2019_mathsMS_olevel_mj_paper1

Extracted Content: 2019_mathsMS_olevel_mj_paper1

Page 1

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4004/01 JUNE 2019 SOLUTION GUIDE

[ QUESTION

SOLUTION

MARK | ADDITIONAL
GUIDANCE

Evidence of accurate
division by 25.

1 (a) 12
ae 0, 5
= 48
(b) 2x8 = 40% I
5 4
(c) 0,0375 ===
10000
a
~ $0

2 T@

(b)

NIP

12:13:15
12+134+15=40

12 100 _ 635
0° 1

£8 a0 $32.50
40° 1 °0""

To express a fraction as

a percentage you
multiply by 100.

Correct reduction of the
numerator and the

denominator using
common factors till they
are in their simplest
form.

Study the pattern and
identify that these are
perfect squares.

1 Study the sequence and
identify that these are
square roots of prime
numbers.

Study the sequence and
identify that the next
term is obtained by
dividing by 2 or
multiplying by

1 3
3 the previous term.

The candidate should
find the total of the
ratios and then calculate
each ration over the

101

total ratio and multiply
by $100

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QUESTION | SOLUTION

MARK

15 100
n° = $37.50

ADDITIONAL
GUIDANCE

4 |@@ | 434, = 4x52 4 3x5! 4 4x50 Bases | Recognition of
to base | expanded index format
ten of numbers in different
= 4x25 + 3x5x4x1 bases when converting
from other.
=100+15+4
Alternatively the
= 119,, l candidate can use the
method of repeated
multiplication as shown
434 below.
x5
204+3
= 23
x5
115+4=119,,
i
(ii) 7510 to base two Understanding of
conversion of numbers
2 78 in base ten to other
2 atenil 1001011, bases. The remainders
are written from bottom
2 [8 rem | l to top.
2 9Orem0
2 4rem |
2 2rem 0
“2. IremO0
Orem |
1 Understanding of the
general rules of addition
102

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QUESTION

SOLUTION

1010,

(b)(i)

Kite

ADDITIONAL
GUIDANCE

of members in base
eight.

Knowledge of the
properties of different
shapes.

Use of the theorem “the

angle in a semi-circle is
aright angle.”

(P'NR)U(R'NQ)

103

Understanding of the set
notation or meaning of,
compliment of a set,
intersection and union
was required for the
candidate to be able to
share the set (P'N R) U

'NQ)

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i (a) 647 cents to dollars Conversion of cents to dollar you divide by 100 647+100 = $6,47 (b) US$11:R13,80

US$75,90: ? More

75,90

x 13,80

=R1047,42

Sx—-2y=26 (1x2

3x+4y=0 (2)x1

10x —4y = 52
3x+4y=0
Add 13x =52
x =4
Substiture 4 for x in (2)
3x4+4y=0
12+y=0
4y = -12
=-3
x=4andy=-

Any correct method of
solving simultaneous
equations is acceptable.
[elimination as shown
substitution, matrix and
graphical methods of
which candidate may ask
for graph paper.]

=a 2%
6) =z2(298)
() = )

x=4andy=-3

104

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SOLUTION

MARK | ADDITIONAL GUIDANCE

10cm

Area of the parallelogram = 10 x 8x
sin 150°

= 10x8x0,5
= 80x0,5

= 40cm?

Probability of picking a yellow sweet =

\w/

8
20

Recall of the formula for
calculating area of a
parallelogram given two sides
and an angle.

Alternatively the candidate can
find the perpendicular height
and then use it to calculate the

i area.
ne
sin 30° = 3
I h= sin30°x8
I h=0,5 x8
h=4cm

-. Area of parallelogram
= 10x4

= 40cm?

Use the knowledge that the
probability of picking a yellow
sweet is equal to the number of
yellow sweets over the total
number of sweets in the box.

Probability of picking two sweets of the
same colour

~8 7 {12 1
~ 20°19" 20°19

_ 56 , 132
~ 380 380

105

Apply the law of probability of
picking two sweets of the same
colour without replacement.
Alternatively the candidate can
use the tree diagram to find the
probability.

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QUESTION | SOLUTION MARK | ADDITIONAL |

GUIDANCE

Identify circles in the
diagram.

Identify the number of

lines of symmetry in the
diagram. Understanding
of the concept of line of
symunetry is important.

(c)

6 if Identifying the number

of angles through which
the shape can be rotated
to its original position
gives the order of
rotational symmetry.

