2005 - WAEC Mathematics Past Questions and Answers - page 3

2x - 3y = 22 ---- eqn I
3x + 2y = 7 ---- eqn II
multiply eqn I by 2
4x - 6y = 44 ---- eqn III
multiply eqn II by 3
9x + 6y = 21 ---- eqn IV
Adding eqn III and IV
=> 13x = 65 => x = 5
Substituting 5 for x in eqn II
3x5 + 2y = 7 => 2y = -8
y = -4

(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 12 8y - 4 - 9y + 3 = 12 <br /> \Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13)

(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27)

(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}<br /> 25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}<br /> 45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10)

Substituting for x in the equation
(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4)

mean (=\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5)

T.S.A of s closed cylinder = (2\pi r(r+h)<br /> =\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right)=\frac{22}{1}\left(\frac{27}{2}\right)=27\time 11=297cm^2)

The perimeter of the sector (=2r+\frac{\theta}{360}\times 2\pi r <br /> \Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28<br /> 73.133+28=101.33cm)

P^I = {1, 7, 15} ; Q^I = {1, 13, 11}
P^I ∩ Q^I = {1}