PiXL Pre Public Examination, June 2016, 3H, Edexcel Style Mark Scheme
Qn
1
Working
Area of trapezium = 16/2 (14+18)
= 8x 32= 256 cm2
Area of two semicircles
𝐴 = 𝜋 x 3.5! = 38.48451001
Answer
Mark
218
4
Notes
P1 For start to process eg. radius = 14 ÷ 4 (= 3.5)
M1 Method to find area of trapezium or semicircle or circle
P1 Process to find area of the shaded region
A1 217.51.. – 38.48..
Area of shaded region = 25638.48451001 = 217.51549 = 218
3sf
2(a)
675 x 1.45075 = 979.25625 =
$979 to nearest dollar
(b)
135 ÷3 × 2 = 30 × 2
Or
60 ÷2 = 30 and 30 × 7 = 210
(c)
3(a)
6th term: 7+11 = 18
7th term: 11 + 18 = 29
8th term: 29 + 18= 47
979
2
shown
2
yes
1
47
1
M1 675 x 1.45075
A1 979 – 979.3
M1 For correct method to convert cost in UK to dollars or vice versa, using
Sanjay;s approximation
C1 Shown with correct calculations
C1For an evaluation
eg. It is a sensible start to the method because he can do the calculations
without a calculator and 1.5 dollars to the £ makes more sense to get
comparisons.
B1 cao
PiXL Pre Public Examination, June 2016, 3H, Edexcel Style Mark Scheme
Qn
3 (b)
(c)
Working
Answer
Mark
4th term: m + 2n+ 2n = m +4n =
5th term: m + 4n + m+2n = 2m
+6n
6th term: 2m + 6n + m+ 4n = 3m
+ 10n
shown
2
M1 Method to show by adding pairs of successive terms shown
C1 m + 4n and 2n + 6n and 3m + 10
m=1 n=2
3
P1 Process to set up two equations
M1 Process to solve equations
A1 m = 1 n = 2
As diagram
3
M1 Begin to interpret given information eg. 3 overlapping labelled ovals
with central region correct
M1 Method to communicate given information eg. 3 overlapping labelled
ovals with all regions inside correct
A1 Fully correct diagram
no
1
C1 No with valid reason- eg 6 is not a factor of 15 so it wouldn’t be in the
overlap
3m +10n = 23
3m + 6n = 15
4n = 8
n=2
Notes
m + 2n = 5 m + 4 = 5 so m = 1
4(a)
(b)
8
PiXL Pre Public Examination, June 2016, 3H, Edexcel Style Mark Scheme
Qn
5
Working
With the water meter
140 litres = 0.140 cubic metres
Cost of 0.140 cubic metre per
day: 0.140 x £1.95 = £0.273
Answer
Mark
Julie should have
a water meter
(from working
with correct
figures)
5
Cost over 365 days: 365 x £0.273
=£99.645
Notes
P1 Process to find number of cubic metres eg. 140 ÷ 1000
P1 Full process to find cost per day
P1 Full process to find total cost of water used per year (accept use of
alternative time period for both options)
P1 Full process with consistent units for total cost of water
A1 Correct decision from correct figures (204.465 or correct figure for
their time period)
Total cost £104.82 + £99.645 =
£204.465 approx £204.47
Since this is less than £393- it is
cheaper for Julie to have a water
meter,
6
11 3
x
4
2
!
4!
3
M1 convert fraction to mixed number
M1 multiply by reciprocal
A1 cao
7(a)
19,36,51,63,73,80
1
B1 cao
(b)
cf graph
2
M1 for at least 5 points plotted at each upper end of the interval (or
consistently plotted not at upper end) and joined
A1 correct graph
(c)
comparable value
and conclusion
3
0.000089
1
(x + 4) (x + 5)
2
M1 for (x ± 4)(x ± 5)
A1 (x + 4) (x + 5)
2
M1 for attempt to isolate terms in x in an inequality or an equation
8
9(a)
(b)
(c)
5x < 19
x<
!"
!
-2, -1, 0, 1, 2, 3, 4
M1 indication of reading taken from cf graph using weight = 3.4kg or find
UQ from 60
A1 for value given between 55 & 57 or 3.6 & 3.8
C1 (dep on M1) for conclusion
B1
A1 x <
2
!"
!
oe
M1 for attempt to divide by 2,
A1 -2, -1, 0, 1, 2, 3, 4
-2 ≤ p < 4.5 oe
10(a)(i)
(ii)
0- 2
-6 - 2
-2
-8
2
B1
B1
(b)
18 = 3x
6
2
M1 form and solve equation 16 = 3x – 2
A1 6
11
12(a)
𝑥=
−3 ± 3! − 4 1 (−6)
2(1)
−3 ± √33
2
-4.372 or 1.372
3
M1 Substitute into quadratic formula – allow sign errors
M1 evaluate as far as
13
!
A1
69
500
4
M1 Begin to interpret given information eg. 3 overlapping labelled ovals
with central region correct
M1 Extend interpretation of given information eg. 3 overlapping labelled
ovals with at least 5 regions correct
M1 Method to communicate given information eg. 3 overlapping labelled
ovals with all regions correct including outside
A1
(b)
!!± √!!
44
172
Shown
2
!""
oe
P1 For correct process to identify correct regions in Venn diagram or
!!
divide by '172' or !
A1
3
!"
!!
!"#
M1 one pair of equal sides or angles with reasons
M1 second pair of equal sides or angles with reasons
C1 proof completed correctly with full reasons and reason for congruence
14(a)
(F(x) = x2 – 7x – 3
F(7) = -3, F(8) = 5
Shown
2
M1 Method to establish at least one root
A1 Change of sign therefore at least one root in interval 7 ≤ x ≤ 8
(b)
x2 = 7x + 3 (÷ x)
Shown
1
C1 for at least one correct step and no incorrect ones
x=7+
!
!
B1 x1 = 7 +
!
!
oe
(c)
x1 = 7.42857....
x2 = 7.40384....
7.4
3
15(a)
5 x 4 = 20
20
2
(b)
P(57) = 1/20
Only one out of a
possible 20
26°
1
A1
5
P1 process to find length CH
A1 √109
P1 process to find angle ECH
M1 tan-1(their EH/CH) (or alternative method using sin or cos correctly)
A1 25 - 26°
16
150÷(3x10) = 5 = EH
CH2 = 102 + 32= 109
Tan ECH =
!
√!"#
ECH = 26°
M1 correct substitution to find x2
A1 cao
M1 5 x 4
A1 cao
!
!"
17(a)
(b)
5
3
!
y=-!+
!"
1
B1
4
B1 -1/”(a)”
P1 process of forming equation y = mx + c using their gradient
M1 finding constant
A1 cao
1
B1
!
18(a)
-9a + 12b
(b)
Shown
4
P1 process to find OM or YN
P1 process to find MN
A1 16b
!
C1 state proof eg MN parallel to XY as MN = !(XY)