grade 10 assessment guide - maths...
TRANSCRIPT
Grade 10 - 2 - Educator’s Guide 2008
Grade 10 Assessment Guide Learning Outcomes 1 and 2
1.1 Assignment : Functions - Rubric 3
1.2 Investigation: Number Patterns – Guide / Memo 4
1.3 Control Test: Products, Factors, Equations - Memo 6
1.4 Project: Finance - Guide 7
1.5 Exam A: Paper 1 - Memo 8
1.6 Exam B: Paper 1 - Memo 12
2 Learning Outcomes 3 and 4
2.1 Assignment: Shape, Space and Measurement - Memo 15
2.2 Investigation: Trigonometry - Memo 19
2.3 Control Test: Trigonometry, Measurement and Analytical Geometry - Memo 21
2.4 Project: Data Handling – Memo/Rubric 24
2.5 Exam A: Paper 2 - Memo 26
2.6 Exam B: Paper 2 - Memo 29
Grade 10 - 3 - Educator’s Guide 2008
Grade 10 Assignment: Functions Marks: 30
6 – 8 4 – 5 2 – 3 0 - 1
Accuracy of graphs
(If technology is used learners must
provide printouts of all graphs)
Accurate and Correct Throughout
Almost all Correct Some correct Mostly incorrect
Completion of Table
All Entries Correct Most Entries Correct Incomplete with Errors
Observations Made
Clear and correct explanations given for
all conclusions
Clear explanations given but conclusions
incomplete
Did not clearly explain the reasoning
No Attempt /Have given a Vague Description of Incomplete Conclusions
Correctness of expressions for linear, parabolic
and rational functions
Accurate and Correct Almost all Correct No attempt / Many
errors
Punctuality
Deadline met Deadline not met
/Negotiated Deadline met
Presentation Acceptable / not
acceptable
Grade 10 - 4 - Educator’s Guide 2008
Grade 10 Investigation: Number Patterns Marks: 100 Option 1
The sequence recurs from the 6th term, provided neither 1T nor 2T is zero.
If zero is chosen as one of the first two terms, in calculating subsequent terms we land up with a zero denominator, which makes it impossible to continue. Proof: Let aT =1 and bT =2
Then a
bT
13
+= and hence
ab
ab
ba
ab
ba
b
T++=×++=
++
= 1111
1
4
( ) ( )
b
a
b
a
ab
bab
b
a
ab
abab
a
bab
ab
T+=
+×+++=
+×+++=
+
+++
= 1
1
11
1
11
11
5
aab
ab
b
ba
ab
abb
a
T =++
×++=++
++
=1
11
11
6
ba
ba
b
aa
T =+
×+=++=
11
11
17
Then terms 3 to 5 are repeated and so on…
Grade 10 - 5 - Educator’s Guide 2008
Option 2 Conjectures:
1. If the first and third digits are the same, the difference is zero (and hence the sum of the difference and the reversed difference is also zero).
2. If the first and third digits differ by 1, the difference is 99 and the difference and the sum of the difference and the reversed difference is 99 + 99 = 198
3. For all other cases, the sum of the difference and the reversed difference is 1 089
Note: if the third digit is zero, this will result in a two digit number on reversal, but the conjectures above still hold. If the first digit is zero, the number will, strictly speaking, not be a three digit number and the given conditions will not be satisfied. Proof: Let the digits of the number be x, y and z
Then the value of the number is 100x + 10y + z ( )zx >
The reversed number has the value 100z + 10y + x
The difference is ( ) ( ) ( )zxxzzx −=−+− 99100
We can assume, without a loss of generality that zx > Conjecture 1: If x = z, then Difference = 0099 =× and the sum of zero and zero is zero! Conjecture 2: If 1+= zx Difference = 99199 =× and the sum = 99 + 99 = 198 Conjecture 3: If 2+= zx , then difference = 198299 =× , reversed difference = 891 and the sum is 1 089. If 3+= zx , then difference = 297399 =× , reversed difference = 792 and the sum is 1 089. If 4+= zx , then difference = 396499 =× , reversed difference = 693 and the sum is 1 089. If 5+= zx , then difference = 495599 =× , reversed difference = 594 and the sum is 1 089. If 6+= zx , then difference = 594699 =× , reversed difference = 495 and the sum is 1 089. If 7+= zx , then difference = 693799 =× , reversed difference = 396 and the sum is 1 089. If 8+= zx , then difference = 792899 =× , reversed difference = 297 and the sum is 1 089. If 9+= zx , then difference = 891999 =× , reversed difference = 198 and the sum is 1 089. There are no other options.
