Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 586–600 of 1462 verified questions
Sort by:
Mathematics
UTME / JAMB 2007
Geometry and Trigonometry
Solution
Find the locus of point equidistant from two straight lines y - 5 = 0 and y - 3 = 0
A
y - 2 = 0
B
y - 4 = 0
C
y - 1 = 0
D
y - 3 = 0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0
Source: Myschool Past Questions Archive (Record #9252)
Mathematics
UTME / JAMB 2007
Geometry and Trigonometry
Solution
Find the size of each exterior angle of a regular octagon
A
51o
B
45o
C
40o
D
36o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Each Exterior ∠ = 360/n
= 360/8
= 45o
Source: Myschool Past Questions Archive (Record #9254)
Mathematics
UTME / JAMB 2007
Geometry and Trigonometry
Solution
Find the value of \(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}\)
A
\(\frac{4}{\sqrt{3}}\)
B
\(\frac{2}{\sqrt{3}}\)
C
1
D
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}= \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\\
=\left(\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}\right)\div \left(\frac{\sqrt{3}\sqrt{3}+1}{\sqrt{3}}\right)\\
=\frac{(3-1)}{(3+1)}\\
=\frac{2}{4}=\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #9255)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Diagram
Solution
(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \(209 cm^{2}\), calculate the height of the cylinder. [Take \(\pi = \frac{22}{7}\)].
(b) In the diagram, O is the centre of the circle and ABC is a tangent at B. If \(\stackrel\frown{BDF} = 66°\) and \(\stackrel\frown{DBC} = 57°\), calculate, (i) \(\stackrel\frown{EBF}\) and (ii) \(\stackrel\frown{BGF}\).
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) TSA of a closed cylinder = \(2\pi r^{2} + 2\pi rh\)
= \(2\pi r (r + h)\)
\(2 \times \frac{22}{7} \times 3.5 (3.5 + h) = 209\)
\(22(3.5 + h) = 209 \implies 77 + 22h = 209\)
\(22h = 209 - 77 = 132\)
\(h = \frac{132}{22} = 6 cm\)
(b) \(< EDB = 90^\circ\) (angle subtended at the circumference by a semi-circle)
\(\therefore < EDF = 90^\circ - 66^\circ = 24^\circ\)
\(< EDF = < EBF = 24^\circ\) (angles in the same segment)
(ii)
\(< DEB = < DBC = 57^\circ\) (Alternate segment theorem)
\(\therefore < BGF = 180 - (57 + 24) = 180 - 81\)
\(< BGF = 99^\circ\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #52987)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
Find the product of 0.0409 and 0.0021 leaving your answer in the standard form
A
8.6 x 10-6
B
8.6 x 105
C
8.6 x 10-4
D
8.6 x 10-5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
0.0409 x 0.0021 = 409 x 10\(^{-4}\) x 21 x 10\(^{-4}\)
= 409 x 21 x 10\(^{-4 * -4}\)
= 8589 x 10\(^{-8}\)
= 8.589 x 10\(^{3}\) x 10\(^{-8}\)
= 8.6 x 10\(^{3-8}\)
= 8.6 x 10\(^{-5}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #36325)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
Simplify \((0.3\times 10^{5})\div (0.4\times 10^{7})\)leaving your answer in standard form
A
\(7.5\times 10^{-4}\)
B
\(7.5\times 10^{-3}\)
C
\(7.5\times 10^{-2}\)
D
\(7.5\times 10^{-1}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((0.3\times 10^{5})\div (0.4\times 10^{7})\\
\frac{3}{10}\times 10^{5}\div \frac{4}{10}\times 10^{7}=3\times 10^{4}\div 4\times 10^{6}\\
\frac{3\times 10^{4}}{4\times 10^{7}}=\frac{3}{4}\times 10^{-2}=0.75\times 10^{-2}\\
7.5\times 10^{-3}\)
Source: Myschool Past Questions Archive (Record #26719)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per meter square
A
N1,575.00
B
N3,150.00
C
N4,734.30
D
N8,268.75
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
N75.00 x 110.25 = N8,268.75
Source: Myschool Past Questions Archive (Record #26730)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the center of the circle
A
2.5cm
B
3.0cm
C
3.5cm
D
4.0cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(5^{2}=3^{2}+x^{2}\)
25 - 9 = x\(^2\)
16 = x\(^2\)
x = 4
Source: Myschool Past Questions Archive (Record #26731)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Diagram
Solution
IN the diagram, |LN| = 4cm, LNM = 90o and tan y = 2/3. What is the area of the ∆LMN?
