2018_mathsMS_olevel_nov_paper1

Extracted Content: 2018_mathsMS_olevel_nov_paper1

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4004/1 NOVEMBER 2018 SOLUTION GUIDE

QUESTION | SOLUTION ADDITIONAL GUIDANCE
1 (a) 2 Understanding that 2? = 2x22 |
5? and 5? = 5x5 respectively is
IxOKE important. The division line is
= Sxs essential.
adlB.
~ 25
Lob @ x 0,24 Evidence of correct division by 25
28 is required.
1 (b) (ii) 0,125 in standard form Knowledge of correct form of
standard form is essential
= 1,25 x 107%
(A x 10” where 1S A < 10 anda
is an integer).
2 (a) [az This question tests knowledge of

3 (b) @

types of numbers.

This question tests knowledge of
types of numbers.

52 ig? 5F g0

4302;
1 0111,
+ 1010

10 0001,

Knowledge of expanded format of
numbers in different bases,
operations in the bases and the
place value system of base five.

General rules of addition of
numbers in base two.

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3 (b) (ii) | 512,

4355
44.

General rules of subtraction of
numbers in base seven.

tz
lz (a) 12.45 a.m.

The correct way of expressing time
in {2 hour notation with only one
dot between hours and minutes and
showing whether it is before noon
(a.m.) or afternoon (p.m.).

wD

4 (b) 212:
- 34

17 38

ei

17 38 or 5.38pm

4(c) 7 1 hectare = 10 000m?

1 km? = 1 000m x 1 000m =

Knowledge of time zones and
meaning of time ahead. Subtract
time ahead from time given.

The correct way of writing 24 hour
notation with no dots in between
and nothing written after.

Conversion of square kilometres to
hectares being tested in this
question.

1 000 000m?
“= 5Skm2 = 5%1000000
49,000 Recall that;
| hectare = 10 000m?
= 500 hectares.
aaa 1 km? =1000m x 1000m =
1 000 000m?
5 (a) 6,07 x10* = 6,07 x 10 000 Conversion from standard form to
ordinary by recognizing that 10* =
= 60700 10.000.
Recognise that the digits move to
the right number of places
equivalent to the index of 10.
rs (b) 2,53x 10+ + 6,1x1074 Convert both numbers to ordinary
form, add and then convert the
=253+0,61 sum to standard form
= 25,91 (Ax10" where I< A < 10 and nis
= 2,591 x10? an mnteEet:)
ee

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6 (a)

PRQ =(180° - 96) + 2
= 84+2

= 42°

The question tests on understanding

of types of angles between two
parallel lines and their special
relationships.

The exterior angle in a triangle is
equal to the sum of the two
opposite interior angles and the fact
that in an isosceles triangle two
angles are equal.

6 (b)

6 (c)

QRB = (180° - 84°) + 42°
= 96° + 42°
= 138°

QSR = 90° - 42°

= 48°

2v+3y=11 (1)x5
3x—Sy=-12 (2)x3
LOx +15y = 55
9x - ISy =- 36
Add 19x =19
x = 1
Substitute 1 for x in (1)
2x14+3y=11
24+3y=11

3y= 11-2

The method of elimination consist

Use of the theorem that allied or
co-interior angles are
supplementary.

Angles PQR and QRS are
alternative hence equal.

The candidate to recall that the sum
of interior angles of a triangle is
180°.

of eliminating one of the variables
with the same coefficient after
multiplying and then substitute to
get the other variable.

Alternatively use the substitution,
matrix and graphical methods.

O-.G 2c
Ce a
Ge ECE)
G=0)

x= Land y=3.

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8 (a) wee Recall the general form of inverse f variation, in this question which is k Ge 7 w x in order to find the equation. k 675 = 90 Use the given values of w and fto find the constant of variation or k= 675 x 90 1 proportionality. k=60 750 60 750 pep 88 1 wet = — 8 (b) f when w = 500 Substitution of given value in the HeRee equation found in (a) to the w= “a numerical value of f. 500 = 80.750 Make /the subject of the equation. ei S00f= 60 750 60750 i= 500 1215 f 10 f= 121,5 L 9 (a) 45,3981 = 45 40.( correct to 4 s.gf) L The zero at the end is very Significant. 9 (b) 75st <85 2 Knowledge of limits of accuracy needed. 10 (a) 4 3x" = 15x ! 1 Need to find the H.C-F of the two terms. 3x (x-5). Brackets are essential. 4 i —} : 2 10 (b) 8ki'm = 2X2x2xkxlxIxm There is need to express all terms 2 EP as products of factors or as sets of = 2 RE m factors so as to find the H.C. F, 28K'lm = 2X2x7xkxkxlxlxlxm = 2x71 Pm l 36?mn = 2X2x3x3xIxlxmxn

