2012 - WAEC Mathematics Past Questions and Answers - page 2
11
If x + y = 2y - x + 1 = 5, find the value of x
A
3
B
2
C
1
D
-1
correct option: b
x + y = 2y - x + 1 = 5
x + y = 2y - x + 1
x + x + y - 2y = 1
2x - y = 1....(i)
2y - x + 1 = 5
-x + 2y = 5 + 1
-x = 2y = 4
x - 2y = -4 .....(ii)
solve simultaneously (i) x 2x - y = 1
(ii) x x - 2y = -4
2x - y = 1
=2x - 4y = -8
3y = 9
y = \(\frac{9}{3}\)
y = 3
substitute y = 3 into equation (i)
2x - y = 1
2x - 3 = 1
2x = 1 + 3
2x = 4
x = \(\frac{4}{2}\)
= 2
Users' Answers & Commentsx + y = 2y - x + 1
x + x + y - 2y = 1
2x - y = 1....(i)
2y - x + 1 = 5
-x + 2y = 5 + 1
-x = 2y = 4
x - 2y = -4 .....(ii)
solve simultaneously (i) x 2x - y = 1
(ii) x x - 2y = -4
2x - y = 1
=2x - 4y = -8
3y = 9
y = \(\frac{9}{3}\)
y = 3
substitute y = 3 into equation (i)
2x - y = 1
2x - 3 = 1
2x = 1 + 3
2x = 4
x = \(\frac{4}{2}\)
= 2
12
The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation
A
12n - 6 = 0
B
3n - 12 = 0
C
2n - 35 = 0
D
5n - 33 = 0
correct option: d
12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3
36 + n - 6n = 3
-5n = 3 - 36
-5n = -33
-5n + 33 = 0
5n - 33 = 0
Users' Answers & Comments36 + n - 6n = 3
-5n = 3 - 36
-5n = -33
-5n + 33 = 0
5n - 33 = 0
13
Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)
A
m \(\leq \frac{5}{4}\)
B
m \(\geq \frac{5}{4}\)
C
m \(\leq - \frac{1}{11}\)
D
m \(\geq - \frac{1}{11}\)
correct option: c
\(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)
= \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\)
12(-2m - 5) \(\geq\) 4(5m - 14)
-24m - 60 \(\geq\) 20m - 56
-24m - 20m \(\geq\) -56 + 60
44m \(\geq\) 4
m \(\leq \frac{4}{-44}\)
m \(\leq \frac{-1}{11}\)
Users' Answers & Comments= \(\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}\)
12(-2m - 5) \(\geq\) 4(5m - 14)
-24m - 60 \(\geq\) 20m - 56
-24m - 20m \(\geq\) -56 + 60
44m \(\geq\) 4
m \(\leq \frac{4}{-44}\)
m \(\leq \frac{-1}{11}\)
14
The curved surface area of a cylindrical tin is 704cm2. If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]`
A
14cm
B
9cm
C
8cm
D
7cm
correct option: a
Curved surface area = 2\(\pi h\)
704 = 2 x \(\frac{22}{7} \times 8 \times h\)
704 = \(\frac{352}{7}\)h
352h = 704 x 7
h = \(\frac{704 \times 7}{352}\)
= \(\frac{4928}{352}\)
h = 14cm
Users' Answers & Comments704 = 2 x \(\frac{22}{7} \times 8 \times h\)
704 = \(\frac{352}{7}\)h
352h = 704 x 7
h = \(\frac{704 \times 7}{352}\)
= \(\frac{4928}{352}\)
h = 14cm
15
The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector
A
360o
B
252o
C
246o
D
234o
correct option: b
Let 0 = angle of the minor sector
angle of the major sector = 360 - \(\theta\)(angle at a point)
2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)
2\(\pi r = 180^o\)
r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)
Lenght of ninor arc
= \(\frac{\theta}{360} \times 2 \pi r\)
54 = \(\frac{\theta}{360} \times 3 \pi r\)
\(\theta = \frac{360 \times 54}{2 \pi r}\)
but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r
\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)
\(\theta = 2 \times 54 = 108^o\)
angle of the major sector = 360 - 108o
= 252o
Users' Answers & Commentsangle of the major sector = 360 - \(\theta\)(angle at a point)
2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)
2\(\pi r = 180^o\)
r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)
Lenght of ninor arc
= \(\frac{\theta}{360} \times 2 \pi r\)
54 = \(\frac{\theta}{360} \times 3 \pi r\)
\(\theta = \frac{360 \times 54}{2 \pi r}\)
but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r
\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)
\(\theta = 2 \times 54 = 108^o\)
angle of the major sector = 360 - 108o
= 252o
16
In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true?
A
/XY/ + /YZ/ = /XZ/
B
/XY/ - /YZ/ = /XZ/
C
/XY/2 = /Y/2 - /XY/2
D
/YZ/2 = /XZ/2 - /XY/2
correct option: c
Users' Answers & Comments17
If cos(x + 40)o = 0.0872, what is the value of x?
A
85o
B
75o
C
65o
D
45o
correct option: d
cos(x + 40)o = 0.0872
x + 40 = cos-10.0872
x + 40o = 84.99o
x = 84.99o - 40o
x = 44.99
x = 45o
Users' Answers & Commentsx + 40 = cos-10.0872
x + 40o = 84.99o
x = 84.99o - 40o
x = 44.99
x = 45o
18
A kite flies on a taut string of length 50m inclined at tan angle 54o to the horizontal ground. The height of the kite above the ground is
A
50 tan 30o
B
50 sin 54o
C
50 tan 54o
D
50 sin 36o
correct option: b
Users' Answers & Comments19
Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores
A
\(\sqrt{10}\)
B
4
C
5
D
\(\sqrt{30}\)
correct option: c
\(\begin{array}{c|c} x & x - x & (x - \bar{x})^2\ \hline 15 & -6 & 36\21 & 0 & 0\17 & -4 & 16\ 26 & 5 & 25 \ 18 & -3 &9 \ 29 & 8 & 64 \end{array}\)
\(E(x - \bar{x})^2\) = 150
N = 6
S.D = \(\sqrt{\frac{(x - x)^2}{N}}\)
S.D = \(\sqrt{\frac{150}{6}}\) = 5
Users' Answers & Comments\(E(x - \bar{x})^2\) = 150
N = 6
S.D = \(\sqrt{\frac{(x - x)^2}{N}}\)
S.D = \(\sqrt{\frac{150}{6}}\) = 5
20
A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours
A
\(\frac{8}{15}\)
B
\(\frac{13}{25}\)
C
\(\frac{11}{15}\)
D
\(\frac{13}{15}\)
correct option: a
Prob(RB + BR) = Total balls = 4 + 6 = 10
= prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\)
= \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)
Users' Answers & Comments= prob(\(\frac{4}{10} \times \frac{6}{9}\)) + prob(\(\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90}\)
= \(\frac{48}{90} = \frac{16}{30} = \frac{8}{15}\)