ORDINARY LEVEL

MATHEMATICS
4004/01
NOVEMBER 2022
EXAMINER REPORT

Introduction
The stakeholder report is for Ordinary Level Mathematics paper one (400401) for the November 2022 session.
The report captures the following:
 General performance of candidates
 Question-by-question analysis
 Highlights on expected solution(s) and method(s) to get the solution.
 Common errors or misconceptions by candidates.
The report ends with a conclusion.
The general performance of the candidates
Very few candidates performed exceptionally well. The majority performance range was from poor to mediocre, showing the barest
mathematical skills.
The report will use the following abbreviation: cwa for: common wrong answer.

Question Expected answer Comments and the method

1(a) 1000001 This was done well. No working was needed
(b) 42 hours This was done well except those who used 12 hours instead of 24 hours.
3
Expected method was 24 + 4 × 24
(c) 110 36/ or Most candidates could not change the decimal part to minutes. They would separate degrees and
11degrees 36minutes minutes by a comma.
6
Expected method was 110 and 10 × 60 minutes
2 (a) 29 377
Expected method was 13
(b) 2 5 3
( ) Expected method was ( ) − ( )
−4 0 4
5 3 −2
Some candidates subtracted ( ) from ( ) resulting in CWA of ( ),
0 4 4

(c) 16 Expected list of square numbers 1;4;9;16

Some candidates confused the concept of square numbers and even numbers resulting in CWA of 4
and 18.
3 (a) 0450 The CWA was 450 or 450 𝑁𝐸. Only three figures should have been used.
90
The correct solution involved 2 and had to convert it to a three-figure bearing

(b) 1;3;5 The expected list of odd numbers was 1;3;5;7. Candidates were expected to choose the first three.
The CWA was 1;2;3. The odd number is a basic maths concept that should not be mistaken for a
prime number.
(c) 0,05 Candidates were to look at the second significant figure (5)
 The CWA was 0.0500. The concept was poorly applied.

4 (a) 8 Some candidates failed to effect the division of a whole number by a fraction.
1 8
7 The expected method was 7 = 1 × 7
8
7
The CWA was .
8
(b) 1 80
The expected method was log 5 16 = log 5 5
Most candidates would divide log 5 80 𝑏𝑦 log 5 16 resulting in the CWA was 5
5 (a) 1 Some candidates obtained the correct answer from the wrong working.
1(2𝑥+3𝑦)
3 Expected method was 3(2𝑥+3𝑦)
(b) 7 49
The expected method was √ 9 after converting to improper fractions
3
7
The CWA was± 3. Candidates were expected to note that the square root was given as positive.

6 −2 The expected method was: factorise and solve (2𝑥 + 1)(𝑥 − 3) = 0 or

3 or 4 5±7
𝑥 = 4 (use of the quadratic formula)
1
The CWAs were 1 and −1 2

7 (a) 0 They were expected to substitute 3 in the function to get 𝑓(3) = 32 − 3 × 3

(b) -2 or 5 3±7
They were expected to form the equation, 𝑑 2 − 3𝑑 = 10 and solve (d-5)(d+2) =0 or 𝑑 = 2 (use
of the quadratic formula)
Most candidates failed to form the required quadratic equation
8 n=13, m=2 Expected method was: Add the 2 equations, remain with 5m=10 or -5n=-65. Solve and substitute or
𝑚 1 1 1 −7
use the matrix method ( )=5 ( ) ( ) or any other correct method of solving simultaneous
𝑛 −2 3 17
equations
9 (a) 53 Most candidates were not familiar with the concept of prime numbers.
Expected list was 47;53;59. Only 53 belongs to the range
The CWA were 51 and 57.
(b) 57 Multiples of 19 are 38; 57;76. Only 57 belongs to the range
(c) 49 Expected list of square numbers was 16;25;36;49;64. Only 49 belongs to the range.
10 (a) 21;28 Analysis of the sequence was required.
The difference between two adjacent numbers was increasing by 1
(b) 1 1 Analysis of the sequence was required.
;
16 64
Candidates were expected to divide the number on the left by 4.
1 1
The CWA were 8 and 32.
11 (a) 8 Most candidates could not expand correctly.
Expected expansion was 1 × 𝑎1 + 1 × 𝑎 0 = 9
𝑎+1=9
The CWA was 10.
(b) 4005 Most candidates did not complete the required steps.
They were expected to convert to base 10 by use of place value:
2× 72 + 0 × 71 + 2 × 70 = 100
Then convert 100 to base 5 by division and writing the remainders
The CWA was 1005 . When they had not converted to base 5
12 (a) 49 Candidates should have added the numerators before squaring.
144 4+3 2 7 2
Expected method was ( 12 ) = (12)
25
The CWA was 144. When they applied the square without first adding the numerators
(b) 5 5 2
Expected working was 4 × 3
6
13 (a) 2 Candidates mistook the rhombus to a square.
The correct diagram was
Where the red lines are lines of
Symmetry.
The CWA was 4.
(b) 36𝑐𝑚2 1 2
Expected method was 48 − (2) × 48
The CWA was 24. Candidates did not realise that they were to use the concept of area of similar
figures where the scale factor is squared.
14 (a) 4(3𝑝2 − 𝑞)(3𝑝2 + 𝑞) Expected method was 4(9𝑝4 − 𝑞 2 ) then factorise using the difference of two squares
Most candidate failed to identify the need to use the difference of two squares.

(b) (2m+3n)(5y+3) Candidates were familiar with the factorisation by grouping

Expected method was 5𝑦(2𝑚 + 3𝑛) + 3(2𝑚 + 3𝑛) or 2𝑚(5𝑦 + 3) + 3𝑚(5𝑦 + 3)
10
15 (a) 36km/h 1000
Expected method was 1
60×60
The CWA was 0,01km/h when most candidates would just convert metres to km.

