2004 - WAEC Mathematics Past Questions and Answers - page 3
21
Find, correct to two decimal places, the mean of 9, 13, 16, 17, 19, 23, 24.
A
23.00
B
17.29
C
16.50
D
16.33
correct option: b
Mean \(= \frac{sum\hspace{1mm}of\hspace{1mm}all\hspace{1mm}the\hspace{1mm}numbers}{Total\hspace{1mm}Number}\
\frac{9+13+16+17+19+23+24}{7}\
\frac{121}{7}=17.2857=17.29\)
22
In the diagram, ∆XYZ is similar to ∆PRQ, |XY| = 5cm, |XZ| =3.5cm and |PR| = 8cm. Find |PQ|
A
5.6 cm
B
11.2cm
C
11.4cm
D
28.0cm
correct option: a
\(\frac{PQ}{PR}=\frac{XZ}{XY}; \frac{PQ}{8}=\frac{3.5}{5}\
PQ=\frac{8\times 3.5}{5}=\frac{28}{5}=5.6cm\)
23
Factorise 27p2x2 - 48y2.
A
9(3px - 4y)2
B
3(3px - 4y)(3px - 4y)
C
9(px - 4y)(3p x + 4y)
D
3(3px -4y)(3px +4y)
correct option: d
27p2x2 - 48y2; 3(32 p2x2-42y2)
3[(3px) 2 - (4y) 2]; 3[(3px-4y)(3px+4y)]
Users' Answers & Comments3[(3px) 2 - (4y) 2]; 3[(3px-4y)(3px+4y)]
24
What is the volume of a solid cylinder of diameter 7cm and height 7cm? (Take \(\pi = \frac{22}{7}\))
A
38.5cm3
B
77cm3
C
269.5cm3
D
1078cm3
correct option: c
Volume of a solid cylinder = Base area x height
Base area \(=\pi r^2 =\frac{22}{7} \times \frac{7}{2}\times \frac{7}{2} = \frac{77}{2}\)
Volume of a solid cylinder = \(\frac{77}{2}\times \frac{7}{1} = \frac{539}{2}cm^3\
=269.5cm^3\)
Users' Answers & CommentsBase area \(=\pi r^2 =\frac{22}{7} \times \frac{7}{2}\times \frac{7}{2} = \frac{77}{2}\)
Volume of a solid cylinder = \(\frac{77}{2}\times \frac{7}{1} = \frac{539}{2}cm^3\
=269.5cm^3\)
25
Find the sum of the roots of the equation 2x2 + 3x - 9 = 0
A
-18
B
-6
C
\(-\frac{9}{2}\)
D
\(-\frac{3}{2}\)
correct option: d
The general quadratic equation
x2 - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\
Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)
Users' Answers & Commentsx2 - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\
Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)
26
Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q
A
{2}
B
{1,9}
C
{315.7}
D
{4, 6, 8, 10}
correct option: b
ξ = {1, 2, 3, . . . . . . ,10}, P= (2, 3, 5, 7) and Q = {1, 3, 5, 9},
Pl∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}
Users' Answers & CommentsPl∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}
27
In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80o. Find ∠QRS.
A
90o
B
80o
C
50o
D
40o
correct option: c
∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 50o
∠QRS = 50o
Users' Answers & Comments∠RPQ = 50o
∠QRS = 50o
28
Find the mean deviation of 6, 7, 8, 9, 10
A
1.2
B
1.5
C
2
D
8
correct option: a
\(\begin{matrix}
x & x-m & |x-m| \
6 & -2 & 2 \
7 & -1 & 1 \
8 & 0 & 0 \
9 & 1 & 1 \
10 & 2 & 2 \
& & 6
\end{matrix}
\)
Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)
Users' Answers & Commentsx & x-m & |x-m| \
6 & -2 & 2 \
7 & -1 & 1 \
8 & 0 & 0 \
9 & 1 & 1 \
10 & 2 & 2 \
& & 6
\end{matrix}
\)
Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)
29
A point X is on the bearing 342o from a point Y. What is the bearing of Y from X?
A
342o
B
252o
C
198o
D
162o
correct option: d
If x is the bearing 342ofrom a point y
= 360o - 342o = 18o;
The bearing of y from x = 180o - 18o = 162o
Users' Answers & Comments= 360o - 342o = 18o;
The bearing of y from x = 180o - 18o = 162o
30
In the diagram, O is the centre of the circle where OS//QR and ∠SOR = 35o
A
35o
B
45o
C
55o
D
70o
correct option: c
∠SOR = ∠ORQ alternate angle.
Also ∠OQR = ∠OQR (Base angle of isosceles)
∠QOR = 108 – 2(35o) = 180 - 70o - 110o ∠QPR = 1/2QOR = 1/2(110o) ∠QPR + 55o (theorem angle at center is twice angle at circumference)
Users' Answers & CommentsAlso ∠OQR = ∠OQR (Base angle of isosceles)
∠QOR = 108 – 2(35o) = 180 - 70o - 110o ∠QPR = 1/2QOR = 1/2(110o) ∠QPR + 55o (theorem angle at center is twice angle at circumference)