Question 1
WAEC 2006A regular polygon has each of it angles at 160. What is the number of sides of the polygon?
- a.36
- b.9
- c.18
- d.20
Real WAEC Mathematics past questions, each with the correct answer and a worked solution. NECO candidates sit the same paper, so these apply there too. Below are 8 to work through free — no account needed.
Sample drawn from papers of 2019, 2006, 2000.
A regular polygon has each of it angles at 160. What is the number of sides of the polygon?
1.3logx+logy =3. Then, y is
X and Y are two events. The probability of X or Y is 0.7 and that of X is 0.4. If X and Y are independent, find the probability of Y.
P (X or Y) = P(X) + P(Y), when they are independent as given. 0.7 = 0.4 + P(Y) P(Y) = 0.7 - 0.4 = 0.30
The simple interest on ₦8550 for 3 years at x% per annum is ₦4890. Calculate the value of x to the nearest whole number.
S.I = PR/T100 ⟹ N 4890 = (8550 × 3 × x)/100 x = (4890 × 100)/(8550×3) x=19.06 x≊19
Simplify 81<sup>(−3/4)</sup> x 25<sup>(1/2)</sup> x 243<sup>2/5</sup>
81<sup>(−34)</sup> x 25<sup>(1/2)<?sub> x 243<sup>(2/5)</sup> = (4√81)<sup>−3</sup> × √25 × (5√243)<sup>2</sup> = 5×3<sup>2</sup>/3<sup>3</sup>= 5/3
Find the value of ((0.5436)<sup>3</sup>)/(0.017×0.219) to 3 significant figures.
= 0.16063/0.017 × 0.219 = 43.1 (to 3 s.f)
If S = (4t + 3)(t - 2), find ds/dt when t = 5 secs.
s=(4t+3)(t−2) ds/dt=(4t+3)(1)+(t−2)(4) = 4t+3+4t−8 = 8t - 5 ds/dt(t=5secs)=8(5)−5 = 40 - 5 = 35 units per sec
The angles of a polygon are given by 2x, 5x, x and 4x respectively. The value of x is
Since there are 4 angles given, the polygon is a quadrilateral. Sum of angle in a quadrilateral = 360° ∴ 2x + 5x + x + 4x = 360° 12x = 360° x = 30°
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