Edexcel Gcse Mathematics 1387

EDEXCEL GCSE MATHEMATICS 1387

HIGHER TIER – MOCK PAPER 5 – MARK SCHEME

NC Ref. / Grade / No. /

Working

/ Answer / Mark / Notes /
NA2a / C
C / 1 / (a)(i)
(ii)
(b) / 2 ×17 ×19 / 0.323
17000
646 / 2
2 / B1 for 0.323
B1 for 17000
M1 for 2 ×17 ×19
A1 for 646
SSM3d / C / 2 / Tick under each of the three boxes
/ 3 / No more than three boxes ticked
B1 for
B1 for
B1 for
NA6a / C / 3 / 4n − 1 / 2 / B2 for 4n − 1
(B1 for 4n + k , where k is any integer)
NA4a / C / 4 / 3000 × 0.82 / 1920 / 3 / M2 for 3000 × 0.82
A1 for 1920
NA5e / C / 5 / (a) / 6x + 2 = 4x − 28
2x = − 30 / − 15 / 2 / M1 for 6x + 2 = 4x − 28 (or for − x = − 7 −)
A1 for − 15
B1 for (x + y)(x + y - 3)
NA5k / B / (b) / 15 − 2x = 4 × 3
15 − 12 = 2x / 1.5 / 3 / M1 for 15 − 2x = 4 × 3
M1 for 15 − 12 = 2x
A1 for 1.5 oe
NA5b / B / (c)(i)
(ii) / (x - 21)(x - 2)
2 and 21 / 3 / B2 for (x - 21)(x - 2)
(B1 for (x ± 21)(x ± 2))
NA5b / B / (d) / (x + y)(x + y - 3) / 1 / B1ft
HD3a / C / 6 / Unbiased suitable question with ‘response boxes’ / 2 / B1 for unbiased suitable question
B1 for suitable ‘response boxes’
SSM4c / C / 7 / Correct construction / 3 / B1 for 60° construction (with construction arcs seen)
M1 for angle bisector construction (intersecting arcs)
A1 for 30° within tolerance
NC Ref. / Grade / No. /

