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Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
From a point P, Q is 5km due West and R is12km due South of Q. Find the distance between P and R.
A
5km
B
12km
C
13km
D
17km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Using Pythagoras theorem
PR\(^2\) = 5\(^2\) + 12\(^2\)
25 + 144 = 169
PR = \sqrt(169)= 13km
Source: Myschool Past Questions Archive (Record #47584)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
Calculate the area of an equilateral triangle of side 8cm
A
8√3
B
16
C
4√3
D
16√3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
An equilateral triangle has all sides equal and all angles equal as 600
Area = \(\frac{1}{2}\) absin\theta
Area = \(\frac{1}{2}\) x 8 x 8 x sin60
= \(\frac{1}{2}\) x 64 x \(\sqrt{\frac{3}{2}}\)
= 16\sqrt3 cm\(^2\)
Source: Myschool Past Questions Archive (Record #47589)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
From a point R, 300m north of P, a man walks eastwards to a place; Q which is 600m from P. Find the bearing of P from Q correct to the nearest degree
A
026\(^o\)
B
045\(^o\)
C
210\(^o\)
D
240\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Cos \theta = \(\frac{adj}{hyp}\)
= \(\frac{300}{600}\)
= 0.5
\theta = Cos - 10.5
= 60
∠ RPQ = ∠ PQs
So the bearing of P from Q is 180 + 60 = 240\(^o\)
Answer is D
Source: Myschool Past Questions Archive (Record #47601)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
If x : y : z = 2 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\)
A
1\(\frac{1}{2}\)
B
2
C
2\(\frac{1}{2}\)
D
3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)
\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)
Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)
\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)
\(T_1\) = 15 and \(T_2\) = 21
Therefore;
x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9 (again)
z = \(\frac{4}{7}\) x 21 = 12
Hence;
\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\)
\(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)
= 1\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #49246)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)]
A
14m
B
10m
C
7m
D
5m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Volume of a cylinder = \( \pi r^2\)h
385 = \(\frac{22}{7}\) x \(r^2\) x 10
385 x 7 = 22 x \(r^2\) x 10
\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)
= 12.25
r = \(\sqrt{12.25}\)
= 3.5m
Hence, diameter of tank = 2r
= 2 x 3.5 = 7m
Source: Myschool Past Questions Archive (Record #49263)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x
A
\(45^o\)
B
\(84^o\)
C
\(85^o\)
D
\(95^o\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)
8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\)
8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\)
8x = 720\(^o\) - 40\(^o\) = 680\(^o\)
x = \(\frac{680^o}{8}\)
= 85\(^o\)
Source: Myschool Past Questions Archive (Record #49345)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
If x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z
A
2:3
B
3:4
C
3:8
D
1:2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)
\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)
But,
x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\)
\(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\)
\(T_1 = 15\) and \(T_2 = 21\)
Thus , x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9
z = \(\frac{4}{7}\) x 21 = 12
Hence; x : z = 6 : 12
= 1 : 2
Source: Myschool Past Questions Archive (Record #49414)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.