Grade 10 - 6 - Educator’s Guide 2008
Grade 10 Test: Products, Factors, Equations, Surds and Number Patterns Time: 1 hour Marks: 50 1.1 224 25309 yyxx +− ����������������
1.2 abba −−+ 326 ����������������
1.3 83 −p ����������������
1.4 662 − ��������
2.1 ( )( )44 22 −+ xx ��������
( )( )( )2242 −++= xxx ��������
2.2 ( )( )132 −− mm ����������������
3.1 Hypotenuse ( ) ( )221212 ++−= ��������
12221222 ++++−=
6= ��������
3.2 ( )( )1−+− yxyx ����������������
One side yx − , other side 1−+ yx
Hence perimeter = 2x-2y+2x+2y-2 24 −= x units ���������������� 4.1 ( ) 35103510 =×+=E ��������
4.2 nEn 35+= ��������
4.3 11635 =+ k �������� 37=∴k �������� 4.4 When n is any odd multiple of 5 ��������
5.1 Any valid set q
p
p
x, and
x
q ��������
Grade 10 - 7 - Educator’s Guide 2008
Grade 10 Project: Finance Marks: 75 Section A 1. R101,70 �������� 2. More: inflation since 2006, cost of shipping, profit for shop… �������� 3. In 2006: exchange rate less: not published in 2004 when rate was even more favourable. �������� 4. US$13,46 (to the nearest cent) ��������� 5. Best in 2005, worst in 2002. Any valid explanation. ��������� 6. Best in 2002, worst in 2005. Any valid explanation. ��������� 7. R100 = A$17 ��������� 8. 1A$=R5,88 (to the nearest cent) ��������� 9. R1=US$0,14 (to the nearest cent) ��������� 10. US$1=A$0,83 (to the nearest cent) ��������� 11. R649,52 In RSA ���������, A$76,36 in Aus.��������� 12. US$1=British Pounds(0,50)(to the nearest penny) or 0,49964…������������ 13. $3,20 = R 22,59 �������� 14. More expensive in USA �������� 15. R1 = $0,2645… or US$ 1 = R4,845375 Rand is undervalued. �������� Section B 1. R650,27 (toi nearest cent) ������������� 2. R684,79 (to nearest cent) �������������������� 3. Monthly payments=R27,79 (to nearest cent) �������������������� 4.1 R375,32 (to nearest cent) ��������������������
4.2 45,53812
08,01170
12
08,01160
12
08,01170
1224
R=
++
++
+ (to nearest cent)
������������������������������������
4.3 ( ) 26,0131067,1890 2 R= (to nearest cent) ���������
4.4 Total available =R913,77 Parents’ offer is better! ��������
Grade 10 - 8 - Educator’s Guide 2008
Grade 10 Mathematics Exam Paper 1 Time: 2 hours Marks: 100 1.1.1 = ��������
����
1.1.2 = ��������
����
����
1.2.1 = ��������
1.2.2 = ����
����
��������
2.1.1
����
or ����
2.1.2
��������
����
2.1.3 ��������
����
����
2.2 Let the distance from Cape Town to Stellenbosch be x km ����
Then time taken on trip to Stellenbosch = hours
Time taken on trip from Stellenbosch to Cape Town = hours ����
Time taken for return trip = hours ����
Grade 10 - 9 - Educator’s Guide 2008
hours ����
Distance covered on return trip = ����
Hence average speed is ����
3. Let the number of boys be and the number of girls be ����
Then (Eqn 1) ����
Then total mass = ����
But total mass = 1 270
(Eqn 2) ����
From eqn. 1: ����
Subst. Into eqn. 2: ����
���� and
Ms Shoko has 18 boys and 20 girls in her class ���� 4.1 The height of the water after 25 minutes is ���� 4.2 Height after 60 minutes = 3 + 12 ��������
