Download Mark scheme - Solve My Maths

For 2m indicates a value between 7.15 and 7.35 inclusive, eg:
 '7.2'
 '7.21'
 '7.211
2
For only 1m shows in working the value 52 , 7 or 52
Accept trigonometric ratios used to calculate a value between
7.15 and 7.35 inclusive.
No marks are awarded for attempts to use trigonometric
rations leading to an answer outside the range 7.15 to 7.35
inclusive.
No marks are awarded where scale drawings have been used to
find an answer.
[2]
-
-
Prism
(a)
Gives a correct justification, eg:
 '62 + 82 = 102'
 '64 + 36 = 100'
 '102 – 64 = 36'
 '62 = 36, 82 = 64, 102 = 100 = 36 + 64'
 'Square the shorter sides and add, square root to get the big side.'
 'It's a 3, 4, 5 triangle.'
 'It's Pythagoras, a2 + b2 = c2' (sides labelled with hypotenuse as c)
 'Cos x = (82 + 62 – 102) ÷ 2 × 8 × 6 = 0, and cos 90 = 0'
Do not accept responses that state only that Pythagoras'
theorem holds, but give no further justification
eg 'It's Pythagoras.'
'It's Pythagoras, a2 + b2 = c2'
(no linking of a, b or c to diagram or given numbers.)
Do not accept incomplete or incorrect working
eg '62 = 36, 82 = 64, 102 = 100'
'62 + 82 = 102, 12 + 16 = 20'
'102 = 100, 62 = 36, 82 = 64, 100 + 36 + 64 = 200'
1
(b)
For 2m indicates 168
and
shows a complete correct method, eg:
 '6 × 8 ÷ 2 × 7 = 168'
 '7 × 6 = 42, × 8 = 336, ÷ 2 = 168'
For 2m or 1m do not accept incorrect methods,
eg '10 + 6 + 8 = 24, × 7 = 168'
'10 × 6 × 8 = 480, ÷ 2 = 240, × 7 = 168'
2
For only 1m indicates 168 but shows no evidence of an incorrect method, eg:
 '168'
 '24 × 7 = 168'
or
Shows a correct method with no computational errors but the answer is omitted, eg
 '6 × 8 ÷ 2 × 7'
 '6 × 8 ÷ 2 = 24, × 7'
or
Shows a complete correct method, including calculating the answer, with not more than
one computational error, eg:
 '6 × 8 ÷ 2 × 7 = 147'
 '6 × 8 = 56, ÷ 2 = 28, × 7 = 196'
Do not accept as a computational error a processing error such
as an incorrect height for the triangle
eg'6 × 10 ÷ 2 × 7 = 210'
or
Correctly finds the volume of the cuboid of which the prism is half, eg:
 '6 × 8 × 7 = 336'
 '336'
(c)
Indicates 120
-
-
1
[4]
(a)
For 2m states value between 242.0 and 242.1 inclusive, eg:
 '242.07436'
 '242'
 '242.1'
2
Accept trigonometric ratios used to calculate value between
242.0 and 242.1 inclusive.
For only 1m substitutes correctly into Pythagoras' theorem but calculates incorrectly, eg:
 '(150² + 190²)
= (30000 + 38000)
= 68000
= 260.8'
Award 1m for the response 240.
(b)
Attempts to use Pythagoras' theorem, eg:
 '150² = 22500
80² = 6400
x² = 80² + 15²
150² + 80² 170'
This mark is awarded if the cupboard diagonal is given as 170
with no working shown.
1
Calculates correct length of cupboard diagonal, eg:
 '170'
Accept trigonometric ratios used to calculate value between 169
and 171 inclusive.
1
Compares correctly the calculated length of cupboard diagonal with width of room,
1
-
eg:
 'Cannot be turned.' (diagonal calculated as 170)
 'No.' (diagonal calculated as 170)
 'Yes.' (diagonal calculated as 81.4)
It is not necessary for 165, the width of the room, to be
mentioned.
For this mark to be awarded a value between 169 and 171
inclusive must have been given as length of cupboard diagonal
of Pythagoras' theorem used in calculating an incorrect length.