Attached Figure / Diagram
A
24cm2
B
12cm2
C
10cm2
D
6cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(tan \hspace{1mm}y = \frac{opp}{adj}=\frac{2}{3}\Rightarrow \frac{2}{3}=\frac{
LN}{MN}=\frac{4}{MN}\\
∴ MN = \frac{12}{2}=6\\
Area \hspace{1mm}of \hspace{1mm}\Delta LMN = \frac{1}{2}\times MN\times LN=\frac{1}{2}\times 6\times 4=12cm\)
Source: Myschool Past Questions Archive (Record #26733)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
Simplify(0.3 x 105) \(\div\) (0.4 \(\times\) 107), leaving you answer in the standard form.
A
7.5 x 10-4
B
7.5 x 10-3
C
7.5 x 10-2
D
7.5 x 10-1
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{0.3 \times 10^5}{0.4 \times 10^7} = \frac{3 \times 10^{-1} \times 10^{5}}{4 \times 10^{-1} \times 10^7}\)
\(\frac{3 x 10^4}{4 x 10^4}\)
\(\frac{3}{4}\) x 104-6 = 0.75 x 10-2
= 7.5 x 10-3
Source: Myschool Past Questions Archive (Record #36231)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
If p \(\alpha \frac{I}{Q}\) which of the following is true?
A
q \(\alpha p^2\)
B
q \(\alpha \frac{1}{p^2}\)
C
q \(\alpha \sqrt{p}\)
D
q \(\alpha \frac{1}{p}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
p \(\alpha \frac{I}{Q}\); p \(\frac{k}{q}\) (where k is constant)
q = \(\frac{k}{p}\)
q \(\alpha \frac{1}{p}\)
Source: Myschool Past Questions Archive (Record #36232)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
Which of the following statement is true for the ste P = {-3.2 \(\leq\) x < 5} where x is an integer?
A
least value of x is -3.2
B
least value of x is -3
C
greatest value of x is 4.9
D
greatest value of x is 5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
p = {-3.2.....4.9}; Since x is an integer
least value of x is -3
Source: Myschool Past Questions Archive (Record #36234)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.
A
2.5cm
B
3.0cm
C
3.5cm
D
4.0cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Distance(d) = d\(^2\) = 5\(^2\) - 3\(^2\)
= 25 - 9 = 16
d = \(\sqrt{16}\) = 4cm
Source: Myschool Past Questions Archive (Record #36263)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per metre
A
N1,575.00
B
N3,150.00
C
N4,734.30
D
N8,268.75
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area = 110.252
f2 = 110.25(square)
l = \(\sqrt{110.25}\) = 10.5m
Since N75.00 per metre
N75.00 = 1 metre; x = 10.5m
x = 787.5 for one side, then four sides = N787.5 x 4
= N3,150.00
Source: Myschool Past Questions Archive (Record #36265)
Mathematics
WAEC / WASSCE 2007
Geometry and Trigonometry
Solution
Each of the interior angle of a regular polygon is 162o. How many sides has the polygon?
A
8
B
12
C
16
D
20
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides)
162n = 180n - 360
162n - 180n = -360
-18n = - 360
n = \(\frac{360}{18}\)
n = 20
Source: Myschool Past Questions Archive (Record #36271)