4.3 Rising at minute ��������
4.4 Height after 5 minutes is ������������
4.5 Height will be 403 :when ���� ���� Time taken = 5 minutes ����
4.6 After 5 hours = 300 minutes: height = ��������
The water will not have overflowed ���� 5.1 NZ$ 1 = R
��������
(correct to the nearest R10)
5.2 Money from investments = ��������������������
She will need to save a further R (to the nearest cent) ����
Grade 10 - 10 - Educator’s Guide 2008
5.3 ��������
= ������������
����
6.1 Parabola: intercepts ������������ shape �������� Straight line: intercepts ����
6.2 ����
����
or
P is the point and Q is the point ��������
6.3 between the point and Q i.e for ��������
6.4 Dotted line ������������
7.1 ��������
��������
Grade 10 - 11 - Educator’s Guide 2008
7.2 lies on the graph (Eqn. 1) ����
. Also lies on the curve . (Eqn. 2) ����
From eqn.1
Subst. Into eqn. 2
����
and ����
7.3 The equation of the horizontal asymptote of is ��������
8.1 ; ����
8.2 period of = ����
8.3 and ��������
8.4 range of = ����
8.5 For ������������
Grade 10 - 12 - Educator’s Guide 2008
Grade 10 Mathematics Exam Paper 1 Time: 2 hours Marks: 100
No Solutions Comments 1.1.1 1.1.2 1.1.3 1.2 1.3.1 1.3.2 1.4 1.5.1
)35)(12( 2 −+− aaa
31192 23 +−+= aaa
3
2
3.2
3.2
3.2
6
3.2
11
11
11
=
=
=
−−−+
−+
−+
aaaa
aa
aa
a
aa
15
31415
10521915
)2(5)73(33
2
5
73
+=
+−+=
−−+=
−−+
x
xx
xx
xx
10
100033
=×=×
NM
NM
32
0)3)(2(
06
042
4)2)(1(
2
2
=−==−+
=−−
=−−−
=−+
xorx
xx
xx
xx
xx
3x = 50 33 = 27 and 34 = 81 ∴ 3 < x < 4 ∴ 56,3≈x
31
622
2422
<≤<≤
<−≤−
x
x
x
�� simplifying � answer
(3) � exponential law (writing as prime bases) � exponential law � answer
(3) �� numerator � denominator � answer
(4) �� dividing both sides by 5 � answer
(3) � multiplying � standard form � factors � answer
(4) �� critical values � answer
(3) �� critical values �� graph
(4 � correct step
1 3
Grade 10 - 13 - Educator’s Guide 2008
1.5.2 1.6
Step 2: )23(2)23(2 +≠+− kmkm 5(2+ 3k) – 2m(3k + 2) ............. (step 1)
= 5(2 + 3k) - 2m(2 + 3k).......... (step 2)
= (5 - 2m)(2 + 3k) .......................(step 3) 3) 3) 3) No: the numberline is dense everywhere so the number of rational numbers on any interval is infinite
� explanation (2)
� �
(2) �� explanation
(2) [30]
2 Let the number of km @ 4 km/h = x ∴ the number of km @ 5 km/h = 22 – x
12
1004885
55
22
4
=∴=−+∴
=−+∴
x
xx
xx
∴ 12 km @ 4 km/h and 10 @ 5 km/h
� �� setting up equation � simplifying � answer
[5]
3.1 3.2 3.3
17 cm
nTn 233−=
33 – 2n = 0 n = 2
116 hours
�� ��� �� �
[8]
4.1 4.2
12 −+= nnTn
2549
15050250
=−+=T
��� � subst. � answer
[5] 5.1.1 5.1.2 5.1.3 5.1.4 5.2.1
848,5 Swiss Francs R311, 91 R4761, 90 In Japan computer will cost R6 000 In South Africa computer will cost R8 500 ∴ import provided cost of importing is less than R2 500
57,27311
)1.1(7000
)1(5
RA
A
iPA n
==
+=
�� (2)
�� (2)
�� (2)
� � �
(3)
� formula �� subst � answer
Grade 10 - 14 - Educator’s Guide 2008
5.2.2
20011
)]12.0(51[7000
)1(
RA
A
niPA
=+=
+=
∴ compound interest wi ll be more profitable
(4) � formula � subst � answer
(3) [16]
6.1 6.2 6.3.1 6.3.2 6.4
