2004 - WAEC Mathematics Past Questions and Answers - page 4
31
Find the average of the first four prime numbers greater than10
A
20
B
19
C
17
D
15
correct option: d
First four prime numbers greater than 10 are 11, 13, 17, 19
The average \(= \frac{11+13+17+19}{4}=15\)
Users' Answers & CommentsThe average \(= \frac{11+13+17+19}{4}=15\)
32
Given that \(\sqrt{128}+\sqrt{18}-\sqrt{k}\), find K,
A
8
B
16
C
32
D
48
correct option: c
\(\sqrt{128} +\sqrt{18}-\sqrt{k}=7\sqrt{2}\
\sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\
8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\
-\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\
=\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)
Users' Answers & Comments\sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\
8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\
-\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\
=\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)
33
In the diagram, PQRW is a circle. Line P, V and QR are produced to meet at M, where ∠WMR = 30o and |WM| = |MR| Find the value of x
A
10o
B
25o
C
35o
D
60o
correct option: b
MPQ = PQM = 3x; ∆MPQ
MPQ + PQM + PMQ = 180o; 3x + 3x + 30o = 180o
6x \times 30o = 180o; 6x + 180o - 30o; 6x + 150o
\(x = \frac{150}{6} = 25^{\circ}\)
Users' Answers & CommentsMPQ + PQM + PMQ = 180o; 3x + 3x + 30o = 180o
6x \times 30o = 180o; 6x + 180o - 30o; 6x + 150o
\(x = \frac{150}{6} = 25^{\circ}\)
34
The table above gives the marks scored by a group of students in a test Use the table to answer the Question below
What is the median mark?
What is the median mark?
A
1
B
2
C
3
D
4
correct option: c
Median = 0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5
Median mark = \(\frac{3+3}{2} = \frac{6}{2} = 3\)
Users' Answers & CommentsMedian mark = \(\frac{3+3}{2} = \frac{6}{2} = 3\)
35
The table above gives the marks scored by a group of students in a test Use the table to answer the Question below
What is the probability of selecting a student from the group that scored 2 or 3?
What is the probability of selecting a student from the group that scored 2 or 3?
A
\(\frac{1}{11}\)
B
\(\frac{5}{22}\)
C
\(\frac{7}{22}\)
D
\(\frac{6}{11}\)
correct option: d
Prob (2 or 3) = prob(2) + prob(3)
Prob (mark 2) \(=\frac{7}{22}\) + Prob(mark 3) = \(\frac{5}{22}\)
Prob(2 or 3) \(=\frac{7}{22}+\frac{5}{22}=\frac{12}{22}=\frac{6}{11}\)
Users' Answers & CommentsProb (mark 2) \(=\frac{7}{22}\) + Prob(mark 3) = \(\frac{5}{22}\)
Prob(2 or 3) \(=\frac{7}{22}+\frac{5}{22}=\frac{12}{22}=\frac{6}{11}\)
36
Find the range of values of x for which\(\frac{x+2}{4}-\frac{x+1}{3}>\frac{1}{2}\)
A
x > 4
B
x > -4
C
x < 4
D
x < -4
correct option: d
\(\frac{x+2}{4}-\frac{x+1}{3}>\frac{1}{2}\
\frac{3(x+2)-4(x+1)>6}{12}; 3x + 6 – 4x – 4 > 6\
-x +2>6; -x > 4; x < -4\)
Users' Answers & Comments\frac{3(x+2)-4(x+1)>6}{12}; 3x + 6 – 4x – 4 > 6\
-x +2>6; -x > 4; x < -4\)
37
A boy walks 800m in 20 minutes. Calculate his average speed in km per hour
A
2.4
B
4
C
16
D
24
correct option: a
\(Average \hspace{1mm}speed = \frac{Distance}{Time\hspace{1mm} taken}\
Distance = 800m = \frac{800}{1000}km = 0.8km\
Time\hspace{1mm}taken = 20\hspace{1mm}minutes = \frac{20}{60}hr=\frac{1}{3}hr\Average \hspace{1mm} speed = \left(0.8\div \frac{1}{3}\right)km/hr = (0.8\times3)km/hr\=2.4km/hr\)
Users' Answers & CommentsDistance = 800m = \frac{800}{1000}km = 0.8km\
Time\hspace{1mm}taken = 20\hspace{1mm}minutes = \frac{20}{60}hr=\frac{1}{3}hr\Average \hspace{1mm} speed = \left(0.8\div \frac{1}{3}\right)km/hr = (0.8\times3)km/hr\=2.4km/hr\)
38
Simplify \(\frac{2}{a+b}-\frac{1}{a-b}\)
A
\(\frac{3}{a+b}\)
B
\(\frac{a-3b}{a^2-b^2}\)
C
\(\frac{3a-b}{a^2 – b^2}\)
D
\(\frac{a-3b}{a^2+b^2}\)
correct option: b
Simplify \(\frac{2}{a+b}-\frac{1}{a-b}; \frac{2(a-b)-1(a+b)}{(a+b)(a-b)}\)
= \(\frac{2a-2b-a-b}{(a+b)(a-b)}\)
= \(\frac{a-3b}{a^2 - ab + ab - b^2}\)
= \(\frac{a-3b}{a^2-b^2}\)
Users' Answers & Comments= \(\frac{2a-2b-a-b}{(a+b)(a-b)}\)
= \(\frac{a-3b}{a^2 - ab + ab - b^2}\)
= \(\frac{a-3b}{a^2-b^2}\)
39
The diagram is a circle centre O. Find the value of x
A
30o
B
50o
C
61o
D
78o
correct option: e
∠PQR (Reflex) = 360° – 68° = 292°
∠PQR = 2∠PQR
292° = 4x + 30°
4x = 292 – 30 = 262
\(x = \frac{262}{4}=65.5°\)
40
Use the graph to answer the Question below
What are the roots of the equation x2 + 3x - 4 = 0?
What are the roots of the equation x2 + 3x - 4 = 0?
A
1, 4
B
-1, -4
C
-1, 4
D
-4, 1
correct option: d
x2 + 3x – 4 = 0 has roots value at the curve makes touches with the x – axis. The curve contact is at x = -4 and +1.
Users' Answers & Comments