Examinations Council of Lesotho
2023 Examiner’s report
Mathematics Paper 1 (0178/01)
2023 – 0178/01 Examiner’s Report
1
Mathematics (0178/01) October/November 2023
Report to Schools
Introduction
This report serves to share information on the examiners’ observations during the marking of
Mathematics paper 1 (0178/01) in the November 2023 examinations. The document gives
comments on the candidates’ general performance, strengths and weaknesses. Problematic topics
and areas on which teachers must put emphasis are also highlighted. It provides detailed
feedback on individual questions by stating the expected responses, candidates’ common
responses (errors), misconceptions and recommendations where necessary.
General comments
Most candidates showed a good performance in this paper, some candidates did not answer
what was asked for in the question. This was seen even when candidates clearly understood
the topic, particularly in question 6 in which depreciation was often given as the answer while
the question asked for the value of the car after one year’s depreciation. In question 4
learners would give one year’s interest instead of the investment period.
Question analysis based on candidates’ responses
Question
Total marks
Performance
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
3
4
4
3
2
2
3
3
3
4
4
4
3
4
4
3
2
2
3
60
Fair
Poor
Poor
Poor
Good
Fair
Good
Poor
Poor
Fair
Poor
Fair
Fair
Fair
Poor
Fair
Fair
Fair
Fair
Total
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1
Evaluate
(a) 3.6 ÷ 0.03
#
#
(b) 3 $ × 3 &
Expected responses: (a) 120
(b) 10
Candidates’ common errors: (a) 0.12, 12, 1.2
(b) un-simplified working
In (a) candidates found it challenging to convert decimals into whole numbers by multiplying
by the powers of 10. For those who managed to convert decimals to fractions struggled to
multiply by the reciprocal of the second fraction. In (b) most candidates managed to convert
mixed fractions into improper fractions. However, they did not simplify the resulting fractions.
#(
The common result was # . Also, some candidates tried to find common denominator even
)&
&
though the operation involved was multiplication, $ × & ×
#*
&
$
× $.
Recommendations: Teachers should give more practice on teaching four basic operations
on all types of numbers, making it clear to learners as to which operations require finding the
LCM.
Conclusion: The question was fairly done.
2
(a)
By rouding each number in the calculation correct to one significant figure,
estimate the value of
(b)
5.62 × 230
0.34
Express 360 as a product of its prime factors.
(b) 23 × 32 × 5
Expected responses: (a) 4000
Candidates’ common errors: (a) 400, 12, 1200
(b) various wrong responses
(a) Most candidates failed to round off 0.34 and 5.62 to one significant figure. As a result,
numerical expressions like
*.(( × )(( *.(( × ) *.(( × )((
(
,
(.,(
,
(.,(
were frequent on the working space.
(b) Numerous
Misconceptions: (a) Most candidates treated division by zero and one equally alike.
Therefore, evaluating
*.(( × )((
(
yielded 1200.
Recommendations: Teachers should teach students place values and rules of counting
significant figures. Besides, teachers should strongly stress that division by zero is
undefined. Teachers should also teach learners to draw the difference between prime and
composite numbers.
Conclusion: Poorly done
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3
3
(a) P = 3.6 × 103 and Q = 1.2 × 102.
Calculate P – Q, giving your answer in standard form.
0
,
(b) Evaluate /3 $ .
#
Expected responses: (a) 3.48 × 103
(b) 1 )
)2
Candidates’ common errors: (a) 3480, 2.4 × 10n (b) $
(a) Some candidates managed to convert the given numbers to ordinary form, however they
could not take the difference back to the standard form as required by the question hence
the result 3480. A sizeable number of candidates subtracted the coefficients irrespective of
powers of 10 having different exponents.
(b) Most candidates managed to change the mixed fraction into an improper one but failed to
apply the third root appropriately.
Recommendations: Teachers should include nth roots of fractions so as to widen the
candidates’ learning scope.
Conclusion: Poorly performed
4
Dlomo invests M8000 in the bank which offers simple interest at the rate 3% per
annum.
Calculate the number of years for which Dlomo will get a total interest of M1200.
Expected responses: 5 years
Candidates’ common errors: Most candidates evaluated an interest for a year, being
M240, but failed to relate M240 with total interest i.e. M1200. For those who managed to
31200
relate, they could not divide correctly 3240 . A few candidates could not differentiate the
principal amount from the total interest on the given information.
Recommendations: Teachers should treat financial mathematics/commercial arithmetic
topic in depth and give learners more practice problems of which the selected examples
should include problems at synthesis level to widen the learners thinking scope.
