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 & Comments
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
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 & Comments
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 & Comments
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 & Comments
36
In what modulus is it true that 9 + 8 = 5?
A
mod 10
B
mod 11
C
mod 12
D
mod 13
correct option: c

9 + 8 = 17

17 + mod 12 = 1 rem 5

it is mod 12

Users' Answers & Comments
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 & Comments
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 & Comments
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 & Comments
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 & Comments
Please share this, thanks: