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Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, O is the centre of the circle ∠PQR = 75o, ∠OPS = yo and \(\bar{OR}\) is parallel to \(\bar{PS}\). Find y
Attached Figure / Diagram
A
15
B
30
C
45
D
75
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram:
Source: Myschool Past Questions Archive (Record #26420)
Mathematics
WAEC / WASSCE 1999
Linear, Simultaneous & Quadratic Equations
Solution
The sides of a right angle triangle in ascending order of magnitude are 8cm, (x-2)cm and x cm. Find x
A
16
B
17
C
34
D
90
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
8\(^2\) + (x - 2)\(^2\) = x\(^2\)
64 + x\(^2\) - 4x + 4 = x\(^2\)
x\(^2\) - 4x + 68 = x\(^2\)
-4x = -68
x = \(\frac{-68}{-4}\) = 17
Source: Myschool Past Questions Archive (Record #26421)
Mathematics
WAEC / WASSCE 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
In the diagram, CDE is a triangle, ABEF is cyclic quadrilateral, AB||CD, ∠BAF = 65o and ∠AFE = 85o calculate ∠CDE
Attached Figure / Diagram
A
20o
B
30o
C
50o
D
65o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #26422)
Mathematics
UTME / JAMB 1999
Inequalities & Linear Programming
Diagram
the shaded portion in the graph is represented by
Attached Figure / Diagram
A
y + x - x \(\leq\) 0, y - x \(\leq\) 0
B
y - x + x, 3 \(\leq\) 0, y - x \(\geq\) 0
C
y + x - 3 \(\geq\) 0, y + x \(\leq\) 0
D
y - x + x3 \(\geq\) 0, y + x \(\leq\) 0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #33084)
Mathematics
UTME / JAMB 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, EFGH is a cyclic quadrilateral in which EH || FG, EG, and FH are chords. If < FHG = 42º and < EFH = 34º, calculate < HEG.
Attached Figure / Diagram
A
34o
B
42o
C
52o
D
76o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
< EFH = < EGH = 34º (angles n the same segment)
< GHF = < GEF = 42º(angles in the same segment)
< FOG = 42 + 34 = 76º (exterior angle)
< FOG = < EOH = 76º (vertically opposite angle)
< EDO = 90º, < DOE = \(\frac{76}{2}\) = 38º
< HEG = 90º - 38º = 52º
Source: Myschool Past Questions Archive (Record #33085)
Mathematics
UTME / JAMB 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, PQRS is a circle with ST||RQ. Find the value of x if PT = pS
Attached Figure / Diagram
A
70o
B
55o
C
40o
D
35o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
o (sum of o - 110o = 70o
in o
2x + 70 = 180o
2x = 180o - 70o = 110o
x = \(\frac{110^o}{2}\) = 55o
Source: Myschool Past Questions Archive (Record #33086)
Mathematics
UTME / JAMB 1999
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, TZ is tangent to the circle QPZ. Find x if TZ = 6 units and PQ = 9 units
Attached Figure / Diagram
A
3
B
4
C
5
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\bigtriangleup\)WOP, b2 = r2 - (4.502...........(i)
In \(\bigtriangleup\)WOT, b\(\bigtriangleup\) + (4.5 + x)2 = a2...........(ii)
In \(\bigtriangleup\)ZOT, a2 = 62 + r2 ..........(iii)
subst. equation (iii) and (i) in eqn (ii)
r2 - (4.5)2 + (4.5 + x)2 = 36 = 20.25
= (4.5 + x)2 = \(\sqrt{56.25}\) = (4.5 + x)
= 56.25
= 4.5 + x = 7.5
x = 3.0m
x =
Source: Myschool Past Questions Archive (Record #33087)
Mathematics
UTME / JAMB 1999
Fractions, Decimals, Percentages & Approximations
Solution
The sectoral allocations to various ministries in a state budget are as follows:
Agriculture - N25,000,000.00
Education - N20,000,000.00
Women affairs - N35,000,000.00
Commercial and Industries - N20,000,000.00
Use a pie chart to represent this information. The corresponding angle to Agriculture is
A
90\(^0\)
B
72\(^0\)
C
60\(^0\)
D
45\(^0\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Total allocation = ₦25m + ₦20m + ₦35m + ₦20m = ₦100,000,000
Fraction for Agriculture = ₦25,000,000 / ₦100,000,000 = 0.25 (or 25%)
Angle for Agriculture = 0.25 \times 360^\circ = 90^\circ
Other sectors:
- Education: (20/100) \times 360^\circ = 72^\circ
- Women Affairs: (35/100) \times 360^\circ = 126^\circ
- Commercial and Industries: (20/100) \times 360^\circ = 72^\circ
Source: Myschool Past Questions Archive (Record #33088)
Mathematics
UTME / JAMB 1999
Trigonometry, Elevation & Bearings
Diagram
Solution
find the length XZ in the triangle
Attached Figure / Diagram
A
\(\sqrt{7m}\)
B
\(\sqrt{6m}\)
C
\(\sqrt{5m}\)
D
\(\sqrt{3m}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
xz2 = xy2 + yx2 - 2(xy) (yz) cos 120o
= 22 + 12 - 2(2) (1) cos 1202
= 4 + 1 - 4x - cos 60 = 5 - 4x - \(\frac{1}{2}\)
5 + 2 = 7
xz = \(\sqrt{7}\)m
Source: Myschool Past Questions Archive (Record #33089)
Mathematics
UTME / JAMB 1999
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
Solution
The grades of 36 students in a class test are as shown in the pie chart. How many student has excellent?
