2019_mathsMS_olevel_mj_paper2

Extracted Content: 2019_mathsMS_olevel_mj_paper2

Page 1

Extracted Text

4004/2 JUNE 2019 SOLUTION GUIDE

QUESTION | SOLUTION

MARK | ADDITIONAL GUIDANCE

1

4415
6+1 7

(a)

Realise that both numerator and denominator
is increasing by |.

(b) 10=3+7

Key is the concept of a prime number and
these have to be less than 10.

()i)

112
05x —— =
$105x Too $117,60

4
(443)

(ii)

Tendai gets = x$10
= $60

. 3
Chipo gets = ae «$105

= $45

Finding 12% of 105 $12,60 then adding to
$105.

Either calculating Chipo’s share then
subtracting it from $105 so as to get Tendai’s
share or the other way round still acceptable.

5

=50 130

6 = 90 — 40 and 180 — 50

Alternatively Cos 40 = 0,7660

Therefore Sin“1(0,7660) = 50

Recalling that Sine of an obtuse angle is also
positive.

(b)(i)

AP = sin37
94°"

- AP = 9,4Sin 37

= 5,657
AC 9,4

Sin 37 Sin 48

AC = 9,4 Sin 37
“sin 48

= 7,612

Identifying the correct trig ratio to use in the
right angled triangle or 9,4 Cos 53 (ii)

Realising that the shortest method would be
using Sine rule.

However using the calculated value of AP in
triangle APC would be applicable giving
AP AP

or ——

i Cos42 Sin 48
je 22857. 9p 5:887
Cos 42 Sin 48

Four significant figures should then be used.

149

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Page 2

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

(i) 360 — 3y or 2(4y + 4) i Recall of angle property of a point as ab
amount of turning of 360°. And that angle at

(c)

OR 7x0,99 7 — 1% one would get 6,93

the centre is twice that subtended on the
circumference.
(ii) La +4) = 360-3y I Appreciation of circle geometry theorems.
8y + 8 = 360 —3y Formulation of the equation and solving it.
lly = 360-8
1ly = 352
3 (a)(i) | 7:91,70 Appreciation of Ratio and expressing this as
a fraction.
* 1:? (less)
$/cent conversion and the importance of the
91,7 zero.
= = 13,1
7
*. 1:13,10 |
(ii) $7 —7x0,01 = $6,93 2 1% means 0,01
Even on the calculator

(b) 84%: $210 3 Original price is always 100%, it is the
selling price, as a percentage, that differs
16% means selling price as a percentage

becomes 84%.

* 100%: ? (more)

_ $210x100

84
= $250
(ce) 2010x100 PRT

P= 3x5 Recalling the formula I = —
= $13 400 1 Making P the subject
1001 = PRT
1001 _
‘a=
——_t_
150

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QUESTION

SOLUTION

MARK | ADDITIONAL GUIDANCE

4 (a)

(b)(i)

Amount = $600x1,043
= $674,918

= $674,92

3@—1)—-(@+2)

x—1
_ Bk B- B= 2
7 ea
_ 2x-5
~ ed

x7 4+3x-8=3x41

x?-8=1
x?7=9
x=343
~ | a) [2% =0,25
A
4
2% =2-2
x=-2
(c) ax+b=d?

Correct substitution and simplification.

Use of the formula

axe
Amount (Yield) = P (1 ++) followed by
substitution and correct simplification.

Alternatively one can calculate simple
interest yearly remember that interest accrued
is added onto the principal amount as one
goes to the next year.

Find the L.C.M of the denominators and
express both fractions under that common
denominator.

The division line is important. Pay particular
attention to change of signs as brackets are
removed.

Interpreting the functional notation.
Formulating the equation.
Solving the quadratic equation.

Using any of the methods of solving
quadratic equations.

Expressing 0,25 as a common fraction.

Appreciating that SAME BASE implies
EQUAL POWERS.

Removal of the square root sign by squaring
both sides.

Isolating the term in x on one side.

Making x the subject by dividing both sides
by a.

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

Division line is essential and should cover
d?—b.

| Show that ruler and compasses have been
used through evidence of construction arcs
and lines.

