2014 - WAEC Mathematics Past Questions and Answers - page 3
21
If x : y = 3 : 2 and y : z = 5 : 4, find the value of x in the ratio x : y : z
A
8
B
10
C
15
D
20
correct option: c
Users' Answers & Comments22
Given that cos x = \(\frac{12}{13}\), evaluate \(\frac{1 - \tan x}{\tan x}\)
A
\(\frac{5}{13}\)
B
\(\frac{5}{7}\)
C
\(\frac{7}{5}\)
D
\(\frac{13}{5}\)
correct option: c
cos x\(\frac{12}{13}\)
132 = 122 + a2
169 = 144 + a2
a2 = 169 - 144
a2 = 25
a \(\sqrt{25}\)
a = 5
tan x = \(\frac{5}{12}\)
\(\frac{1 - \tan x}{\tan x} = \frac{1 - \frac{5}{12}}{\frac{5}{12}}\)
\(\frac{\frac{1 - \frac{5}{12}}{12 - 5}}{12} = \frac{\frac{7}{12}}{\frac{5}{12}}\)
= \(\frac{7}{2} \div \frac{5}{12}\)
= \(\frac{7}{12} \times \frac{12}{5} = \frac{7}{5}\)
Users' Answers & Comments132 = 122 + a2
169 = 144 + a2
a2 = 169 - 144
a2 = 25
a \(\sqrt{25}\)
a = 5
tan x = \(\frac{5}{12}\)
\(\frac{1 - \tan x}{\tan x} = \frac{1 - \frac{5}{12}}{\frac{5}{12}}\)
\(\frac{\frac{1 - \frac{5}{12}}{12 - 5}}{12} = \frac{\frac{7}{12}}{\frac{5}{12}}\)
= \(\frac{7}{2} \div \frac{5}{12}\)
= \(\frac{7}{12} \times \frac{12}{5} = \frac{7}{5}\)
23
approximate 0.0033780 to 3 significant figures
A
338
B
0.338
C
0.00338
D
0.003
correct option: c
Users' Answers & Comments24
Simplify \(\frac{8^2 \times 4^n + 1}{2^{2n} \times 16}\)
A
16
B
8
C
4
D
1
correct option: c
\(\frac{\sqrt{8^2 \times 4^{n + 1}}}{2^{2n} \times 16}\)
= \(\frac{\sqrt{2^{3(2)} \times 2^{2(n + 1)}}}{2^{2n} \times 2^4}\)
= \(\frac{\sqrt{2^6 \times 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^6 + 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^{2n + 8}}}{2^{2n} + 4}\)
= \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\)
= \(\sqrt{2^{2n - 2n} + 8 - 4}\)
= \(\sqrt{2^4}\)
= \(\sqrt{16}\)
= 4
Users' Answers & Comments= \(\frac{\sqrt{2^{3(2)} \times 2^{2(n + 1)}}}{2^{2n} \times 2^4}\)
= \(\frac{\sqrt{2^6 \times 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^6 + 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^{2n + 8}}}{2^{2n} + 4}\)
= \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\)
= \(\sqrt{2^{2n - 2n} + 8 - 4}\)
= \(\sqrt{2^4}\)
= \(\sqrt{16}\)
= 4
25
If \(\frac{2}{x - 3} - \frac{3}{x - 2}\) is equal to \(\frac{p}{(x - 3)(x - 2)}\), find p
A
-x - 5
B
-(x + 3)
C
5x - 13
D
5 - x
correct option: d
\(\frac{2}{x - 3} - \frac{3}{x - 2}\) = \(\frac{p}{(x - 3)(x - 2)}\)
\(\frac{2(x - 2) - 3(x - 3)}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)
= \(\frac{2x - 4 - 3x + 9}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)
\(\frac{-x + 5}{(x - 3)(x - 2)} = \frac{p}{(x - 3)(x - 2)}\)
p(x - 3)(x - 2) = -x + 5(x - 3)(x - 2)
p = -x + 5 or p = 5 - x
