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Mathematics
WAEC / WASSCE 2006
Indices, Logarithms & Surds
Diagram
Solution
In the diagram, |QR| = 5cm, PQR = 60o and PSR = 45o. Find |PS|, leaving your answe in surd form.
Attached Figure / Diagram
A
4\(\sqrt{5}\)cm
B
3\(\sqrt{7}\)cm
C
4\(\sqrt{6}\)cm
D
5\(\sqrt{6}\)cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
tan 6o = \(\frac{|PR|}{|QR|}\)
\(\sqrt{3} = \frac{|PR|}{5}\)
= |PR| = \(5 \sqrt{3}\)cm
sin 45 = \(\frac{|PR|}{|PS|}\)
\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)
|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)
= 5\(\sqrt{6}\)cm
Source: Myschool Past Questions Archive (Record #36203)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, \(\frac{PQ}{RS}\), find xo + yo
Attached Figure / Diagram
A
360o
B
300o
C
270o
D
180o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
PQR = QRS = Y(alt. amgles)
PQR + x = 180o(angles on a straight line)
y + x = 180o
Source: Myschool Past Questions Archive (Record #36204)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58o. Find < QRS
Attached Figure / Diagram
A
112o
B
116o
C
122o
D
148o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
PRQ = 90 - 58 = 32o(angle in a semi-circle)
Since PRS = 90o(radius angular to tangent)
QRS = 90 + 32
= 122o
Source: Myschool Past Questions Archive (Record #36205)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?
Attached Figure / Diagram
A
260o
B
130o
C
100o
D
80o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Draw a line from P to Q
o
Also, o
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
Source: Myschool Past Questions Archive (Record #36206)
Mathematics
WAEC / WASSCE 2006
Trigonometry, Elevation & Bearings
Diagram
Solution
In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find angle PQR.
Attached Figure / Diagram
A
60o
B
70o
C
120o
D
160o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Q1 = 50(alternate angle)
Q2 = 180o - 110o (straight line angle)
Q2 = 70
PQR = Q1 + Q2
= 50o + 70o
= 120o
Source: Myschool Past Questions Archive (Record #36207)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, the two circles have a common centre O. If the area of the larger circle is 100\(\pi\) and that of the smaller circle is 49\(\pi\), find x
Attached Figure / Diagram
A
2
B
3
C
4
D
6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
area of larger circle = 100\(\pi\)
\(\pi r^2 = 100\pi\)
R2 = 100
R = 10(Radius)
Area of smaller circle = \(49 \pi\)
\(\pi r^2 = 49\pi\)
r2 = 49
r = 7(radius)
Since R = x + r
x = R - r
x = 10 - 7 = 3
Source: Myschool Past Questions Archive (Record #36209)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQR is a straight line, QRST is a parallelogram, < TPQ = 72o and < RST = 126o. What type of triangle is \(\bigtriangleup\) PQT?
Attached Figure / Diagram
A
An equilateral triangle
B
A right-angled triangle
C
An isosceles triangle
D
A scalene triangle
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
< TSR =
Source: Myschool Past Questions Archive (Record #36210)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
How many pupils took the test?
Attached Figure / Diagram
A
36
B
38
C
39
D
40
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No. of pupils = 1 + 3 + 4 + 7 + 8 + 6 + 6 + 4 + 1 = 40
Source: Myschool Past Questions Archive (Record #36211)
Mathematics
WAEC / WASSCE 2006
Fractions, Decimals, Percentages & Approximations
Diagram
Solution
If the pass mark was 4. What percentage of the pupils failed the test?
Attached Figure / Diagram
A
20.0%
B
37%
C
62.5%
D
80.0%
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No. of pupils that failed = 1 + 3 + 4 = 8
% failed = \(\frac{8}{40} \times 100\)% = 20.0%
Source: Myschool Past Questions Archive (Record #36212)
Mathematics
WAEC / WASSCE 2006
Fractions, Decimals, Percentages & Approximations
Diagram
Solution
The diagram shows an arc MN of a circle, centre O, with 10cm. If o. calculate the length of the arc, correct to three significant figures. [Take \(\pi = \frac{22}{7}\)]
Attached Figure / Diagram
A
6.28cm
B
12.6cm
C
22.6cm
D
62.8cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Length of arc = \(\frac{\theta}{360 \times 2 \pi r}\)
= \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 10\)cm
= 12.57
= 12.6cm(3 s.f)
Source: Myschool Past Questions Archive (Record #36213)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram what is x + y in terms of z?
