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Mathematics
WAEC / WASSCE 2009
Permutations, Combinations & Probability
Solution
A box contains black, white and red identical balls. The probability of picking a black ball at random from the box is \(\frac{3}{10}\) and the probability of picking a white ball at random is \(\frac{2}{5}\). If there are 30 balls in the box, how many of them are red?
A
3
B
7
C
9
D
12
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total no of balls = 30
Let x = no. of red balls
Pr(red) = \(\frac{x}{30}\)
Pr(black) = \(\frac{3}{10} = \frac{9}{30}\)
Pr(white) = \(\frac{2}{5} = \frac{12}{30}\)
No. of black balls = 9
No. of white balls = 12
9 = 12 + x = 30
x = 30 - 21
x = 9
No. of red balls = 9
OR
Let w = no of white balls
B = no of black balls
R = no of red balls
Pr(black balls) = \(\frac{3}{10}\)
Pr(white balls) = \(\frac{2}{5}\)
But Pr(white balls) = \(\frac{2}{5}\) = \(\frac{W}{30}\)
cross multiply
W = no of white balls = 12
Pr(black balls) = \(\frac{3}{10}\) = \(\frac{B}{30}\)
cross multiply
B = no of black balls = 9
Total number of balls in the basket = W + B + R = 30 ( given)
= 12 + 9 + R = 30
R = 30 - 12 - 9 = 9
Therefore, the no of red balls = 9.
Source: Myschool Past Questions Archive (Record #36472)
Mathematics
WAEC / WASSCE 2009
Indices, Logarithms & Surds
Solution
Simplify; \(\frac{1}{2}\sqrt{32} - \sqrt{18} + \sqrt{2}\)
A
zero
B
\(\sqrt{2}\)
C
2\(\sqrt{2}\)
D
4\(\sqrt{2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{1}{2}\sqrt{32} - \sqrt{18} + \sqrt{2}\) = \(\frac{1}{2} (4\sqrt{2}) - 3\sqrt{2} + \sqrt{2}\)
= 2\(\sqrt{2} - 3\sqrt{2} + \sqrt{2}\)
= 3\(\sqrt{2} - 3\sqrt{2} - 3\sqrt{2}\) = 0
Source: Myschool Past Questions Archive (Record #36473)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Solution
The angles of triangle are (x + 10)o, (2x - 40)o and (3x - 90)o. Which of the following accurately describes the triangle?
A
it is a scalene triangle
B
it is right angled isosceles triangle
C
t is an equilateral triangle
D
It is an isosceles triangle but not right angled
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(x + 10)o + (2x - 40)o + (3x - 90)o = 180
6x - 120 = 180
6x = 180 + 120
6x = 300
x = \(\frac{300}{6}\)
x = 50
x + 10o = 50o + 10o = 60o
2x - 40 = 2(50o) - 40 = 60o
3x - 90 = 3(50o) - 90o = 60o
Hence, it is an equilateral triangle
Source: Myschool Past Questions Archive (Record #36474)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify (x - 3y)2 - (x + 3y)2
A
2(x + 3y)
B
(2x - 3y)
C
-12xy
D
6xy
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(x - 3y)2 - (x + 3y)2 = [(x - 3y) - (x + 3y)]
[(x + 3y + x + 3y)] = [-6y] [2x]
= -12xy
Source: Myschool Past Questions Archive (Record #36475)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
The following is the graph of a quadratic friction, find the co-ordinates of point P
Attached Figure / Diagram
A
(0, 4)
B
(4, 0)
C
(0, -4)
D
(-4, 0)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #36481)
Mathematics
WAEC / WASSCE 2009
Linear, Simultaneous & Quadratic Equations
Diagram
Solution
The following is the graph of a quadratic friction, find the value of x when y = 0
Attached Figure / Diagram
A
1, 2
B
1, 4
C
2, 3
D
1, 6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
when y = 0, value of x = roots of the equation
x2 - 5x + 4 = 0
x = 1, 4 (see graph)
Source: Myschool Past Questions Archive (Record #36482)
Mathematics
WAEC / WASSCE 2009
Trigonometry, Elevation & Bearings
Diagram
Solution
PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate the perpendicular distance between the parallel sides
Attached Figure / Diagram
A
3.35cm
B
3.73cm
C
4.50cm
D
4.62cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
h = ?
\(\frac{h}{5cm}\) = sin 45o
h = 5cm x 0.6691 = 3.35cm
Source: Myschool Past Questions Archive (Record #36483)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
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 )
Attached Figure / Diagram
A
27cm2
B
30cm2
C
36cm2
D
37cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
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.)
Source: Myschool Past Questions Archive (Record #36485)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP
Attached Figure / Diagram
A
120o
B
90o
C
60o
D
45o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Tan \(\theta = \frac{6}{6} = 1\)
\(\theta\) = tab - 1(1) = 45o
o = 90o
Source: Myschool Past Questions Archive (Record #36486)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, IG is parallel to JE, JEF = 120o and FHG = 130o, fins the angle marked t
Attached Figure / Diagram
A
40o
B
70o
C
80o
D
100o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Now, t + 60o + 50o = 180o
t = 180o - 110o
t = 70o
Source: Myschool Past Questions Archive (Record #36487)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
Find the value of x in the diagram
Attached Figure / Diagram
A
50o
B
60o
C
70o
D
80o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
30 + x = 100
x = 100 - 30
= 70o
Source: Myschool Past Questions Archive (Record #36488)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x
Attached Figure / Diagram
A
5.6
B
6.5
C
6.6
D
6.8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\)
\(\frac{x}{4 + x} = \frac{7}{7 + 5}\)
\(\frac{x}{4 + x} = \frac{7}{12}\)
(2x = 7)(4 + x); 12x = 28 + 7x
12x - 7x = 28
5x = 28
x = \(\frac{28}{5}\)
x = 5.6
Source: Myschool Past Questions Archive (Record #36491)
Mathematics
WAEC / WASSCE 2009
Trigonometry, Elevation & Bearings
Diagram
Solution
Using the diagram, find the bearing of X from Y
Attached Figure / Diagram
A
300o
B
240o
C
120o
D
60o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The bearing of X from Y
= 360 - 120 = 240
Source: Myschool Past Questions Archive (Record #36502)
Mathematics
WAEC / WASSCE 2009
Trigonometry, Elevation & Bearings
Diagram
Solution
If /XY/ = 50m, how far cast of X is Y?
Attached Figure / Diagram
A
25.5m
B
40.6m
C
40.8m
D
43.3m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Required distance = x
where \(\frac{1}{50m}\) = sin 60 x 50m x 0.866
x = 43.3m
Source: Myschool Past Questions Archive (Record #36503)
Mathematics
WAEC / WASSCE 2009
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
in the diagram, angle 20º is subtended at the centre of the circle, find the value of x
Attached Figure / Diagram
A
30o
B
50o
C
80o
D
100o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2(30 + x) = 360 - 200
60 + 2x = 160
2x = 160 - 60
2x = 100
x = \(\frac{100}{2}\)
x = 50
Source: Myschool Past Questions Archive (Record #36504)