level 8 ans
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 + 82 = 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 + 82 = 102 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 • • •
• •
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62 + 82 36 + 64
1
State or imply that the triangle is an enlargement of a 3, 4, 5 right-angled triangle eg
(b)
•
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 Do not accept: incomplete explanation eg It’s like a 3, 4, 5 triangle
Gives a correct justification
1
eg •
6 .9 × 8 = 9.2 6
•
8 × 1.15 = 9.2
•
9.2 ÷ 1.15 = 8
•
6.9 ÷ 9.2 = 6÷8=
•
3 4
3 4
6 6.9 is a 15% increase 8 × 0.15 = 1.2 8 + 1.2 = 9.2
•
8 tan–1 = 53.1... 6 6.9 × tan 53.1… = 9.2
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2
Accept: minimally acceptable explanation eg • •
6.9 ×8 6 8 × 1.15
6.9 6 = 9.2 8 Do not accept: incomplete explanation eg • 9.2 ÷ 1.15 Do not accept: explanation attempts to use Pythagoras’ theorem eg •
• (c)
6.92 + 9.22 = 11.52
Shows the digits 115
1
eg •
1.15 × 108
•
115 000 000
•
11.5
Shows the correct value in standard form, 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
1.150 × 108 [4]
3.
Gives an integer value between 16 500 and 17 000 inclusive
2
eg •
17 000
•
16 700
•
16 667 !
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Gives a non-integer value within the correct range eg • 16 666.(…) Condone
3
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
•
100 × 5000 30
•
5000 ÷ 30 = 200 (premature rounding), 200 × 100 = 20 000 [2]
question
y5 ...................... 4 . ( a ) ( + ) 2 0 a n d – 2 0 , i n e i t h e Hertfordshire La Schools
4
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 eg
1
The denominator is zero, and fractions with denominators of zero are not defined
60 isn’t defined 0 Accept minimally acceptable explanation eg The denominator would be zero You can’t divide by 0 There’s nothing to divide 60 by
!
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60 0
Use of ‘infinity’ Condone eg, accept The closer the denominator gets to 0, the more the fraction tends towards infinity 5
Anything divided by 0 = infinity
60 = 0
Do not accept incomplete or incorrect explanation eg
It’s
60
and that’s impossible
0
(c)
Because 10 – 10 = 0 You cannot divide by zero and you cannot find the square root of zero The denominator would be zero but 60 = 60 0 60 =0 0
Gives a value less than 10 Accept correct set of values described eg x < 10
1
Less than 10 [3]
5.
(a)
Draws the correct triangle in any orientation
1
eg ▪
(b)
Draws a correct shape in any orientation, ie
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1
6
or
or
or
! !
Lines not ruled or accurate Accept provided the pupil’s intention is clear Side lengths labelled Ignore, even if incorrect [2]
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7
6.
(a)
Gives a correct explanation that shows the correct application of Pythagoras’ theorem eg
1
(52 + 52) = (25 + 25) 5 × 5 + 5 × 5 = 50, so y = 50 y2 = 52 + 52 y2 = 50 y = 50
50 y
5
25
5 25
so y 2 = 50 and y = 50
It’s an enlargement of a 1, 1, 2 triangle, so it’s 52 and 52 = 50 Accept minimally acceptable explanation eg
!
(52 + 52)
52+ 52 (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
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y2 = 52 + 52 y2 = 50 y2 (error) = 50 5 × 5 + 5 × 5 = 50, so length is 50 = 7.5 (error)
8
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 eg
1
250 = 4 × 50 = 200 The sides would be 10cm (102 + 102) = 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 Accept minimally acceptable explanation eg 2 50 = 4 × 50
(102 + 102)
10 2 + 10 2 10 × 10 = 100, 100 × 2
Do not accept incomplete or incorrect explanation eg
200 = 2 50 10 2 100 = 10 50 × 2 then × 2 again Area = 10 × 10 × 2 = 200 [2]
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9
7.
Expressions (a)
For 2m indicates a correct simplified expression, eg:
18x3
18 (x3) For 2m do not accept x3 not expressed as a power eg:
2
‘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
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’
Simplifies fully but makes one computational error, eg:
(b)
(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.
Indicates 90 Accept values of 90 multiples of 360, provided 90 is also shown.
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1
10
(c)
For 2m indicates 5
2
For only 1m shows a correct value for x3, eg:
or
Makes one computational error then correctly follows through to solve for x, eg:
or
x3 = 125 For 1m, accept implied values eg: ‘500/0.5 = 1000, ‚ 8 = 125’
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.
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:
31
31.11
32
1
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:
(c)
1
85
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).
