2018_mathsMS_olevel_nov_paper2

Extracted Content: 2018_mathsMS_olevel_nov_paper2

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4004/2 NOVEMBER 2018 SOLUTION GUIDE

I(a) i) | ax—x(a— b) + 2bx

= ax—ax+bx+2bx

Evidence of correct removal of
2 brackets and directed numbers
-xXX—b = +bx

= 3bx
(ii) | @—2)? — x?

=x? 4x 4+ 4—y?

| ecaree by grouping like terms
(@— 2) - 2)

Grouping and simplifying like

Expansion of the square bracket i.e.

_ (@~3)@+3)— («+ 2)(x-2)
~ (x - 2)(% +3)

(x —2)(x +3)

_ x? —9— (x? — 4)
= (x — 2)(x + 3)
_ x? -9-x7 44
~~ @-2G@+43)
=5
(-—2)@+3)

_ Xi +3x-3x-9- (4? -2x42x—4)

4x44 terms
l(b) pre +E 7 é 1)] Correct substitution followed by
~ 212 + 3 a correct removal of inner brackets
and getting 2.

1/1 1

=3(5+15) ae a eee ee: eS a
2\20 "2. 2 Or multiplying 2% 2 +5%5 giving

=5 (2) 44

(c) x-3 x42 — Evidence of correct denominator |
x—-2 x43 with all the essential brackets.

1 Understanding of “difference of
two squares” can be applied in
expanding (x — 3)(x + 3)and
(x —2)(x + 2).

Brackets are very important on
- (+ 2)(x — 2) so that the pitfall
L of getting —x? — 4 is avoided.

2(a)(i)
$50

5kg

= $10

Understanding that ‘per’ means
“divided by” dividing the total cost
by 5 kg

[==

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(ii) 115% : $50 ~. 100% : (less)

$50x100
115

= $4348

-. VAT = $50 - $43,48

= $6,52

Any price i.e Cost Price is always
100%, so adding Vat means selling
price becomes 115%

115% : $50

15% (VAT)

$50x15% . z
pads 7]
Tism giving $6,52

iii)

10
Too * 2x50 +30+5x28 + 28+ 3x50)

10
= Too (100 + 30+ 140 + 28+ 150)

=" Gee
apg a)

ig eres
= Too *

= $36,40
3
(b) Amount = $400 (1 + =)

$400 x 1,037

I

= $437,09

3(a\G) | 4x -2<5x+2and5x+2<2x+8

—2-2<5x—4x and5x-2x < 8-2
—4<sx 3x <6
x<2

~4<ex<2

Evidence of calculating the total
cost and then finding 10% of the
total cost as the discount offered.

Recall and application of the
formula for compound interest.

A=P(1+2)'

Alternative method would be to
calculate yearly i.e. interest for year
1 and add it, becomes principal for
ear 2 and so on.

Ability to split the inequalities into
two and working on each and
combine the result.

(ii)

67

A number line with solution line
closed a + —4 and opena + +2

Evidence of “open and closed
intervals.

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[ ®) [pos

Q+bx
R?(Q + bx) = ax—p
QR? + bxR? = ax—p
QR? + p = ax — bxR2
ax — bxR? = QR? +p

= 2 2,
x(a oR ) — QR?+p
a=bR? a—bR2

p+ QR?
x= ———_—_
a—bR?

Squarring both sides so as to
temove the square root sign.

Removal of fractions by
multiplying both sides by Q + br

Collecting terms in x on one side
and factorise to make x the subject.

NB: All signs can change in the
ptQR?
a-bR2

expression

(c) m(2m?n? + 3mn — 2)

= m[2mn(mn + 2) — 1(mn + 2)]
= m[(2mn — 1)(mn + 2)]

m(2mn — 1)(mn + 2)

Realising that there is a common
factor of m.

Factoring out m leaves a quadratic
expression that can be factorised
using any method like the ring

number of —4m?n?
___|

Show that ruler and compasses
have been used through evidence of
construction arcs and lines. If
construction space is not enough or
a mistake has been made which
cannot be erased ask for plain paper
and construct on the plain paper
and attach it as additional material,

(a) Triangle ABC with sides AB = 6 cm, BC =6.5|4
em and ABC = 45° with correct construction

arcs.

A complete circle centre A with a tadius of

4cm

First construct an angle of 90° then
bisect the angle to get the 45°
angle.

The construction ares are part of the
solution so should be clear.

Use a ruler for all straight lines,
Circle to be clear and complete.