106

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QUESTION

SOLUTION MARK

—

ADDITIONAL GUIDANCE

log 30 = log(6x5)

=log6 +log5S
= 0,8881 + 0,6990

= 1,4771

Application of the law of
logarithm log ab = loga + logb

log 1 200 000 = log(1,2
x1 000 000)

= log1,2 +
log 1 000 000
= log (2) +6
= log6 —
log5 + 6
=0,0791 + |!
6
= 6,0791
360° I
18
= 20°

(7 — 2)x180°

= 5x180°

= 900°

Application of the laws of
logarithms;

log ab = loga + logb
a
logy = loga —logb

Thee was also need to know that
the log 1000 000 = 6

Recall of the theorem of the sum
of exterior angles of a polygon is
360°.

Understanding of a regular
polygon would assist the
candidate to know that the
exterior angles are of the same
size since the interior angles will
also be of the same size.

Recall of the formula for
calculating the sum of interior
angles of any polygon.

[(@m — 2)x180°or (2n — 4)x
90°]

107

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QUESTION | SOLUTION

MARK | ADDITIONAL
GUIDANCE

The general form
of direct
variation is to be
used to find the
equation. [N o
5].

Use the given
values of N and S
to find the
constant of
variation.

ail

14 | (a) N«Ss
N=ks
40 = 1000k
= 10
~ 10000
a
25
Ree S
~ 25
(b) _1
N= sae
180 =— S
~ 25
S = 180x25
S = 4500
15 | (a)

I2cm

Substitution of

L given value of N
in the equation
found in (a) to
find the
numerical value
of 3.

Make 3 the
subject.

Sem

tan ACD ==
12

12cm

Knowledge of
trigonometric
ratios is required
for the candidate
to express tan
ACD asa
common fraction.

Sem

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QUESTION | SOLUTION

(b) AC? = 57 + 12?

AC? = 254144
AC? = 169
AC? = ¥169

AC = 13

x 5
CosDAC =

MARK

in BDC =—
sin = 13

Modal mass = 10kg

ADDITIONAL
GUIDANCE

Using the triangle in (a).
Use the Pythagoras
Theorem to find the length
of AC. Use trigonometric
ratio for cosine to find

cos DAC.

Use the triangle in (a) and
knowledge of the
trigonometric ration for sin
to find sin BD C.

Modal class is the class
with the highest frequency.

qi T0410
Median mass = ——

Mean mass =

= (p+ 2)(p—2)

(ii) 2p? + 7p +6

=2p+4p+3p+6

54+5+10+10+10+20

6

The median is the entry in
the middle if data is
arranged in order. In this
case the median is the mean
of the 3% and 4" entries.

The mean is calculated by
finding the sum of the
entries and then divided by
total frequency.

Use the concept of
difference of two squares to
factorise the expression.
Brackets are essential.

Knowledge of factorising

quadratic expression either

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= 2p) +2) + 3(p + 2)

= (2p + 3)@+2)

p?-4= (p+ 2)(p—2)

2p? + 7p +6 = (2p+3)(p+2)

The H.C-P of p? — 4 and 2p? +
7p+6isp+2

by grouping in pairs or the
ring number method.

Use the factorised
expressions in (a) to find
the highest common factor.

Sem

Vertical height h = VI5?—12? | ,

h=V225—144

h= V81
h=9cem l

110

There is need to visualise
the diagram or to draw a
sketch diagram.

Use the Pythagoras
Theorem to calculate the
vertical height.

The candidates should
know that the vertical
height is perpendicular to
the diameter,

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UESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

Substitution of value of r (radius) and h
(height) in the given formula and
simplify accurately.

The candidate should remember not to
substitute the value of 7 as the answer
is required in terms of 7.

f(x) =3x?-2x-8 Understanding of function notation
5 where f (—4) means that substitute —4
F(-4) = 3x(-4) for x in the function.
—2x(-4)

-8

f(-4) = 3x16+8-8
f(-4) =484+8-8
f(-4) = 48

f(x) =0 The candidate can solve the quadratic
equation by factorisation or use the
quadratic formula.

3x? -—2x-8=0
3x + 4)(x—2) =0
les allie x =24/(-2)74x@)x(—8)
either 3x + 4 =0or
x-2=0

orx=2

111

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

—— 5 =
20 | (a) (x7) la = (4)%/ Use of law of indices x“/b = 2/¢4 js ]
required to solve the equation.
3
x = (v4) I
x= £28
x=8o0r-—8 I |
(b) 2 3 Knowledge of method of solving equations
x-2 x+2 involving fractions by first multiplying
Very term by the L.C.M of th
2x + 2) = 3(x ae we a
2/1 denominators to remove the denominators.
The equation that remains is a linear
2x+4=3x-6 equation in one variable.
2x —3x-6-—4
—x=-10

no |

112

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QUESTION

21

SOLUTION

~ | MARK | ADDITIONAL GUIDANCE

(a)