Conclusion: poorly done.
5
The interior angle of a regular polygon is 108°.
Calculate the number of sides of the polygon.
Expected response: 5
Candidates’ common errors: Most of the candidates got 72°. For those who went further
failed to go extra mile in using 72° to find number of sides of such given polygon.
2023 – 0178/01 Examiner’s Report
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Recommendations: Teachers should encourage candidates to regularly attempt problems
involving interior and exterior properties.
Conclusion: Well performed.
6
A car is valued at M320 000.
The car depreciates by 5% every year.
Calculate the new value of the car at the end of the first year.
Expected responses: 304 000
&%
Candidates’ common errors: 16000, 1600 or 320000 × #((
&
Most of the candidates failed to multiply out or simplify 320000 × #(( on which they, in large
numbers, got 1600.
Recommendations: Teachers should emphasise the importance of simplifying before
multiplying so as to minimise the chances of working with large numbers.
Conclusion: Fairly done
7
−1
A=8
2
3:
3 2:
and B = 8
5
1 0
Find
(a)
3B,
(b)
A – B.
Expected responses: (a) ;<, *(=
Candidates’ common errors: (a) ;<, *,=
(b) ;–?# &#=
(b) ;?# #&=
Most candidates did not manage to multiply or add correctly.
Misconceptions: Some candidates treated 0 and 1 equally alike in their operations and
thereafter got challenged by directed numbers. Candidates missed 3 × 0 and – 1 − 3 which
let to loss of marks.
Recommendations: Teachers should deal with directed numbers frequently in their
calculations.
Conclusion: Fairly done
8
Solve
2@ A 1
3
=
@B2
4
.
Expected responses: 2
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Candidates’ common errors: Most candidates still had divided by 12 even at the stage the
denominators are no more expected
?()@ A #) E ,(@ B ))
#)
or partial multiplication of numerator
terms by the LCM which led into incorrect responses.
Recommendations: Equations and expressions should be well differentiated.
Conclusion: Poorly done
Expected responses: (a) 6
,
(b) ?
Candidates’ common errors: Most candidates got sin 6 as their answer simply because
they failed to relate 10 and 5 as hypotenuse from the given ratio or Pythagorean triples to
find the wanted side.
Recommendations: Teachers should raise awareness into the learners thinking on the
application of Pythagorean triples from the basic one, 3, 4, and 5.
Conclusion: Poorly done.
10
Factorise completely.
(a) 8𝑥 ) −50
(b) 𝑥 ) −5𝑥 −6
Expected responses :(a) 2(2𝑥 − 5)(2𝑥 + 5)
Candidates’ common errors: (a) 2(4𝑥 ) − 25)
2023 – 0178/01 Examiner’s Report
(b) (𝑥 − 6)(𝑥 + 1)
(b) −6, −5, 1
6
In (a), most candidates factorised partially and did not realise that the resulting bracket
consisted of difference of two squares which needed further factorisation, hence 2(4𝑥 ) − 25)
was common. In (b), most candidates failed to choose the correct pair on the factors of
whose sum is −5 instead they only focused on the pair of numbers whose sum is −5 or
product of −6.
Misconceptions: Most of the candidates got a variety of incorrect answers as an implication
that they did not factorise the given expression correctly.
Recommendations: Better emphasis is needed in factorisation but more especially on the
difference of 2 squares when the coefficient of 𝑥 2 is greater than one. Also, teachers should
caution candidates on the best way to choose correct factors which can be used in grouping
the terms of the expressions before engaging in factorisation process.
Conclusion: Fairly done
11
Mr Phela has 120 animals which are cows, horses and goats.
20% of the animals are cows.
(a) This information is drawn on a pie chart.
Calculate the sector angle which represents the number if cows.
(b) One animal is chosen at random from Mr Phela’s animals.
Calculate the probability that the animal chosen is either a horse or a goat.
Expected responses: (a) 720
(b)
?
&
Candidates’ common errors: 24
(a) Most candidates calculated the number of cows instead of the sector angle for the
number of cows as a result 24 was common.
Recommendations: for (a), teachers should emphasise that the frequency is directly
proportional to the sector angle. While for (b), teachers should show the importance of
linking percentages, frequencies, and sector angles to probability.
Conclusion: Poorly done
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Expected responses: (a)
90
360
×
22
7
×7×7 =
1
4
× 22 × 7 =
11 × 7
)
=
77
2
=38.5 cm2 (AG)
(b) 45.5 cm2
Candidates’ common errors: In (a) most candidates substituted correctly into the sector
area equation but did not manage to simplify to the given answer. In (b), most candidates got
the area of a rectangle correct earning a part mark. There were a handful of candidates who
got incorrect answer due to failure to subtract decimals from whole numbers. 46.5cm2 was
common.
