2019_mathsMS_olevel_nov_paper2

Extracted Content: 2019_mathsMS_olevel_nov_paper2

Page 1

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4004/2 NOVEMBER 2019 SOLUTION GUIDE

UESTION | SOLUTION | MARK | ADDITIONAL GUIDANCE
1 4.(2 4. =) Rules of precedence or order
43 of operation
1 Simplify brackets first by
a (=) adding the two fractions in the
brackets.
41 Remember the minus (— )sign.
=4-—
12
_ 48-41
~ 12
= + or 0,583
12 1
(b)) | =5,25x4,15 1 Appreciation of upper and
lower limits (bounds)
= 21,787 1 decimal place means the 2"
decimal place is the key.
= 21,79 I YZ (minimum) When both y and z
i take values on the lower limits.
als eis iti Be
(OM = x2928 I Application of ratio, 3:5
1 830 1
(ii) 66 612 — 19x1 830 I Find number of Executive
2928 — 1830 1 rooms.
Find the total amount realised
$29 1 from General rooms and
subtract it from $66 612.
Divide by the number of
Executive rooms.

235

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Page 2

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2 (ay@

3(¢+2)-3x14<7

QUESTION | SOLUTION MARK | ADDITIONAL
GUIDANCE
Ie Lee

Find the determinant of
Matrix A, in terms of x.
Given that this
expression in less than 7,
write an inequality, in x
and solve it,

On a number line what
would be the largest
integer value of x,

3x+6-42<7
3-3x14<7
3x < 43
x< 14s
2
14
Det A = 48 — 42
=6

173 -14
a
: =5(5, 16

Substitute 14 for y and
find the determinant of
Matrix A, hence write
down the inverse of
Matrix A.

—

— 2
PN = 3 Gx + Gy)

=2x+4y

(b)(i) QR = by QW: WR is 1:5, therefore
QR is 6 times QW.
(ii) PR = 3x + 6y I PR = PG + OR. Use the
triangle law of vector
= addition.
I Ratio PN:NR is 2:1

therefore PN is : of PR.

(iv)

—

QN = —3x+ 2x + 4y

Srx+ 4y

|

QN = QP + PN. Use the
triangle law of vector
addition.

|
F

(ay) | sOP 1 Equal ares/chords
a subtend equal angles.
ii) | P@s = 90°— 57° Angle in a semi-circle is
equal to 90°. therefore
=. 33° if

236

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Page 3

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QUESTION

SOLUTION

MARK | ADDITIONAL

= 55,59 cm?

237

GUIDANCE
SPQ and SOP are
complimentary.
QRs = 180° — 57° PSRQ is a cyclic
quadrilateral opposite
= [23° 1 angles are
supplementary.
(iv) | QSR = 180° — (90° + 66°) t In a cyclic quadrilateral,
opposite angles add up to
= 180° - 156° 180°.
= 24° I
(b)(i) =_ 37 1 Angle QPS = 90° —
ames 22,3° = 67,7°
=9.751 em I 3,7
cos67,7° = a therefore
3,7 PS
PS = Gos67,7° sinisa6
SIN22,3
pg = 52Sini34.6 _
sin22,3
9,757cm
ct1)) i. -1 (3:7 it
QPR = Cos (=)
= 446° Ll Find the angle QPR.
. Subtract angle RPQ from
(ili) | SPR = 67,7° — 44,6° Il angle SPQ getting angle
SPR.
= 23,1° it
4 (aw Vv 22 % (4) x49 it Application of the
7 2 formula for volume of a
r cylinder = mr?.

Substitute for 7, radius r
and length /.
Simplify.

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Page 4

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——

QUESTION | SOLUTION
}+————_——
55,6 x0,63

=35¢

(b)@)

Area = 15 X6 + 5 x20x15

= 90 + 150

= 240 cm?

ADDITIONAL
GUIDANCE

Use of formula for
density to be used.

mass |
volume|’
hence mass = density x
volume.

[density =

|

The arrow is made up of
a rectangle measuring 15
by 6 and an isosceles
triangle with a base of 20
and height 15.
Alternatively find the
area of the large
rectangle 30 by 20.
Subtract the areas of two
smaller rectangles and
two right angled
triangles.

| Perimeter = 2x7 + 2x15 +6+

| 2V15? + 102

(ii)
= 144+ 3046+ 2V325
= 86,06

L+1

Perimeter refers to
distance along the edges,
as such there is only one
6 cm edge.

Application of
Pythagoras Theorem
Appreciation of
composite shapes and
finding their perimeters.

238

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| QUESTION | SOLUTION

(d)@)

ADDITIONAL
GUIDANCE
Divide finally by the
common factor of x.

in 32
“p= 32t 3 orp =

Introduce a constant of
proportional, k.

| Find the value of k by
substitution.

