2019_mathsMS_olevel_nov_paper1

Extracted Content: 2019_mathsMS_olevel_nov_paper1

Page 1

Extracted Text

2019 SOLUTION GUIDE

ADDITIONAL GUIDANCE

ON
2 460 = 1 000

Knowledge of conversion of cm to

2,460 litres (1 000 cm’ = | litre)

(b) 10000m? 100 Knowledge of conversion of hectares to square metres

250000 m2 * “t and square kilometres to square metres.

=4% 1 hectare = 10 000 m?
1 Km? = | 000 000 m?

2 0,25 Km* = 250 000 m?
2 | fa) (~8)3 Application of the law of indices (x)i = (bvx)* or bV¥x@
=——, 2
(3v=8 ) a

=3/(—8)?

= (-2)? <i

_ = 3V64
a4

(b) Vi47 + Vi08

= V49 x3 + V36x3
= 7V¥3 + 6v3

= 13V3

Knowledge of rules of surds and ability to identify perfect
Squares is essential in simplifying the expression.

3x-y=2(1)x2
Sx-2y=0(2)x1

6x-2y=4
Sx-2y=0
Subtract x =

Any correct method of solving simultaneous equations is
acceptable (elimination as illustrated in the working,
substitution or the graphical method.

Alternatively use the matrix method as shown below.

G=(3 D®)

a O)= (Go)
y= 10 x=4and y= 10
x=4 andy=10
a eee es ia
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QUESTION | SOLUTION

MARK

ADDITIONAL GUIDANCE

(b)

qt-r
= (-6)x2 — (—1)
= -12+1

=-11

Substitution of given values and simplify accurately.
Ability to manipulate directed numbers.

Substitution of given values and simplify accurately.

()

(q+r)°
=(-6-1)?

=(-77

=49

Substitution of given values and simplify accurately.

Understanding the concept of order of rotational symmetry
being the number of times the shape has to be rotated
through the same angle to get to its original position.

(b)

20x = 1800
x — 90°

x+x+x+x L
+ 2x + 2x + 2x
+ 2x 4+ 2x 4+ 2x
+ 2x+ 2x

= (12 — 2)x180°

4x + 16x = 1800

Use of the formula for finding the sum of interior angles
of a polygon.

[(n — 2)x180°]
Ability to formulate a linear equation and solve it.

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MARK | ADDITIONAL GUIDANCE

_
UESTION _| SOLUTION
[QUESTION

6 | (a) : of 54 Km

I

Ability to simplify accurately,

(b) 53:7
7 is to 35
15 is to --

15 35
pave

74
| =75

4

Find the total of the ratios and then use simple proportion
to calculate the total number of sweets.

+

7 /@@ P! = {1; 2; 7; 8; 9; 10}

|

Understanding of the meaning of compliment of a set and
intersection was key for the candidate to list the elements

Q = {4;5; 6; 7; 8} in the intersection.
iz a P'NQ = {7;8} 1
(ii) (PU Q)' = {1;2;9; 10} Ability to identify the elements in the compliment set and
then state the number.
“n(PUQ)=4 1
es ee —t
vA (KN 1L)' or K'uL! 1 Identify the unshaded region which is

K OL. The shaded region is not K 0 L.
hence in set notation it is (Kn L)',

8 | (a) Knowledge of the difference of two squares.
4 Brackets are essential.
1 1
=(x+ 5 (x =|
(b) x(x — 2) - 2xy + 4y Knowledge of factoring using the method
of grouping pairs.
=x(% — 2) — 2y(« — 2) 1 Brackets are essential.
L_| = (% = 2y)(@ — 2) l

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

9 | (a) 2214, = 2x53 + 2x5? +1x5? + 4x59

~

There is need for recognition of expanded
index format of numbers in base 5.

(b) 101, = 3719 Use of the expanded index format of a
number in base x and then formulate a
1xn? + Oxn! + 1xn° = 37 1 quadratic equation.
Solve the quadratic equation.
n?+0+1=37 Take note that the base cannot be negative
hence n= 6.

n*+1-37=0
n? —36=0
(n+ 6)(n-6) =0

Eithern+6=0orn—6=0

n=-6or6
n=6 I
EE ee
10 | (a) Pis a 2x3 matrix Knowledge of order of matrices and that
Qisa 3x1 matrix if two matrices are multiplied the product
PQ=H has number of rows equal to the first
matrix and number of columns equal to
2X3 product 3x1=2x1 the second matrix. [p by g X q by r=p by
r matrix]
Order of H is 2X1 or 2 by | L
10 | (b) A= C 1 ) ( 4 ) Knowledge of multiplication of matrices
3 -3/\3 -3

that is row by column and the ability to
manipulate directed numbers.
Brackets are essential.

