Download 1. (a) 64 2 213/27 M1 for 2(5+8) or 26 A1 cao (b) 1 1 B1 cao 2. (i) 1 1

1.
(a)
64
2
213/27
M1 for 2(5+8) or 26
A1 cao
(b)
1
1
B1 cao
[3]
2.
(i)
1
1
B1 cao
(ii)
1
16
1
B1 cao accept 0.0625
(iii)
64
B1 cao condone  64
1
[3]
3.
S, P, R, Q
2
B2 all correct
(B1 for 2 or 3 correct)
[2]
4.
(1, 1)(0, 1)
(1, 1) (0, 0)
(1, 0) (1, 1)
3
B3 for 6 points correct
(B2 for 3 points correct)
(B1 for 1 point correct)
NB: B1 for each additional point over six
[3]
5.
(a)
(10, 4, 0)
1
B1 cao
Glyn Technology School
1
(b)
(10 ÷ 2, 4 ÷ 2, 0)
(5, 2, 0)
M1 for two correct coordinates or for two of “10” ÷ 2, “4” ÷ 2,
“0” ÷ 2, ft from (a)
A1 ft from (a)
If the answer to (a) is correct, ie (10, 4, 0), then in part (b);
(5, 2, 0) gets 2 marks.
(5, 2, 4), (5, 4, 0), (10, 2, 0) all get 1 mark for two correct
coordinates.
(5, 4, 4), (10, 2, 8), (10, 4, 0) all get 0 marks.
If the answer to (a) is incorrect, for example (4, 10, 8), then in
part (b)
(2, 5, 4) gets 2 marks, following through; ie dividing each of the
coordinates by 2
(2, 5, 0), (4, 5, 4), (2, 6, 4) all get 1 mark for two “correct”
coordinates.
(5, 4, 4), (2, 2, 8), (4, 5, 0) all get 0 marks.
2
[3]
6.
(a)
2x + 6 +3x +18
5x + 24
2
M1 for 2 × x + 2 × 3 or for 3 × x + 3 × 6
A1 for 5x + 24 cao
(b)
3y(y – 4)
2
M1 for 3y(ay – b) or for 3(ay2 – by) or for y(3y – 12)
A1 for 3y(y – 4) cao
(c)
(t – 4)(t + 4)
1
B1 for (t – 4)(t + 4) oe
[5]
7.
(a)
15p – 6
1
B1 for 15p – 6
(b)
6x + 3 + 6x – 2
= 12x + 1
B2 for 12x + 1
(B1 for 12x or + 1 or 6x + 3 or 6x – 2)
Glyn Technology School
2
2
(c)
8.
(a – 8)(a – 8)
B2 for (a – 8)(a – 8) or (a – 8)2
(B1 for a in both brackets and two numbers
multiplying to 64 or –64)
Condone the missing trailing bracket.
x
2 x  3
2
3
x2 x  3
2 x  32 x  3
x
2x  3 
[B1 for x(2x + 3) seen
AND B1 for (2x  3)(2x + 3) seen]
B3 for
[3]
9.
(i)
76
2
180 – 2 × 52
B1 cao
(ii)
tgts to a circle are equal in length
B1 for mention of equal tgts to a circle from a point
[2]
10.
(a)
90
1
B1 cao
(b)
140
2
M1 for sight of 20°
or 360 – 90 – 90 – 40
A1 for 140°
SC: Award B1 for an answer or 220
(c)
2
B1 for Angle between tangent and radius = 90°
or Tangents from a point to a circle are equal
B1 for Isosceles triangle POQ so angle OQP = 20°
or Angles in a triangle add up to 180°
[5]
Glyn Technology School
3
11.
(a)
30
3
angle TPO = angle OPN = angle PNT
2x + x + 90 = 180
M1 for angle TPO = angle OPN = angle PNT
M1 (dep) for 2x + x + 90 = 180 (oe)
A1 cao
(b)
PS = SN
3
M1 for identifying triangle NOP as isosceles
M1 for identifying angle NSO or angle PSO as 90
A1 for use of symmetry together with result
OR M2 for 3 relevant statements
[M1 for 2 relevant statements]
A1 for use of congruent triangles together with result
OR M1 for tan ONS =
OS
OS
and tan OPS =
oe
SN
SP
[use of sine or cos accepted if isosceles triangle identified]
OS
OS
M1 for SN =
and SP =
tanONS
tanOPS
A1 for result
[6]
12.
(i)
27
4
Tangent 90 to diameter /radius/ line from (through) centre
B1 for 27 cao
B1 for reason
(ii)
63
180 (90+ “27”)
angle in semicircle (is 90)/Alternate segments
/angle at centre twice at circumference
B1 ft for 90”27” if not 63
B1 for reason
[4]
13.
1
(180 – 86) = 47
2
90 – 47 = 43
2
1
1
(180 – 86) or 47 or for 90 – ‘47’ or (180 – “94”)
2
2
A1 for 43 cao
M1 for
[2]
Glyn Technology School
4