[5]
Triangles
(a)
Correct explanation
The most common correct explanations are:
1
Showing or implying that 62 is added to 82, with either 102 or the use of , eg
 62 + 82 = 102
 62 + 82 = 100, 100 = 10
 10 is 100, and 100 = 64 + 36
 AB2 + BC2 = 100,  = 10
 AB2 + BC2 = AC2
Do not accept answer found through scale drawing
Do not accept in any part of this question.
Do not accept incomplete explanation that does not refer to either 102 or 
eg AB2 + BC2 = 100
36 + 64 = 100 = 10
or
Referring to the 3, 4, 5 triangle
 Each side is double 3, 4, 5
 It’s the 3, 4, 5 triangle.
 The 3, 4, 5 triangle must have a right angle.
Minimally acceptable explanation
eg 6, 8, 10 triangle
Do not accept incomplete explanation
eg Because of Pythagoras.
If it wasn’t 10 it wouldn’t be right-angled
a2 + b2 = c2 (without linking to the diagram).
(b)
136 or 11.7 or 11.6(..)
or
For only 1 Complete correct method, eg
 (62 + 102)
 136 = (incorrect)
!
Answer 12 or 11
Accept only if a valid method, or more accurate response, seen.
Accept use of tangent to find an angle, then correct use of sine
or cosine.
Do not accept Partial Method
If Pythagoras is used, the square root must be seen or implied.
Do not accept AD2 = 136 as sufficient.
(c)
6 or 5.9(..)
or
For only 2 Both angles correct, ie
x as 37 or 36.9 or 36.8(..), and
y as 31 or 30.9(..)
2
-
-
3
or
Complete correct method, eg
 tan-1 0.75 – tan-1 0.6
 cos-1 0.8 – cos-1(10  their b)
or
For only 1 One angle correct.
or
For only 1 Correct trigonometric ratios for both angles identified
 tan y = 0.6, tan x = 0.75
 tan = 6  10, tan = 6  8
-
!
Incorrect evaluation of angles
For 2, accept provided correct trigonometric ratios are shown,
and the angles are subtracted
eg tan x = 6  8, x = 43°
tan y = 6  10, y – 34°
43 – 34 = 9
Do not accept misunderstanding of trigonometric ratio
eg tan x = 6  8, x = 0.75°
[6]
(a)
For 2m indicates a value between 8.9(0) and 8.91 inclusive.
For only 1m shows a complete correct method
eg
 ‘(4.8² + 7.5²)'
 '79.29'
 '79'
 'tan–1 7.5 ÷ 4.8 = 57.3, 7.5 ÷ sin 57.3'
 '4.8 ÷ cos 57.4'
Tbroughout the question do not accept answers found through
scale or scaled drawings.
For 2m accept 9 only if a correct value or correct working is
seen.
For 1m if Pythagoras is used, the intent to square both, add and
square root must be shown, even if there are computational
errors. Do not accept 79(.29 .... ) without the intent to square
root shown or implied.
2
(b)
For 2m indicates a value between (0)56 and (0)56.5 inclusive
eg
 '056'
 '56'
Ignore any subsequent correction for magnetic north.
2
For only 1m shows a correct method,
eg
 'tan  = 6 ÷ 4'
 'tan–1 6/4'
 '62 + 42 = 7.22, cos x = 4 ÷ 7.2'
 'sin = 6 ÷ 7.211'
 '6/sin  = 7.2/sin 90
For 1m do not accept a wrong method, eg 'tan  = 4 ÷ 6' unless
it is clear, eg by labelling on the diagram or subsequent
subtraction from 90°, that the angle referred to is the one at
Bargate. If this angle is then correctly found as between (0)33.5
and (0)34 inclusive, award 1m.
(c)
For 3m indicates a value between 1.4(0) and 1.41 inclusive.
For only 2m shows the total distance north is a value between 5.4(0) and 5.41 inclusive.
or
Shows a complete correct method for the distance north of Bargate
eg
 '6 ÷ tan 48 – 4'
 '6 × tan 42 – 4'
c
N
a
42
b
4 km

