2009 - WAEC Mathematics Past Questions and Answers - page 4
31
Find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3
A
2x2 - 2x + 3 = 0
B
2x2 - 2x - 3 = 0
C
2x2 - 5x + 3 = 0
D
3x2 - 5x - 3 = 0
correct option: d
x = \(\frac{-1}{2}\)
x = 3
2x = -1, x = 3; 2x + 1 = 0
x - 3 = 0
(2x + 1)(x - 3) = 0
2x2 - 6x + x - 3 = 0
2x2 - 5x - 3 = 0
Users' Answers & Commentsx = 3
2x = -1, x = 3; 2x + 1 = 0
x - 3 = 0
(2x + 1)(x - 3) = 0
2x2 - 6x + x - 3 = 0
2x2 - 5x - 3 = 0
32
The following is the graph of a quadratic friction, find the co-ordinates of point P
A
(0, 4)
B
(4, 0)
C
(0, -4)
D
(-4, 0)
correct option: a
Users' Answers & Comments33
The following is the graph of a quadratic friction, find the value of x when y = 0
A
1, 2
B
1, 4
C
2, 3
D
1, 6
correct option: b
when y = 0, value of x = roots of the equation
x2 - 5x + 4 = 0
x = 1, 4 (see graph)
Users' Answers & Commentsx2 - 5x + 4 = 0
x = 1, 4 (see graph)
34
PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate the perpendicular distance between the parallel sides
A
3.35cm
B
3.73cm
C
4.50cm
D
4.62cm
correct option: a
h = ?
\(\frac{h}{5cm}\) = sin 45o
h = 5cm x 0.6691 = 3.35cm
Users' Answers & Comments\(\frac{h}{5cm}\) = sin 45o
h = 5cm x 0.6691 = 3.35cm
35
PQRS is a trapezium. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate, correct to the nearest cm2, the area of the trapezium (h = 3.35cm2 )
A
27cm2
B
30cm2
C
36cm2
D
37cm2
correct option: a
Area = \(\frac{1}{2}(a + b)\)h
where a = QR = 6cm
b = PA = 10cm
Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm2
= \(\frac{1}{2}(16)\) x 3.35cm2
= 26.8cm2
= 27cm2 (approx.)
Users' Answers & Commentswhere a = QR = 6cm
b = PA = 10cm
Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm2
= \(\frac{1}{2}(16)\) x 3.35cm2
= 26.8cm2
= 27cm2 (approx.)
36
In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP
A
120o
B
90o
C
60o
D
45o
correct option: b
Tan \(\theta = \frac{6}{6} = 1\)
\(\theta\) = tab - 1(1) = 45o
< top = 20
= 2 x 45o = 90o
Users' Answers & Comments\(\theta\) = tab - 1(1) = 45o
< top = 20
= 2 x 45o = 90o
37
In the diagram, IG is parallel to JE, JEF = 120o and FHG = 130o, fins the angle marked t
A
40o
B
70o
C
80o
D
100o
38
Find the value of x in the diagram
A
50o
B
30o
C
22o
D
17o
correct option: d
Sum of exterior angles of a polygon = 360o
(x + 64) + 2x + (3x - 54) + 5x + 4x + 3x + (2x + 10) = 360o
20x + 20 = 360o
20x + 360 - 20 = 340o
x = \(\frac{340}{20}\)
x = 17o
Users' Answers & Comments(x + 64) + 2x + (3x - 54) + 5x + 4x + 3x + (2x + 10) = 360o
20x + 20 = 360o
20x + 360 - 20 = 340o
x = \(\frac{340}{20}\)
x = 17o
39
In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x
A
5.6
B
6.5
C
6.6
D
6.8
correct option: a
\(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\)
\(\frac{x}{4 + x} = \frac{7}{7 + 5}\)
\(\frac{x}{4 + x} = \frac{7}{12}\)
(2x = 7)(4 + x); 12x = 28 + 7x
12x - 7x = 28
5x = 28
x = \(\frac{28}{5}\)
x = 5.6
Users' Answers & Comments\(\frac{x}{4 + x} = \frac{7}{7 + 5}\)
\(\frac{x}{4 + x} = \frac{7}{12}\)
(2x = 7)(4 + x); 12x = 28 + 7x
12x - 7x = 28
5x = 28
x = \(\frac{28}{5}\)
x = 5.6
40
Using the diagram, find the bearing of X from Y
A
300o
B
240o
C
120o
D
60o