I
ys 2x y<-—x+300
_|¥ zai

Line that passes through
the points (0;0) and

(100;200)

Gradient of the line =22-°
100-0

= 2

4

=2

Equation of the liney =
mx +C

0=2x0+¢
c=0

-. Equation of the line is
y=2x

Inequality is y < 2x line
that passes through the
points (300;0) and (0;300)

Gradient of the line = 222° | |
0-300
=-1
Equation of the line y =
mx +c
0=1x300+c
c= +300

Equation of the line is y =
—x + 300

Inequality is y < —x +
300

The two inequalities are

113

Realization that equation of a straight line
is in the form y = mx + c (where m is
the gradient and c is the y—intercept).
The gradient can be found using different
methods. (measurement or two points on
the line). Knowledge of the fact that if the
line is broken it will be < or > and when
itis continuous it will be <or >.

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2 Identify one value of x and y respectively from the shaded region that give x + y its

maximum value.

—__——_|_

618 000 = 6,18x105

1

Knowledge of the standard form [A
x10" where | < A < 10 nis an integer]
required.

0,000423= 4,23x 1074

l

nN

(8,76x10-7) + (7,89
x1077)

= 0,0876 + 0,0789
= 0,1665

= 1,665 x1071

Expressing a decimal number in standard
| form.
2. Candidate can convert both numbers to

ordinary form, add and convert the sum
to standard form,

Alternatively, the candidate can use the
factorisation method

(8,76x10~?) + (7,89x1072)
= 1077(8,76 + 7,89)

= 1665x107?

= 1,665x101x1072

= 1,665x1071

Ley)

@@ | 4¥=4o+ ox

Knowledge of the triangular law when
adding vectors.

Knowledge of the triangular law when
adding vectors and application of the
ratio given.

114

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QUESTION | SOLUTION

ADDITIONAL
GUIDANCE

- AX is parallel to YB

- YB is 3 times AX[YB = 32K]

The bearing of Dube village from
Moyo Village is 200° or
320° or 20°W of S.

Let the distance be hd
d? = 5? + 6? — 2x5x6x cos 40°
d? = 25 + 36 — 60x0,77
d? = 61 — 46,20

d* = 14,80

d= ./14,80

115

Use the cosine rule to

Understanding of the
properties of the vectors.

Use of a sketch diagram
is required to assist the
candidate to visualise
what is on the ground.

The sketch diagram is
used to find the bearing.

find the distance as two
sides and the included
angle are given.

Substitute the values in
the formula and simplify
correctly.

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SOLUTION

8 is to 16

|,44cm? is to ? More
ey 1,44
a) E

_4, 144
nd

= 5,76cm?

as

8 is to 16

Jisto L6cm*

Ea
16. 1

ADDITIONAL
GUIDANCE

Candidate should use the
knowledge that the area
factor is equal to the square
of the scale factor, that is;
scale factor is |:k area factor
is ?:k*

Candidate should use the
knowledge that the volume
factor is equal to the square
of the cube of the scale

factor, that is; scale factor is

[:k volume factor is :k°

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QUESTION SOLUTION

Rotation

270° clockwise or 90°
anticlockwise

Centre of rotation (0;0)

(b) A(-1;1) and A,(1;-2)
Translation vector = (4) -
Gr

a (4)

117

ADDITIONAL
GUIDANCE

Knowledge of
constructing a
perpendicular bisector
is essential.

To find the centre of
rotation the candidate
has to construct
perpendicular bisector
of BB, and DD, and
where they intersect is
the centre of rotation in
this case it is (0;0).

The candidate should

subtract the position
vector of A, from A in
order to get the
transaction vector.

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QUESTION | SOLUTION

] MARK

Total distance covered

= 150x5+ 150
= 750 + 150

= 900m

= + (90 + 60)10 +5 x5 x60 |!

ADDITIONAL
GUIDANCE

The area of the shape is the
distance covered. The area
can be divided into a
trapezium and a triangle or
into a rectangle and two
triangles.

If the shape is divided into a
rectangle and two triangles it
will be as follows.

Total distance covered

= 10x60 +5x10x30 +
+x5x60
2

= 600+ 150+ 150

= 900m

(b) Velocity = 208

15

= 60m/s

To calculate velocity, the |
candidate should divide the
total distance covered by the
total time taken.

; ; 60-0
(c) Deceleration = ——
15-0

The deceleration is equal to
the gradient of graph from

t=10tt=15.|-=
~12].

The candidate should take
note that the question is
asking for deceleration so it
will be positive unless if it
was acceleration.

118

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