48
a = tan 48 × 4 = 4.44, b = 6 – 4.44 = 1.56
c = tan 42 × 1.56'
 'd/sin 8.3 = 7.21 ÷ sin 48'
 'x ÷ sin 8 = 7.21 ÷ sin 48'
For only 1m shows a correct method for finding the total distance north
eg
3
 '6 ÷ tan 48'
 '6 × tan 42'
or
Using the sine rule, makes one error
eg
 'x ÷ sin 8 = 7.21 ÷ sin 56'
or
Using similar triangles, makes one error or correctly finds the values of a and b as shown
above.
[7]
Solving x
For 2m indicates a correct value, eg:
 '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
2x
D
5x
3x
2x
B
3x
C
or
Forms a correct equation, eg:
 '8x = 180'
 '2x + 3x + 3x = 180'
The angles may be shown on the diagram or written elsewhere.
2
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]
(a)
For 2m states correct value in standard form, eg:
 '4.3953 × 1047'
 '4.4 × 1047'
2
For only 1m shows in work that indices area added together, eg:
 'value is 5.98 + 7.35 × 10 24 + 22 = 13.33 × 1046'
 '5.98 × 1024 × 7.35 × 1022 = 43,953 × 1046 = 43.95346 = 4.3953 47'
Accept numbers rounded or truncated to one or more decimal
places.
For only 1m states correct answer but not in standard form eg
'43953 × 1046'
'4395.3 × 1044'
'43953 followed by 43 noughts'
'4.3953E + 47'
(b)
For 2m states correct value in standard form, eg:
 '1.5132 × 1011'
 '1.5 × 1011'
For only 1m shows in work that squaring 105 gives 1010, eg:
2
 '3.89 × 3.89 × 105 × 105 =3.89 × 2 × 1010 =7.78 × 1010 = 77800000000'
 '3.89 × 10 5 squared = 3.895 squared =15.132110 =1.5132111'
Accept numbers rounded or truncated to one or more decimal
places
For only 1m states correct answer but not in standard form eg
15.1321 × 1010
151321000000
1.51321E + 11
(c)
For 2m states correct value in standard form, eg:
 '3.5276888 × 1019'
 '3.5 × 1019'
2
For only 1m 3.5276888 or 3.5276889 or 3.5276882 (or one of these numbers rounded or
truncated to one or more decimal places) shown in working or given in answer, eg:
 '6.67 × 10–20 × 1.19 × 1055 = 7.973 × 10 35 ÷ 2.25 × 1016 = 3.5276888 × 1051 =
3.53 × 1051'
Accept numbers rounded or truncated to one or more decimal
places.
For 2m accept 3.5276889 × 1019 or 3.5276882 × 1019
For only 1m states correct answer but not in standard form eg
35276889 × 1012 , 35276882 followed by 12 noughts,
3.5276888E + 19,35.3 × 1018
[6]
(a)
Indicates a value between 3.67 and 3.68 inclusive, or 3.6 or 3.7, eg:
1
 '3.6781609'
 '3.7'
 '3.6'
 '3.67 × 10°'
 '0.3678 × 10'
 '0.3678 × 101'
Award the mark if answer is given as 4 (or about 3½ or about 4) where a
value between 3.6 and 3.7 inclusive is seen as the answer in the working.
(b)
For 2m indicates a value between 1496 and 1521 inclusive, eg:
 '1519.5312'
 '1.519 × 103'
2
 '1500'
For only 1m shows in working the use of a correct pair of ratios, eg:
x
50


10
389
.  10
128
.  109

x
389
.  1010

50 128
.  10 9
Award 1m if value given has as its first three non-zero digits
149, 15(0), 151 or 152
eg 151
0.15195312
15
14.96
Do not accept incorrect pairs of ratios eg
x
50