=( 2x2+1x3 2x1+1x(-3) )
~ \3x2+ (-3)x3 3x14 (—3)x(—3)

4 i.
=(575 seal
a 2

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QUESTION | SOLUTION

MARK | ADDITIONAL GUIDANCE

Identify that DAC = CBD angles
subtended by the same chord DC.

ODB = CBD [alternate angles are equal].

Realise that triangle AOD is isosceles,
Using the knowledge that the angle ina

angle, DAC and DéA are complimentary.
DCA = ABD [ angles are subtended by the

[angles subtended by the same chord AB

11 | (a) ODB=30° 1
=e a
(b) ABD = 60° 1
hence DAO = ABO.
semi-circle is a right
same chord AD.]
(ce) ACB = 60° Realize that ADB = ACB
are equal].
12 | (a) @) | Locus of points 4 cm from O. {i

Key words are “points 4 cm from 0”.

Locus of points equidistant from point A
and point O.

The key words are equidistant from A and O

Ability to identify and shade the region
which suits the given
description of P.

47

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QUESTION

SOLUTION

ADDITIONAL GUIDANCE

USSL is to 12 Rands
USS$5,40 is to -- 1

= R64,80

Use simple proportion to calculate the amount
in Rands using the exchange rate.

ye 2 000x20x2 1
an 100

1 = 20x20x2
1 = 800
Total amount payable

= $2 000 + $800 =$2 | 1
800

Use of the formula for calculating Simple

PTR
Interest [7 = "Taal:

Substitute the numerical values in the formula
and simplify accurately.

Add the Principal to the Simple Interest to get
the amount payable after 2 years.
Alternatively the Simple Interest can be
calculated yearly and then add the interest

for the 2 years to the Principal.

The bearing of B from | 1

Key word is “from C” where the angle to give

C is 210° or S30° W the bearing should be calculated.
or 30° West of South. Drawing a line at point C parallel to the north
_| line can assist to calculate the angle.
BC 4 1 Use the Sine Rule to calculate BC as the
Sin10° Sin50° length of one side is given.
Use knowledge of sum of interior angles to

BC_ 4 find the third angle which is required in the
0,2 0,8 calculation.

_ 4x0,2 1
BC= i

An 8.
BC= ia
BC=1Km 1
—t
15 | (a) ee i =1 4a 1 Use of the law of logarithms
093 343 ~ ‘09335 logM? = plogM and the knowledge that
log33 = 1.

=log,37>
=—5logs3
=-5 1

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7] —— (b) log381 = (2x — 1)

(b)

3@2-D = g1
3x1) _ 34
2¥=1=4
2x=5

Use knowledge of log? = x means a* = 5, Realise the need
to have same bases in order to equate the powers, hence
gex-1 3

2x-1=4

Solve the linear equation.

The general equation of direct variation is to be used to
find the equation.

Use any set of the given values of h and V to find the
constant of variation.

Value of q = 6

Substitution of the numerical value of Vin the
equation found in (a) to find the numerical value of
h which is represented by q in the table.

Solving for h is by finding the cube root on both sides.

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QUESTION | SOLUTION

"79 1 2G)

ADDITIONAL GUIDANCE

Multiply the transformation matrix by the position
vector of A to get the coordinates of A,.
Coordinates to be given in the correct form and not

_ ( 1x4 + 0x2 ) as a column vector.
—3x4 + 1x2 Brackets are essential
_( 4
- (40)
A,(4; —10)
(b) Shear The invariant line is x = 0 or y - axis, The distance

Invariant line x = 0 from point A to A, is 12 units. Note that it is

negative since it is counted downwards shear factor
Shear factor —3

acer Wiles
—
18 | (@) 3x-652x-3<4x4+1 Split the inequality in two parts.
3 Solve each inequality separately.
3x-6<52x-3 : 2x- Combine the solutions.
3<4x+1
3x—-2x<56-3 : -3-
1<4x—-—2x
x<3 1 4<
2x
—2<x
—2<x<3
(b) ons ig Realise that — 2 is not part of the solution set hence
ASOuga fae the circle above — 2 is not shaded whilst the one
above 3 is shaded because 5 is part of the solution

set.

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— 19 (a) hod Substitute 20 for h in the expression and simplify. a= Realize that 16 and 25 are perfect squares hence the St+h —_—- 35 can be found, _ {20-4 l 9™* |5+20 _ {16 4~ 125 4 g = 5or 0,8 1 (b) a i | Realise that the square root can be removed by (g)? = h-4 raising both sides to the power 2. S+h Clear fractions and remove brackets. Collect terms with h on one side of the equation. h-4 Make / the subject of the formula. g= eax I Alternatively it can be worked depending on how

grouping of like terms is done having the answer
with all signs changed.

gE+h=h=4
Sg? +g?h=h—4
Ph-h=-4—5g?
h(g? — 1) =-4—5g2

—4-59? 5g7+4
h= or g >
_g*-1 1-9 1

a

20 (a)

AB =AO+0OB Use the triangular rule to add the vectors.

