation eg:

(x = 35, y = 20) [7]

23. Gives all three correct expressions, ie 2

y + 15

2y

y + 3a

or Gives two correct expressions 1

U1

! Expressions unsimplified or use unconventional

notation

eg, for the third expression

• y + a + a + a

• 1y + 3 × a

Condone [2]

24. (a) Indicates missing values for Capacity of bucket are 24 and 30 1

Indicates first two missing values for Number of buckets 1 are 300 and 240

Indicates that missing value for Number of buckets corresponding to 1 capacity of 15 is 160

(b) Indicates an equation equivalent to T = BN or 2400 = BN, eg: 1

T = B × N

B = T

N

B × N = 2400

t = n × b = 2400

Expression eg:

BN


Equations with words eg:

T = B times N

T equals B × N

The use of letters other than those given in the question, apart from their lower case versions.

(c) For 2m indicates 5 hours 20 minutes. 2

For only 1m shows required computation is 4000 divided by 12.5 (or any equivalent division).

or

Shows in working the value 320

It is not necessary for the computation to be attempted.

For 1m Attempts to multiply 12.5 by particular values to obtain 4000 unless the value 320 is shown.

(d) Indicates 2 hours 30 minutes. 1

21

2 hours or 150 minutes.

(e) For 2m indicates 3 hours 20 minutes. 2

For only 1m shows a correct computation in working, or shows 3.33 or 31

3 in

working but incorrectly converts this to hours and minutes, eg:

2

3 × 5

5 ÷ 1.5

5 × 100

150

For 2m accept 200 minutes.

For 1m accept a response given as 3 hours 33 minutes,

or 3 hours 34 minutes, or 31

3 hours as evidence of correct

working.


(f) For 2m indicates 10 hours. 2

For only 1m shows in working that the given time is multiplied by 8, eg:

1hr 15 × 8

1.25 × 2 × 2 × 2

1.15 × 8

For 2m accept 600 minutes.

For 1m accept 9.2 or equivalent [11]

25. (a) For 2m indicates a value in standard form between 2.9 × 105 and 3.2 × 105 2

inclusive, eg:

3.055 × 105

3.1 × 105

For only 1m indicates a correct value between 290000 and 320000 not in standard form, eg:

0.0003055 × 109

305500

305555.56

For 1m accept a response using the E notation with a value between 2.9 and 3.2 inclusive. eg:

3.055 E 5

or Accept a response involving a value between 2.9 and 3.2 inclusive with (+)5 or (+)05 eg:

3.15

(b) For 2m indicates a value in standard form as x × 105 where x is a value 2 between 8.4 and 9.9 inclusive with no non-zero digits given for thousandths and below or

Indicates 1(.0) × 106, or 106

or Indicates a value as a number between 840000 and 1000000 inclusive with no non-zero digits given for hundreds and below, eg:

9.17 × 105

1.0 × 106

917000

1000000

For 2m accept a response not given in standard form where the equivalent answer in digits would have no non-zero digits for hundreds and below eg:


0.917 × 106

For only 1m indicates a value in standard form as x ×105 where x is a value between 8.4 and 9.9 with non-zero digits given for thousandths and below

For 1m accept a response not given in standard form where the equivalent answer in digits would have non-zero digits for hundreds and below eg:

0.9167 × 106

or

Indicates a value as a number between 840000 and 1000000 with non-zero digits given for hundreds and below, eg:

9.167 × 105

8.415 × 105

916666.6667

916700

For 1m accept a response using the E notation with a value between 8.4 and 9.9 inclusive eg:

9.16667 E + 5

or Accept a response involving a value between 8.4 and 9.9 inclusive with ( + )5 or ( + )05 eg:

9.16667 05

(c) For 2m indicates a value between 7.9 and 8.7 inclusive, eg: 2

8

81

3

For only 1m shows in working a correct computation for distance, or a correct computation relating to the distance travelled by sound in 1 second, eg:

1200

3600 × 25

12 10

60 60

3.

× 25

1.2 × 103 ÷ 3600

1200 ÷ 60 ÷ 60

20 ÷ 60 [6]

[2] - Amazon Web Servicesverulam.s3.amazonaws.com/resources/ks3/maths/level 8 ans.pdf ·...

Documents

Transcript of [2] - Amazon Web Servicesverulam.s3.amazonaws.com/resources/ks3/maths/level 8 ans.pdf ·...