Grade 10 Assessment Guide - Maths Excellencemathsexcellence.co.za/papers/tutorials/Grade_10_Memo_-_Topic... · Grade 10 - 2 - Educator’s ... Grade 10 Project: Finance Marks: 75

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 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 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…


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 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 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 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 ��


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) ��


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 ��

��


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 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


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


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]


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 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

�

�

�

�


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

=

++= �

�

�


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)


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 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

��


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 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

�

�


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(

≠∴=

−+−=

=∴

=

=

−+−= �

�

�

�

�


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 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)


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


(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)


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)


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 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

=

××=∆

==

−+−−−=

�

�

�

�

�

�

�

�

� �

�

�

�

�

�

� �

� �

�

�

� �


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

=×=

∆×=

�

�

�

�

�


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

�

�

�

�

�

�

� �

�

�

�


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

=

=

�

�

�

� �

�

� �

�


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'

=

=

� �

�

�

�

�

�

�

�

� �

�

�

� �

� �

�

�

� �


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

�

�

�

�

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