) __|

Gi)

pt? = 32

1
2x5 xt° = 32x2

1 Substitute for p

Isolate # and solve by
finding the cube root of
64.

t3 = 64
t=4
Ks (a) | 85% : $55
The selling price as a
-- 100% : -- more percentage is 85%.
Original price as a
= $55x100 percentage is 100%.
85 {+1 Use simple proportion to
Ee $64,71 I find the original price.
bli 1 enti
(b) (i) ie I Use of percentages
100
$4,12

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE “li 6 (a) i» __ *)?K3° Express 81 as an 3h = Bil exponent of 3. [ Appreciation of a 38x35 laws/rules of indices aS Qu [Ga]? = x and x4 + x? = xe] gk = 313-11 Equal bases implies 3k = 32 equal powers. { k=2 = (b)(i) | 2 y? — Sy +2) There is a common factor 2 and the quadratic factor 2(y —2)(y-1) 2 that remains can be factorised using any method of quadratic expressions factorisation. Brackets are essential. (ii) | axt+atb+bx Regroup the terms. Factorise in two parts. =a+1)+b(1+x) Common bracket (factor) identified as (x + 1) ms 6. 4 (a+b)@+1) | (e) 6 a x(2-x) x _ 6—3(2=x) ~ x(2—x) L _ 8 64+3x 1 ~ x(2=x) _ 3x Factorise 2x — x?. © x(2=x) Find the L.C.M of the denominators and = = I express both fractions

with that common
denominator. The
division line is essential.
Pay particular attention
to change of signs as
brackets are removed.

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QUESTION | SOLUTION

8 (a)@) d= 0,8(12 + 1,5)
2

= 0,4«13,5

=54

MARK

$e

ADDITIONAL
GUIDANCE

(ii) h(12+b)=2A

2A

~ 12+b

(b)G) | (2x)? = x? + 44 —2x4xxC0s120

-1
Ax? = x? + 4? —8xx F)
4x? =x? 4164 4x

3x? -4x-16=0

243

Accurate substitution and
evaluation to a numerical
auswer.

Clear fractions by
multiplying by the
common denominator 2.
Make h the subject by
dividing both sides by 12
+b.

Application of cosine
tule.

Remove the brackets on
(2x)?

Cos120 = —Cos(180
— 120)

= —Cos60

—2x4xx(—Cos60)
simplying to + 4x

Correct reduction with no
wrong working seen.

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Page 8

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QUESTION | SOLUTION

L

(ii)

412,00
- 4,12

$407.88

it

MARK _ | ADDITIONAL
GUIDANCE

Understanding of
Crediting and Debiting.

) |

(iii)

407,88
~ 292,88
$115,00

——]

Study the pattern.
Add or subtract.

(ce) | 600x4x3 i Formula for Simple ‘|
100 Interest [1 = ==]
100
=$72 l Substitute in the formula
and simplify.
(ii) | 600x1,043 — 600 1 Formula for Compound
Interest as Amount (of
= 674,92 — 600 I investment) =P (I+
t
=
=74,92 1 =)
Then Interest = Amount
— Principal OR
Calculate interest for Y 1
using the formula for
simple interest then add
that interest to Principal
for Y, and so on until the
end of three years.
Getting $24; $24.96 and
$2596
Add to get $74,92.
r
(ili) | $74,92 i Simple subtraction
-72.00 realising that interest
$2,92 from Option B is higher
than interest from Option
A.
iL | |

242

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[ SORHION

SOLUTION

9

(a)

x = 360 — (180 + 120)

= 360 — 300

MARK | ADDITIONAL
GUIDANCE

Angle property of a point
as an amount of turning
of 360°.

—

(b) (i) Simple proportion.
= 60
| (ii) 80 | Simple proportion.
qe Zz
=40
(iii) zo 80x90 1 Simple proportion.
(c) Class centres: 1; 3; 5;7 Fr class centres using
the formula sum of class
Mea notrsxotrsxcoixen L boundaries divided by
240 two.
_ 80+180+200+420
= 240 Estimate mean:
There is need to multiply
BBO I the class centre of each
ae class with its
2 . corresponding frequency.
= 3,6 or 3,67 or 3 zor 3 hours 40 minutes ‘| Add the products and
divide by total frequency.
7 r. i
(d) 100 99 1 Independent events apply
240 a 239 the Product Law.
Since there is no
2 0,1726 l replacement both
a8 numerator and
denominator should be
less by L.
It is normal to give
fraction in its lowest
terms though answer as a
decimal fraction is
_| acceptable.