Working

/ Answer / Mark / Notes /
NA5j / C / 8 / (a) / −
/ −2, −1, 0, 1, 2,3 / 3 / M1 for −
A1 for − 2 and 3.
A1 for − 1, 0, 1, 2
NA5j / B / (b) / 5 + 11 > 5x − x / x < 4 / 2 / M1 for 5 + 11 > 5x − x oe isolation
A1 for x < 4
NA2c / B / 9 / (a) / Distance =
= / / 3 / M1 for
A2 for
(A1 for either or for , 1≤k<10)
B / (b) /
= 120 000 − 90 000 / 30 000 / 2 / M1 for either or 120 000 or 90 000 seen
A1 for 30 000
HD4a / B
B / 10 / (a)(i)
(ii)
(b) / 25 − 13 / 13
12
correct box plot / 2
3 / B1 for 13
B1 for 12
B1 for median marked at 20
B1ft for box ends at lq (13) and uq (25) (ft on (a)
B1 for end of whiskers at 4 and 34
NA5b / B / 11 / (a)(i)
(ii) / LHS =
= /
Printed result / 5 / B1 for
M1 for
A1 for a convincing completion to printed result
(iii) / a = 102.5, b = 97.5
a + b = 200 , a − b = 5
Require / 20 012.5 / M1 for either a + b = 200 or a − b = 5
A1 for 20 012.5
SSM2h / B
B / 12 / (a)
(b) / Angle PQR = 90° − 56° = 34° / Explanation
146° with reasons / 1
3 / B1 for PQT = 90° (radius is perpendicular to tangent)
B1 for PRQ = 90°, angle in a semi-circle
B1 for QRT = 56°, TR = TQ tangents from a point to a circle, so RQT is an isosceles triangle
B1 for 146°
NB A solution without any reasons can only get a maximum of 1 mark for 146°.
NA5h / A / 13 / (a) /
25 = 16k / / 3 / M1 for
M1 for 25 = 16k
A1 for
A / (b) / 6.25 / 1 / B1 for 6.25
A / (c) / / ± 2.4 / 2 / M1 for
A1 for ± 2.4 (accept either value)
NA3a / A / 14 / (a)(i) / 1 / 4 / B1cao
(ii) / / B1cao
(iii) / / B1cao
(iv) / 5 / B1cao
NA5b / A* / (b) / / 1 / B1
HD4a / A
A / 15 / (a)
(b) / 15
60
bar ht 31 width 20
bar ht 23 width 10 / 2
2 / B1 for 15
B1 for 60
B1 for bar ht 31 width 20
B1 for bar ht 23 width 10
SSM3c / A / 16 / Correct triangle drawn; vertices at
(3,−1); (4.5, −1);
(3,3.5) / 3 / B1 for lengths of all sides
B1 for a negative enlargement
B1 for correct position
NA3n / A* / 17 /
/ 3.5 / 3 / B1 for
M1 multiplying numerator and denominator by √8 oe
(oe eg reduces √2 ÷ √8 to ½)
A1 for 3.5
NA6e / A* / 18 / / − 0.5, 1.5 / 7 / M1 for denominator(s) (x+2)(x−1) or x2+x−2
M1(dep) for numerator(s) 7(x−1)+x+2
/ A1 for oe
8x − 5 = 4(x+2)(x−1) / M1 for multiplying through by (x+2)(x−1)
/ M1(dep) for correct attempt at expansion of terms and collection on one side to give a quadratic expression
/ A1 for
A1for − 0.5, 1.5
SSM2i / A / 19 / (a) / / 128p. / 2 / M1 for
A1 for 128p.
A / (b) / “128p. ” −
= 128p − 16p = 112p. / Printed result / 2 / M1 for “128p. ” −
A1 printed result convincingly found
A* / (c) / Area of top of frustum = 16p.
Area of bottom of frustum = 64p.
Slant length of small cone is 5 cm
Slant length of large cone is 10 cm
Curved surface area of frustum
= p × 8 × “10”− p × 4 × “5”
= 60p.
Total surface area = 16p.+64p.+60p. / 140p. / 5 / B1 for either (circular area) 16p. or 64p.
B1 for either slant length
M1 for p × 8 × “10”− p × 4 × “5”
A1 for 60p.
A1 for 140p.
HD4g / A* / 20 / P(green and 2 colours the same) = P(GRR)+P(GBB)=
P(blue and 2 reds) = P(RRB)+P(RBR)+P(BRR)
=
P(red and 2 blues) = P(BBR)+P(BRB)+P(RBB)
=
P(exactly 2 colours the same) = / / 5 / M1 complete relevant branches of a tree diagram oe
(can be implied by equivalent work)
M1 sum of at least 2 double products
A1 for oe
M1for either oe
or for
A1 for oe
NB Award alternative methods equivalent marks
SSM2g / A*
A* / 21 / (a)
(b)(i)
(ii) / When x = 0, y = 6 p + q = 6
When x = 180, y = 2 p − q = 2 / p = 4, q = 2
(240,3), (480,3) / 2
2 / M1 for either p + q = 6 or p − q = 2
A1 for p = 4, q = 2
B1 for (240,3)
B1 for (480,3)
NA5b / A / 22 / 2n and 2n + 2 , where n is an integer

=
is always a multiple of 8 so is never a multiple of 8. oe / Printed result proved algebraically / 4 / B1 for either 2n or 2n + 2
M1 for correct expansion
A1 for correct simplified algebraic expression with some factorisation
A1 convincing conclusion to the mathematical argument

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UG012657 – Edexcel GCSE Mathematics 1387: Mock Papers with Mark Schemes Higher Tier Paper 5 Mark Scheme