WAEC 2024 General Mathematics (OBJ/ESSAY) WASSCE (PC 2ND) Questions and Answers

2024 WAEC Mathematics (MATHS) Objective (OBJ) Theory (ESSAY) Questions and Answers is out: 

The West African Examinations Council (WAEC) has given students helpful materials to understand what's needed for the Mathematics final exam.

These materials include comments, analysis of how students have done in the past, things students have done well, things they've struggled with, and possible answers to questions.

These resources are really useful for students getting ready for the exam. They show how to answer questions effectively and what's expected.

Students should go through these resources carefully to do better in the exam.


2024 WAEC General Mathematics Paper 2, WASSCE (PC 2ND)


Question No 1:

(A) Kojo invested $4,000.00 and Alidu invested $6,000.00 in a business. After one year, they made a profit of $4,800.00. After deducting expenses of $1,280.00, they shared the remaining profit based on their investments. Find how much each person received.

(B) In an Arithmetic Progression (A.P.), the 8th term is -8 and the 3rd term is 12. Find:

(i) the common difference;

(ii) the first term.

Solutions:

(A) Profit Sharing:

Kojo's investment = $4,000.00
Alidu's investment = $6,000.00

Total investment = Kojo's investment + Alidu's investment = $10,000.00

Profit to be shared = Total profit - Expenses = $4,800.00 - $1,280.00 = $3,520.00

Kojo's share = (Kojo's investment / Total investment) * Profit to be shared = ($4,000.00 / $10,000.00) * $3,520.00 = $1,408.00

Alidu's share = (Alidu's investment / Total investment) * Profit to be shared = ($6,000.00 / $10,000.00) * $3,520.00 = $2,112.00

(B) Arithmetic Progression:

Given: 8th term (a8) = -8
3rd term (a3) = 12

To find the common difference (d):

For a8, we have: -8 = a1 + 7d

For a3, we have: 12 = a1 + 2d

Now, solve these equations simultaneously to find d, and then find a1.

Observation:

The Chief Examiner noted that many candidates were able to successfully tackle part (b) of the question. However, some still struggled with understanding the question's requirements. In part (a), while the majority of candidates provided the correct answer, some failed to indicate the decimal place for the monetary value.

IN PART (A), candidates appropriately found the ratio as 2:3. After deducting expenses, the remaining amount was calculated as $3,520.00. Kojo's share was $1,408.00 and Alidu's share was $2,112.00.

FOR PART (B) (I), candidates correctly found the equation for the 8th term (a + 7d = -8) and the equation for the 3rd term (a + 2d = 12). Solving these equations yielded the common difference (d) as -4.

IN PART (B) (II), candidates correctly substituted -4 for d in either equation (1) or (2) to find the first term (a) as 20.

Question No 2:

The distance between two ports, M and N, is 2,100 km. Two ships are traveling towards each other. One ship leaves port M at 20 km/h, while simultaneously, another ship leaves port N at 15 km/h.

(a) How long will it take the two ships to meet?

(b) How far will they be from port M when they meet?

Solutions:

(a) Let t be the time it takes for the ships to meet. The total distance traveled by both ships when they meet is the sum of the distances each ship travels. Using the formula:

distance = speed × time

The equation for the total distance traveled by both ships is:

20t + 15t = 2100

(b) Let d be the distance from port M when the ships meet. Since both ships are traveling towards each other, the distance traveled by one ship is 20t and by the other ship is 15t. So, the equation for the distance from port M is:

d = 20t.

Observation:

The Chief Examiner reported that the majority of candidates attempted this question, but many struggled with forming the required equation involving speed, distance, and time.

In part (a), candidates were expected to find the total distance traveled by both ships, which should lead to the equation 35t = 2100, where t represents time in hours. Solving this equation gives t = 60 hours.

In part (b), candidates were expected to substitute 60 hours for t in the equation to find the distance from port M, which should be 20 * 60 = 1200 km. Therefore, the distance from port M is 1200 km.

Question 3:

The curved surface area of a cone is 242 cm2. If the slant height is 4 cm more than the radius, calculate, correct to one decimal place, the:

  1. (a) radius;
  2. (b) height;
  3. (c) volume, of the cone.

[Take π=22/7]

Solutions

Given:

Curved surface area of cone = 242 cm2

Slant height = Radius + 4 cm

(a) Radius (r):

Let's denote:

  • r = radius of the cone
  • l = slant height of the cone
  • h = height of the cone

We have the following formulas:

  1. Curved surface area of a cone: CSA = πrl
  2. Slant height of a cone: l = r + 4
  3. Height of a cone using Pythagoras theorem: h = √(l2 - r2)

First, let's find the radius (r):

Given that CSA = 242 cm2 and l = r + 4:

CSA = πrl = πr(r + 4) = 242

πr2 + 4πr = 242

r2 + 4r - 242/π = 0

Now, let's solve this quadratic equation to find r.

