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Mathematics
UTME / JAMB 2015
Geometry and Trigonometry
Solution
The area of a circle of radius 4cm is equal to (Take π = 3.142 )
A
10.3cm2
B
15.7cm2
C
50.3cm2
D
17.4cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of a circle (A) = πr2
Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)
Source: Myschool Past Questions Archive (Record #41285)
Mathematics
UTME / JAMB 2015
Geometry and Trigonometry
Solution
The sum of two numbers is 5; their product is -14. Find the numbers.
A
x = 5 or 1
B
x = 7 or 1
C
x = 7 or 0
D
x = 7 or − 2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let x represent the first number;
Then, the other is (5 − x) , since their sum is 5 and their product is 14
x(5 − x) = 14
5x − x2 = 14
x2 − 5x − 14 = 0
(x2 − 7x) + (2x − 14) = 0
x(x − 7) + 2(x − 7) = 0
(x − 7)(x + 2) = 0
Either (x − 7) = 0 or (x + 2) = 0
x = 7 or x = − 2
The two numbers are 7 and − 2
Source: Myschool Past Questions Archive (Record #41293)
Mathematics
UTME / JAMB 2015
Geometry and Trigonometry
Solution
How many sides has a regular polygon whose interior angles are 120°each
A
4
B
5
C
6
D
3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Each exterior angle = 180 − 120=60
Exterior angle =360/n
60 = 360/n
n =360/6
= 6
Source: Myschool Past Questions Archive (Record #41295)
Mathematics
UTME / JAMB 2015
Geometry and Trigonometry
Solution
A public car dealer marked up the cost of a car at 30% in an attempt to make 20% gross profit. Due to the value of dollar, he now placed 20% discount on the car. What profit or loss will he make?
A
3%
B
2%
C
4%
D
1%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let assume the cost price is 100%
Marked up price + cost price = 20 + 100 = 120%
Discount at 20% = 20/100 × 120 % of cost price
Selling price = cost price − gain
= (120 − 24)% of cost price
= 96% of cost price
Loss = (100− 96)% of cost price
= 4% of cost price
∴ He will wake 4% loss
Source: Myschool Past Questions Archive (Record #41298)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Solution
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in Naira?
A
N250,000.00
B
N200,000.00
C
N150,000.00
D
N100,000.00
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
N1.00 = $0.075
N X = $15,000
X = \(\frac{1.00 \times 15000}{0.075}\)
= N200,000.00
Source: Myschool Past Questions Archive (Record #45696)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre. If PQ//RS and ∠ONS = 140^\circ, find the size of ∠POM.
Attached Figure / Diagram
A
40o
B
50o
C
60o
D
80o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In the diagram above,
∠MNO = 140^\circ and angles on a straight line is 180^\circ
: ∠NMO = (180 - 140)^\circ = 40^\circ
Hence; ∠POM = 40^\circ ( alternate angle ∠S)
Source: Myschool Past Questions Archive (Record #45754)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the diagram, PTR is a tangent to the centre O. If angles TON = 108^\circ, Calculate the size of angle PTN
Attached Figure / Diagram
A
132o
B
126o
C
108o
D
102o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram; 108^\circ + x + x = 180^\circ (sum of angle in a triangle)
108^\circ + 2x = 180^\circ
x = 180^\circ - 108^\circ
= 72^\circ
x = \(\frac{72^o}{2}\)
= 36^\circ
(Angle between tangent and a chord through the point of contact)
Hence, angle PTN = 90 + 36
= 126^\circ
Source: Myschool Past Questions Archive (Record #45757)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQ//RT, QR//SU,
Attached Figure / Diagram
A
134o
B
132o
C
96o
D
48o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram; a = b = 48º (alternate < S)
x = 180 ^\circ - b (angles on a str. line)
x = 180º - 48º
= 132º
Source: Myschool Past Questions Archive (Record #45767)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and
Attached Figure / Diagram
A
73o
B
67o
C
57o
D
37o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram given, < PRT = 3\(^o\) (Change in same segment)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)
Source: Myschool Past Questions Archive (Record #45775)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the diagram, O is the centre of the circle, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and < TOS = 76\(^o\). Using the diagram, find the value of angle PTR.
Attached Figure / Diagram
A
76o
B
71o
C
57o
D
33o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
PTR = 180º - (90 + 33)º
= 180º - 123
= 57º
Source: Myschool Past Questions Archive (Record #45785)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Solution
\(\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let: (\(x_1, y_1\)) = (0, 3)
(\(x_2, y_2\)) = (\(\frac{5}{4}, \frac{11}{2}\))
Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)
= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)
= \(\frac{11 - 6}{2} \div \frac{5}{4}\)
= \(\frac{5}{2} \div \frac{5}{4}\)
= \(\frac{5}{2} \times \frac{4}{5}\)
= 2
Source: Myschool Past Questions Archive (Record #45788)
Mathematics
UTME / JAMB 2015
Geometry and Trigonometry
Solution
Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.
A
\(\frac{2\sqrt{13}}{13}\)
B
\(\frac{3\sqrt{13}}{13}\)
C
\(\frac{4\sqrt{13}}{13}\)
D
\(\frac{6\sqrt{13}}{13}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)
thus m2 = 22 + 32
= 4 + 9 = 13
m = \(\sqrt{13}\)
Hence, 2sin x = 2 x \(\frac{2}{m}\)
2 x\(\frac{2}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}}\)
= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)
= \(\frac{4\sqrt{13}}{13}\)
Source: Myschool Past Questions Archive (Record #45797)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Solution
PQRS is a square. If x is the midpoint of PQ, calculate correct to the nearest degree,
A
53o
B
55o
C
63o
D
65o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram given,
tan\(\alpha\) = \(\frac{1}{0.5}\) = 2
\(\alpha\) = tan - 1(2) = 63.43º
= 63º
Source: Myschool Past Questions Archive (Record #45809)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Solution
The angle of elevation of an aircraft from a point K on the horizontal ground 30\(\alpha\). If the aircraft is 800m above the ground, how far is it from K?
A
400.00m
B
692.82m
C
923.76m
D
1,600.00m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In \(\bigtriangleup\)KBC, sin 30 = \(\frac{800}{IKCI}\)
IKCL = \(\frac{800}{sin30^o}\)
= \(\frac{800}{0.5}\)
= 1600m
Source: Myschool Past Questions Archive (Record #45810)
Mathematics
WAEC / WASSCE 2015
Geometry and Trigonometry
Diagram
Solution
In the given diagram, \(\bar{QT}\) and \(\bar{PR}\) are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.
Attached Figure / Diagram
A
35o
B
40o
C
55o
D
60o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In the diagram, QOR + 2m(vertically opposite angles)
So, m + 90^\circ + 2m = 180^\circ
(angles on str. line)
3m = 180^\circ - 90^\circ
3m = 90^\circ
m = \(\frac{90^o}{3}\)
= 30^\circ
substituting 30^\circ for m in
2m + 4n = 200^\circ gives
2 x 30^\circ + 4n = 200^\circ
60^\circ + 4n = 200^\circ
4n = 200^\circ - 60^\circ
= 140^\circ
n = \(\frac{140^\circ}{4}\)
= 35^\circ
Source: Myschool Past Questions Archive (Record #45861)