MATHEMATICS – SEC LEVEL
MS06
MARKING SCHEME – PAPER IIA
MAY 2006 SESSION
Question
Number
1(a)
(b)
Answer
Mark
4.5% --- Lm382.5
100% --- 382.5 ÷ 4.5 × 100
= Lm8500
M1
3km/h --- 30min
track is 1.5km long
5km --- 60min
1.5km --- 60 ÷ 5 × 1.5
= 18min
Notes
A1
Dividing by 4.5 and multiplying by 100
[or use formula and substitute values]
Accept 8500 [or 850 000c, unit given]
M1
1.5 (or equivalent)
M1
A1
Dividing by 5 and multiplying by 1.5
Total: 5 marks
2
AD̂E = AB̂E (angles on same arc)
ED̂F = CB̂E (Exterior angle = interior
opposite angle of a cyclic quadrilateral)
AB̂E = CB̂E (BE bisector)
AD̂E = ED̂F ; hence DE bisects AD̂F
M1
M1
Reason must be given
Reason must be given
A1
M1
No need for reason
Up to AD̂E = ED̂F enough
Total: 4 marks
3(i)
f(−6) = 2 − −36 = 2 + 2 = 4
(ii)
y = 2 − 3x
x
3
M1
= 2− y
Making
x
3
subject [not − 3x ]
M1
A1
Multiplying by 3 (or LCM method)
Or equivalent (using correct function
notation)
M1
Substitution of both functions
[×3] 9(2 − x) + 6 − x = 4
18 − 9x + 6 − x = 4
18 + 6 − 4 = 9x + x
20
20 = 10x; x = 10
M1
Multiplying by common denominator [or
LCM method]
x=2
A1
x = 3(2 − y)
f−1(x) = 3(2 − x)
(iii)
B1
f−1(x) + f(x) =
4
3
3(2 − x) + 2 −
x
3
=
4
3
M1
M1
Collecting like terms
Making x subject
No ft marks awarded here
Total: 9 marks
Page 1 of 5
MATHEMATICS – SEC LEVEL
4(i)
(ii)
AĈB common
A1
CÊD = CB̂A (or CD̂E = CÂB) (corresponding
angles)
A1
Reason must be given
M1
Establish relationship between
corresponding sides
x+1=6
M1
Simplification [removing
denominators and multiplication]
x=5
A1
AB = 6(5) = 30cm
BC = x + 1 = 6cm
DE = 2x = 2(5) = 10cm
A1ft
A1ft
A1ft
ft for incorrect value of x in (ii)
ft for incorrect value of x in (ii)
ft for incorrect value of x in (ii)
M1
M1
Substitution
Making CD subject from previous
statement
DE
CE
2x
2
(iii)
(iv)
MS06
=
=
AB
CB
6x
x +1
CD
CE
=
AC
CB
CD
2
=
27
6
CD = 2 × 27 ÷ 3
CD = 9cm
A1
Total: 11 marks
5(a)
x 3 − 7 x − 24 = 0
[x = 3.5] 3.53 − 7(3.5) − 24 = −5.625
[x = 3.6] 3.63 − 7(3.6) − 24 = −2.544
[x = 3.7] 3.73 − 7(3.7) − 24 = 0.753
[x = 3.66] 3.663 − 7(3.66) − 24 = −0.59
Hence root lies between 3.66 and 3.7
x = 3.7 (1 d.p.)