128
.  109 389
.  1010
[3]
(a)
For 2m indicates a value in standard form between 2.9 × 105 and 3.2 × 105
inclusive, eg:
 '3.055 × 105'
 '3.1 × 105'
2
-
-
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
between 8.4 and 9.9 inclusive with no non-zero digits given for thousandths and
below or
2
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:
 '8'
2
1
 '8 '
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
.  10 3
 '
× 25'
60  60
 '1.2 × 103 ÷ 3600'
 '1200 ÷ 60 ÷ 60'
 '20 ÷ 60'
[6]
(a)
Indicates the top statement
1
and
gives a correct explanation eg:
 '4 × 103 = 4000, 43 = 64'
 '4 × 103 means 4 × 10 × 10 × 10 which is greater than 4 × 4 × 4'
 'Just the 103 on its own is more than 43'
Accept any indication that 4 × 103 > 43, even if no statement is
ticked.
Provided the top statement is indicated, accept evaluation of 4
× 103 or 103 as sufficient
eg 4 × 103 = 4000
103 means × 1000
Do not accept an incorrect statement ticked, even if the
explanation is correct.
The explanation need not compute values for 4 × 103 or 43.
However, any computation must be correct hence do not accept
incorrect computations
eg 43 = 162 = 256
(b)
Indicates the last statement, ie 0.36 × 105
1
(c)
Indicates the first statement, ie 25 × 10–4
1
(d)
Indicates a correct simplified response eg
1
-
 '6'
 '6 × 100'
Indicates a correct simplified response eg
 '3 × 104'
 '30000'
1
The question does not ask for a response in standard form,
hence accept a correct number × 10 to the appropriate power
eg, for the second part
300 × 102
Accept a correct value written as a fraction with the
denominator 1, otherwise do not accept partial simplification
eg, for the second part
'
6 10 4
'
2
[5]
(a)
For 2m indicates correct probability, eg:
 '7/25'
 '0.28'
 '56 out of 200'
 '28 in 100'
2
For only 1m shows in working that the probability is calculated by adding the numbers in
the box together and dividing the result by 200, but gives an incorrect answer
or
gives an answer where the numerator or denominator involves fractions or decimals, eg:
 '5.6/20'
Accept for 2m equivalent fractions or percentages
eg 14/50
28%
Do not accept for 2m or only 1m answers in ratio form
eg 56 : 144
56 to 144
7 : 18
7 to 18
Do not accept for 2m or only 1m 28 without percentage sign.
(b)
For 2m indicates a value between 332 and 340 inclusive, eg:
 '336'
2
or
indicates the value obtained by correctly multiplying the probability given in part (a) by
1200, eg:
 '360' if 0.3 given as answer to(a)
or
correctly multiplies incorrect total of matchsticks calculated in part (a) by 6, eg:
 '330' if 55 matchsticks calculated in(a).
For only 1m shows in working that the value is calculated by multiplying the probability
given in part (a) by 1200, but performs this calculation incorrectly
or
shows in working that the value is calculated by multiplying the total number of
matchsticks found in part (a) by 6, but performs this calculation correctly.
Accept for 2m 'About ....'for any value within the range
specified.
Allow follow through from part (a) for 2m or only 1m.
[4]
(a)
Indicates correctly which statements are true and which are not true, eg:

Accept any indications.
Do not accept responses where one or more spaces are left
blank
eg
1
(b)
Indicates correctly which statements are true and which are not true, eg:

1
Accept any indications.
Do not accept responses where one or more spaces are left
blank
eg
(c)
States median number of marks between 36 and 38 inclusive, eg:
 '37'
 '361/2'
1
Accept 36 or 38
(d)
States 43
1
(e)
States 116
1
-
[5]
(a)
Indicates a correct probability eg:
 69
'
100
1
'
(b)
Indicates a correct probability eg:
 27
'
57
1
'
Accept decimals or percentages which round to 0.47 or 47%
(c)
Indicates the correct calculation eg:
1
 _ _ _ _
____
____
27 26
____

