WAEC Mathematics Past Questions & Answers - Page 3

$log_3 x + log_3 (x - 8) = 2$

$log_3 x(x - 8) = log_39$ since 2 =   $log_39$

$log_3$ cancels out

⇒ x(x - 8) = 9

⇒ $x^2 - 8x = 9$

⇒ $x^2 - 8x - 9 = 0$

⇒ $x^2 - 9x + x - 9 = 0$

⇒ x(x - 9) + 1(x - 9) = 0

⇒ (x - 9)(x + 1) = 0

⇒ x = 9 or x = -1

Since we can't have a log of negative numbers,

∴ x = 9.

∠RMN = 90° (angles in a semicircle is a right angle)
∠SRN = ∠RMN + ∠MNR (ext. angle of a ∆ is equal to the sum of two opp. int. angles)
⇒ 5x + 20 = 90 + x
⇒ 5x - x = 90 - 20
⇒ 4x = 70

x = $\frac{70}{4}$

 = 17.4°

therefore, 2x = 2$\times17.5$ = 35°

Let x = original price

A decrease of 22% = (100-22)% = 78% of the original price.

78% of x = 27.30

= $\frac{78x}{100} = 27.30$

78x = 2730

then, x = $\frac{2730}{78}$

x =  35

Therefore, the original price of the shoe was  $35.00.

 Let A = (1, -3), B = (6, 2) and C = (0,4)

$d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}$
|AB| = $\sqrt{(2 - (-3))^2 + (6 - 1)^2}$ 
= $\sqrt{(2 + 3)^2 + (6 - 1)^2}$ 
= $\sqrt{5^2 + 5^2}$ 
= $\sqrt{25 + 25}$
= $\sqrt50$
= 5$\sqrt2$ units

|BC| = $\sqrt{(4 - 2)^2 + (0 - 6)^2}$
= $\sqrt{2^2 + (-6)^2}$
= $\sqrt{4 + 36}$
= $\sqrt40$ 
= 2$\sqrt10$  units

|AC| = $\sqrt{(4 - (-3)^2 + (0 - 1)^2}$
= $\sqrt{(4 + 3)^2 + (0 - 1)^2}$
= $\sqrt{7^2 + (-1)^2}$ 
= $\sqrt{49 + 1}$ 
= $\sqrt50$
= 5$\sqrt{2}$  units

Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.

The symbolic form for the statement "If Siah is from Foya, then Foya is in Lofa" is:

p ⇒ q

This is read as "p implies q". It means that if p is true, then q must also be true.

In other words, if Siah is from Foya, then Foya must be in Lofa.