If construction space is not enough ora
mistake has been made, ask for plain paper
and construct on the plain paper and attach it
as additional material.

Remember to write your candidate details.

(a)(i) | Triangle ABC with two sides
7cm and an angle of 120° with
correct construction arcs.
Bisector of ABC with correct
and clear construction arcs.

Construction of 120° can be done by
constructing an angle of 60 externally.

(ii)

The bisection arcs should show evidence of
use of relevant mathematical instruments.

(iii) Perpendicular bisector of side 2, Points of intersection of the two sets of arcs
BC with correct and clear must be clear.

construction ares.

in

(b)@ | BAD correctly constructed with | 2 Emphasis is on construct of an angle of 45° at
clear and correct construction A.
arcs.

Could be done by first constructing an angle
of 90° then bisect the angle to get 45°.

OR realising that ABC is an isosceles triangle
with angle at A = 30°, falling short of 15° to
make it 45, therefore constructing 60° and
bisecting it gives 30° angle which can be

bisected also giving an angle of 15°.

152

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Page 5

Extracted Text

QUESTION

SOLUTION

6 (a)(i)

Point D clearly marked and
labelled.

ADDITIONAL GUIDANCE

In form of small cross or clear dot.

Region correctly identified and

clearly shaded.
n(PNQ)= 0

(b) | @

n(PUQ)=35+415

=50

(a)(i)

5x—-13 <x-—bandx-—6<
9+ 4x

4x < 7 and —15 < 3x

Understanding of the intersection of the two
Loci,
Note that 59 — (15+ 35+ 9) =0

Suggesting that set P and Q are disjoint sets.

Following the conclusion that set P and Q are

disjoint.

The key is the understanding that the two
probabilities on the two attached branches
should add up to 1,a CERTAINITY.

P(G and G) requires Product law.

1 Pe and F) requires Product law.

The use of the tree diagram will simplify the
problem and where to apply addition law.

The inequality is to be split into two and
solve the two separately then combine the
results.

153

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Page 6

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| QUESTION

SOLUTION

-—
x <3/,and-5 <x

-h<x<i 4,

—

MARK | ADDITIONAL GUIDANCE

—-

| ! A number line closed at 1: and open at -5,

Directed numbers and positions on a number
line and key is the term integer, a whole
number positive or negative including zero.

F ——}— ———_
(b) = Rb 2 { Recap on trig ratios in right angled triangles.
Sin AC B=
2x+3
Ti x+2 _ a tr Deductive reasoning.
(ii) | 16(x + 2) = 9(2x + 3) Remove fractions by multiplying by 16(2x +
3) both sides or cross multiplying
16x + 32 = 18x +27
Collecting like terms
§= 2%
Solving for x
L
2 2 =x 1
(iv) ae =x : is I Substitution for x.
‘- = 8cm |
(v) BC = f8?-4,52 l Application of Pythagoras theorem.
= /43,75 Using the calculate to evaluate,
= 6,614 , ¥8?-4,5?

[ wa tp= 1744)

154

Substitute for x and simplify.

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Page 7

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QUESTION | SOLUTION

ADDITIONAL GUIDANCE

Find the value of y when x = 5 in y = x? —
4x

(b)(i) Plots should be visible and the graph should
be drawn using free hand, not grossly thick

and passing through the correct points.

155

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Page 8

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

s
H

eae Os: es ee
4

(ii) The line y = 3 — x correctly 2 In general lines drawn should be at least 3em
drawn to cut the curve in two unless stated otherwise.
places.
(OW | x=—0,8 1 These values are found where the straight
line intersects the curve.
x= 3,8 I
(i) x=2 2 Mirror line for bi-lateral symmetry.
| Parallel to Y axis (x = 0).
9 (a) 1 ENDS A 1 The formula for finding area of a triangle
2 *26x24x Sin 40 given two sides and an included angle.
= 20 2 { 1
Caen [5 ab Sin C, where C is the included angle
a

(b) | AD? = 26? + 242 I Application of Cosine rule.
— 2x26
x24 Cos 40 Correct substitution in the formula

