2014 - WAEC Mathematics Past Questions and Answers - page 4
31
If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime
A
\(\frac{5}{9}\)
B
1\(\frac{4}{9}\)
C
\(\frac{1}{3}\)
D
\(\frac{2}{9}\)
correct option: c
p = { 1, 2, 3}
Q = {2, 3, 5}
prob(prime number) = prob(1 and 2) or
= prob(2 and 3) or
= prob(3 and 2)
There are three possibilities of the sum being prime
Total possibility = 9
probability(sum being prime) = \(\frac{3}{9}\)
= \(\frac{1}{3}\)
Users' Answers & CommentsQ = {2, 3, 5}
prob(prime number) = prob(1 and 2) or
= prob(2 and 3) or
= prob(3 and 2)
There are three possibilities of the sum being prime
Total possibility = 9
probability(sum being prime) = \(\frac{3}{9}\)
= \(\frac{1}{3}\)
32
If log 5.957 = 0.7750, find log \(3 \sqrt{0.0005957}\)
A
4.1986
B
2.9250
C
1.5917
D
1.2853
correct option: b
\(3 \sqrt{0.0005957}\)
\(\begin{array}{c|c} \log No(0.0005957) & \frac{1}{3} \log \frac{1}{3} \ \hline (0.0005957)^{\frac{1}{3}} & 4.7750 \times \frac{1}{3} \ & 6 + 2.7750 \\hline & 3 \\hline & 2.9250\end{array}\)
Users' Answers & Comments\(\begin{array}{c|c} \log No(0.0005957) & \frac{1}{3} \log \frac{1}{3} \ \hline (0.0005957)^{\frac{1}{3}} & 4.7750 \times \frac{1}{3} \ & 6 + 2.7750 \\hline & 3 \\hline & 2.9250\end{array}\)
33
The probability of an event P happening is \(\frac{1}{5}\) and that of event Q is \(\frac{1}{4}\). If the events are independent, what is the probability that neither of them happens?
A
\(\frac{4}{5}\)
B
\(\frac{3}{4}\)
C
\(\frac{3}{5}\)
D
\(\frac{1}{20}\)
correct option: c
prob(p) = \(\frac{1}{5}\)
prob(Q) = \(\frac{1}{4}\)
Prob(neither p) = 1 - \(\frac{1}{5}\)
\(\frac{5 - 1}{5} = \frac{4}{5}\)
prob(neither Q) = 1 - \(\frac{1}{4}\)
\(\frac{4 - 1}{4} = \frac{3}{4}\)
prob(neither of them) = \(\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}\)
= \(\frac{3}{5}\)
Users' Answers & Commentsprob(Q) = \(\frac{1}{4}\)
Prob(neither p) = 1 - \(\frac{1}{5}\)
\(\frac{5 - 1}{5} = \frac{4}{5}\)
prob(neither Q) = 1 - \(\frac{1}{4}\)
\(\frac{4 - 1}{4} = \frac{3}{4}\)
prob(neither of them) = \(\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}\)
= \(\frac{3}{5}\)
34
Each exterior angle of a polygon is 30o. Calculate the sum of the interior angles
A
540o
B
720o
C
1080o
D
1800o
correct option: d
number of sides = \(\frac{360^o}{\theta} = \frac{360^o}{306o}\)
n = 12o
Sum of interior angle = (n - 2) 180o
(12 - 2) 180v = 10 x 180o
= 1800o
Users' Answers & Commentsn = 12o
Sum of interior angle = (n - 2) 180o
(12 - 2) 180v = 10 x 180o
= 1800o
35
Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141.
