WAEC 2024/2025 Further Maths OBJ/ESSAY Questions and Answers


Complete 2024 WAEC May/June Further Maths (OBJ) Objectives  and (ESSAY) Theory Questions and Answers EXPO Room for School Candidates & private|WAEC Further Mathematics (OBJ/ESSAY) Questions and Answers. 2024 WAEC FREE QUESTION AND ANSWER ROOM FOR Further Mathematics OBJ THEORY [School Candidates





2024 WAEC Further Maths (OBJ) Objectives Answers:

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2024 WAEC Further Mathematics (ESSAY) Theory Answers:







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Past WAEC 2023 further (Maths) Mathematics Questions And Answers

No. 4)
(a) To find the number of terms in the series, we can use the formula for the sum of an arithmetic progression:
Sum = (n/2) * (first term + last term)
where "n" represents the number of terms in the series.
Given:
  • First term (a₁) = -8
  • Last term (aₙ) = 52
  • Sum (S) = 286
Using the formula:
286 = (n/2) * (-8 + 52)
Simplifying the equation:
286 = (n/2) * 44
Dividing both sides of the equation by 44:
286/44 = n/2
6.5 = n/2
Multiplying both sides of the equation by 2:
13 = n
Therefore, the number of terms in the series is 13.
(b) To find the common difference (d), we can use the formula:
Last term = First term + (n - 1) * common difference
Given:
  • First term (a₁) = -8
  • Last term (aₙ) = 52
  • Number of terms (n) = 13
Using the formula:
52 = -8 + (13 - 1) * d
Simplifying the equation:
52 = -8 + 12d
Adding 8 to both sides of the equation:
60 = 12d
Dividing both sides of the equation by 12:
5 = d
Therefore, the common difference in the arithmetic progression is 5.
WAEC Further Mathematics answers objectives loading.......

WAEC 2023 further Mathematics  Questions Papers:











Continuation of the Wace 2023 further maths answers

(6)
We are given:
P(n) = 1/3 and P(n') = 1 - 1/3 = 2/3
P(T) = 1/5 and P(T') = 1 - 1/5 = 4/5
We want to find the probability that only one of them will solve the question. This can be calculated as:
= (1/3 × 4/5) + (1/5 × 2/3)
= 4/15 + 2/15
= 6/15
= 2/5
So the probability is 2/5.
(7)
We are given:
m = 3i - 2j
n = 2i + 3j
p = i + 6j
We need to simplify the expression:
4(3i - 2j) + 2(2i + 3j) - 3(-i + 6j)
Expanding the expression, we get:
12i - 8j + 4i + 6j + 3i - 18j
Combining like terms, we get:
12i + 4i + 3i - 8j + 6j - 18j
Simplifying further, we get:
19i = 20j

 2018 WAEC MAY/JUNE FURTHER MATHS OBJ AND THEORY

(1)
Solve the matrix equation:
x-3 -4 3
Expanding the equation, we get:
(x-3)[2(6-x) + 8] + 4[5(6-x) - 4] + 3(-20 - 4) = -24
Simplifying the equation, we get:
(x-3)[-2x + 20] + 4[-5x + 26] + 3(-24) = -24
Further simplifying, we get:
-2x^2 + 26x - 60 - 20x + 104 - 72 = -24
Combining like terms, we get:
-2x^2 + 6x - 4 = 0
Factoring the quadratic equation, we get:
x^2 - 3x + 2 = 0
x^2 - 2x - x + 2 = 0
x(x - 2) - 1(x - 2) = 0
(x - 1)(x - 2) = 0
Therefore, x = 1 or x = 2.
(7)
Given:
m = 3i - 2j
n = 2i + 3j
p = i + 6j
Find the value of:
4(3i - 2j) + 2(2i + 3j) - 3(-i + 6j)
Expanding the equation, we get:
12i - 8j + 4i + 6j + 3i - 18j
Combining like terms, we get:
19i - 20j
Therefore, the result is 19i = 20j.
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