WAEC GCE 2025 CBT MATHEMATICS QUESTIONS BELOW
WAEC GCE 2025 CBT MATHEMATICS SOLUTIONS BELOW
(10ai)
Let’s define the events:
X: Team X qualifies
Y: Team Y qualifies
P(X) = 3/2 is not possible (probabilities can’t be greater than 1). Assuming it’s 3/4 and 3/4 for X and Y respectively doesn’t match the question, let’s assume P(X) = 2/3 and P(Y) = 3/4.
P(only one team qualifies) = P(X and not Y) + P(Y and not X)
= P(X) × P(not Y) + P(Y) × P(not X)
= (2/3) × (1/4) + (3/4) × (1/3)
= 2/12 + 3/12
= 5/12
(10aii)
P(at least one team qualifies) = 1 – P(neither team qualifies)
= 1 – P(not X and not Y)
= 1 – (1/3) × (1/4)
= 1 – 1/12
= 11/12
(10b)
First, find the mean:
Mean = (11 + 15 + 8 + 10 + 14 + 12) / 6
= 70 / 6
= 11.67
Next, find the deviations from the mean:
|11 – 11.67| = 0.67
|15 – 11.67| = 3.33
|8 – 11.67| = 3.67
|10 – 11.67| = 1.67
|14 – 11.67| = 2.33
|12 – 11.67| = 0.33
Mean deviation = (0.67 + 3.33 + 3.67 + 1.67 + 2.33 + 0.33) / 6
= 12 / 6
= 2
(5)
3 < x < 8 and 8 < y < 15
The mean of the numbers is 7, so we can set up the equation:
(2 + 3 + x + 8 + y + 15) / 6 = 7
Combine like terms:
(28 + x + y) / 6 = 7
Multiply both sides by 6:
28 + x + y = 42
Subtract 28 from both sides:
x + y = 14
Since x < 8 and y > 8, let’s try to find a combination that works:
x = 6, y = 8 doesn’t work (y > 8)
x = 5, y = 9 works!
So, x = 5 and y = 9.
The product of x and y is:
xy = 5 × 9 = 45
OBJECTIVE QUESTIONS AND ANSWERS BELOW