A
11
B
12
C
13
D
14
correct option: d
T1, T2, T3
2, -9, -20 .... -141
l = a + (n - 1)d
first term, a = 2
common difference d = T3 - T2
= T2 - T1
= -20 - (-9) = -9 -2
= -20 + 9
= -9 -2
= -20 + 9
= -11
-11 = -11
d = -1
last term l = -141
-141 = 2 + (\(\cap\) - 1) (-11)
-141 = 2 + (-11 \(\cap\) + 11)
= 2 - 11\(\cap\) + 11
-141 = 13 - 11\(\cap\)
-141 - 13 = -11\(\cap\)
-154 = -11\(\cap\)
\(\cap\) = \(\frac{-154}{-11}\)
\(\cap\) = 14
Users' Answers & Comments2, -9, -20 .... -141
l = a + (n - 1)d
first term, a = 2
common difference d = T3 - T2
= T2 - T1
= -20 - (-9) = -9 -2
= -20 + 9
= -9 -2
= -20 + 9
= -11
-11 = -11
d = -1
last term l = -141
-141 = 2 + (\(\cap\) - 1) (-11)
-141 = 2 + (-11 \(\cap\) + 11)
= 2 - 11\(\cap\) + 11
-141 = 13 - 11\(\cap\)
-141 - 13 = -11\(\cap\)
-154 = -11\(\cap\)
\(\cap\) = \(\frac{-154}{-11}\)
\(\cap\) = 14
36
In what modulus is it true that 9 + 8 = 5?
A
mod 10
B
mod 11
C
mod 12
D
mod 13
37
The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q?
A
2.5cm
B
5.0cm
C
7.5cm
D
20.0cm
correct option: a
volume of cylinder = \(\pi r^2h\)
volume of cylinder p = \(\pi r^2 \times 10\)
= 10\(\pi r^2\)
volume of cylinder Q = \(\pi (2r)^2 h\)
= 4\(\pi r^2\)h
4\(\pi r^2 = 10 \pi r^2 h\)
h = \(\frac{10}{4} = 4.5cm\)
Users' Answers & Commentsvolume of cylinder p = \(\pi r^2 \times 10\)
= 10\(\pi r^2\)
volume of cylinder Q = \(\pi (2r)^2 h\)
= 4\(\pi r^2\)h
4\(\pi r^2 = 10 \pi r^2 h\)
h = \(\frac{10}{4} = 4.5cm\)
38
The bar chart shows the scores of some students in a test. How many students took the test?
A
18
B
19
C
20
D
22
correct option: c
score 0 = 4 students
score 1 = 2 students
score 2 = 5 students
score 3 = 6 students
score 4 = 3 students
= 20 students
Users' Answers & Commentsscore 1 = 2 students
score 2 = 5 students
score 3 = 6 students
score 4 = 3 students
= 20 students
39
The bar chart shows the scores of some students in a test. If one students is selected at random, find the probability that he/she scored at most 2 marks
A
\(\frac{11}{18}\)
B
\(\frac{11}{20}\)
C
\(\frac{7}{22}\)
D
\(\frac{5}{19}\)
correct option: b
at most 2 marks = 5 + 2 + 4 students = 11 students
probability(at most 2 marks) = \(\frac{11}{20}\)
Users' Answers & Commentsprobability(at most 2 marks) = \(\frac{11}{20}\)
40
In the diagram, the value of x + y = 220o. Find the value of n
A
20o
B
40o
C
60o
D
80o
correct option: b
y + z = 180o
z = 180o - y
x + k = 180o
k = 180o - x
the sum of angles in the \(\bigtriangleup\) are: n + z + k = 180
but n + 180o - y + 180o - x = 180o
n + 180o - y - x = 0
but we want to find n
n + 180o = x + y
but x + y = 220o
n + 180o = 220o
n = 220o - 180o
n = 40o
Users' Answers & Commentsz = 180o - y
x + k = 180o
k = 180o - x
the sum of angles in the \(\bigtriangleup\) are: n + z + k = 180
but n + 180o - y + 180o - x = 180o
n + 180o - y - x = 0
but we want to find n
n + 180o = x + y
but x + y = 220o
n + 180o = 220o
n = 220o - 180o
n = 40o