Users' Answers & Comments\(\frac{2(x - 2) - 3(x - 3)}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)
= \(\frac{2x - 4 - 3x + 9}{(x - 3)(x - 2)}\) = \(\frac{p}{(x - 3)(x - 2)}\)
\(\frac{-x + 5}{(x - 3)(x - 2)} = \frac{p}{(x - 3)(x - 2)}\)
p(x - 3)(x - 2) = -x + 5(x - 3)(x - 2)
p = -x + 5 or p = 5 - x
26
Subtract \(\frac{1}{2}\)(a - b - c) from the sum of \(\frac{1}{2}\)(a - b + c) and \(\frac{1}{2}\)
(a + b - c)
(a + b - c)
A
\(\frac{1}{2}\) (a + b + c)
B
\(\frac{1}{2}\) (a - b - c)
C
\(\frac{1}{2}\) (a - b + c)
D
\(\frac{1}{2}\) (a + b - c)
correct option: a
\(\frac{1}{2}\)(a - b + c) + \(\frac{1}{2}\)(a + b - c) - [\(\frac{1}{2}\) (a - b - c)]
\(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}(a + b + c)\)
Users' Answers & Comments\(\frac{1}{2}a - \frac{1}{2}b + \frac{1}{2}c + \frac{1}{2}a + \frac{1}{2}b - \frac{1}{2}c - \frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c\)
= \(\frac{1}{2}(a + b + c)\)
27
A man's eye level is 1.7m above the horizontal ground and 13m from a vertical pole. If the pole is 8.3m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes.
A
33o
B
32o
C
27o
D
26o
correct option: c
tan \(\theta = \frac{opp}{adj} = \frac{6.6}{13}\)
tan \(\theta = 0.5077\)
\(\theta\) = tan-1 0.5077
\(\theta = 27^o\)
Users' Answers & Commentstan \(\theta = 0.5077\)
\(\theta\) = tan-1 0.5077
\(\theta = 27^o\)
28
A chord subtends an angle of 120o at the centre of a circle of radius 3.5cm. Find the perimeter of the minor sector containing the chord, [Take \(\pi = \frac{22}{7}\)]
A
14\(\frac{1}{3}\)cm
B
12\(\frac{5}{6}\)cm
C
8\(\frac{1}{7}\)cm
D
7\(\frac{1}{3}\)cm
correct option: a
perimeter of minor sector
2r + \(\frac{\theta}{360} \times 2 \pi r\)
= 2 x 3.5 + \(\frac{120^o}{360^o} \times 2 \frac{22}{7} \times 3.5\)
= 7 + \(\frac{154}{21}\)
= 7 + 7.33
= 14.33
= 14\(\frac{1}{3}\)cm
Users' Answers & Comments2r + \(\frac{\theta}{360} \times 2 \pi r\)
= 2 x 3.5 + \(\frac{120^o}{360^o} \times 2 \frac{22}{7} \times 3.5\)
= 7 + \(\frac{154}{21}\)
= 7 + 7.33
= 14.33
= 14\(\frac{1}{3}\)cm
29
In parallelogram PQRS, QR is produced to M such that |QR| = |RM|. What fraction of the area of PQMS is the area of PRMS?
A
\(\frac{1}{4}\)
B
\(\frac{1}{3}\)
C
\(\frac{2}{3}\)
D
\(\frac{3}{4}\)
correct option: c
There are 3 triangles: \(\bigtriangleup\)MRS, \(\bigtriangleup\) and \(\bigtriangleup\) PRQ
Area PRMS has 2 triangles: \(\bigtriangleup\) MRS and \(\bigtriangleup\) RSP
the fraction = \(\frac{2}{3}\)
Users' Answers & CommentsArea PRMS has 2 triangles: \(\bigtriangleup\) MRS and \(\bigtriangleup\) RSP
the fraction = \(\frac{2}{3}\)
30
In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years?
A
15%
B
35%
C
65%
D
85%
correct option: c
Users' Answers & Comments