NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICS P3
FEBRUARY/MARCH 2011
MEMORANDUM
MARKS: 100
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QUESTION 1
1.1
1.2
1.3
Mean
3,2 + 3,2 + 3,2 + 4,2 + 4,5 + 4,9 + 8,3 + 9,5 + 11,7 + 12,2 + 12,5
=
11
77,4
=
11
= 7,03
Median = 4,9
Mode = 2,3
Mode
This is the lowest value and will indicate that the increases are
very poor.
Mean.
This is the highest value and can be used to indicate that
increases are good.
9 Mean
9 Median
9 Mode
(3)
9 mode
9 reason
(2)
9 Mean
9 Reason
(2)
[7]
QUESTION 2
2.1
2.2
90 − 65
2
σ = 12,5
σ=
University A:
78 − 65 = 13
Her result lies just over 1 standard deviation from the mean.
University B:
x + σ = 54
x + 2σ = 59
Her result lies just over 2 standard deviations from the mean.
Her result for University B is better.
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9 method
9 answer
(2)
9 1 sd from the
mean
9 2 sd from the
mean
9 University B.
(3)
[5]
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QUESTION 3
3.1
36% Fall
63%
Rain
37%
(Rain & Fall)
9 63% Rain
9 36% Fall
64%
(Rain & Not Fall)
Not Fall
12%
99 structure of
the tree diagram
Fall
9 64% Not fall
(No Rain & Fall)
No Rain
88%
(No Rain & Not Fall)
Not Fall
3.2
3.3
⎛ 37 88 ⎞ ⎛ 63 64 ⎞
×
=⎜
×
⎟
⎟+⎜
⎝ 100 100 ⎠ ⎝ 100 100 ⎠
407 252
=
+
1250 625
911
=
1250
= 0,7288
37 12
P(Dry & Fall) =
×
100 100
111
=
2500
= 0,0444
P(Not Fall)
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9 88% Not Fall
(6)
37 88
×
100 100
63 64
9
×
100 100
9 answer
9
(3)
37 12
×
100 100
9 answer
9
(2)
[11]
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QUESTION 4
Average of trial
examination
80
68
94
72
74
83
56
68
65
75
88
Final examination mark
72
71
96
77
82
72
58
83
78
80
92
4.1
3 marks:
All points
plotted
correctly.
Final Examination Mark
Scatter Plot showing the trial examination mark vs final examination
mark
110
100
2 marks:
7 – 10
points
plotted
correctly
90
80
70
60
50
1 mark:
3–6
points
plotted
correctly
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90
(3)
100
Trial Examination Mark
99 a
9b
4.2 a = 25, 38
(25, 38342009…)
b = 0,71
(0, 7069044703…)
yˆ = a + bx
yˆ = 25,38 + 0,71x
4.3
9 answer
(4)
Scatter plot showing the trial examination mark vs final examination
mark
Final Examination Mark
110
100
90
9 slope
9 accurate
drawing
(2)
80
70
60
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90
100
Trial Examination Mark
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4.4 r = 0,74
4.5
5
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99 answer
(2)
9
substitution
9 answer
(2)
[13]
(0, 7391817008…)
yˆ = 25,38 + 0,71x
yˆ = 25,38 + 0,71(75)
= 78,63 %
If the original values of a and b then yˆ = 78,401
QUESTION 5
Broken a limb
Not broken a limb
TOTAL
463
b
782
Female
a
c
d
TOTAL
913
617
1 530
Male
5.1
a = 450
b = 319
c = 298
d = 748
5.2
5.3
P(Female who has not broken a limb)
298
=
1530
149
=
765
P(Female & broken a limb)
450
=
1530
5
=
17
= 0,2941176471...
= 0,29
P(Female) × P(Broken a limb)
748 913
=
×
1530 1530
= 0,29
The events of being female and having broken a limb are independent.
9 answer for a
9 answer for b
9 answer for c
9 answer for d
(4)
9 298
9 answer
(2)
9
463
1530
99
782 913
×
1530 1530
9 independent
(4)
[10]
If a candidate answers not independent due to the fact that the answers are
not accurate to more than 2 decimal places, award full marks.
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QUESTION 6
6.1
6.2
Number of different ways the shirts and trousers can be arranged
= (7 + 4)!
= 11!
= 39 916 800
9 11
9 11!
(2)
Number of ways so that the shirts are together and trousers are together
= 7!.4!.2
= 241 920
9 7!
9 4!
9×2
(3)
6.3
P(Shirt at beginning and trouser at the end)
9!×4 × 7
=
11!
14
=
55
9×4×7
9 9!
9 11!
9 answer
(4)
[9]
QUESTION 7
99 answers
7.1
7
4
–3
–5
–9
–32
–27
–113
–81
–356
(2)
–243
– 113; – 356
9
Tk +1 = Tk − (3) k
9 T1 = 7
9k≥1
7.2 Tk +1 = Tk − (3) k
T1 = 7
k≥1
OR
Tk +1 = Tk − 3(3) k −1 ;
OR
Tk = Tk −1 − (3) k −1 ;
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T1 = 7 ; k ≥ 1
T1 = 7 ;
(3)
[5]
k≥2
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QUESTION 8
A
2
2
D
8.