4 . Note that AO = OA
=C)+@)

Za Ge @

En ie

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QUESTION _| SOLUTION

MARK | ADDITIONAL GUIDANCE

Let the coordinates of P be (x, y).

BP = OA + 20B

()-@=-G)+2@)

(a) @)

6)-O-()+0

()-O=@)

()-O+0)
pao G)=(2) fiat _|

3x+4x0=12
3x = 12
x=4

P(4; 0) l

Consider the coordinates of P to be (x; y).

Express the coordinates of P in column form and
equate to OA + 20B.

(ii)

3x + 4y = 12

When x=0
3x0+4y =12
4y = 12

2 (053) L

Realise that when the line cuts the x-axis the y
coordinate will be equal to zero.

Substitute 0 for y in the equation

3x + 4y = 12 to find the x coordinate,

Note that the coordinates should have brackets.

Realise that when the line cuts the y axis the x
coordinate will be equal to zero.

Substitute 0 for x in the equation

3x + 4y = 12 to find the y coordinate.
Brackets are essential.

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_

QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE
T (bw 3x +4y = 12 Express the given equation in the form
4y =-3x+12 y = mx + c where m is the gradient of the
3 of the line
=—-x +3
4
Gradient of line is —2
= reer: {
(b) (ii) PQ = 32442 1 Realise that OQ = 3 units and OP = 4 units.
PQ =v9416 Use the Pythagoras Theorem to find the
length of PQ.

PQ=V25
PQ =5 units 1 |

22 | (a)(i)_ | The modal class is 40 <h < 50 1 Modal class is the class with the highest

ai

0 | ean = [o 4) +6 2)
1028) +2)

a(A)]/30

4X25+6X35+10xX45+2X554+8x
30

100+2104+450+110+520
ee
30

1390

30

= 462 or equivalent form

|

|i

ay

{

t frequency.

There is need to calculate the class centres

of each class by adding the lower and upper
limit then divide by two. This class centre is
then multiplied by the frequency and all the
products of class centres and frequency are
added. The sum is divided by total frequency.

ae

|

There is need to add the frequencies of the
two classes where the plant can be chosen
from and then divide it by total frequency.

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eae B (a) EBD 1 Understanding of the concept of similar triangles is key. Ability to identify equal angles and corresponding sides is essential such that

the similar triangle is named correctly.

(b) AB AC Use the answer in (a) to identify corresponding
EB ED sides.
Substitute the sides with numerical values and
AB 3,5 then simplify accurately.
42° 21
3,5x4,2

= AB=7cm l

(ii) (2,1)? is to (3,5)?

Square the scale factor () in order to get the

ain?
22,5 is to --- area factor. (3)
Use simple proportion and simplify accurately.
22,5 @)
oe sg (2
1 “\aa
_ 22,5x12,25
aay: I
= 62,5 cm? 1
24 | (c) Acceleration = 24-2 1 The acceleration is equal to the gradient of part of
s=0 the graph from t = 0 tot=6.
ie Correct units of acceleration (m/s?) should be
a stated.
=3 mis? i

Let the velocity after 10 seconds

v Realise that the deceleration which is equal to the
1 gradient is constant. Form an equation in V and
solve it.

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QUESTION

SOLUTION

ADDITIONAL GUIDANCE

(c) 1 1 The area under the graph is the
[5(26+546 +5 x9x54] /15 distance covered.
Divide the shape into a trapezium
a SRO N RAR ORE? and a triangle or into two triangles
as and a rectangle.
2704243 Knowledge of the correct formula
45 to find area of a trapezium,
1
5 [F(a + b)A]
~ 15
3
25 | (a) 2x 22 y 35 Knowledge of the formula for
: calculating circumference of a
_ circle (2mr or rd).
iia Substitute the numerical values of
=22¢m 1 7 and d and then simplify
= accurately,
(b) Perimeter = 14 + 10,5 + [2 x2x 2% x3,5x2] +105 | 1 Understanding of the concept of
3 i perimeter is important.
=24,5+224105 Use the formula for calculating
. ae perimeter of a semi-circle.
1 1
=57¢m [ar or 57d]
Substitute the numerical values in
— the formula and simplify correctly.
(c) Area of the shaded part = 14X10,5 — 3x22 x 35% Calculate the area of the rectangle
: md

3,5
1

= 147-66 x0,5x3,5

= 147-1155

= 31,5 cm?

ABDE and then subtract area of

1 3 circles,

Identify that the two semi-circles
constitute one circle.

Knowledge zd of the formula for
calculating area of a circle. [xr]

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