2013 - WAEC Mathematics Past Questions and Answers - page 3
21
If p = (y : 2y \(\geq\) 6) and Q = (y : y -3 \(\geq\) 4), where y is an integer, find p\(\cap\)Q
A
{3, 4}
B
{3, 7}
C
{3, 4, 5, 6, 7}
D
{4, 5, 6}
correct option: c
p = (y : 2y \(\geq\) 6)
2y \(\leq\) 6
y \(\leq \frac{6}{2}\)
y = \(\leq\) 3
and Q = (y : y -3 \(\geq\) 4)
y - 3 \(\geq\) 4
y \(\geq\) 4 + 3
y \(\geq\) 7
therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....}
P\(\cap\)Q = {3, 4, 5, 6, 7}
Users' Answers & Comments2y \(\leq\) 6
y \(\leq \frac{6}{2}\)
y = \(\leq\) 3
and Q = (y : y -3 \(\geq\) 4)
y - 3 \(\geq\) 4
y \(\geq\) 4 + 3
y \(\geq\) 7
therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....}
P\(\cap\)Q = {3, 4, 5, 6, 7}
22
Find the values of k in the equation 6k2 = 5k + 6
A
{\(\frac{-2}{3}, \frac{-3}{2}\)}
B
{\(\frac{-2}{3}, \frac{3}{2}\)}
C
{\(\frac{2}{3}, \frac{-3}{2}\)}
D
{\(\frac{2}{3}, \frac{3}{2}\)}
correct option: b
6k2 = 5k + 6
6k2 - 5k - 6 = 0
6k2 - 0k + 4k - 6 = 0
3k(2k - 3) + 2(2k - 3) = 0
(3k + 2)(2k - 3) = 0
3k + 2 = 0 or 2k - 3 = 0
3k = -2 or 2k = 3
k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\)
k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))
Users' Answers & Comments6k2 - 5k - 6 = 0
6k2 - 0k + 4k - 6 = 0
3k(2k - 3) + 2(2k - 3) = 0
(3k + 2)(2k - 3) = 0
3k + 2 = 0 or 2k - 3 = 0
3k = -2 or 2k = 3
k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\)
k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))
23
If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9
A
8
B
7
C
6
D
5
correct option: a
y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\)
6 = k\(\sqrt{3 + 1}\)
6 = k\(\sqrt{4}\)
6 = 2k
k = \(\frac{6}{2}\) = 3
y = \(\sqrt{(x + 1)}\)
9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)
\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)
3 = \(\sqrt{x + 1}\)(square both sides)
9 = x + 1
x = 9 - 1
x = 8
Users' Answers & Comments6 = k\(\sqrt{3 + 1}\)
6 = k\(\sqrt{4}\)
6 = 2k
k = \(\frac{6}{2}\) = 3
y = \(\sqrt{(x + 1)}\)
9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)
\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)
3 = \(\sqrt{x + 1}\)(square both sides)
9 = x + 1
x = 9 - 1
x = 8
24
The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the values of k
A
zero
B
-2
C
-4
D
-8
correct option: d
y = x2 + 2x + k at point(2,0) x = 2, y = 0
0 = (2)2 + 2(20 + k)
0 = 4 + 4 + k
0 = 8 + k
k = -8
Users' Answers & Comments0 = (2)2 + 2(20 + k)
0 = 4 + 4 + k
0 = 8 + k
k = -8
25
What is the locus of the point X which moves relative to two fixed points P and M on a plane such that < PXM = 30o
A
thebisector of the straight line joining P and M
B
an arc of a circle with PM as a chord
C
the bisector of angle PXM
D
a circle centre X and radius PM
correct option: b
Users' Answers & Comments26
When a number is subtracted from 2, the result equals 2 less than one-fifth of the number. Find the number
