2010 - WAEC Mathematics Past Questions and Answers - page 4
31
If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k
A
\(\frac{8}{3}\)
B
\(\frac{7}{3}\)
C
\(\frac{5}{3}\)
D
\(\frac{2}{3}\)
correct option: a
x2 + kx + \(\frac{16}{9}\); Perfect square
But, b2 - 4ac = 0, for a perfect square
where a - 1; b = k; c = \(\frac{16}{9}\)
k2 - 4(1) x \(\frac{16}{9}\) = 0
k2 - \(\frac{64}{9}\) = 0
k2 = \(\frac{64}{9}\)
k = \(\sqrt{\frac{64}{9}}\)
k = \(\frac{8}{3}\)
32
If x km/h = y m/s, then y =
A
\(\frac{7}{18}\)x
B
\(\frac{11}{20}\)x
C
\(\frac{4}{15}\)x
D
\(\frac{5}{18}\)x
correct option: d
x kmh-1 = y ms-1
\(\frac{x km}{1 hr}\) = y ms-1
\(x \times \frac{1km}{1hr}\) = y ms-1
\(x \times \frac{1000m}{60 \times 60s}\) = y ms-1
\(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms-1
\(x \times \frac{5}{18} ms^{-1}\)
\(x \times \frac{5}{18} ms^{-1}\) = y ms-1
y = \(\frac{5}{18}\)x
Users' Answers & Comments\(\frac{x km}{1 hr}\) = y ms-1
\(x \times \frac{1km}{1hr}\) = y ms-1
\(x \times \frac{1000m}{60 \times 60s}\) = y ms-1
\(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms-1
\(x \times \frac{5}{18} ms^{-1}\)
\(x \times \frac{5}{18} ms^{-1}\) = y ms-1
y = \(\frac{5}{18}\)x
33
The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x
A
x \(\leq\) 3
B
x \(\geq\) 3
C
x < 3
D
x > 3
correct option: a
mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5
= \(\frac{14 + 2x}{4} \leq 5\)
= 14 + 2x \(\leq\) 5 x 4
14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14
2x \(\leq\) 20 - 14
2x \(\leq\) 6
x \(\leq\) \(\frac{6}{2}\)
x \(\leq\) 3
34
In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events?
A
\(\frac{3}{32}\)
B
\(\frac{7}{3}\)
C
\(\frac{5}{3}\)
D
\(\frac{5}{16}\)
correct option: d
Pr. (winning 100m race) = \(\frac{1}{8}\)
Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)
Pr. (winning high jump) = \(\frac{1}{4}\)
Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump)
= (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\))
= \(\frac{3}{32} + \frac{7}{32}\)
= \(\frac{10}{32}\)
= \(\frac{5}{16}\)
Users' Answers & CommentsPr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)
Pr. (winning high jump) = \(\frac{1}{4}\)
Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump)
= (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\))
= \(\frac{3}{32} + \frac{7}{32}\)
= \(\frac{10}{32}\)
= \(\frac{5}{16}\)
35
In the diagram, < PSR = 220o, < SPQ = 58o and < PQR = 41o. Calculate the obtuse angle QRS.
A
90o
B
100o
C
121o
D
60o
correct option: c
Joining SQ. In \(\bigtriangleup\) SPQ,
(22o + a) + 55o + (41o + b) = 180o
121o + a + b = 180o
a + b = 180 - 121
a + b = 59o.....(1)
In \(\bigtriangleup\) SRPQ; R + a + b = 180o
R + 59o = 180o
(in (1), a + b = 59o)
R = 180 - 59
R = 121o
Users' Answers & Comments(22o + a) + 55o + (41o + b) = 180o
121o + a + b = 180o
a + b = 180 - 121
a + b = 59o.....(1)
In \(\bigtriangleup\) SRPQ; R + a + b = 180o
R + 59o = 180o
(in (1), a + b = 59o)
R = 180 - 59
R = 121o
36
The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test?
A
100
B
85
C
80
D
70
correct option: b
Pass mark = 50%
No. of students that passed = f50 + f65 + f80
= 45 + 25 + 15
= 85
Users' Answers & CommentsNo. of students that passed = f50 + f65 + f80
= 45 + 25 + 15
= 85
37
The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50?
A
55\(\frac{1}{3}\)%
B
60%
C
65%
D
66\(\frac{2}{3}\)%
correct option: d
Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\)
= \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100%
= \(\frac{120}{180}\) x 100%
= 66\(\frac{2}{3}\)%
Users' Answers & Comments= \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100%
= \(\frac{120}{180}\) x 100%
= 66\(\frac{2}{3}\)%
38
What is the value of m in the diagram?
A
20o
B
30o
C
40o
D
50o
correct option: b
4m - 15o = m + 75o
(Vertically opposite angles are equal)
4m - m = 75 + 15
3m = 90
m = \(\frac{90}{3}\)
m = 30o
Users' Answers & Comments(Vertically opposite angles are equal)
4m - m = 75 + 15
3m = 90
m = \(\frac{90}{3}\)
m = 30o
39
In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75o. Find the value of y
A
105o
B
110o
C
130o
D
150o
correct option: a
In \(\bigtriangleup\) PQR,
Q = S = 75o (Corresponding angle)
R = Q = 75o (Base angles of an isosceles \(\bigtriangleup\))
But, y + 75o = 180o (Sum of angles in a straight line)
y = 180 - 75
y = 105o
Users' Answers & CommentsQ = S = 75o (Corresponding angle)
R = Q = 75o (Base angles of an isosceles \(\bigtriangleup\))
But, y + 75o = 180o (Sum of angles in a straight line)
y = 180 - 75
y = 105o
40
In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\)
A
\(\frac{KL}{JI}\)
B
\(\frac{HK}{JK}\)
C
\(\frac{JI}{KL}\)
D
\(\frac{HK}{LK}\)
correct option: a
\(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ
< HKL = HJI = xo
Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\)
\(\frac{LH}{JH} = \frac{KL}{JI}\)
Users' Answers & Comments< HKL = HJI = xo
Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\)
\(\frac{LH}{JH} = \frac{KL}{JI}\)