xay = + q 39 a= + 1
a3 = 8 a = 2 q = 1 y = 1
12
1)( +
=x
xh or 12)( += −xxh
12)( −−= xxh
422
16
1712
−=
=+
x
x
∴ x = – 4
�� value of a �� value of q
(4) �� equation
(2) �� equation
(2) �� equation
(2) �� setting up equation � answer
(3) [13]
Grade 10 - 15 - Educator’s Guide 2008
7.1 7.2 7.3.1 7.3.2 7.3.3
72
7)( −= xxf
P(0 ; 9) R(3) PR2 = 92 + 32
= 90 ∴PR = 9,49 – 3 < x < 3 The graph will be narrower
�� equation (2)
� shape of graph � asymptote � x - intercept
(3)
� coordinates of P and Q � � �
(4)
�� (2)
�� (2)
[13] 8.1 8.2 8.3
a = 2 ; q = 1 x = 0o ; 180; 360o
180o
� � �� �
[5] 9.1 9.2 9.3
Motorist M1 travels at 150 km/h Motorist M2 travels at 100 km/h M1 : y = 150x The co-efficient of x (the gradient)
� � �� formula �
[5]
-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7
-5
-4
-3
-2
-1
1
2
3
4
5
x
y
Grade 10 - 16 - Educator’s Guide 2008
Grade 10 Assignment: Shape, Space and Measurement Marks: 75 NOTES FOR TEACHERS: 1. Learners require compasses, ruler, eraser, pencil and at least two unlined sheets of
paper. 2. Where the question requires learners to plot points on a grid, they should either use the
grid provided on the question paper or should be provided with grid paper. The grid must be square otherwise learners cannot make conjectures based on visual perceptions.
3. It is suggested that learners discuss their definitions of polygons with the teacher before proceeding with 4.1 – 4.8.
4. Where learners are asked to prove conjectures, any valid proof is acceptable. It does not have to match the proof in the memorandum.
1. Construction AB = 9cm and DC = 9cm; �������� AB//DC ���� (2)
1.1. parallelogram ���� (1)
1.2. AD=BC, DB,CA == (4)
Let diagonals intersect at O, then AO = OC, DO = OB (any four) ����������������
(Note: conjectures about alternate angles are not conjectures about ABCD)
1.3. 1 mark for each equal pair measured accurately��������������������
2. Construction of circle, radius 5 cm ���� (1)
2.1. 1 mark for each diameter. NB the diameter MUST be drawn accurately through circle centre. �������� (2)
2.2. PQ = 10cm.���� Diameter is twice radius. ���� (2)
2.3. RK, SK, QK.�������� All radii of circle and radii equal. ���� (2)
2.4. Isosceles.�������� PK = RK (from 2.3)���� (2)
2.5. Isosceles.���� PK = KS (from 2,3)���� (2)
2.6. x== SPKPSK ……….. base angles isos triangle
y== PRKRPK ……….. base angles isos triangle
°=∴
°=+∴°=+∴
90RPS
90
∆ofangles...........18022
yx
yx
(4)
P
K
S
Q
R
�
�
�
�
Grade 10 - 17 - Educator’s Guide 2008
2.7. SRQR,QSQ,SP �������� (2)
2.8. The proof is not complete, but the fact that SRQR,QSQ,SP are all right angles
can be proved in the same way that it was proved that °= 90RPS , in which case it would have been proved that PRQS is a rectangle (all angles °= 90 ). (2)
3. Plot F and G. ���� (1)
Note: Two possible solutions in given quadrant.
3.1. 5 units ���� (1)
3.2. Plotting E correctly ���� Explanation: 3,4,5 Pythagorean triangle used. �������� (3)
3.3. H plotted correctly. ���� (1)
3.4. All sides equal.���� (1)
3.5.