100 99
Accept any indication.
[3]
(a)
States a value between 30.50 and 32.00 inclusive, eg:
 '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
1
(b)
States a value between 82 and 87 inclusive, eg:
 '85'
1
(c)
Shows on the grid vertical lines drawn down from the cumulative
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.
1
-
-
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.
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
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
(d)
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]
Tests
(a)
For 2m completes the graph within the tolerance as shown on the overlay.
Note that the line from (0, 0) to (0, 10) has already been drawn,
ie is not part of the pupil's response.
For 2m or 1m the graph may be a curve or a series of straight
lines.
For only 1m marks all points correctly but does not join them or does not
join them correctly.
-
-
or
Marks, and joins, at least 3 points correctly.
The other point may be incorrect or omitted.
2
or
Makes an error in plotting one point then follows through correctly,
including joining the points.
or
Draws a series of columns or bars of the correct height
-
Ignore any column or bar drawn over (0, 0) to (10, 3)
(b)
For 2m completes the frequency graph, eg:
For only 1m shows 4 columns of the correct height, eg:
or
Shows 3 columns of the correct height with a total frequency of 30, eg:
or
Shows all columns of constant but incorrect height.
Points or lines need not be accurate provided the pupil's
intention is clear.
For 2m or 1m the horizontal lines connecting the columns need
not be drawn but may be inferred by correct points
eg, for 2m
-
2
[4]
(a)
Indicates 250
Do not accept responses which include operations
eg 6 ÷ 6 × 60
Indicates (0).1 or
1
10
1
1
Accept equivalent fractions or decimals.
(b)
Indicates a = 20
1
Indicates b = –5
1
Indicates c = 6 or –6 or both.
Do not accept redundant information, ambiguous responses or
partial evaluation
eg a = × 20
a = 100 ÷ 5
b = 5–
c = 6²
1
For 3m indicates x = 20 and y = 5
For 3m accept embedded values only if they are embedded in
both the given equations
eg 20 + 8 × 5 (= 60); 4 × 20 – 4 × 5 (= 60)
3
(c)
For only 2m gives only one of the correct values for x or y.
or
Gives values for x and y that satisfy one of the original equations and shows a complete
and correct algebraic method with only one error.
For 2m accept embedded values given in only one of the
equations provided no other embedded values are also
indicated.
For 2m values for x and y need not be integers. The other
positive integral solutions for x, y that satisfy x + 8y = 60 are:
4, 7
12, 6
28, 4
36, 3
44, 2
52, 1
60, 0
Solutions for 4x – 4y = 60 are such that
y = x – 15
For only 1m shows the correct multiplication of one of the equations by any constant eg:
 'x + 32y = 240'
 '8x – 8y = 120'
 'x – y = 15'
or
Gives values for x and y that satisfy one of the original equations.
or
Correctly rearranges one of the given equations eg:
 'x = 60 – 8y'
 '4x = 60 + 4y
 'x = (60 + 4y) ÷ 4'
 '4y = 4x – 60'
or
Correctly equates both equations and simplifies to 2 terms eg:
 '12y = 3x'
 '–4y + x = 0'
[8]
Rearrange
(a)
Rearranges correctly to make e the subject
eg
p –2f
 e 
2
1
 e  ( p – 2 f )
2
p
 e  – f
2
1
 e  – f  p
2
2
Accept Minimally acceptable correct rearrangement
eg e = (p – 2f) ÷ 2
e=p÷2–f
Do not accept For 2, incorrect equation
1
eg e  p – 2 f
2
or
Expands the brackets correctly
eg
 p = 2e + 2f seen
Do not accept p incorrectly multiplied by 2 at the same time as
the brackets expanded
eg 2p =2e + 2f
1
or
Divides by 2 throughout
eg
p

 e  f seen
2
or
Expands incorrectly to give p = 2e + f, then follows through correctly
eg
 p = 2e + f (error)
p– f
and so e 
2
p– f
Do not accept e 
without previous working shown
2
As there is no way of knowing how many errors were made, do
not accept
(b)
Rearranges correctly to make d the subject
eg
 d = c – 2r
Accept Minimally acceptable correct rearrangement
eg d = (2c – 4r) ÷ 2
2
d= c–
d=c–
r
0.5
r
1
2
or
Shows 2r – c = – d or
1
1
d= c–r
2
2
1
or
As a correct first step, multiplies by 2, or divides by a half, throughout
eg
 2r = c – d seen
r