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11
Indicates a value between 9.50 and 11.50 inclusive. 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)
1
For 2m draws a graph passing through the points (25,0), (30,1), (35,3), (40,6), (45,10), (50,20), (55,27) and (60,30)
2
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. Do not allow follow through from an incorrect graph drawn as the correct information is given in the tables.
1
[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.
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12
For only 1m finds, in terms of x, at least 2 of the missing angles as shown below: A 2x
D
5x 3x 2x
3x
B
or
C
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 or equivalent probability eg
1
660 [0.66] 1000
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13
(b)
Gives a value between 0.5 and 0.61 inclusive or equivalent probability eg
160 [0.5517…] 290
150 [0.5172…] 290
160 [0.5333…] 300
1
[2]
11.
(a)
9.43 × 1012 ! Zero(s) given after the last decimal place within standard form notation eg for part (a)
1
• 9.430 × 1012 Condone
(b)
7.35(54) × 1013 or 7.36 × 1013 or 7.4 × 1013 ! For part (b), follow through Accept 7.8 × their (a) provided this is written correctly in standard form to at least 2 s.f.
1
[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. For 3m, do not accept equivalent fractions or decimals
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1.6 3
14
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 –
20.97 2.34 100 49.87
20.97 × 2.34 = 49.07 49.87 – 49.07 = 0.8 0.8 49.87 (
49.87 20.97 – 2.34) × × 100 20.97 49.87
21.(...) – 20.97 49.87 = 21.(…), 21.(...) 2.34
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 eg
1
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
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15
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 algebraic manipulation, eg
a = 4b
b = a/4
8b = 2a
Forms a correct equation for the perimeter of the triangle, and simplifies, eg
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 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
1
1
[]
............................... 1 mark
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16
(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
The most common correct explanations: 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 Accept minimally acceptable explanation 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 Accept minimally acceptable explanation eg The lines in the shaded bit are 3 apart The lines in the boxes are the medians
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Arrows indicating both medians on the diagram
17
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
!
(b)
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
Indicates A and gives a correct explanation The most common correct explanations:
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
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18
Do not accept incomplete or incorrect explanation eg 5 is the larger inter-quartile range 31 – 27 is less The inter-quartile range for A is 4 cm and 3.2 cm [scale ignored]
for B is
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 Accept minimally acceptable explanation 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
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19
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.
or
1 or equivalent probability 36 ! 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 Shows or implies a complete correct method, even if values are rounded or truncated eg
1
6 1 1 6 6 6
1×
2
1 1 6 6
1 1 6 6 3
1 ×6 6
0.17 × 0.17 0.02
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20
or Shows or implies a correct method to find the total number of possible outcomes eg 216 6×6×6 1 6
3
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 1 1 match it, so it’s then 6 6 ! 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
1 , or 6
the percentage equivalents For 1m, do not accept 0.2 for
1 unless a 6
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
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21
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]
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22
17. 100
0 0
or
10
Draws a complete correct curve within the tolerance as shown above For 3m, do not accept points joined with straight lines for a curve
3
Draws a curve within the tolerance as shown above between (2, 50) and (5, 20), even if the curve is incorrect or omitted elsewhere
2
or Indicates at least 5 correct points on the graph, even if the points are not joined or joined with straight lines
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23
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
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24
Accept: minimally acceptable explanation 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 Do not accept: incomplete explanation 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
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25
!
(c)
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 Accept: minimally acceptable explanation 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 Do not accept: incomplete explanation eg • The lower quartile taken away from the upper quartile gives 23 [no indication of quartiles on graph]
Indicates False and gives a correct explanation The most common correct explanations:
1 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 the median on paper 1, accept 44 to 45 inclusive For the median on paper 2, accept 51 to 52 inclusive 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
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26
Accept: minimally acceptable explanation 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 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°
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3
27
4.
Justifies why the hexagon will not fit eg 135 120
135 + 120 + 90 360
135 135 135 120
45
90
90 + 45 = 135° which is 15° too big
135 240
135 + 90 = 225 but it should be 240 !
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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
28
or Shows at least one correct value from each of the following three sets of angles, even if it is not clear to which shape the angle belongs
2
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 angles, even if it is not clear to which shape the angle belongs
1
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]
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29
20.
Gives all three correct values, ie
2
3 6 9 or
Gives two correct values ! 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
1
3 , 2 provided this does not give a value of 0 or 3 their second value ×
[2]
21.
Births (a)
1920
1 Accept unambiguous indication eg
(b)
1.13 106
4.5 104
2
Shows or implies the value 45 000 eg
1
or
45 000
45 103
0.45 105 Do not accept incorrect value eg
45 104
4.54 [3]
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30
22.