156

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Page 9

Extracted Text

MARK | ADDITIONAL GUIDANCE

7 =b? +c? —2be CosA

Use of the calculator or relevant
mathematical tables.

d = 12,77 Sin40

= 8,208cm

QUESTION | SOLUTION
AD? = 295,977
AD = 295,977
= 17,2em |
(ce) BC 24 2
Sin30 Sin(110)
_ 24 Sin 30 i
~ Sin 110
= 12,77cm !
@ [dad
Es = Sin 40

BCD = 180° — (30° + 40°)
= 110°

Use of Sine rule.

The shortest distance is one that makes a
right angle with BD and passing through C.

Use the trig ratio Sin 40° to find the shortest
distance.

165 <A < 180

(ii) 165 <h< 180

(iii) 160 <h<180

(b) [s (150 + 160)

9 (160 + 165)
2

165 + 180
yg G68 + 180)

2
9 (180 + 190)

7 |

ed

+42

That class interval with highest frequency.

Stated as it is in the table.

Median at > (42 + 1)"

1
3x43" ie.21,5°%

Lower quartile at ; (43)" = 10,75%"
Find the class centres for each class or
interval.

Multiply the class centre with its
corresponding frequency.

Add the products.

Divide by total frequency.

157

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Page 10

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QUESTION | SOLUTION MARK | ADDITIONAL GUIDANCE

= (5x155 + 9x162,5

+18%x172,5 +
10185) + 42
_ 7192,5
~ 42

= 171,25
LL :
(c)

_ 27 26 Since there is no replacement subsequent
“Ra probabilities should be less by | for both
numerator and denominator,

~ class width

=< or 0,407 3
287 Independent events hence apply the product
| law.
_——s
(d) Correct histogram drawn with 3 For histogram, height is found by calculating

frequency densities (heights) frequency density,
0,5: 1,8: 1,2 and 1.

Using the formula:
| : | requenc
noe cs at nO and fra = Femency

Find area of ACDE

Find area of rectangle ABCE

Add the two areas.

= Sm? 3
(b) V=5x3 Volume = Cross-sectional Area X length

= 15m3 2
(c) 23m? Cannot buy a fraction of the tin therefore
4,5m2 round up!
aE

158

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QUESTION

SOLUTION MARK | ADDITIONAL GUIDANCE

(b) (Gi)

=5,111

= 6tins 2
(d)@) | pg =./72 4 0,62 Use of Pythagoras Theorem

= 1,166 &

(ii) 1,166x3x2 Area of the rectangular sides

= 6,996x$6,40 DEX3mx2 sides.

= $44,77 3 Material costs $6.40 for every square metre,
(a\@ | (3 3 =3 2 Scalar multiplication means each entry is

(3) aa ( 1 ) multiplied by the scalar.

= (3) 7 (3) Addition / subtraction of vectors.

~\9 3

Brackets are essential.

-_ ee
~\6
12? + 62

= v180

Magnitude/modulus of the vector as size of
the line segment representing vector (2)
Apply Pythagorus theorem

= 13,42

2(p + q) Scalar multiplication of a vector.

=2p+2q

OM = 0A + AM

Use the triangle law of vector addition.

AC =AB+BC
Use the triangle law of vector addition.

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Extracted Text

QUESTION | SOLUTION

MARK | ADDITIONAL GUIDANCE

(iv) > 1 I — Scalar multiplication.
OT =k(p+ 3 q)
=kp+ Ska
) | OF = p+ he 2hq I AT = hAC
= (1+h)p + 2hq) = h(p + 2q)
=hp+ 2hq

Therefore OT = OA + AT

Use the triangle law of vector addition.

——
(vi)

A1th=kuw()

il
(1+h)p + 2hq = kp +5kaq

=

Two equations are to be formed and solved
l simultaneously.

Use any method of solving simultaneous
equations.

[+1

wh aah 2
Roel)
1
a=
4 1
k= -orl1=
(vii) + ‘
ViL
hors

!
other.
i

Parallel vectors imply scalar multiple of each

160

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