= c – d seen
0.5

r
1
= c – d seen
2
[4]
Graphs
(a)
Gives a correct justification by referring to at least 2 points on the line
and indicates how at least one of the points relates to the equation, eg:
 '(0, 8) is on the line and 0 + 8 = 8, (2, 4) is also on the line and 2 × 2 + 4 = 8'
 
x
0
4
y
8
0
'2 of the x + the y = 8'
 'x = 1, y = 6, so 2x + y = 8, similarly for x = 2, y = 4'
 '6 + 2 = 8, 2 + 6 = 8' (with the points (3, 2) and (1, 6) marked on the diagram).
or
Gives a correct justification by referring to all points on the line, using one
point as an example, eg:
 'For every point on the line, 2x + y = 8, eg 2 × 5 – 2 = 8'
or
Gives a correct justification by referring to the negative gradient and to the
intercept of the line, eg:
 'The line has a gradient of – 2 and the intercept is at 8'
 'The line goes down 2 for every 1 across, and it crosses at 8'
Accept omission of brackets for co-ordinates.
-
1
The points may be identified either through the use of coordinates, or through the use of x and y, or through linking to
the diagram.
Do not accept responses which do not relate numbers to points
on the line
eg '6 + 2 = 8, 2 + 6 = 8' (with no indication of co-ordinates,
or use of x and y to link to the values 3 and 2 or 1 and 6,
or marking of (3, 2) and (1, 6) on the diagram).
Do not accept reference to points on the line without indication
of how at least one of the points relates to the equation
eg '(2, 4) and (4, 0) are on the line.'
'The line goes through (4, 0) and (0, 8)'
'x
0
4
y
8
0
'x = 1, y = 6, and x = 2, y = 4'
Do not accept explanations that refer to only one point without
reference to all points on the line
eg '(0, 8) is on the line and 0 + 8 = 8'
'For example, (3, 2) is on the line and 6 + 2 = 8'
Do not accept explanations which restate the question without
demonstrating the relationship
eg 'For each point, 2x + y = 8'
'Every point it's twice the x value added to the y value to give 8'
Do not accept an incomplete description of the line
eg 'Gradient = 2, intercept = 8'
'It goes through 4 on the x axis, 8 on the y axis.'
'It's going through x4 and y8 so 2x + y = 8'
-
-
(b)
Indicates a correct equation, eg:
 'x + y = 8'
 'y = 8 – x'
 'x = 8 – y'
 '2x + 2y = 16'
1
(c)
Draws a correct line within the tolerance as shown on the overlay.
The line should be at least 5cm in length.
Accept lack of label provided there is no ambiguity about which
is the correct line.
1
(d)
For 3m indicates x = 1½ and y = 4
For 3m accept x given as an improper fraction, otherwise do
not accept incomplete processing.
3
-
For only 2m gives only one of the correct values for x or y with no evidence of
an incorrect method or errors in finding that value.
For 2m or 1m accept incomplete processing, eg:
'x = 3 ÷ 2'
or
Gives values for x and y that satisfy one of the original equations and shows a complete
and correct algebraic method with not more than one error, eg:
 '3 × (2x + 1) = 6x + 3 so 6x + 3 = 4x + 6
so 2x = 3, answer x = 3, y = 7'
or
Draws the line 3y = 4x + 6, and correctly follows through from their intersection
for x and y.
The line should be drawn within an accuracy of ±2mm, be at
least 3cm in length and intersect with their line for part (c).
-
For only 1m shows in working the correct multiplication of one of the given
equations by any constant, eg:
 '2y = 4x + 2'
 '3y = 6x + 3'
 '6y – 12 = 8x'
 '6y – 12x = 3y – 4x'
 '3y = 8x'
or
Substitutes correctly for one variable, eg:
 '3(2x + 1) = 4x + 6'
 '6x + 3 = 4x + 6'
 '3y = 4(y – 1) ÷ 2 + 6'
or
Draws the line 3y = 4x + 6
-