Attached Figure / Diagram
A
7
B
8
C
9
D
12
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
degree of excellent in the pie-chart = 360o - (90o + 120o + 80o) = 360o - 290o = 70o
= 360o = 36 students
70 = \(\frac{70^o}{360^o}\) x 36 = 7
Source: Myschool Past Questions Archive (Record #33090)
Mathematics
WAEC / WASSCE 1999
Indices, Logarithms & Surds
Solution
(a) Simplify \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
(b) Evaluate \(\log_{5} (\frac{3}{5}) + 3 \log_{5} (\frac{5}{2}) - \log_{5} (\frac{81}{8})\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
= \((\frac{25}{4})^{\frac{1}{2}} \times 2^{4} \div (\frac{2}{15})^{2}\)
= \(\frac{5}{2} \times 16 \times \frac{225}{4} = 2250\)
(b) \(\log_{5} (\frac{3}{5}) + 3\log_{5} (\frac{15}{2}) - \log_{5} (\frac{81}{8})\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 15 - 3\log_{5} 2 - \log_{5} 81 + \log_{5} 8\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} (3 \times 5) - 3\log_{5} 2 - \log_{5} 3^{4} + \log_{5} 2^{3}\)
= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 3 + 3\log_{5} 5 - 3\log_{5} 2 - 4\log_{5} 3 + 3\log_{5} 2\)
= \(2\log_{5} 5 \)
= \(2\)
Source: Myschool Past Questions Archive (Record #54376)
Mathematics
WAEC / WASSCE 1999
Sets, Logic & Venn Diagrams
Solution
(a) If \(\varepsilon\) is the set \({1, 2, 3,..., 19, 20}\) and A, B and C are subsets of \(\varepsilon\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;
(b) Find : (i) \(A \cap B\) ; (ii) \(A \cap C\) ; (iii) \(B \cup C\).
(c) Using your results in (b), show that \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) A = {5, 10, 15, 20}
(ii) B = {4, 8, 12, 16, 20}
(iii) C = {3, 6, 9, 12, 15, 18}
(b) (i) \(A \cap B = {20}\)
(ii) \(A \cap C = {15}\)
(iii) \(B \cap C = {12}\)
(c) \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\)
\((A \cap B) \cup (A \cap C) = {15, 20}\)
\(A \cap (B \cup C) = {5, 10, 15, 20} \cap {3, 4, 6, 8, 9, 12, 15, 16, 18, 20}\)
= \({15, 20}\)
\(\therefore (A \cap B) \cup (A \cap C) = A \cap (B \cup C)\)
Source: Myschool Past Questions Archive (Record #54414)
Mathematics
WAEC / WASSCE 1999
Trigonometry, Elevation & Bearings
Solution
ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37^\circ and |BC| = 5 cm. Calculate :
(a) |AC|, correct to 3 significant figures ;
(b) < PBA.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Let |PC| = x cm; Hence, |AC| = 2x cm
\(\tan 37^\circ = \frac{x}{5}\)
\(x = 5 \tan 37 \)
\(x = 3.768 cm\)
\(\therefore |AC| = 2 \times 3.768\)
= \(7.536 cm\)
\(\approxeq 7.54 cm\) (3 sig. figs)
(b) From \(\Delta ABC\),
\(\tan < ABC = \frac{7.536}{5} = 1.5072\)
\(< ABC = \tan^{-1} (1.5072) = 56.436^\circ\)
\(\therefore < PBA = < ABC - < PBC\)
= \(56.436^\circ - 37^\circ\)
= \(19.436^\circ \approxeq 19.44^\circ\)
Source: Myschool Past Questions Archive (Record #54417)
Mathematics
WAEC / WASSCE 1999
Mensuration (Areas, Perimeters & Volumes)
Solution
In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :
(a) |AB| ;
(b) the area of the triangle BCD ;
(c) |CD| ;
(d) perimeter of the trapezium.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(|AB| = \sqrt{|BD|^{2} - |AD|^{2}}\)
= \(\sqrt{17^{2} - 15^{2}} = \sqrt{289 - 225}\)
= \(\sqrt{64} = 8 cm\)
(b) Area of triangle BCD = \(\frac{1}{2} \times 9 \times 8 \)
= \(36 cm^{2}\)
(c) \(|EC| = 6 cm\)
\(|CD| = \sqrt{8^{2} + 6^{2}} = \sqrt{100}\)
= 10 cm.
(d) Perimeter of the trapezium = 15 cm + 8 cm + 9 cm + 10 cm
= 42 cm.
Source: Myschool Past Questions Archive (Record #54418)
Mathematics
WAEC / WASSCE 1999
Permutations, Combinations & Probability
Solution
(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(3y - 2x = 21 ... (1)\)
\(4y + 5x = 5 .... (2)\)
Multiply (1) by 4 and (2) by 3 so we have,
\(12y - 8x = 84 ... (3)\)
\(12y + 15x = 15 ... (4)\)
(3) - (4) : \(-8x - 15x = 69 \implies -23x = 69\)
\(x = \frac{69}{-23} = -3\)
Put x = -3 in (1),
\(3y - 2(-3) = 21\)
\(3y + 6 = 21 \implies 3y = 21 - 6 = 15\)
\(y = \frac{15}{3} = 5\)
\(x, y = -3, 5\).
(b)(i) Probability of losing
Probability of winning = \(\frac{1}{6}\)
\(\therefore\) Probability of losing = \(1 - \frac{1}{6} = \frac{5}{6}\).
(ii) Losing $20.00 after two picks = picking the card numbered 1 twice.
= \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
Source: Myschool Past Questions Archive (Record #54420)