(a)
Indicates y = –10 Accept equivalent equations. Do not accept responses not given as an equation.
1
(b)
Indicates, in either order, A and B. Accept any unambiguous indication of the correct line eg: through ‘(0, 15)’ and ‘(15, 0)’
1
(c)
Indicates, on the diagram, the line y = x 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.
1
(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.
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31
(f)
Indicates (35, 20) Allow follow through from part (e) for numerical values only. Do not accept unconventional notation eg:
1
(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)
(b)
Indicates missing values for Capacity of bucket are 24 and 30
1
Indicates first two missing values for Number of buckets are 300 and 240
1
Indicates that missing value for Number of buckets corresponding to capacity of 15 is 160
1
Indicates an equation equivalent to T = BN or 2400 = BN, eg:
1
T=B×N
B=
B × N = 2400
t = n × b = 2400 Expression eg: BN
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T N
32
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 2 hours or 150 minutes. 2
1
(e)
For 2m indicates 3 hours 20 minutes.
2
For only 1m shows a correct computation in working, or shows 3.33 or 3
1 in 3
working but incorrectly converts this to hours and minutes, eg:
2 ×5 3
5 ÷ 1.5
5×
Hertfordshire La Schools
100 150 For 2m accept 200 minutes. For 1m accept a response given as 3 hours 33 minutes, 1 or 3 hours 34 minutes, or 3 hours as evidence of correct 3 working.
33
(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:
(b)
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
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:
Hertfordshire La Schools
34
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:
(c)
9.16667 05
For 2m indicates a value between 7.9 and 8.7 inclusive, eg:
8
8
2
1 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 × 25 3600
12 . 103 × 25 60 60
1.2 × 103 ÷ 3600
1200 ÷ 60 ÷ 60
20 ÷ 60 [6]
Hertfordshire La Schools
35
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 + 82 = 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 + 82 = 102 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 • • •
• •
Hertfordshire La Schools
62 + 82 36 + 64
1
State or imply that the triangle is an enlargement of a 3, 4, 5 right-angled triangle eg
(b)
•
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 Do not accept: incomplete explanation eg It’s like a 3, 4, 5 triangle
Gives a correct justification
1
eg •
6 .9 × 8 = 9.2 6
•
8 × 1.15 = 9.2
•
9.2 ÷ 1.15 = 8
•
6.9 ÷ 9.2 = 6÷8=
•
3 4
3 4
6 6.9 is a 15% increase 8 × 0.15 = 1.2 8 + 1.2 = 9.2
•
8 tan–1 = 53.1... 6 6.9 × tan 53.1… = 9.2
Hertfordshire La Schools
2
Accept: minimally acceptable explanation eg • •
6.9 ×8 6 8 × 1.15
6.9 6 = 9.2 8 Do not accept: incomplete explanation eg • 9.2 ÷ 1.15 Do not accept: explanation attempts to use Pythagoras’ theorem eg •
• (c)
6.92 + 9.22 = 11.52
Shows the digits 115
1
eg •
1.15 × 108
•
115 000 000
•
11.5
Shows the correct value in standard form, 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
1.150 × 108 [4]
3.
Gives an integer value between 16 500 and 17 000 inclusive
2
eg •
17 000
•
16 700
•
16 667 !
Hertfordshire La Schools
Gives a non-integer value within the correct range eg • 16 666.(…) Condone
3
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
•
100 × 5000 30
•
5000 ÷ 30 = 200 (premature rounding), 200 × 100 = 20 000 [2]
question
y5 ...................... 4 . ( a ) ( + ) 2 0 a n d – 2 0 , i n e i t h e Hertfordshire La Schools
4
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 eg
1
The denominator is zero, and fractions with denominators of zero are not defined
60 isn’t defined 0 Accept minimally acceptable explanation eg The denominator would be zero You can’t divide by 0 There’s nothing to divide 60 by
!
Hertfordshire La Schools
60 0
Use of ‘infinity’ Condone eg, accept The closer the denominator gets to 0, the more the fraction tends towards infinity 5
Anything divided by 0 = infinity
60 = 0
Do not accept incomplete or incorrect explanation eg
It’s
60
and that’s impossible
0
(c)
Because 10 – 10 = 0 You cannot divide by zero and you cannot find the square root of zero The denominator would be zero but 60 = 60 0 60 =0 0
Gives a value less than 10 Accept correct set of values described eg x < 10
1
Less than 10 [3]
5.
