2005 - WAEC Mathematics Past Questions and Answers - page 3
21
Solve the equation 2x - 3y = 22; 3x + 2y = 7
A
-5
B
-4
C
4
D
5
correct option: b
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
Users' Answers & Comments3x + 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
22
Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\)
A
-8
B
-13
C
13
D
19
correct option: b
\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 12 8y - 4 - 9y + 3 = 12 \
\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)
Users' Answers & Comments\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)
23
If \(log_9x= 1.5\),find x
A
36
B
27
C
24.5
D
13.5
correct option: b
\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)
Users' Answers & Comments24
A sequence is given by \(2\frac{1}{2}, 5, 7\frac{1}{2}, .....\) if the nth term is 25, find n
A
9
B
10
C
12
D
15
correct option: b
\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\
25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\
45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\)
Users' Answers & Comments25 = \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}\
45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\)
25
Given that the root of an the equation \(2x^2 + (k+2)x+k=0\) is 2, find the value of k
A
-4
B
-2
C
-1
D
\(-\frac{1}{4}\)
correct option: a
Substituting for x in the equation
\(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\)
Users' Answers & Comments\(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\)
26
Find the mean of the numbers 1, 3, 4, 8, 8, 4 and 7
A
4
B
5
C
6
D
7
27
What is the total surface area of a closed cylinder of height 10cm and diameter 7cm? [Take \(\pi = \frac{22}{7}\)]
A
77cm2
B
227cm2
C
297cm2
D
374cm2
correct option: c
T.S.A of s closed cylinder = \(2\pi r(r+h)\
=\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right)=\frac{22}{1}\left(\frac{27}{2}\right)=27\time 11=297cm^2\)
Users' Answers & Comments=\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\)
28
An arc of a circle of radius 14cm subtends angle 300o at the center. Find the perimeter of the sector formed by the arc
A
14.67cm
B
42.67cm
C
101.33cm
D
543.33cm
correct option: c
The perimeter of the sector \(=2r+\frac{\theta}{360}\times 2\pi r \
\Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\
73.133+28=101.33cm\)
Users' Answers & Comments\Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\
73.133+28=101.33cm\)
29
which of the following statements describes the locus of a point R which moves in a plane such that its equidistant from two intersecting lines?
A
the bisector of the angle formed by the lines
B
the point of intersection of the two lines
C
A cone, with two intersecting lines as slant height
D
A circle, with the point of intersection of the two lines as the center
correct option: a
Users' Answers & Comments30
P = {3, 9, 11, 13} and Q = {3, 7, 9, 15} are subset of the universal set ξ = {1, 3, 7, 9, 11, 13, 15} find PI ∩ QI
A
{3, 9}
B
{5, 7, 9}
C
{1}
D
{1, 11}