A
11
B
\(\frac{15}{2}\)
C
5
D
\(\frac{5}{2}\)
correct option: c
Let the number be y, subtract y from 2 i.e 2 - y
2 - y = 4 < \(\frac{1}{5}\) x y, 2 - y = 4 < \(\frac{y}{5}\)
2 - y - 4 < \(\frac{y}{5}\)
2 - 4 - y \(\frac{x}{5} - 4\), multiplying through by 5
5(2 - x) = 5(\(\frac{x}{5}\)) - 5(4)
10 - 5x = x - 20
-5x - x = -20 - 10
-6x = -30
x = \(\frac{-30}{-6}\)
= 5
Users' Answers & Comments2 - y = 4 < \(\frac{1}{5}\) x y, 2 - y = 4 < \(\frac{y}{5}\)
2 - y - 4 < \(\frac{y}{5}\)
2 - 4 - y \(\frac{x}{5} - 4\), multiplying through by 5
5(2 - x) = 5(\(\frac{x}{5}\)) - 5(4)
10 - 5x = x - 20
-5x - x = -20 - 10
-6x = -30
x = \(\frac{-30}{-6}\)
= 5
27
Express \(\frac{2}{x + 3} - \frac{1}{x - 2}\) as a simple fraction
A
\(\frac{x - 7}{x^2 + x - 6}\)
B
\(\frac{x - 1}{x^2 + x - 6}\)
C
\(\frac{x - 2}{x^2 + x - 6}\)
D
\(\frac{x - 27}{x^2 + x - 6}\)
correct option: a
\(\frac{2}{x + 3} - \frac{1}{x - 2}\) = \(\frac{2(x - 2) - (x - 3)}{(x + 3) (x - 2)}\)
= \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\)
= \(\frac{x -7}{x^2 + x - 6}\)
= \(\frac{x - 7}{x^2 + x - 6}\)
Users' Answers & Comments= \(\frac{2x - 4 - x - 3}{x^2 - 2x + 3x - 6}\)
= \(\frac{x -7}{x^2 + x - 6}\)
= \(\frac{x - 7}{x^2 + x - 6}\)
28
An interior angle of a regular polygon is 5 times each exterior angle. How many sides has the polygon?
A
15
B
12
C
9
D
6
correct option: b
Let the interior angle = xo
interior angle = 5xo (sum of int. angle ann exterior)
(angles = angle or straight line)
6x = 180
x = \(\frac{180}{6}\)
x = 30o
no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\)
= \(\frac{360}{30}\) = 12
Users' Answers & Commentsinterior angle = 5xo (sum of int. angle ann exterior)
(angles = angle or straight line)
6x = 180
x = \(\frac{180}{6}\)
x = 30o
no. of sides = \(\frac{\text{sum of exterior angles}}{\text{exterior angle}}\)
= \(\frac{360}{30}\) = 12
29
Given that P = x2 + 4x - 2, Q = 2x - 1 and Q - p = 2, find x
A
-2
B
-1
C
1
D
2
correct option: b
P = x2 + 4x - 2, Q = 2x - 1
Q - p = 2, (2x - 1) - (x2 + 4x - 2) = 2
2x - 1 - x2 - 4x + 2 = 2
-2x - x2 + 1
-x2 - 2x - 1 = 0
x2 + 2x + 1 = 0
x2 + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1)(x + 1) = 0
x + 1 = 0 or x + 1 = 0
x = -1 or x = -1
x = -1
Users' Answers & CommentsQ - p = 2, (2x - 1) - (x2 + 4x - 2) = 2
2x - 1 - x2 - 4x + 2 = 2
-2x - x2 + 1
-x2 - 2x - 1 = 0
x2 + 2x + 1 = 0
x2 + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1)(x + 1) = 0
x + 1 = 0 or x + 1 = 0
x = -1 or x = -1
x = -1
30
A pyramid has a rectangular base with dimensions 12m by 8m. If its height is 14m, calculate the volume
A
322m3
B
448m3
C
632m2
D
840m2
correct option: b
Volume of pyramid = \(\frac{1}{3}\) x base area x height
= \(\frac{1}{3} \times 12^4 \times 8 \times 14\)
= 4 x 8 x 14 = 448m3
Users' Answers & Comments= \(\frac{1}{3} \times 12^4 \times 8 \times 14\)
= 4 x 8 x 14 = 448m3