(a)
Draws the correct triangle in any orientation
1
eg ▪
(b)
Draws a correct shape in any orientation, ie
Hertfordshire La Schools
1
6
or
or
or
! !
Lines not ruled or accurate Accept provided the pupil’s intention is clear Side lengths labelled Ignore, even if incorrect [2]
Hertfordshire La Schools
7
6.
(a)
Gives a correct explanation that shows the correct application of Pythagoras’ theorem eg
1
(52 + 52) = (25 + 25) 5 × 5 + 5 × 5 = 50, so y = 50 y2 = 52 + 52 y2 = 50 y = 50
50 y
5
25
5 25
so y 2 = 50 and y = 50
It’s an enlargement of a 1, 1, 2 triangle, so it’s 52 and 52 = 50 Accept minimally acceptable explanation eg
!
(52 + 52)
52+ 52 (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
Hertfordshire La Schools
y2 = 52 + 52 y2 = 50 y2 (error) = 50 5 × 5 + 5 × 5 = 50, so length is 50 = 7.5 (error)
8
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 eg
1
250 = 4 × 50 = 200 The sides would be 10cm (102 + 102) = 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 Accept minimally acceptable explanation eg 2 50 = 4 × 50
(102 + 102)
10 2 + 10 2 10 × 10 = 100, 100 × 2
Do not accept incomplete or incorrect explanation eg
200 = 2 50 10 2 100 = 10 50 × 2 then × 2 again Area = 10 × 10 × 2 = 200 [2]
Hertfordshire La Schools
9
7.
Expressions (a)
For 2m indicates a correct simplified expression, eg:
18x3
18 (x3) For 2m do not accept x3 not expressed as a power eg:
2
‘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
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’
Simplifies fully but makes one computational error, eg:
(b)
(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.
Indicates 90 Accept values of 90 multiples of 360, provided 90 is also shown.
Hertfordshire La Schools
1
10
(c)
For 2m indicates 5
2
For only 1m shows a correct value for x3, eg:
or
Makes one computational error then correctly follows through to solve for x, eg:
or
x3 = 125 For 1m, accept implied values eg: ‘500/0.5 = 1000, ‚ 8 = 125’
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.
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:
31
31.11
32
1
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:
(c)
1
85
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).
Hertfordshire La Schools
11
Indicates a value between 9.50 and 11.50 inclusive. 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)
1
For 2m draws a graph passing through the points (25,0), (30,1), (35,3), (40,6), (45,10), (50,20), (55,27) and (60,30)
2
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. Do not allow follow through from an incorrect graph drawn as the correct information is given in the tables.
1
[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.
Hertfordshire La Schools
12
For only 1m finds, in terms of x, at least 2 of the missing angles as shown below: A 2x
D
5x 3x 2x
3x
B
or
C
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 or equivalent probability eg
1
660 [0.66] 1000
Hertfordshire La Schools
13
(b)
Gives a value between 0.5 and 0.61 inclusive or equivalent probability eg
160 [0.5517…] 290
150 [0.5172…] 290
160 [0.5333…] 300
1
[2]
11.
(a)
9.43 × 1012 ! Zero(s) given after the last decimal place within standard form notation eg for part (a)
1
• 9.430 × 1012 Condone
(b)
7.35(54) × 1013 or 7.36 × 1013 or 7.4 × 1013 ! For part (b), follow through Accept 7.8 × their (a) provided this is written correctly in standard form to at least 2 s.f.
1
[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. For 3m, do not accept equivalent fractions or decimals
Hertfordshire La Schools
1.6 3
14
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 –
20.97 2.34 100 49.87
20.97 × 2.34 = 49.07 49.87 – 49.07 = 0.8 0.8 49.87 (
49.87 20.97 – 2.34) × × 100 20.97 49.87
21.(...) – 20.97 49.87 = 21.(…), 21.(...) 2.34
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 eg
1
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
Hertfordshire La Schools
15
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 algebraic manipulation, eg
a = 4b
b = a/4
8b = 2a
Forms a correct equation for the perimeter of the triangle, and simplifies, eg
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 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
1
1
[]
............................... 1 mark
Hertfordshire La Schools
16
(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
The most common correct explanations: 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 Accept minimally acceptable explanation 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 Accept minimally acceptable explanation eg The lines in the shaded bit are 3 apart The lines in the boxes are the medians
Hertfordshire La Schools
Arrows indicating both medians on the diagram
17
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
!