245

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ADDITIONAL GUIDANCE

[ QUESTION | SOLUTION MARK

Correct substitution in
the quadratic formula

= —(-4) + /16 — 4x3x(—16)

Gi)

= 2 —btvb?—4ac
WAM T sc 2a
= 4x16 +192 +1 Round off the answers to
6 3 significant figures.
Completing the square.
- 4+208
~ 6 i 3x? 4x 16 0
3 3 3 3
= 3,07 or— 1,74 +1
2_ 4x _ 16
aa a
‘What must be added to
make LHS a perfect
square.
4. 4.16 4
eee a 8
( 2) _ 4844
a hoa:
( ay _ 52
~~ 3) ~9
2 52
= -4+ [=
aan
2, v52
x= >+—
3 3
2
eal V2
3 3
_ 2+v52
a

244

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE (b) Triangle C correctly drawn with vertices at | | + 1+ 1 | Mark point (2; —1), the

1

(2; 25), 7) and ( 7.92

2

)

centre of enlargement.
Enlargement factor of
=1 ; means that the
image is inverted and on
the other side of the
centre of enlargement.
Join point on B to the
centre of enlargement
and produce that line.
Use a ruler and short
calculation to mark the
three vertices of triangle
Cc.

Join the three vertices to
have triangle C.

(c)

(d)

Translation vector got
by:

Final point — initial point
expressed as column
vectors not in coordinate
form. Brackets are
essential.

On the graph join point
(—2; 2) and (6;—2) bya
line segment.

Express the line segment
as a column vector.

Le ia 2 a
=(6 7 ry

(2;12), (—2; 6), (—2; 8)

Express coordinates of A
in matrix form.
Pre-multiply by the
operator.

Express in coordinate
system.

Brackets essential.

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QUESTION

10 ‘| (a)

Lt |

SOLUTION

Graph with points (1:80), (3:60) (5:40),
(7:60) and (9:0).

Correctly plotted and joined by straight
lines.

p=3—4(3)? +4
=35

ADDITIONAL |
GUIDANCE

Find the class centres.
Class centres used for
drawing the frequency
polygon using class
centres and
corresponding
frequencies as the
coordinates of the
vertices.

Time cannot be negative,
hence the polygon cannot
be closed on the negative
time axis but on the
positive at (9:0).

Substitute +3 for x in the
function to find the value
of p.

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SOLUTION

Rotation

Centre (—1;-1)

Through 90° clockwise or 270
anticlockwise

252

ADDITIONAL
GUIDANCE

Take any two points on
triangle A and join them
to their corresponding
points on triangle B.

If the two lines are not
parallel it shows that the
transformation is neither
Reflection nor
Translation, leaving it as
Rotation.

Find the centre of
rotation using a ruler and
compass.

Establish the angle of
rotation by joining a
point on A to the centre
of rotation and a
corresponding point on B
also to the centre of
rotation.

Then state the direction
in terms of clockwise or

anticlockwise direction,

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L

QUESTION

SOLUTION

MARK | ADDITIONAL

GUIDANCE

all

—t

(iii)

—\—

S10 small squares:--
When 100 small square: 1 unit?

510x1
100

5,1 unit?

|

The region should be
correctly identified as
evidenced by marking by
straight lines without
necessarily shading the
tegion.

Area should be found by
counting squares and
multiplying the number
of squares by area of one
square.

The area of one square
should be found using
the scale.

Using cm would result in
having to multiply by 4
giving an answer of 20,4
cm? . Units of em?
essential.

(iv)

2<x<34

+—

1

The question is: For what
range of values of x is
the graph, curve above y

PE

il ‘Tes "eae

—

I

——

[==

Constraints or
restrictions in
formulating inequalities
in linear programming.
Key words are “not more
than” and “‘at least”

(b)

|
2x + 4y < 16

x+2y<8

1

I

What are the limitations?
Express as an inequality.

(c)

x24
yal

+

Appreciation of “at least”
meaning “equal to” or
“greater than”.

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QUESTION | SOLUTION

ADDITIONAL
GUIDANCE

Profitimax) = 2X30 + 3x40 Substitution in the
equation

Profit = 30 x + 40y

= 60 + 120
= 180

al

Use of correct scale for
each axis is necessary so
that all diagrams can fit
| on the given grid. Make
ai iiss ees are ere ges aipaes sure all vertices are

| | | clearly marked and
joined by straight lines.

12

ers ees |
Seepage HED Ser al ES ferret paneer eae
Make sure all vertices are
(a)@) | Triangle A correctly drawn I clearly marked and

joined by straight lines to
make it a complete
triangle.

|

+

(ii) | Triangle B correctly drawn 1 Make sure it is a
complete triangle.

250

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