Recommendations: Teachers should include formulae of different polygons, both regular
and irregular, in order to develop a deeper understanding on the candidates to calculate
area.
Conclusion: Fairly done
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13
Solve the simultaneous equations.
𝑥 + 2𝑦 = 7
2𝑥 + 3𝑦 = 8
Expected responses: 𝑥 = −5
𝑦=6
Candidates’ common errors: Most candidates did not multiply both sides of the equation
when eliminating variables of their choice. However, for those who opted for substitution
method could not proceed beyond substitution stage.
Recommendations: It should be drawn to learners’ attention that in solving equations
whatever is done to the left hand side should also be done to the right.
Conclusion: Fairly done
Expected responses:
(a) Angle CBT 60°, reason: Tangent to the circle is perpendicular to diameter
(b) Angle BAC 60°, reason: Angles in a triangle
Candidates’ common errors: Most candidates were able to find wanted angles but failed to
link those angles to their corresponding properties.
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Misconceptions: Instead of stating circle theorem, they showed how those angles are
calculated.
Recommendations: Strong emphasis is needed on angle properties.
Conclusion: Fairy done
15
The table shows the ages of 30 boys.
Age
10
11
12
13
14
Frequency
18
2
5
2
3
(a) Find the modal age.
(b) calculate the mean age.
Expected responses: (a) 10
Candidates’ common errors: (a) 18,12
(b) 11
(b) 10,12
In (b), few candidates managed to get a correct answer. Most candidates got incorrect
products of variable and frequency.
Misconceptions: (a) Majority of candidates chose mode as 18 simply because 18 is the
largest number in the table.
(b) Few candidates could not calculate the mean correctly from the frequency table. They
treated each variable as having a frequency of 1 which led to dividing the sum of variables
#( B ## B #) B #, B #?
by 5. Thus,
&
= 12.
Recommendations: Teachers should emphasise that mode is the variable that has the
largest frequency. Teachers should include calculations of measures of central tendency
from frequency tables regularly.
Conclusion: Poorly performed.
16
A straight line passes through the points P(0,3) and Q(−2,1).
(a) Find the gradient of the line PQ.
(b) Write the equation of the line.
Expected responses: (a) 1
(b) 𝑦 = 𝑥 + 3
Candidates’ common errors: for (a) There were multiple responses. Most candidates were
challenged by the use of gradient formula with directed numbers involved. In (b) some
candidates went on to calculate 𝑦 −intercept even though it was already evident on one of
the coordinates.
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Recommendations: Teachers should encourage students to pay more attention to the way
they substitute values of 𝑥 and 𝑦 into the gradient formula. Additionally, teachers should
expose learners to the use of different commanding words in questions.
Conclusion: Fairly done
Expected responses: 300
Candidates’ common errors: 60 and 120
Recommendations: Teachers should integrate angle properties with bearings. They should
also emphasise the direction in which the bearing is measured.
Conclusion: Fairly done.
18
f(𝑥) = 3 −𝑥.
Find f(−2).
Expected responses: 5
Candidates’ common errors: Most candidates got challenged by 2 consecutive negative
signs and ended up with subtraction 3 – 2 = 1 or multiplying as opposed to adding which let
them to 6.
Recommendations: Teachers should deal with directed numbers in many calculations.
Apart from that, teachers should teach learners the difference between evaluating a function
at a given constant (f(𝑘)) and f(𝑥)= 𝑘.
Conclusion: Teacher should dig deeper on evaluating functions, not only using positive
integers but even the negative ones.
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Expected responses:
(a) 910 ± 1
(b) Angle bisection with construction of arcs
Candidates’ common errors:
(a) Numerous angles out of expected range of answer
(b) Inaccurate angle bisection, bisection without arcs
Recommendations: In (a), teachers should instil the skill of accurately measuring the
angles into learners. For (b), teachers should enforce neat non free hand work in
construction.
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Conclusion: Fairly performed
Conclusion and recommendations
To succeed in this paper, the entire core syllabus ought to be covered. Candidates should
remember formulae and apply them appropriately and give answers in the form required.
Candidates should use suitable level of accuracy where necessary. Teachers should
complete the syllabus timeously so that learners can have adequate practice and exposure
to examination type questions. Candidates in preparation for examination should be made
aware that more than 1 mark generally indicates that progress towards a correct solution
may be credited and as such workings should be clearly shown.
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