(b) Height (h):

Once we find the radius (r), we can use the formula for height (h) to find the height of the cone.

(c) Volume:

Finally, once we have the radius (r) and height (h), we can use the formula for the volume of a cone to find the volume.

Volume \( V = \frac{1}{3} \pi r^2 h \)

Observation:

The Chief Examiner reported that this question was popular among candidates, indicating a good understanding of the question.

Part (a):

In this part, a good percentage of candidates that attempted it could not apply the Pythagoras theorem where necessary.

They were expected to substitute the given values into the curved surface area formula to get \( 242 = r(r + 4) \).

Simplifying gives \( r^2 + 4r - 77 = 0 \). Factorizing gives \( (r - 7)(r + 11) = 0 \).

Solving gives \( r = -11, r = 7 \). Therefore, \( r = 7.0 \) cm.

Part (b):

In this part, some candidates were able to apply the trigonometry ratio correctly, but a good number of them could not get the trigonometry ratio correctly.

They were expected to recall that the radius, slant height, and the height form a right-angled triangle.

Using the Pythagoras theorem gives \( h^2 + r^2 = (r + 4)^2 \).

Then substituting the value of the radius gives \( h^2 + 7^2 = 11^2 \).

Simplifying gives \( h^2 = 72 \). Taking the square root of both sides gives \( h = \). (The calculation for \( h \) is missing in the observation.)

Part (c):

In this part, they did as expected and substituted the given parameters into the volume relation to obtain \( \text{Volume} = 72 \).

Simplifying gives \( \text{Volume} = 435.576 \) cm³.

Therefore, \( \text{volume} = 435.6 \) cm³ (1 d.p.).

In the diagram, \( UR \) and \( RS \) are tangents to the circle with center \( O \). If \( URS = 80° \), find \( UTS \).

Given that point \( P \) is 270 m from \( Q \) on a bearing of \( 328° \) and point \( R \) is 420 m from \( Q \) on a bearing of \( 058° \), find the bearing of \( P \) from \( R \).

Observation:

The Chief Examiner noted that this question was unpopular among the candidates, with only a few attempting it satisfactorily.

Part (a):

Candidates were expected to draw straight lines to join point U and O, and also to join point O and S. They should have observed that ∠OUR = ∠OSR = 90° since quadrilateral RUTS forms a quadrilateral. It means ∠O = 360°. Substituting gives 900° + 900° + 800° = 3600°. Simplifying gives 3600° - 2600° = 1000°. It implies ∠OUS = 1000°. From the angle at the center and circumference theorem, ∠OUS = 500°.

Part (b):

Candidates were expected to draw the bearing diagram as shown.

We observe that the triangle formed by the diagram is a right-angled triangle. Then tan(θ) = US / OU. Solving gives θ = arctan(US / OU) = arctan(2380 / 2720) = 32.740°. Therefore, the bearing = 32.740° + 238° = 271°.

WAEC Mathematics OBJ ESSAY Questions and Answers

See also : WAEC Physics Paper 1, May/June 2008-2024 (Questions and Observation)

General Comments:

Overall, the Chief Examiner noted that the quality of the paper was good and similar to previous years. The questions were clear, straightforward, and evenly distributed, with no confusion in the instructions. However, despite this, many candidates struggled with certain questions, indicating a lack of preparation.

The Chief Examiner also mentioned that candidates did better compared to previous years.

Candidates' Weaknesses:

  • Many candidates didn't include units or provide answers in monetary value when required.
  • Some candidates didn't follow the instructions properly.
  • Many candidates made approximations too early or inaccurately.
  • There was a lack of understanding in interpreting questions and applying mathematical principles correctly.
  • Candidates struggled with Bearing and Circle Geometry, Ratio, Proportions, Rates, and Probability.
  • There was overreliance on calculators, leading to skipped essential steps.
  • Candidates spent too much time on one part of the questions, leaving insufficient time for others.

Suggested Remedies:

  • Candidates should focus more on understanding topics related to diagrams, such as bearing, circle geometry, and graphs.
  • Candidates should carefully read and understand the requirements of each question before attempting them.
  • Encourage candidates to cover the entire examination syllabus during their preparation to improve performance.
  • Parents or guardians should provide modern textbooks to aid in studying.
  • Candidates should allocate more time to learning fundamental principles and practicing mathematical problem-solving.
  • Candidates should be encouraged to read questions carefully and understand their requirements before attempting them.

Candidates' Strengths:

  1. Candidates demonstrated a strong understanding of Numbers and Algebra.
  2. There was noticeable improvement in handling questions related to Statistics.
  3. Candidates showed good knowledge in tackling questions on Mensuration.
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