=2?x PP mn

—_

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H.CF of 8kl"m, 28k7Pm and 36Fmn is 1 2x2xlxlxm = 41m? A (6;2) and B (8;5) Use of formula for gradient of a Gradient of line AB = = given two points on the line. 3 =3. L increase in 7 (Gradient = —_———) increase in x (b) y-b = Use of one of the points and the x-a calculated gradient to find the y-2_ 3 equation. x-6 2 l Make y the subject to find the 2(y — 2) = 3(x—6) equation in the required form. 2y—4 =3x-18 Alternatively use one ps the points and the calculated gradient to find 2y =3x— 1844 the value of c =3x- 3 2y =3x-14 (y= Sx+c} 2 y= Sx -7 I 12. aat+6 _at3 Factorise the expression then a-3 " a-—2a-3 inverts the divisor. Common _ 2(a+3) — (a+1) (a3) factors can be divided to remain a ae gas 141 with the answer in that form where

=2(a+1) or 2a+2

13 (a) 3,5kg: 800g
3 500: 800

35:8

brackets are essential or expanded
form with no brackets.

Express the two masses in the same
unit (grammes) and reduced to its
simplest form.

A simplified form with no common
factor in the two terms.

13 (6) | 120% is to 457

100% is to? less

100 4,5
“420 4
=25
~ 2

Z = 3,75 tonnes.

| Recall that 4,5 tonnes represent
120% hence there was need to find
100% that would represent the
harvest in 2015.

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14 (a)

4-5x< 2x +8

4-8<2x+5x

ai

—

Solution of inequalities by
grouping like terms and dividing
both sides by 7.

a
Log Va = log a2

wie

x

RIw

=1

1
-4<7x
—4 z
7 SX
14 (b) The smallest integer that satisfies the | Understanding of real number line
inequality and position of numbers on the real
number line. Recall that 0 is an
4-5x< 2x +8is 0. L integer.
15 (a) log a=3 and log b=7 The use of law of logarithms,
log ab = log a+ log b.
log ab = log a + log b
=34+7
=10 L
15 (b) log —_ log 1 —log b Use of law of logarithms,
‘ log ¢ = log a—log b.
=0-7 Recall that log 1 = 0
Jes =~—7. L
15 (c)

There was need to express Va =
a

as
t

Hence log Va =log a3 = 5loga.

[16 @)

F@) = (+4) 2x-1)
f3)=G4+4 23-1)

=7x5

i
{ 3
Understanding of function notation

where /(3) means that substitute 3
for.x in the function.

16 (b)

12(3m) — 12(m) _ 2(5)
4 3B 2

Knowledge of method for solving
linear equations involving fractions
by first multiplying every term by
L.C.M. of the denominators to
remove the denominators.

The equation that remains is a
linear equation in one variable.

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9m — 4m = 30 1

5m = 30
m=6 1
17 (a) p-q= (°) -_ (7) Subtract corresponding elements.
Vector brackets are essential and
there is not supposed to be a
_ (c al division line.
= (4
17 (b) (—x)? +(—4)? = 5? 1 Understanding of the modulus or

e+ 16=25
x= 25-16 L
x =V9

magnitude sign and how to find the
magnitude of a vector.

Knowledge of formation of a
quadratic equation and solving it is
required.

The clue is the word regular which
means all sides are equal and all

interior angles are also equal.

115°+ 89° +x+x4+x4+x=

(6 —2)x 180°

4x+204° = 720°

The concept of sum of interior
angles of a hexagon to be used to
form an equation and solve it to
find the value of x.