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(c)

Bisector of BCA with correct and clear
construction arcs

The bisection arcs should be very
clear to show ability to use relevant
mathematical instruments.
Appreciation of the intersection of

There should be marks to show the
position of the two points with

correct labelling.

The key words are “points
equidistant” which show locus

Explanation that might not use
these words is still acceptable

rovided the meaning in there.

The key is the ability to identify
square numbers and knowing the

Being able to use the set symbols

The intersection of set is the set of

commion elements in the given sets.

The ‘n’ refers to a number.

(d) The points X1 and X2 clearly marked on the | 2
intersection of the circle and bisector of BCA. the two Loci.
(e) Locus of points equidistant from sides CA and | |
CB.
language.
S(ay) | A= (1; 459} 2
cut off.
and notation.
(i) | ANB= {4} L
(iii) | n(AUB)=4 L

(b)Ga)

(ii)

The number of elements in the
union of sets A and B should not be
in set brackets.

PCR

iD

69

The correct Venn diagram should
show that P is contained in Q and Q
is contained in R.

Any shapes e.g. rectangles or
squares may be used.

Correct labelling of sets is essential.

The set symbol should show that
set P is a subset of set R or R
contains P.

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Mor)

R
&
9
R
tie
10 3
9 B
R
3 2
10 9
B
2
9 B

The hint on completion of the tree |
diagrams is that the two
probabilities on the two attached
branches should add up to la
certainity.

There is no replacement therefore
subsequent probabilities
denominators should decrease by |.

Mean = (10x2,5 + 12x7,5 + 37x12,5 + 51x

17,5 + 10x22,5) + (10 + 12 + 37 +51 + 10)
— 254+90+462,5+892,5+225
= 284904462,5+892,54225

120
_ 1695

120
= 14,125 or 142

oat

L

n= 22+37
=59

(ii) lp (R and R)= = xs I The application of the tree diagram.
Us
oe 1 Product law applies because these
LL two events are independent. =|
(iii) P (at least R) = P(RR)or P(RB)or P(BR) Meaning of “at least one is Red”
eget By Big 7 Or is translated to mean addition
wt io*%stin%s t Alt ;
7 7 9 ernatively 1 — p(Not R and R)
“as "30+ 30 = 1—p(BB)
— 28
30 32
es = I =i- to x 5
= it
=
6(a) Mid-point are 2,5; 7,5: 1235517555225: if [ Understanding of the idea of mid-

points or class centres in Grouped
Data.

Mid-point= “+4
the Upper limit and

LLL. is the Lower Limit of that

class

where ULL. is

Finding Mean of grouped data i.e.
Adding all frequency x class centre
and divide by the total of all
frequencies,

“Cumulative” means “adding on”

This is got by adding frequencies
Up to the required class. [10 + 12 +
37]

—

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(b)(ii)

(c)(ii)

Very smooth curve passing through
the correct 5 points.

Use of correct scale is a
prerequisite as it shapes the curve.
The principle used in graphs is that
one large square is of length 2 cm.
the points should be plotted with
visible plots. The curve should be
hand drawn and smooth.

There should be evidence of using the graph with
dash lines from cumulative frequency 60,5 point
to intersect with curve and from this curve to
intersect with the age (horizontal) axis.

Median = 15,2

Median at (120 + 1) position on
Cumulative Frequency axis.

A dashed line from cumulative frequency 90,75
point to intersect with the curve and from the
curve intersection to the age (horizontal) axis.
Upper quartile= 17,5

Upper quartile is at + (120 + 1)

position on the C.F. that is the 90,75
position

BAC = 140°— 75°
= 65°

AC 9

Bearing is with reference from the
North and measured in the
clockwise direction.

Identify that this question requires

(a)(ii) Sin80° Sin65° use of “sine rule.
_ 95in80° Correct application of sine rule.
~_Sin65°
= 9.780 km Giving answer to the required
degree of accuracy. _
(a)(iii) | Let the shortest distance be d km Realising that the shortest distance
i _ Sin 35° is one that makes an angle of 90°
9 with AC and passing through B.
d = 9Sin35°
= 5,162 km Use Sine trig ratio of angle ACB
which is calculated as 180° - (80°
+(65°) = 35°
7(b)(i) Pso = 180° — 72° Ability to see the isosceles triangle
~ 2 PQS.
= 54°
Angles opposite the equal sides
being equal, “base angles”.
SRP = 72° Angles subtended by same are are
(b) (ii) equal.

ae

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a

Angles SRP and POS are “|
subtended by arc PS.