The diagonals have the same midpoint and therefore they bisect each other. (3)
�
�
H(10;4)
E(4;5)
F(1;1)
E(5;4)
H(9;5)
G(6;1)
)2;5(
215
;2
64EG ofMidpoint
=
++=
)2;5(
251
;2
91FH ofMidpoint
=
++=
)5,2;5,5(
214
;2
65EG ofMidpoint
=
++=
)5,2;5,5(
241
;2101
FH ofMidpoint
=
++= �
�
�
Grade 10 - 18 - Educator’s Guide 2008
4. Three or more sides; all sides equal; all angles equal. 4.1. No. ���� No equal sides. No equal angles.���� (2)
4.2. No. ���� Equal sides. Opposite angles equal.���� (2)
4.3. No. ���� One pair of opposite sides equal. Two pairs of equal angles.���� (2)
4.4. No.���� Two equal sides. Two equal angles.���� (2)
4.5. Yes.���� All sides equal. All angles = 90o���� (2)
4.6. No.���� Two pairs of equal sides. 1 pair of equal angles.���� (2)
4.7. No.���� Opposite sides equal. Opposite angles equal.���� (2)
4.8. Yes.���� All sides equal. All angles equal.���� (2) 5.
5.1. Isosceles (1)
A(-1;2)
B(0;-5)
C(4;7)
Grade 10 - 19 - Educator’s Guide 2008
5.2.
AB=AC therefore ∆ ABC is isosceles. (5)
5.3.
One solution above. Drawing: 1 pair equal sides;�� 2 pairs equal sides�� (4)
5.4. Marks awarded as follows: AD� AB� BC� DC� Both pairs of adjacent sides =, therefore a ABCD a kite� (5)
160
)75(4)-(0BC
50
)72(4)-(-1AC
50
))5(2(0)-(-1AB
22
22
22
=
−−+=
=
−+=
=
−−+=
�
�
� �
�
C(10;2)
D(1;4)
B(3;8)
A(0;7)
Grade 10 - 20 - Educator’s Guide 2008
Grade 10 Investigation: Trigonometry Marks: 85 1.1 ������������( for each error or omission)
1.2 ��������������������
1.3 ��������������������
��������������������
��������������������
1.4 ��������������������
��������������������
��������������������
2.1 x co-
ordinate y co-
ordinate
1,6 1,2 0,8 0,6 0,75
2,4 1,8 0,8 0,6 0,75
4 3 0,8 0,6 0,75
1,6 0,8
2,4 0,8
-3 4 0,8
1
1
1
����������������������������������������
Grade 10 - 21 - Educator’s Guide 2008
2.2 Each ratio is the same for the three points on the terminal ray of the same angle.
and etc… ������������������������������������
3.1 ������������������������
3.2 They are the same �������� 3.3.1 , , no minimum or maximum for ���������������������������������������� 3.3.2 any correct values �������� 3.3.3 Any correct values ���������������� 3.4 Signs of the three ratios in the three quads
for all and for all
����������������
Grade 10 - 22 - Educator’s Guide 2008
Grade 10 Test: Trigonometry, Measurement and Coordinate Geometry Time: 1 hour Marks: 50 1.
1.1. (2)
1.2. (2)
1.3. (2) 2. (4) 3.
3.1. (2)
3.2. (1)
3.3. (2)
4.
4.1.
(2)
°=
=
23.32158
sin
θ
θ
°=
=
20,6829
2cos
θ
θ
�
�
�
�
°=
=
70,363
5tan
θ
θ �
�
m
PQ
PQ
33,3140cos
24
2440cos
=°
=
=° �
��
�
mm
XY
6600
10066
=×= �
�
mm
XZ
500
24002900
=−=
°=
=
33,4ˆ6600
500ˆtan
ZYX
ZYX
�
�
�
A(-5;0)
B(3;4)
C(5;0)
M
O
)2;1(
204
;2
35
−=
++−=M
�
�
Grade 10 - 23 - Educator’s Guide 2008
cmh
h
h
hblv
3,81215
1500
12151500
=∴×
=∴
××=∴××=
8.025.1
0.1
83.05.1
25.1
=
=
4.2.
(3)
4.3.
102:
10
100
2)0;5(.
+−=∴=
+−=+−=
xyBCeqn
c
c
cxyintosubst
(4)
4.4. gradientsequalMOBC K// (2)
5. (5) 6.
6.1.