1
4x = 720 — 204
4x = 516°
x= 129° L
19 (a) Cos 45° = 86 Knowledge of trigonometric ratios
aa was required to answer this
BC = Cos 45° x 12 question.
-2,2 . Use of cosine 45° in surd form to
21 calculate the length of BC.
=6v2

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19 (b)

5 A 6v2
Sin ACD Sy

at)

Knowledge of the fact that the sine
of an obtuse angle is equal to sine
of its supplement.

19 (c)

Tan ACD =-]

N

Knowledge of the fact that the
tangent of an obtuse angle is equal
to the negative of the tangent of its
supplement.

20 (a) (i)

Modal height 6< h <

Modal class is the class with the
highest frequency.

(ii)

Median height 6< h < 8

The median is the entry in the
middle of data when arranged in
order. In this case the total
frequency is 50 hence the median is
the mean of the 25" and 26"
entries.

Both are found in the same class

6<h<B

20 (b)

(4) +16 9)+ 2 9 0 OZ

12 30

6(2 + 6) + 8(6 +8) + 6(8 + 10) + S(10 + 12)
50

48+112+108+110
——__

50

378

50

728
50

Calculate the class centres of each
class by adding the lower and upper
limits then divide by two.

This class centre is then multiplied
by the frequency and all the
products of class centre and
frequency are added and the sum
divided by total frequency.

=
21 (a)

Application of the law of
probability for both which is to
multiplying the probabilities of
both not scoring.

21 (b)

1

Applying the same law as above
but now multiplying the
probabilities that they do not score.

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21 (©) |

This is the probability that one
scores and the other does not score.
There are two scenarios hence

4, 16
“i212 addition of two products.
- 1+6
4

awe

~ 42,

22 (a) 12 . 1 Conversion involves changing 12m
1000 60x60 to km by diving by a | 000 and
__12 3600 converting the seconds to hours by
=To00~ 4 diving by 3 600.
= 2
~ 10
= 43,2 km/h

22 (b) @ 12x5 +4%3x12 The area of the trapezium is the

2 distance covered. The candidate
= 60+ 80 can divide the shape into a triangle
and a rectangle. Alternatively use
= 78 metres the formula for calculating area of a
trapezium.
1
2 (8+ 5)x 12
= +x13x12
2
= 78 metres.

(b) (ii) 0-12 The acceleration is equal to the
8-5 gradient of the graph from ¢=5 tot
== “és
~ 3

=— 4m/s?
=
23 (a) ACB =1x60° Use of the theorem “angle at the
; centre is twice angle at the
= 30° circumference”.
|

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23 (b) “Todp = : (180° — 60° Angles in an isosceles triangle

because sides OA and OB are equal
A being radii of the same circle. Since
angle is 60° then the triangle is

equilateral.
23 (c) Length of minor arc AB Knowledge of the formula for

calculating length of an arc

= 60°

60° slog esa L
360° 1" 7 “7 oe
(js %2nr)
= 22
~ 3

ul
)
'
fe}
]

Substitute the values 7 and r

simplified correctly. _|
23 (d) Area of the minor sector AOB The candidate should know the for
calculating area of a sector
60° 22 7 x 7 1
XK xo
360°° 7°15 141 nue
lee Er )
et
~ 3 Substitute the values and h simply
correctly.
= 252 cm? I y

24 (a) 25335:7 2 Clear understanding of prime

numbers. Prime numbers are
numbers with only two factors 1
and that number. 1 is not a prime

number.
24 (b) P ={2;3:5:7} 1 The three sets have no common

elements hence their intersection is

S = {1:49} <innliy

M = {3;6;9}

PNSNM isan empty set

n(PASNM)=0
The Venn diagram should be

completed with all elements as
shown. No elements should be
repeated. The format of the

diagram shows that there is no

intersection between the sets P and
Ss.

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25 (a)

Translation

Translation vector =(;)) - (

3)

25 (b) (i)

The translation vector can be found
by using corresponding vertices
and calculating the movements in
the x and y direction.

Join the vertices on X to a
corresponding vertices on Y.

Observe parallel lines of same
Jength and that helps to find the
transformation.

The matrix for an enlargement is
the factor multiplying the identity
matrix of a 22 matrix.

25 (b) (ii)

gO
ee
@ 90-03

Vertices of triangle z are

= (8:2) and (12;10).

41

To get vertices of triangle Z
multiply the enlargement matrix by
the vertices of triangle ¥

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