(b)Gii) SPR = 90° — 72° PR is the diameter of the circle
= 18° hence angle PSR is 90°
The other two angles are
complimentary meaning they add
up to 90° _|
(b)(iv) PTS = 180° — 2x72° Use of the following circle
= 180°- 144° theorems namely:
= 36° The angle formed by a tangent and
a chord at the point of contact is
equal to the angle in the opposite
segment
Tangents from an external point to
the same circle are equal
lm 1 k Ability to express the statement asa
Vilas AE, y= ale general equation involving a
ak =yvx constant of variation.
k=2V9 Finding the particular equation/law
— from the general equation by
6 substitution.
y F a
(i) 1 6 | Substitution for y
2 vx Solving for x by removing the
(vx)? = (12)? | square root sign by squaring on
x= 144 fk both sides.
(b)(i) Log(3x +1) + log(x-3) = log 10. Recall and apply the laws of

-.log(3x + 1) + log(x3) — log10 =0

song [SE*DE=B) .g

(3x + 1)(x -— 3)

as a 10

= (3x + 1)(x — 3)

4 10

+10 = (3x +. 1)(x-3)

3x*7-8x-13=0

Logarithms i.e.

Log(MN) = log M+ log N, to do the
correct reduction showing all stages
clearly with no wrong working
seen.

10! is based on the assumption that
when the logarithm base is not
given, the base is 10.

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8 (b) Gi)

3x7 —8x— 13 =0
> a=3, b=-8,c=-13

—bivb2—4ac
But x= oe

— ~(-8)ty (-8)?—4x3x(-13)

al

Gi)

Application of the quadratic
formula and giving answers to one
decimal place.

All stages to be clearly stated,
however method of completing the
square may still be used.

(iii)

2(3) 8 13
xt-ox= 7
= ——e What must be added to make
x? ox a perfect square ?
_ B4V220
: (7 al _ 13, 16
=-L,1 or 3,8 3) 3 "9
4)?_ 55
G==>
= ty
37 3
_ 44v55
3
=-1,1 or 3,8
BC =BA+AC I Appreciation of addition law on
=-at+b vectors.
: BC =BA+AC
BN =2BC Ratio of BN : NC
3 .
12,
so
=5 (-a+b)
=-sat=b
AN = AB +BN 2 Use of then triangle law of vector

vat (atts

2 1
=+a+=b

addition.

BM =BA+AM

1
=—-a+=b
2

i | Sai law of vector addition.

| | |

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(>) | a¥=AB +BY y2 Express BY in terms of scalar h. . Xpress BX in terms of scalar h =a+h(—a+=b) d Express a -hb in simple / factorised 1 form =—a-hat+=-hb 2 1 =(1-ha+=hb 2 9 (c) AX =k@a++b) nl Use the fact that 7 8 AX = KAN - 2 ka 4 kb Substitute for AN. ————__— 9 (d) (l-ha hb = ha + tkh 4 Equate the coefficients of the 2 3 3 vectors, coming from 2 Ailton 2 (1) (-h)a=$k. 3 Solve the two equations 1 1 simultaneously using any method ghe 3k (1) of solving simultaneously equations though in this case substitution 5 1 Sp 2% method method is the shortest. = = 3 3 23 phe ex, 4 tS 10(a) p = 2(-1)+3-(-1)? Correct substitution of -1 and +1 it for x in the function y = 2x +3 — =-24+3-1 x? | p=0 q=2(1)+3-12 q=24+3-1 q=4 a Li

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The use of correct scale will make

Plots should be visible and the
graph should be drawn using free
hand, not grossly thick and passing
through the correct points.

In general lines drawn should be at
least 3 cm long unless stated

These values are the x-coordinates
where the straight line cuts the

There should be evidence of
marking of the correct region. Area

(b)G) | Smooth curve passing through the correct 7
points. the graph fit.
4
(ii) The line y = —x correctly drawn to cut the
curve in two places.
otherwise.
(c)@) | x=0,8
x = 3,8 L
L curve.
Gi) | Area = 4,2 units?
3

should be found by counting
squares and multiplying the number
of squares, by area of one square, in
this case its 1 units?

Where one decides to work out the
area in cm’, getting 4 om?
equivalent to 1 unit®. In this case
| the units must be stated.

