CML 2017 Marks EH1E PDF
CML 2017 Marks EH1E PDF
CML 2017 Marks EH1E PDF
Mathematics
Paper 1 (Non-Calculator)
Higher Tier
1 e.g. C S V
5 : 4 5 : 4 = 15 : 12
3 : 2 3 : 2 = 12 : 8 P1
Giving
C S V
15 : 12 : 8 Choc to Vanilla = 15 : 8 P1 A1 Total 3
√ 26 + 1.98 5+2
2 2 ≈ B1
(5.9) − 8.3 36 − 8
7
= M1
28
1
= or 0.25 A1 Total 3
4
4 . 6 6 1... P2
6 2 2 8 9 . 0 0 0...
2 4 8
4 1 0
3 7 2
3 8 0
3 7 2
8 0 £4.66 (to the nearest penny) A1 Total 4
4 (a) e.g. The median of 5 numbers will be the 3rd one when they are
arranged in order of size so it will be one of the numbers.
As all the numbers are odd, the median will be odd. C1
(b) e.g. No. For example, if we have 2, 4, 6, 8, 12, the total of the
numbers is 32 and the mean is 32 ÷ 5 = 6.4 which is not
an even number. C2 Total 3
3 5
5 1 – 8 = 8 of income not on rent
6 5
1 – 11 = 11 of rest of income is saved P1
5 5 25
Fraction saved = 11 × 8 = 88 P1 A1 Total 3
1
7 (a) 7
of 140 = 20
Let the no. of non-fiction paperbacks be x
The number of fiction hardbacks must be 140 – (80 + 20 + x)
= 40 – x
So 80 + (40 – x) = 80 + x + 10
120 – x = 90 + x
30 = 2x
x = 15 [Intuitive methods are easier and fine!]
ξ 20
P F
(20 and 80) C1
20 4
(b) 45
[= 9 ] M1 A1 Total 5
(b) C1
x
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9
x ≤ –4 or x ≥ 4 M1 A1
11 + 1
9 (a) f(11) = =6 B1
2
3
(b) =9 M1
x
3 = 9x
3 1
x= 9 = 3 A1
1 3
1 2
+1 2 3
(c) f( 2 ) = = 4=
2 2
1 3 3 4
gf( 2 ) = g( 4 ) = 3 = 3 × 3 = 4 M1 A1 Total 5
4
(b) e.g. Each year the previous year's value is multiplied by 0.63 M1
So new value = 63% of previous value
Annual % decrease = 100 – 63 = 37% A1 Total 4
4.2 10 4
(b) = ×
1.4 10−6
= 3 × 1010 M1 A1 Total 4
[Note, quick method: Gethin must have covered 10 miles in 12 minutes] Total 4
14 25x = (52)x = 5 2x B1
1 1
− −
3
125 3
= (5 ) 3
= 5 –1
7
So, 5 2x = 5 2 × 5 –1
7 5
+ (−1)
5 2x = 5 2 = 52 M1
5
2x = 2
5
x= 4 A1 Total 3
1.0
0.5
C1
0
0 10 20 30 40 50 60 Number
of sales
(d) e.g. It assumed that the days in the 30 to 40 class were split
evenly with half being below 35 and half being 35 or over C1 Total 7
18 Rearrange: 3x + 2y = 26
2y = 26 – 3x
3
y = 13 – 2 x
3
Gradient of tangent = – 2 P1
−1 2
Gradient of radius perpendicular to tangent = = 3 P1
( )
−
3
2
2
Radius passes through origin so equation is y = 3 x
2
Sub y = 3 x into 3x + 2y = 26 to find point on circle:
2
3x + 2( 3 x) = 26 P1
9x + 4x = 78
13x = 78
x=6
2
When x = 6, y = 3 × 6 = 4
Hence (6, 4) is point where tangent touches circle
Let radius be r : r 2 = 62 + 42 P1
r 2 = 36 + 16 = 52
Equation of circle is x 2 + y 2 = r 2
So, x 2 + y 2 = 52 A1 Total 5
19 (4x + a)(x – 2) = 4x 2 – 8x + ax – 2a P1
(2x + 1)2 = 4x 2 + 4x + 1
So, 4x 2 – 8x + ax – 2a ≡ 4x 2 + 4x + 1 + b
Hence: –8 + a = 4 P1
a = 12 A1
And: –2a = 1 + b
–24 = 1 + b
b = –25 A1 Total 4