 # 2023/2024 May-June WAEC General Maths Exams: Authentic questions 2023/2024 May-June WAEC General Maths Exams: Authentic questions are for those serious students preparing for general mathematics in this forthcoming examination. So, are you one of those seriously preparing for this examination? Do you want to make good grades in mathematics? If YES, then this post is for you

These questions are the standard questions that compare favorably with those of the previous years. That is why we project it as tentative questions for this year. In fact, it has good coverage of the different sections of the syllabus and it is within the ability of every good and average candidate. The rubrics were also clear and unambiguous.

## Completefmc.com Exams info:

Here, you can find standard questions that are alike to those of the previous years. In fact, it has a different section which is why you should read through and understand the questions before answering. Completefmc.com gives you access to doing so.

## The Questions:

Here are the questions. You should read through and understand the questions before turning to the answers. In fact, it will be better to attempt to solve the questions before turning to the answers. Good luck

Qts 1.

• Factorize completely: m2 – 2mn + n2 – 9r2
• Solve simultaneously, the equations:

5x – 4y = 6

33(y – x)= .

Notes:

In part (a), candidates must recognize that the expression was that of difference of two squares and so you must be able to factorize the expression as required. You re expected to factorize m2 – 2mn + n2 to obtain (m – n)2. This, therefore, meant that the expression m2 – 2mn + n2 – 9r2 was equivalent to (m – n)2 – 9r2 which by the method of difference of two squares factorized further as

((m – n) + 3r)((m – n) – 3r) i.e. (m – n + 3r)(m – n – 3r).

In part (b), candidates should obtain the second equation. You re expected to express both sides of the equality in index form with the same base and equate the indices.   expressed in index form was 3-3. The second equation would now be 33(y – x) = 3-3 which implied that 3(y – x) = -3 or y – x = -1. Solving this equation with the first one simultaneously gave x = 2 and y = 1.

### Question 2 – 2023/2024 May June WAEC General Maths Exams: Authentic questions

A man has a wife and 6 Children and his total income in a year was GH¢ 850.00. He was given the following tax-free allowances:

Personal                      GH¢ 120.00;

Wife                           GH¢ 30.00;

Children                       GH¢ 25.00 per child, for a maximum of 4 children;

Medical                        GH¢ 40. 00.

The rest was taxed as follows:

First GH¢ 200.00 at 10%;

Next GH¢ 200.00 at 15%;

Next GH¢ 200.00 at 20%;

Remainder at 25%.

Calculate his:

• taxable income;
• monthly tax.

### 2023/2024 May-June WAEC General Maths Exams: Authentic questions

Notes:

For this question, candidates re encouraged to read the questions carefully before answering so that they can appreciate its demand(s).

You re expected to respond as follows:

Total tax free allowance = 120 + 30 + (25 × 4) + 40  =GH¢ 290.00

Therefore, taxable income = GH¢ (850 – 290) = GH¢ 560.00.

The tax would be deducted as follows:

From the first GH¢ 200.00, tax = × 200 = GH¢ 20.00. From the next GH¢200.00, tax = × 200 = GH¢ 30 .00. Remaining taxable income = GH¢(560 – 400) =GH¢ 160.00. Therefore, tax on GH¢ 160.00 = × 160 = GH¢ 32.00. Total annual tax = 32 + 30 + 20 =GH¢ 82.00. The monthly tax was then obtained as = GH¢ 6.83

Question 3

1.

In the diagram, PQ and QR are straight lines, |PS| = 6cm, |QS| = 4 cm,

|QT| = 5 cm and ∠QTS = ∠RPQ. Calculate |TR|.

If sin x = , 0o ≤x ≤ 90o, evaluate, without tables or calculator,

_____________________________________________________________________________________________________

### 2023/2024 May-June WAEC General Maths Exams: Authentic questions

Observation

This was another question that was reported to be very popular among the candidates. According to the report, while the majority of the candidates performed well in part (b), their performance in part (a) was poor, indicating a poor knowledge of geometry.

In part (a), the majority of the candidates wer reported not to apply the concept of similar triangles correctly. From the diagram, candidates were expected to recognize that triangle QTS was similar to triangle QPR. Hence,  .  = . Solving this equation gave |TR| = 3 cm.

In part (b), the majority of the candidates wer reported to have applied the concept of right-angled triangles to obtain the ratio of cos x and tan x as and respectively. They were also reported to have substituted these ratios into the given fraction to obtain = = =.

Question 13

Observation

The Chief Examiner reported that this question was the most unpopular question and the majority of those who attempted it performed poorly.

In part (a), candidates were expected to respond as follows:

x + y =  +  =. Therefore, |x + y| =  = 5 = 7.1.
In part (b), candidates wer expected to show that if M(x, y) was the midpoint of
, then x =  = 2 and y =  = 1. Hence, the required point was M(2, 1). Similarly, if S(x, y) was the midpoint of, then, x =  = 5 and y =  = -1. Therefore, the midpoint was S(5, -1).  =  –  = . Similarly,  =  – = . By comparing the two vectors they would conclude that  = 2. To find the equation, candidates would first find the gradient of as  =. The required equation was y – 1 =(x – 2) which simplified to 2x + 3y = 7.

Question 4

### 2023/2024 May-June WAEC General Maths Exams: Authentic questions

Without using tables or a calculator, evaluate log10( ) – 2log10( ) + log10( ).

• :
•  Y;
•  Y).

Observation

The Chief Examiner reported that this question was attempted by the majority of the candidates and they performed well in it.

In part (a), candidates wer reported to apply the laws of logarithms correctly. They were able to express the given problem as a single logarithm i.e. log10( ) – 2log10( ) + log10( ) = log10( ) = log10( ) = log10( ) = log1010 = 1.

In part (b), the majority wer reported to have obtained the elements of sets X and Y as X = {10, 11, 12, 13, 14} and   Y = {10, 12, 14, 16}. X Y = {10, 12, 14}. X’ = {15, 16, 17, 18, 19, 20}. X’ Y = {16}. Therefore, n(X’ Y) = 1.

### Candidates’ weaknesses/remedies

The Chief Examiner noticed that some candidates did not close some of the elements of sets with curly brackets. Other areas where candidates’ weaknesses notice included:

• Premature approximation and omission of essential details
• Writing answers to the required degree of accuracy
• Inability to apply mathematical principles correctly
• Circle Geometry and its applications
• Geometrical construction
• Change of subject of the formula
• Mensuration
• Interpretation/solution to  word problems
• Factorization
• Similar triangles
• Commercial Arithmetic
• Functions

### SUGGESTED REMEDIES

The following remedies were suggested by the Chief Examiner:

• Candidates wer encouraged to cover all the topics in the syllabus while preparing for the examination
• Teachers as well as candidates were encouraged to put more effort into studying geometry.
• Teachers were encouraged to use instructional materials during lessons so as to re-enforce the learning of mathematical concepts
• Candidates wer encouraged to adhere to the rubrics of the question, especially with regard to the use of calculators and the degree of accuracy
• Candidates were encouraged to avail themselves of past WASSCE General Mathematics questions.

### Candidates’ strengths

The Chief Examiner observed that there was an improvement in representing given information with diagrams in those areas where they were necessary.  The Chief Examiner also commended candidates’ performance in the following areas:

• Completion of the table of values and drawing of graphs of quadratic equations
• Geometric Progression
• Application of the theories of logarithms
• A solution to simple linear inequalities
• Drawing of bar charts
• Solution of basic trigonometry