B
1
22°
1
C
O
9S
AO = OB
AO = BC
OB = BC
Ô1 = 22°
(radii)
(given)
B̂ 2 = 44°
(ext ∠ ∆ = sum int opp)
 = 44°
AOˆ D = 66°
(∠s opp = radii)
(ext ∠ ∆ = sum int opp)
9S
9 S/R
9S
(∠s opp = radii)
9S
9 answer
[5]
QUESTION 9
P
T
2
•
O1
R
S
9.1
9 construction
Join PO and OS
Let Ô 1 = 2 x
(∠ at circ centre = 2 ∠ at circumference)
(∠s round a point)
9 S/R
R̂ = 180° − x
(∠ at circ centre = 2 ∠ at circumference)
T̂ + R̂ = x + 180° − x
9 S/R
T̂ = x
Ô 2 = 360° − 2 x
= 180°
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9S
9S
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A
D
1
4
2 3
C
2
E
1
1
2
2
F
1
B
9.2.1(a)
9.2.1(b)
D̂ 4 = Ĉ
(∠s opp = sides)
Ĉ = x
DÊB = 180° − x
(∠ sum ∆)
(opp ∠ cyclic quad supp)
 = 180° − 2 x
(ext ∠ cyclic quad = int opp ∠)
9 S/R
9S
9 S/R
(3)
9S
9R
(2)
9.2.2
D̂ 1 + Â = Ê 1
(ext ∠ ∆ = sum int opp)
9 S/R
(∠ sum ∆) OR proved above
9 statement
(corres ∠s =)
9 Reason
D̂ 1 = x
Ĉ = x
D̂ 1 = Ĉ = x
DE || CB
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(3)
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QUESTION 10
A
36
24
F
48
D
40
E
B
G
40
C
10.1
10.2
EG 24
=
(DE || FG)
48 36
48 × 24
EG =
36
EG = 32 cm
BC AC
=
DE AE
120 × 40
BC =
48
= 100 cm
OR
AB AC
=
AD AE
120 × 36
AB =
48
AB = 90
∆ABC ||| ∆ADE
BC AB
=
DE AD
90 × 40
BC =
36
BC = 100 cm
OR
∆ABC ||| ∆ADE
BC AC
=
DE AE
120 × 40
BC =
36
BC = 100 cm
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9 S/R
9 answer
(2)
9 statement
99
substitution
9 answer
(4)
9S
(∠∠∠)
9S
(sides in proportion)
9 90
9 answer
(4)
(∠∠∠)
(sides in proportion)
9S
9S
9 substitution
9 answer
(4)
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QUESTION 11
DBE/Feb. – Mar. 2011
A
3
2
1
2
C
1
Let  1 = x
In ∆ABC and ∆ADT
 1 = B̂ 2 = x (tan ch th)
1.
1
2
B
B̂ 2 = Â 3 = x (AC || BD alt ∠s)
 1 =  3
11.2
(ext ∠ cyclic quad)
2.
T̂3 = BĈA
3.
B̂1 = D̂ 1
(3rd ∠ on triangle)
∆ABC ||| ∆ADT
(∠∠∠)
 1 = Ĉ 2 = x (tan ch th)
T̂1 = Ĉ 2 = x
(AC || BD; alt ∠s)
2
3
2
11.1
D
1
1
4
T
P
9 statement
9 reason
9 statement
9 statement
9 reason
9 statement
(6)
9 S/R
9 S/R
∴ T̂1 = Â 1 = x
T̂4 = x
T̂4 = Â 1
(= x)
PT is a tangent
(conv tan ch th)
OR
 1 = B̂ 2 =  3 = x
(AC || BT)
 3 = T̂1 = T̂4 = x
(∠s in same segment)
 1 = T̂4 = x
PT is a tangent
(conv tan ch th)
OR
B̂1 = T̂2
(∠s in same seg)
B̂1 = D̂1
(||| Δs)
D̂1 = T̂2
PT is a tangent
11.3
(vert opp angles)
(conv tan ch th)
In ∆APT and ∆TPD
1.
P̂ is common.
T̂4 = Â 1
(proven)
2.
AT̂P = D̂ 2
(3rd ∠ on triangle)
3.
∆APT ||| ∆TPD (∠∠∠)
9 Reason
(3)
9 S/R
9 S/R
9 Reason
(3)
9 S/R
9 S/R
9 Reason
(3)
9 S/R
9 S/R
9S
(3)
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11.4
AP PT
(∆APT ||| ∆TPD)
=
PT PD
AP.PD = PT.PT
1
AP. AP = PT 2
3
AP 2 = 3PT 2
11
NSC – Memorandum
DBE/Feb. – Mar. 2011
9 statement
9 reason
1
9 DP = AP
3
9 substitution
(4)
[16]
TOTAL: 150
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