WAEC GCE 2025 MATHEMATICS ANSWER


Mathematics-Obj

Pls kindly trace it from your system. The options are being reshuffled. You might see number 1 as any number, so trace and be fast. You might also see option A as option C, B or D. Just use the answers in words and and trace properly. Kindly Trace it carefully to avoid errors!!!!

1: (C) 3
2: (A) -4 7/12
3: (C) C. 21?, 43?
4: (A) (p - 9)(p + 6)
5: (D) -2
6: (D) x ? -2
7: (C) 4x² - 4x - 15 = 0
8: (A) 5
9: (B) 2xy
10: (B) 71
11: (B) (PUQ)
12: (D) 4
13: (D) 4.8
14: (C) 12
15: (D) x = 1/2, -1
16: (A) 2%
17: (A) 15cm
18: (A) 9
19: (C) 145.20
25: (B) 7/24
26:(C) 5 13/24 m
27: (B) 7/24
28: (C) 60°
29: (C) 110°
30: (B) 11
31: (C) 71.72 m
32: (D) (u²) / (u - 2)
33: (A) 56°
34: (A) 40cm
35: (D) 7cm
36: (D) 45°
37: (C) m = (np)/(n + p)
38: (D) 52°
39: (D) 103°
40: (C) b + y
41: (A) 32
42: (D) 35 cm.
43: (C) 3.
44: (D) (140 - 7x)°
45: (D) 10cm
46; (B) 4.
47; (C) 2.75.
48; (B) 19/35
49; (C) If Fati does not pas her examination then she had not studied hard
50: (A) -3.


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(1)


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(2)


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(3)


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(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

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(6ai)
w = kn²/p + 30
where k is a constant.
We are given that a worker earned $250 for producing 15 items and being absent for 3 days:
250 = k(15)²/3 + 30

Subtract 30 from both sides:
220 = k(225)/3
220 = 75k
Divide both sides by 75:
k = 220/75
k = 2.93 (approximately)
So, the equation becomes:
w = 2.93n²/p + 30

(6aii)
w = 2.93(20)²/4 + 30
= 2.93 × 400 / 4 + 30
= 2.93 × 100 + 30
= 293 + 30
= 323

The worker earned $323.

(6b)
Let the distance be d km.

Time taken to travel from Bontokrom to Kramokrom = d/5 hours
Time taken to return = d/4 hours

Total time = 4.5 hours
d/5 + d/4 = 4.5

Multiply both sides by 20:
4d + 5d = 90
9d = 90

Divide both sides by 9:
d = 10km

The distance between Bontokrom and Kramokrom is 10 km.

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(8a)
The perimeter of a rhombus is 52 cm, so each side is:
52 / 4 = 13 cm

Let's call the diagonals d1 and d2. We know d1 = 24 cm.

The diagonals of a rhombus bisect each other at right angles. Using the Pythagorean theorem:
(13)² = (d1/2)² + (d2/2)²
169 = (24/2)² + (d2/2)²
169 = 12² + (d2/2)²
169 = 144 + (d2/2)²
(d2/2)² = 25
d2/2 = 5
d2 = 10

The area of a rhombus is:
Area = (d1 × d2) / 2
= (24 × 10) / 2
= 120

The area of the rhombus is 120 cm².

(8b)
Let the integers be x and x + 2 (since they're consecutive odd integers).

The sum of their squares is 290:
x² + (x + 2)² = 290
x² + x² + 4x + 4 = 290
2x² + 4x - 286 = 0
x² + 2x - 143 = 0

Factoring the quadratic equation:
(x + 13)(x - 11) = 0

x = -13 (not possible since x is positive)
x = 11

The integers are 11 and 13.

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(9)


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(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

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(11)


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(12)


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