Attached Figure / Diagram
A
180o - z
B
180o + z
C
z - 180o
D
z
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x + y + z = 180(sum of angles in a \(\bigtriangleup\))
x + y = 180 - z
Source: Myschool Past Questions Archive (Record #36214)
Mathematics
WAEC / WASSCE 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram |XY| = 12cm, |XZ| = 9cm, |ZN| = 3cm and ZY||NM, calculate |MY|
Attached Figure / Diagram
A
3cm
B
4cm
C
5cm
D
6cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{|YM|}{|YX|} = \frac{|ZM|}{|ZX|}\)
\(\frac{|YM|}{12} = \frac{3}{9}\)
|YM| = 3 x \(\frac{12}{9}\)
|MY| = 4cm
Source: Myschool Past Questions Archive (Record #36215)
Mathematics
WAEC / WASSCE 2006
Indices, Logarithms & Surds
Solution
(a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
(b) By rationalising the denominator, simplify : \(\frac{7\sqrt{5}}{\sqrt{7}}\), leaving your answer in surd form.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
\(4\frac{2}{9} - 1\frac{13}{15} = \frac{38}{9} - \frac{28}{15}\)
= \(\frac{190 - 84}{45}\)
= \(\frac{106}{45}\)
\(2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3} = \frac{11}{5} + (\frac{4}{7} \times \frac{7}{3})\)
= \(\frac{11}{5} + \frac{4}{3}\)
= \(\frac{33 + 20}{15}\)
= \(\frac{53}{15}\)
\(\therefore \frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}} = \frac{106}{45} \div \frac{53}{15}\)
= \(\frac{106}{45} \times \frac{15}{53}\)
= \(\frac{2}{3}\)
(b) \(\frac{7\sqrt{5}}{\sqrt{7}}\)
= \(\frac{7\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\)
= \(\frac{7\sqrt{35}}{7}\)
= \(\sqrt{35}\)
Source: Myschool Past Questions Archive (Record #53010)
Mathematics
WAEC / WASSCE 2006
Fractions, Decimals, Percentages & Approximations
Solution
(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?
(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \(\pi = \frac{22}{7}\)].
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Let the cost of the car to the dealer be x.
Sales of car to a man with 15% profit : (115% of x = 1.15 x\)
Sale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00)
\(0.95 \times 1.15x = N120,175.00\)
\(1.0925x = N120,175.00\)
\(x = \frac{120,175.00}{1.0925}\)
\(x = N110,000\)
The original cost of the car to the dealer = N110,000.00.
(b) Circumference of a circle = \(2\pi r\)
Distance covered by the wheel = 1.05 km
Radius of the wheel = 18cm
Circumference of the wheel = \(2 \times \frac{22}{7} \times 18\)
= \(113.14 cm\)
Note : 1.05 km = 105,000 cm
Number of revolution = \(\frac{105,000}{113.14}\)
= \(928.05 revs\)
\(\approxeq 928 revs\)
Source: Myschool Past Questions Archive (Record #53011)
Mathematics
WAEC / WASSCE 2006
Linear, Simultaneous & Quadratic Equations
Solution
(a) Solve for x and y in the following equations :
\(2x - y = \frac{9}{2}\)
\(x + 4y = 0\)
(b)
In the diagram, TA is a tangent to the circle at A. If \(\stackrel\frown{BCA} = 40^\circ\) and \(\stackrel\frown{DAT} = 52^\circ\), find \(\stackrel\frown{BAD}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(2x - y = \frac{9}{2} .... (1)\)
\(x + 4y = 0 ....... (2)\)
From (2), x = - 4y. Put into (1), we have
\(2(- 4y) - y = \frac{9}{2}\)
\(-8y - y = \frac{9}{2} \implies - 9y = \frac{9}{2}\)
\(y = \frac{\frac{9}{2}}{-9} = -\frac{1}{2}\)
\(x = - 4y = -4(-\frac{1}{2})\)
\(x = 2\)
\((x, y) = (2, -\frac{1}{2})\)
(b)
In \(\Delta ACD, \stackrel\frown{ACD} = 52^\circ \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\)
\(\therefore \stackrel\frown{BCD} = 40^\circ + 52^\circ = 92^\circ\)
Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180^\circ
\(\therefore \stackrel\frown{BAD} = 180^\circ - 92^\circ = 88^\circ\)
Source: Myschool Past Questions Archive (Record #53016)