The line should be drawn within an accuracy of ±2mm and be at
least 3cm in length.
[6]
Rectangles
Multiplies out one of the pairs of brackets correctly
eg,
 (y + 5) (y + 1) = y2 + 6y + 5 seen
 (y + 10)(y – 3) = y2 + 7y – 30 seen
1
!
Expansion is not simplified
eg, for (y + 5)(y + 1)
y2 + 5y + 1y + 5
Accept unsimplified expansions for the first two marks, but do
not accept for the third mark
Forms an equation equating the two areas and multiplies out the other pair of
brackets correctly
eg
 (y + 5)(y + 1) = (y + 10)(y – 3)
 y2 + 6y + 5 = y2 + 7y – 30
Accept Implicit equating
eg A = (y + 5)(y + 1)
A = (y + 10)(y – 3)
1
Simplifies their equation by at least removing terms in y2
eg
 6y + 5 = 7y – 30
 30 + 5 = y
 y = 35
!
Follow through from their incorrect equation
Accept provided it has terms in both y2 and y
1
35, with no incorrect algebra shown
Do not accept For this mark, do not follow through
!
35 with no supporting algebra
If there is no incorrect algebra, this final mark may be awarded.
Do not accept 35 from incorrect algebra
eg y2 + 6y + 5 = y2 + 7y – 30
1
-
-
7y + 5 = 8y2 – 30 (error)
35 = y
[4]
(a)
1m stated 325
Award 1m if 25 given but working shows this corresponds
to triangular number 325.
2
1m states 351
Award 1m if 26 given but working shows this corresponds
to triangular number 351.
Accept responses in either order.
Also award only 1m if both 25 and 26 are given but not
corresponding to 325 and 351.
(b)
1m states 2701
Award 1m if 73 given but working shows this corresponds to
triangular number 2701.
2
1m states 2775
Award 1m if 74 given but working shows this corresponds to
triangular number 2775.
Accept responses in either order.
Also award only 1m if both 73 and 74 are given but not corresponding
to 2701 and 2775.
[4]
-
-
(a)
For 2m indicates x = 3
For only 1m shows evidence of simplifying the equation correctly to fewer
than four terms, eg:
 '2x – 5 = 1'
 '4x = 2x + 6'
 '2x = 6'
 'x = 6/2
2
(b)
For 1m forms correct equation, eg:
 '2y + 5 = 23 – y'
 '23 – y = 2 × y + 5'
 '23 – y = y2 + 5'
1
Accept symbols other than y provided the same symbol is used
throughout the equation formed.
For 1m indicates 6
Accept 17 given as Aled's number provided the working clearly
shows 6 as the value chosen.
Accept follow through only from the incorrect equation
2y + 5 = y – 23, with y = –28 as answer.
1
For 3m indicates a = 4 and b = 7
3
(c)
For only 2m gives only one of the correct values for a or b
For only 1m shows in work the correct multiplication of one of the equations
by any constant, eg:
 '2a + 6b = 50'
 '6a + 3b = 45'
 '15a + 45b = 50a + 25b'
 '3a + 9b = 10a + 5b
or
Shows in work on variable correctly expressed in terms of other variable from
one or both of the equations, eg:
 'a = 25 – 3b'
 'b = 15 – 2a'
 'a = 2b – 10'
 'b = (a + 10) ÷ 2'
or
Uses an appropriate algebraic method to solve for a and b, making only one error
throughout
A variable correctly expressed in terms of the other variable
must be in the form a = ... or b = ...
Do not accept , for example, 2b = a + 10.
[7]
Simplify
(a)
Correct explanation
eg
a 2 – b 2 (a – b)(a  b)


a–b
a–b
Accept Minimally acceptable explanation
eg a2 – b2 = (a – b)(a + b)
!
Numerical substitution
Ignore if accompanying a correct algebraic explanation,
otherwise do not accept
1
(b)
a
1
Accept
(c)
a1
or
a1b0
a–b
2
or
Shows a correct partial simplification
eg
a 2 b – ab 2

(dividing through by ab)
ab
a 3 – a 2b

(dividing through by b2)
a2
a 2b 3
 a – 2 2 (partial fractions, first term simplified)
a b
1
-
-
Do not accept Incorrect simplification
a – a 2b3
eg
a 2b 2
[4]