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——— H(@q@ [rx2?x15 + nx3?xi d+ axa?xig]

V=

—mx1?x3x14

= 62 + 13,50 + 240 — 4,50

l
it

Realise that it’s the volume of the

three discs minus the volume of the
hole.

= 63 301,9 m2
= 63 301.9 + 10000

= 6,330 ha

= 397
= 122,5cm*
i
—— B
(ii) | 289 04925 em? | l Use of formula for density to be
cm3 2 m
used. D = >
343g L
“m=Dxv
i mw] Cost = $7,50 x 343
2
= $2 572,50
(b) Area = + x400x440x Sin46° Recall of the formula for finding
a area of a triangle given two sides
4 and an included angle then

converting the area to hectares by
dividing by 1 0000.

For those who cannot recall the
formula
Area = ; abSinc
Follow the longer route of first
finding a perpendicular height, h
h :
rrr Sin46
h = 400S5in46
= 287,7

Area = > basexheight

= 2%440x287,7
= 63 294

= 63 300 + 10 000

= 6,330 ha

hy

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12(a)

Triangle P correctly drawn

Triangle Q correctly drawn with vertices at
(—2; —4); (—2; -8) and (—4; —8)

Triangle R correctly drawn at
(—2; —3), (—2;—-1) and (—1; —1)

I Make sure all vertices are clearly
marked and joined by straight lines.

The coordinates of the vertices of
the triangle can be found by
multiplying by matrix for

enlargement on 1.88

—2 OV¢l 2 2
( 0 ») G 4 iy
OR by drawing straight lines from
each vertex passing through the
centre of enlargement, in this case

the origin and then measuring using
the factor of 2, mindful that the

image is inverted.
Translate each point by the given

vector

(2) + (C3) (4) + (Ce) ana

GeO)
OR using the graph to displace each

point by (2) i.e. 3 units to the left
followed by 5 units downwards.

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(d)

CREE
giving the points

(3; 1), (5; 3) (6; 2)

Matrix multiplication

Triangle S drawn correctly.

———<strong></strong>
One-way stretch

Mark all points clearly and join
them using a ruler

y-axis (x = 0) invariant.
Stretch factor 2

i?

There is need to describe fully the
single transformation by stating its

name and properties.

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ZIMBABWE SCHOOL EXAMINATIONS COUNCIL

General Certificate of Education Ordinary Level

MATHEMATICS 4004/1
PAPER |
JUNE 2019 SESSION 2 hours 30 minutes
Additional materials:
Candidates answer on question paper
Geometrical instruments

Allow candidates 5 minutes to count pages before the examination.

TIME 2 hours 30 minutes
This booklet should not be punched or stapled, and pages should not be removed.

INSTRUCTIONS TO CANDIDATES

Write your name, Centre number and candidate number in the spaces at the top of this page.
Write your centre number and candidate number in the box on the top right corner of every page
of this paper.

Check that all the pages are in the booklet and ask the invigilator for a replacement if there are
duplicate or missing pages.

Answer all questions.

Write your answers in the spaces provided on the question paper using black or blue pens.
If working is needed for any question, it must be shown in the space below that question.
Omission of essential working will result in loss of marks.

Decimal answers which are not exact should be given correct to three significant figures
unless stated otherwise.

Answers in degrees should be given correct to one decimal place.

Mathematical tables or electronic calculators should not be brought in the examination
room.

INFORMATION FOR CANDIDATES
The number of marks is given in brackets [ ] at the end of each question or part question.

Copyright: Zimbabwe School Examinations Council, N2018.
© ZIMSEC

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Answer all questions.

NEITHER MATHEMATICAL TABLES NOR SLIDE RULES NOR
CALCULATORS MAY BE USED IN THIS PAPER.

Express
2
#) = as a decimal fraction,
i)
b) 2
5 as a percentage,
TREND) ecrerpamntics
c) 0,0375 as a fraction in its lowest terms.
Answer(c)

Write down the next term in each of the following sequences.

a) 15459; 16;25;36;__

Answer(a)

2 3 VS VT __

Answer(b)

©) 16;8;4;2;1;_

Answer(c)

4004/1 J2019

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3. Three girls aged 12 years, 13 years and 15 years share $100,00 in the ratio of their ages.

Calculate the amount of money that each girl receives.

4. a) Convert

i) 434, to base ten,

Answer (a)(i) [1]

ii) 75,, to base two.

Answer (a)(ii) [1]

b) Evaluate 377, + 411, leaving the answer in base 8.

Answer(b) (]

4004/1 J2019

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