A
27m
B
13.5 \(\sqrt{3m}\)
C
13.5 \(\sqrt{2m}\)
D
9\(\sqrt{3m}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
From the diagram,
tan 30\(^o\) = \(\frac{h}{27}\)
h = 27 tan 30\(^o\)
= 27 x \(\frac{1}{\sqrt{3}}\)
= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)
= \(\frac{27 \sqrt{3}}{3}\)
= 9\(\sqrt{3m}\)
Source: Myschool Past Questions Archive (Record #49443)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Diagram
Solution
Find the value of t in the diagram
Attached Figure / Diagram
A
63\(^o\)
B
117\(^o\)
C
126\(^o\)
D
234\(^o\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In the diagram,
\(\alpha\) = 180\(^o\) - 177\(^o\) (angles on a straight line)
\(\alpha\) = 63\(^o\)
t = 2 x 63\(^o\) (angle at centre = 2 x angle at circum)
= 126\(^o\)
Source: Myschool Past Questions Archive (Record #49478)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Diagram
Solution
In the diagram, PR is a tangent to the circle at Q, QT//RS, < SQR = 35º and < RSQ = 50º. Find the value of
Attached Figure / Diagram
A
40\(^o\)
B
65\(^o\)
C
85\(^o\)
D
95\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In the diagram,
a = 50\(^o\) (alternate angles)
b\(_1\) + a + 35\(^o\) = 180\(^o\) (sum of angles on a straight line)
i.e; b\(_1\) + 50\(^o\) + 35\(^o\)
= 180º
b\(_1\) + 180\(^o\) - 85\(^o\) = 90\(^o\)
But b\(_2\) = \(b_1\) = 95\(^o\) (angles in alternate segment)
Source: Myschool Past Questions Archive (Record #49487)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Diagram
In the diagram, WXYZ is a rectangle with dimension 8cm by 6cm. P, Q, R and S are the midpoints of the sides of the rectangle as shown. Using this information, what type of quadrilateral is the shaded region?
Attached Figure / Diagram
A
Trapezium
B
Prism
C
Rectangle
D
Rhombus
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #49491)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Diagram
Solution
In the diagram, WXYZ is a rectangle with diamension 8cm by 6cm. P, Q, R and S are the midpoints of the rectangle as shown. Using this information calculate the area of the part of the rectangle that is not shaded
Attached Figure / Diagram
A
25cm\(^2\)
B
24cm\(^2\)
C
16cm\(^2\)
D
12cm\(^2\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area of shaded part = 2 x area of \(\bigtriangleup\)SPQ
= 2 x \(\frac{1}{2}\) 8 x 3 =24cm
Hence, area of the unshaded part of the rectangle = area of rectangle - area of shaded part
= 8 x 6 - 24
= 48 - 24
= 24 cm\(^2\)
Source: Myschool Past Questions Archive (Record #49520)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Diagram
Solution
In the diagram, PQ//RS. Find x in terms of y and z
Attached Figure / Diagram
A
x = 240\(^o\) - y - z
B
x = 180\(^o\) - y - z
C
x = 360\(^o\) + y -z
D
x = 360\(^o\) - y - z
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In the diagram,
a = z (alternate angles)
b = 180\(^o\) - a (angles on a straight line)
b = 180\(^o\) - z
c = 180\(^o\) - x (angles on a straight line)
y = b + c (sum of oposite interior angles)
y = 180\(^o\) - z + 180\(^o\) - x
y = 360\(^o\) - z - x
x = 360\(^o\) - y - z
Source: Myschool Past Questions Archive (Record #49564)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
A piece of thread of length 21.4cm is used to form a sector of a circle of radius 4.2cm on a piece of cloth. Calculate, correct to the nearest degree, the angle of the sector. [Take \(\pi = \frac{22}{7}\)]
A
170\(^o\)
B
177\(^o\)
C
182\(^o\)
D
192\(^o\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Length of arc, L = 21.4 - 2 x 4.2cm
= 21.4 - 8.4
= 13cm
But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)
i.e 13 = \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 4.2
= 13 x 360\(^o\) x 7
= \(\theta\) x 2 x 22 x 4.2
\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)
= \(\approx\) 177.27\(^o\)
\(\approx\) 177\(^o\) (to the nearest degree)
Source: Myschool Past Questions Archive (Record #49598)
Mathematics
WAEC / WASSCE 2018
Geometry and Trigonometry
Solution
If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x
A
\(\frac{1}{10}\)
B
\(\frac{1}{5}\)
C
\(\frac{5}{12}\)
D
1\(\frac{2}{5}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x - cos x
= \(\frac{4}{5} - \frac{3}{5}\)
= \(\frac{1}{5}\)
Source: Myschool Past Questions Archive (Record #49606)