The scale factors are similar, but not exactly equal. (3)
6.2. 3150015005.1 cmmll ==
31250125025.1 cmmll ==
3100010000.1 cmmll == (3)
6.3. (3)
�
�
�
�
�
�
�
�
�
21
2
−=−
=MOofgradient
xyMOofEquation 2: −=
�
��
253
04
−=−−=BCofgradient
�
�
�
� �
�
ABAC
AC
AB
AB
31
22
31
22
13
)69()35(
34
343
306
)621()312(
≠∴=
−+−=
=∴
=
=
−+−= �
�
�
�
�
Grade 10 - 24 - Educator’s Guide 2008
22.484
7,71027,71321013
cm
Area
=
××+××+×=
3
2
3
1
4
))(2)(2(
))()((
x
xxxV
x
xxxV
=
==
=
3
22
22
3
221
)()(
4
)()(
y
yV
y
yV
y
y
ππ
ππ
=
=
=
=
6.4. length = 14cm; breadth = 11cm (2)
6.5. (3) 7. Cube Cylinder
In each case, the volume has increased by a factor of 4, so volumes will be equal.
� �
��
�
�
�
�
�
�
(5)
Grade 10 - 25 - Educator’s Guide 2008
Grade 10 Project: Introduction to Data Handling Marks: 55 Note to teachers Before the learners embark on this project, it may be worthwhile doing a class work exercise where they decide on the standard questions they will ask and develop a questionnaire. They will also need to warn their sample group in advance about their intentions so that they can measure their distances. Part 1
5 4 3 2 1 0
Logical method of recording, grouped
well, clear distinction between
subjects
Clear records, good, accurate
information collected, well
presented
Good records made
Records are present but randomly presented
Disorganized recording, messy,
incomplete
No attempt
Part 2
• Modal transport ����� Mean Time���� Median Time���� Mean Distance���� Median Dist���� (6)
• Not numeric, any reasonable explanation �������� • Data very varied, any reasonable explanation �������� • Suitable intervals �������� Correct groupings �������� Title of histogram ���� Labelling of axes ���� Correct representation of data ������������������������
• Correct answer read off histogram �������� • Good explanation, discussing possibly availability of public transport, wealth of learners, school environment, degree to which school is attended by those in the community etc ���������������� (28)
Grade 10 - 26 - Educator’s Guide 2008
Part 3 Scatter plot: Title ���� Axes �������� Plotting of points ���������������������������� Discussion: Correctly interpreting scatter plot, making mention of type of correlation, if any and good explanation for the reason ���������������� (10) Part 4
4 3 2 1 0
Well presented, accurate calcultions with accurate chart
Correct calcultions, correct
chart
Correct calculations, inaccurate chart
Inaccurate, incomplete
No attempt
Part 5
4 3 2 1 0
Good conclusion, tying up different
findings well, clearly understood
Good explanations, using some
findings
Discussion without using findings
Attempted, but inconclusive
No attempt
Grade 10 - 27 - Educator’s Guide 2008
(half distance accross base)
h
4
5
Grade 10 Mathematics Exam Paper 2 Time: 2 hours Marks: 100 Question 1
1.1 45
69
−−−=PQm ���� 1.2 3=∴⊥ PSmPSPQ ����
9
3
−= ���� 3
5
39 =−−
−∴x
��������
3
1−= ���� (3) ( )x−−=∴ 536
x3156 −−=∴ ����
7−=∴ x ���� (5)
1.31 3
1
17
13 −=+−
−=SRm �������� (2) 1.3.2 PQSR ����, same gradient ���� (2)
1.4.1 ( ) ( ) 409357 22 =−++−=SP ���� 1.4.2 SRPSPRSArea ⋅⋅=∆2
1
( ) ( ) 401317 22 =−++−=SR ���� 40402
1 ⋅⋅= ��������
SRSP ⊥ )( SRPQproven ���� 20= ���� (3)
PRS∆ is a right angled, isosceles triangle �������� (5) Question 2
2.1 26488 cmbaseArea =×= �������� (2) 2.2 20582
1 =××=∆Area ��������
��������
214464204 cmTSA =+×=∴ ���� (5)
2.3 2.4 3
heightbaseAreaV
×= ����
3643
364cm=×= �������� (3)
��������
cmh 31625 =−=∴ �������� (4)
Grade 10 - 28 - Educator’s Guide 2008
CB
A
25m
50°°°°
Question 3 3.1 L ���� 3.2 C ���� 3.3 G �������� 3.4 D �������� (6) ���� ���� ���� ���� 3.5 reflected about the x-axis followed by a reflection about the y-axis (or other way round)
or rotate through the origin by o180 (4) 3.6 ( ) ( )yxyx ;; −→ �������� (2)
���� ���� 3.7 reflection about xy = followed by reflection in the y-axis �������� or rotate through the
origin o90 anti-clockwise (4) Question 4
4.1.1 13
5sin =θ ���� (1) 4.1.2 12=AC (pythag) ����
4.1.3 169
144
13
12cos
22 =
=θ �������� (2) 6
5
12
52tan2 =⋅=∴ θ �������� (3)
4.2.1 30sin3
=BC �������� 4.2.2 30cos
3=CD
����
unitsBC 5,130sin3 ==∴ �������� (4) unitsCD 60,230cos3 ==∴ �������� (3)
4.2.3 60,2=BE (rect BCDE) ����
5,1=ED (rect BCDE) ����
20tan60.2
=AE ����
95,020tan60,2 =⋅=∴ AE ����
unitsAD 45,2=∴ ���� (5)
Question 5
o50ˆ =BCA (alt angles) ����
o50tan25 =BC
���� o50tan25 ⋅=∴ BC ����
BC=∴o50tan
25 ����
mBC 98,20=∴ ����
���� ∴shark is 15,98 m from swimmer Question 6 6.1 21770287 =−=range �������� (2)
Grade 10 - 29 - Educator’s Guide 2008
6.2.1 8
27528370270283283268287 +++++++=x ��������
375,252= ���� (3)
6.2.2 287,283,283,283,275,270,268,70 6.2.3 283mod =e ���� (1)
2792
283275 =+=∴median �������� (2)
6.3 mean has greatest increase ���� mode unaffected and median only slightly ���� although sum drops by 70, divisor drops too and therefore answer significantly changes ���� (3) Question 7 7.1
�������������������� (5) 7.2 Cell phone usage over one week ����
80 100 120 140 160 180 time in minutes ���� (5) 7.3 modal class: 140 – 159 ���� (1)
7.4 %83.7010024
17 =× �������� (2)
7.5 any acceptable comment based on data displayed �������� (2)
Time in minutes Frequency
80 - 99 4
100 - 119 3
120 - 139 6
140 - 159 9
160 - 179 2
F r e q u e n c y ����
��������
Grade 10 - 30 - Educator’s Guide 2008
Grade 10 Mathematics Exam Paper 2 Time: 2 hours Marks: 100
Question 1
1.1 (3)
1.2
(4)
1.3 (3)
1.4 (5)
1.5 )1;1()12;78( −=−+−=E (2)
1.6 BEAC , translation results in gradients
xinchange
yinchangebeing equal. (2)
1.7 Transformation from A to B: down five units, left five units Transformation from C to E: down five units, left five units. (2)
1.8 ACEB is a parallelogram because both pairs of opp. sides are parallel. (2)
A(-3;7)
C(4;6)
B(-8;2)
D(-2;k)
3
112
4
)8(4
26
=
=
−−−=BCofGradient
4
37
13
7
1)3(2
7
3
1
1
=−=−
−=−
−=−−−
−×
−=×
k
k
k
k
mDAmBC
65,12
160
)26()8(4( 22
==
−+−−=BCLength
2
22
20
101605,0
16,3
10
)47())2(3(
units
ABCArea
ADheightLength
=
××=∆
==
−+−−−=
�
�
�
�
�
�
�
�
� �
�
�
�
�
�
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�
�
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Grade 10 - 31 - Educator’s Guide 2008
Question 2
2.1
2.1.1 (3)
or
2.1.2 (2)
2.2
2.2.1 unitsbaseofPerimeter 36= (1)
2.2.2 The slanted triangles are congruent to the base. (1)
2.2.3 (3)