(b)(i)
Line A: y = 3
Line B: gradient =1; intercept = 0
y=x
Line C: gradient = 52 ; intercept = 3
y=
(ii)
2
5
x+3
y≥3
y≥x
y ≤ 52 x + 3 [ft from equation]
M1
M1
M1
Substituting any value 3 < x < 4
Substituting to obtain a better
approximation
Substituting to obtain a better
approximation
A1
Conclusion
B1
M1
A1
[Or equivalent]
[Or equivalent]
M1
A1
[Or equivalent]
B1
B1
A1ft
[Or equivalent]
[Or equivalent]
[ft for incorrect equation in (i)]
Total: 12 marks
Page 2 of 5
MATHEMATICS – SEC LEVEL
6(i)
(ii)
MS06
A = −5
C = 10
E = 23 or 0.66 or 0.67
B = −10
D=5
F = 12
[Refer to graph]
B1
B1
B1
For any two values
For any other two values
For remaining two values
A1
M4
For scale on both axes
1 mark for each 3 points plotted correctly
according to table of values of student
Overall shape obtained [Ends must not be
joined]
Accept 0
A1
(iii)
(iv)
y will approach 0
B1
when x = 0: y = 3
when y = 0: 4x + 3 = 0
4x = −3; x = − 34
[Refer to plotting of straight line]
(v)
From graph: x = −1; x =
1
4
M1
For attempting to obtain the coordinates of
at least 2 points on line [e.g. table of
values]
A1
A1
A1
For coordinates of x intercept
For coordinates of y intercept
For correct plotting of line on same axes
[Correct values only]
M1
Use graph to determine x coordinate of
points of intersection [or algebraic method
– obtaining quadratic equation]
A1
Accept either exact values only for
algebraic solution or −1.1 ≤ x ≤ −0.9 and
0.2 ≤ x ≤ 0.4 for graphical solution
[Accept in coordinate form – ignore y
coordinate]
Total: 16 marks
7(i)
Area ∆PBS =
1
2
× 6× 6
2
= 18cm
(ii)
Volume =
1
× 18 × 6
3
M1
A1
Using 6cm for base [or height]
M1
A1
For formula and substitution using ans. (i)
[accept verification method]
B1
M1
For 8 × volume of 1 pyramid
= 36cm3
(iii)
Volume of cube = 123 = 1728cm3
Volume of 8 pyramids × 36
= 8 × 36 = 288cm3
Hence 1728 − 288
= 1440cm3
M1
Subtracting volume of 8 pyramids from
volume of cube
A1
(iv)
6 + 8 = 14 faces
B1
B1
8
14
Total: 10 marks
Page 3 of 5
MATHEMATICS – SEC LEVEL
8(i)
a=9
c=2
e=2
(ii)
P(B, B, B) =
=
(iii)
(iv)
b=7
d=8
f=8
B3
1 mark for each 2 correct values
7
10
× 96 × 85
M1
A1
For multiplying respective probabilities
Or equivalent [accept 3 d.p or more]
7
10
× 96 × 83
M1
A1
Multiplying the 3respective probabilities
Or equivalent
M1
Using P(A) = 1 − P(A*)
M1
A1
For subtracting P(B, B, B) from 1
7
24
P(B, B, Y) =
=
MS06
7
40
P(at least 1 Y) = 1 − P(No Yellow)
= 1 − P(B, B, B)
7
= 1 − 24
=
17
24
Total: 10 marks
9(i)
In ∆ABC: AC2 = 232 + 262
AC =
(ii)
1205 = 34.71 = 34.7 cm
23 − 18 = 5cm
AD2 = 52 + 262
AD =
701 = 26.47 = 26.5 cm
tan 35 = DY
AD
DY = 701 × tan 35
(iii)
(iv)
M1
A1
For Pythagoras and substitution
B1
M1
A1
For (23 − 18) or 5
For Pythagoras and substitution
M1
For using tangent ratio and substitution,
including AD
Accept also 18.6cm [ft for incorrect AD in
(ii)]
DY = 18.54 = 18.5cm
A1ft
In ∆ACX: AĈX = 90 AX2 = AC2 + CX2 = 34.712 + 18.542
AX = 39.35 = 39.4cm
M1
M1
A1ft
For using ∆ACX as a right angled ∆ at C
For Pythagoras and substitution
[ft for incorrect AC and/or CX in (i)/(ii)]
M1
For any trigonometric ratio and respective
substitution
[ft for incorrect CX and/or AC in (i)/(ii)]
.54
tan XÂC = CX
= 18
AC
34.71
XÂC = 28.11 = 28.1
A1ft
Total: 12 marks
Page 4 of 5
MATHEMATICS – SEC LEVEL
MS06
10(a)
x
x +1
− x ( x2+1) +
2
x
=
x ( x )− 2 + 2( x +1)
x ( x +1)
=
x 2 − 2+ 2 x + 2
x ( x +1)
(
)
= xx (xx++12)
=
M1
M1
x+2
x +1
M1
M1
LCM and all numerators correct [If
denominator is crossed out award first M1
only and no other marks]
Expand all numerators [do not award for
trivial cases due to incorrect previous
step]
Factorise numerator [do not award if there
is no common factor due to previous
errors]
Dividing numerator and denominator by
common factor
A1
5th term = 5(5) + 3 = 28
B1
(ii)
2n + 4 = 3(28)
2n = 80
n = 40
M1
M1
A1
For establishing relationship
Making 2n subject
(iii)
216 = 2n + 4
2n = 212
n = 106
M1
For establishing relationship
(b)(i)
A1
Total: 11 marks
Page 5 of 5