(b)
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
Indicates A and gives a correct explanation The most common correct explanations:
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
Hertfordshire La Schools
18
Do not accept incomplete or incorrect explanation eg 5 is the larger inter-quartile range 31 – 27 is less The inter-quartile range for A is 4 cm and 3.2 cm [scale ignored]
for B is
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 Accept minimally acceptable explanation 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
Hertfordshire La Schools
19
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.
or
1 or equivalent probability 36 ! 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 Shows or implies a complete correct method, even if values are rounded or truncated eg
1
6 1 1 6 6 6
1×
2
1 1 6 6
1 1 6 6 3
1 ×6 6
0.17 × 0.17 0.02
Hertfordshire La Schools
20
or Shows or implies a correct method to find the total number of possible outcomes eg 216 6×6×6 1 6
3
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 1 1 match it, so it’s then 6 6 ! 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
1 , or 6
the percentage equivalents For 1m, do not accept 0.2 for
1 unless a 6
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
Hertfordshire La Schools
21
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]
Hertfordshire La Schools
22
17. 100
0 0
or
10
Draws a complete correct curve within the tolerance as shown above For 3m, do not accept points joined with straight lines for a curve
3
Draws a curve within the tolerance as shown above between (2, 50) and (5, 20), even if the curve is incorrect or omitted elsewhere
2
or Indicates at least 5 correct points on the graph, even if the points are not joined or joined with straight lines
Hertfordshire La Schools
23
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
Hertfordshire La Schools
24
Accept: minimally acceptable explanation 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 Do not accept: incomplete explanation 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
Hertfordshire La Schools
25
!
(c)
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 Accept: minimally acceptable explanation 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 Do not accept: incomplete explanation eg • The lower quartile taken away from the upper quartile gives 23 [no indication of quartiles on graph]
Indicates False and gives a correct explanation The most common correct explanations:
1 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 the median on paper 1, accept 44 to 45 inclusive For the median on paper 2, accept 51 to 52 inclusive 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
Hertfordshire La Schools
26
Accept: minimally acceptable explanation 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 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°
Hertfordshire La Schools
3
27
4.
Justifies why the hexagon will not fit eg 135 120
135 + 120 + 90 360
135 135 135 120
45
90
90 + 45 = 135° which is 15° too big
135 240
135 + 90 = 225 but it should be 240 !
Hertfordshire La Schools
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
28
or Shows at least one correct value from each of the following three sets of angles, even if it is not clear to which shape the angle belongs
2
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 angles, even if it is not clear to which shape the angle belongs
1
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]
Hertfordshire La Schools
29
20.
Gives all three correct values, ie
2
3 6 9 or
Gives two correct values ! 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
1
3 , 2 provided this does not give a value of 0 or 3 their second value ×
[2]
21.
Births (a)
1920
1 Accept unambiguous indication eg
(b)
1.13 106
4.5 104
2
Shows or implies the value 45 000 eg
1
or
45 000
45 103
0.45 105 Do not accept incorrect value eg
45 104
4.54 [3]
Hertfordshire La Schools
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22.
(a)
Indicates y = –10 Accept equivalent equations. Do not accept responses not given as an equation.
1
(b)
Indicates, in either order, A and B. Accept any unambiguous indication of the correct line eg: through ‘(0, 15)’ and ‘(15, 0)’
1
(c)
Indicates, on the diagram, the line y = x 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.
1
(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.
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(f)
Indicates (35, 20) Allow follow through from part (e) for numerical values only. Do not accept unconventional notation eg:
1
(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)
(b)
Indicates missing values for Capacity of bucket are 24 and 30
1
Indicates first two missing values for Number of buckets are 300 and 240
1
Indicates that missing value for Number of buckets corresponding to capacity of 15 is 160
1
Indicates an equation equivalent to T = BN or 2400 = BN, eg:
1
T=B×N
B=
B × N = 2400
t = n × b = 2400 Expression eg: BN
Hertfordshire La Schools
T N
32
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 2 hours or 150 minutes. 2
1
(e)
For 2m indicates 3 hours 20 minutes.
2
For only 1m shows a correct computation in working, or shows 3.33 or 3
1 in 3
working but incorrectly converts this to hours and minutes, eg:
2 ×5 3
5 ÷ 1.5
5×
Hertfordshire La Schools
100 150 For 2m accept 200 minutes. For 1m accept a response given as 3 hours 33 minutes, 1 or 3 hours 34 minutes, or 3 hours as evidence of correct 3 working.
33
(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:
(b)
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
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:
Hertfordshire La Schools
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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:
(c)
9.16667 05
For 2m indicates a value between 7.9 and 8.7 inclusive, eg:
8
8
2
1 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 × 25 3600
12 . 103 × 25 60 60
1.2 × 103 ÷ 3600
1200 ÷ 60 ÷ 60
20 ÷ 60 [6]
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