( b) 6 and -6 The question required two values as a solution.

Expected method was 𝑚 × 2𝑚 − 6 × 12 = 0
𝑚2 − 36 = 0
Some did not see that it is a quadratic equation hence it needs two solutions by factorisation
16(a) 1 1
𝑦> 𝑥 Some gave the inequality 𝑦 > leaving out the 𝑥.
3
3
(b) 400 ≤ 2𝑥 + 2𝑦 < 560 . Candidates were required to find the perimeter in terms of 𝑥 and 𝑦 .
or equivalent when The CWA was 400 ≤ 𝑝 < 560
dividing by common
factor
17(a) 12 Most candidates only solved the inequality.
Some failed to interpret. 𝑛 > 11
The CWA were 11 but the inequality does not have equality
(b) 7cm
Candidates could have done well had they drawn a rough sketch.
Expected method was the use of the Cosine formula
𝑥 2 = 32 + 52 − 2 × 3 × 5 × cos 1200
1
𝑥 2 = 32 + 52 − 2 × 3 × 5 × (− 2)
𝑥 = √49
 1
The CWA was 19. Most candidates used for the value of cos 1200 .
2

18(a) -4 and 4 The formulation of an equation was necessary.

Expected method was 𝑥 2 + 32 = 52
(b)(i) 1;4;9 Most candidates had 4 and 9 skipping 1.
Candidates did not realise that 1 is a perfect square.

(b)(ii) 4 Candidates could not list the factors of 8

Expected factors of 8 were 1;2;4;8
The cwa was the inclusion of 3

(b)(iii) 𝐵 ⊂ 𝐴 or A⊃B Candidates were not familiar with set notation

19(a) 180 C Wrong units were used

Candidates were expected to select the number with highest frequency
Some gave the 18° without the C.
(b) 210 C 20+22
Expected method was 2 for the median

(c) 110 C The concept of range was not well understood by candidates.
Expected method was 28-17
(d) 21,20 C Some candidates failed to add all the numbers given.
18+20+22+23+26+28+22+18+18+17 212
Expected method was = 10 or any other correct method
10
20 (a)() 2𝑑0 or 180−4𝑑 2 The cwa was 450 .
The answer should have been in terms of d.
Candidates could have used either alternate angles or sum of opposite interior angles equal to
exterior angle in a triangle.
(a)(ii) 300 Candidates were required to form the correct equation.
Expected equation was 4𝑑 + 2𝑑 = 180 or 180 − 4𝑑 + 𝑑 = 3𝑑 or 𝑑 + 180 − 3𝑑 = 4𝑑 then solve
to find the value of 𝑑
(b)(i) 380 Expected method was 1800 − 1420 ( triangle OBA is an isosceles triangle)
Some candidates failed to identify the theorem to apply

(b)(ii) 760 Expected method was 2 × 380 (B𝑂̂𝐶 𝑖𝑠 𝑡𝑤𝑖𝑐𝑒 𝐶𝐴̂𝐹)

Most candidates failed to identify the correct theorem to apply
21 (a)( i) 120 Expected method was 13𝑥 + 2𝑥 = 180
The CWA was 24. Some candidates equated the equation to 3600 instead of 1800 .

15 3600 3600
Expected method was 2×120 or 1800 −13×120
(a)(ii)
The CWA was 30. Most candidates divided 3600 by 12 instead of 24°.

21(b) 178 .
3 3 5 5 12 12
400 Expected method was with replacement 20 × 20 + 20 × 20 + 20 × 20
The question was not answered by many. Most of those who attempted it worked without
replacement

22 (a)( i) 8m/s Substitution was familiar to candidates.

Expected method was 𝑣 = 5 + 4 × 3 − 32
(a)( ii) 5 Only the positive was the expected answer since time is positive
−4±6
Expected method was: factorise and solve (5 − 𝑡)(1 + 𝑡) = 0 or 𝑡 = −2 (use of the quadratic
formula)
Some included the negative value of time in the final answer when they were expected to use the
positive only
(b) 𝐴 − 2𝜋𝑟 2 Expected method was 2𝜋𝑟ℎ = 𝐴 − 2𝜋𝑟 2.
ℎ= Some few candidates could not isolate h.
2𝜋𝑟

23 (a) 8cm Candidates were expected to equate the equal sides to x; 5+x+17+x=42. X=10. The equal sides are
10 cm long. 5
 10 h 10

17
Using Pythagoras theorem; ℎ = 102 − 62 . ∴ ℎ = 8
2

The CWA was 10 just copying one of the sides. Most candidates could not proceed to find the
height.

1
(b) 88𝑐𝑚2 Expected method was 2 (5 + 17)8 or the isosceles trapezium could be divided into parts.
The CWA was 110. Candidates used 10 as the height
24 (a) $150 100
Expected method was 120 × 180.
The CWA was $144, 00. Most candidates subtracted 20% of $180 from $180.

(b) $120
80 80 20
Expected method was 100 × 150 or 120 × 180 or 150 - 100 × 150

The CWA was $115, 20. Most candidates subtracted 20% of $144 from $144.

Conclusion
The report highlighted the performance of the candidates in the component.
Question by question analysis shows that the following topics need attention by Teachers in preparing Candidates for future examinations:
Bearing
Theory of logarithms
Functional notation
Sequences
Number bases
Area of similar shapes
Units of measurement
Inequalities
Cosine rule,
Range
Exterior angles
Probability
Profit and loss