234,62
39,10125,0
units
Area
=
××=∆
39,10
108
612
:
,
222
==
−=
⊥∴∆
h
h
PythagorasUsing
RQofbisectorishlequilateraisPQR
12
12
12
R Q
P
h
cm
h
h
39.10
60sin12
60sin12
=°=
°=
�
�
�
�
�
�
236,249
34,624
4
units
AreaareasurfaceTotal
=×=
∆×=
�
�
�
�
�
Grade 10 - 32 - Educator’s Guide 2008
Question 3
3.1.1 On diagram (2) 3.1.2 On diagram (2) 3.1.3 On diagram (2) 3.1.4 On diagram (4) 3.1.5 A is moved six units right and six units down and then reflected about the x-axis. (2) Question 4 4.1 (4) 4.2
4.2.1 125
tan =A (1)
4.2.2 53ˆsin =DBC (1)
4.2.3 (3) 4.3 (2)
12
10
8
6
4
2
-2
-4
-6
-8
-10
-12
-15 -10 -5 5 10 15
A
B
C
D
E
EF
F
��
��
��
��
��
��
��
A B
D
C
13
12
3
units
BD
units
CB
4
925
5
1213 22
=−=
=−=
5
165
4
5
12ˆcosˆtan
=
+=+ CBDBCA
°=
= −
6,22
13
12cos 1A
�
�
�
�
�
�
� �
�
�
�
Grade 10 - 33 - Educator’s Guide 2008
Question 5 5.1 5.1.1 (3) 5.1.2 (3) Question 6
D1
D2 y x
3 m
yx
6.1 (3)
N
K
M
L
18cm
20o
20o
cmKN
KN
KN
linessaltNMK
16,6
20sin18
20sin18
).(20
=°=
°=
∠°=
22,95
16,6)1491,16(5,0
91,16
20cos18
units
KLMNArea
cm
NM
=
×+×==
°=
yD
Dy
tan
3
3tan
2
2
=
=
xD
Dx
tan
3
3tan
1
1
=
=
�
�
�
� �
�
� �
�
Grade 10 - 34 - Educator’s Guide 2008
6.2 Distance covered in 1 second:
m
Dist
51,20
6tan3
5,3tan3
=°
−°
=
(3) 6.3 (2) Question 7 7.1
AB Ashwell
15 5
25 60
30 70
40 5
30 5
10 10
30 60
Sum 180 215
N 7 7
Mean 25.71 30.71
Mode 30 5
7.1.1 (4)
7.1.2 AB’s mode = 30 Ashwell’s mode = 5 (2) 7.1.3 Would advise coach to select AB. Although his average is lower, his modal number of runs is higher, indicating that he is more reliable. Ashwell’s mean number of runs has been affected by a few high scores, but he is not consistent. (2)
Question 8 8.1 128 men and 243 women took part in the survey. (2) 8.2 Multiply the midpoint of each class by the class frequency. Sum these numbers and divide by the number of men surveyed (128). (3) 8.3 90 – 119 (1)
8.4 %3,2410024359 =× (2)
hrKm
hrKm
Kmm
/43,77
/360002151,0
sec/02151,0sec/51.21
=×=
=
runs
meansAshwell
71,307
215'
=
=
runs
meansAB
71,257
180'
=
=
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�
�
�
�
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�
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Grade 10 - 35 - Educator’s Guide 2008
8.5 90 – 119 (2) 8.6 The statement does not take into account the fact that significantly more women than men visit the mall overall. If the frequency for the class 60 – 89 minutes is converted to a percentage for both men and women, then the figures can be compared directly. The percentages are similar, as shown below. (3) 8.7 8.8 Accept any valid and substantiated observations, for example:
• The most number of people spend either 30 – 59 minutes or 90 – 119 minutes at the mall (the data is effectively bi-modal).
• Very few people spend more than 180 minutes at the mall. • There is not a significant difference in the number of people visiting the mall in the time
range from 30 minutes to149 minutes.
%62,15100128
20%
%87,16100243
41%
=×=
=×=
men
women
Length of time spent at the mall
0
10
20
30
40
50
60
70
80
0 - 29 30 - 59 60 - 89 90 - 119 120 - 149 150 - 179 180 +
Time (minutes)
Nu
mb
er o
f p
eop
le
�
�
�
