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 7921–7935 of 14893 verified questions
Sort by:
Biology
UTME / JAMB 2003
Reproduction in Plants & Animals
The type of reproduction that leads to variation animal and plant population is
A
budding
B
sexual
C
vegetative
D
asexual
medium Difficulty
Verified Correct Answer: Option B
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #7452)
Biology
UTME / JAMB 2003
Variety of Organisms & Classification of Living Things
An insect with a mandibulate mouth part will obtain its food by
A
biting and chewing
B
chewing and sucking
C
chewing
D
sucking
medium Difficulty
Verified Correct Answer: Option A
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #7453)
Biology
UTME / JAMB 2003
Adaptation & Evolution
An argument against Lamarck's theory of evolution is that
A
disuse of body part cannot weaken the part
B
disused part is dropped off in the offspring
C
acquired traits cannot be passed on to the offspring
D
trait cannot be acquired through constant use of body parts
medium Difficulty
Verified Correct Answer: Option C
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #7454)
Biology
UTME / JAMB 2003
Respiration & Gaseous Exchange
The presence of sunken stomata and the folding of leaves are adaptations to
A
Prevent entry of pathogens
B
Reduce water loss
C
Remove excess water
D
Prevent guttation
medium Difficulty
Verified Correct Answer: Option B
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #7455)
Biology
UTME / JAMB 2003
Adaptation & Evolution
Spines and shells on animals are adaptation for
A
camouglage
B
chemical defence
C
physical defence
D
mimicry
medium Difficulty
Verified Correct Answer: Option C
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #7456)
Biology
UTME / JAMB 2003
Adaptation & Evolution
An example of cryptic colouration is the
A
Mottled colours on the moths that rest in lichens
B
bright colour of an insect pollinated flower
C
Green colour of a plant
D
Bright marks on poisonous tropical frog on veriegated leaves
medium Difficulty
Verified Correct Answer: Option A
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #7457)
Biology
UTME / JAMB 2003
Adaptation & Evolution
Which of the following requires the use of carbon dating to prove that evolution has occurred?
A
Comparative anatomy
B
Biochemical similarities
C
molecular records
D
Fossil records
medium Difficulty
Verified Correct Answer: Option D
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #7458)
Biology
UTME / JAMB 2003
Adaptation & Evolution
the inactive state exhibited by an animal during hot dry seasons is termed
A
Aestivation
B
Dormancy
C
Resting
D
Hibernation
medium Difficulty
Verified Correct Answer: Option A
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #7459)
Biology
UTME / JAMB 2003
Variety of Organisms & Classification of Living Things
In the egg of a bird, the embryo is located in the
A
Yolk
B
Embryo disc IV
C
Chalaza
D
Albumen
medium Difficulty
Verified Correct Answer: Option A
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Official Exam / Past Question Archive
Biology
UTME / JAMB 2003
Variety of Organisms & Classification of Living Things
In the egg of a bird, the embryo is located in the
A
Yolk
B
Embryo disc IV
C
Chalaza
D
Albumen
medium Difficulty
Verified Correct Answer: Option A
Biological Concept & Syllabus Guide
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #41438)
Mathematics
UTME / JAMB 2003
Binary Operations, Matrices & Determinants
Solution
If \(\left| \begin{array}{cc} -x & 2 \\ 4x &1 \end{array} \right|\) = \(\left| \begin{array}{cc} 3 & 3x \\ 4 & -5 \end{array} \right|\), find the value of x
A
5
B
2
C
-2
D
-5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\left| \begin{array}{cc} -x & 2 \\ 4x &1 \end{array} \right|\) = \(\left| \begin{array}{cc} 3 & 3x \\ 4 & -5 \end{array} \right|\),
Means that the determinants of both Matrices are equal
- x - 8x = -15 - 12x
-9x = -15 - 12x
12x - 9x = - 15
3x = -15
x = \(\frac{-15}{3}\) = -5
Source: Myschool Past Questions Archive (Record #76304)
Mathematics
UTME / JAMB 2003
Binary Operations, Matrices & Determinants
Solution
A matrix P has an inverse P\(^{-1}\) = \(\begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}\). Find P
A
\(\begin{pmatrix} -1 & 3 \\ 0 & -1 \end{pmatrix}\)
B
\(\begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix}\)
C
\(\begin{pmatrix} 1 & -3 \\ 0 & -1 \end{pmatrix}\)
D
\(\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let \( A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} \) (given as \( P^{-1} \)).
Determinant of \( A \) = \( (1)(1) - (-3)(0) = 1 \).
Inverse of \( A \) is \( \frac{1}{1} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \), which is exactly \( P \).
Source: Myschool Past Questions Archive (Record #76305)
Mathematics
UTME / JAMB 2003
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram above, O is the centre of the circle, POM is a diameter, and \(\angle\) MNQ = 42º. Calculate \(\angle\) QMP.
Attached Figure / Diagram
A
42º
B
48º
C
132º
D
138º
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
POM is a diameter, so arc MP (lower semicircle via Q) = 180^\circ.
∠MNQ = 42^\circ is an inscribed angle subtending arc MQ.
Therefore, arc MQ = 2 \times 42^\circ = 84^\circ
Arc MQ + arc QP = lower semicircle:
arc QP = 180^\circ − 84^\circ = 96^\circ
- ∠QMP is an inscribed angle subtending arc QP.
Therefore, ∠QMP = \(\frac{1}{2}\) \times 96^\circ = 48^\circ.
(Alternatively: in right triangle PQM, ∠PQM = 90^\circ (angle in a semicircle), so the angles at M and P sum to 90^\circ; the arc method directly gives the result.)
Source: Myschool Past Questions Archive (Record #76318)
Mathematics
UTME / JAMB 2003
Plane Geometry, Polygons & Circle Theorems
Solution
Find the value of p, if the line that passes through (-1, -p) and (-2p, 2) is parallel to the line 2y + 8x - 17 = 0
A
\(\frac{7}{6}\)
B
\(\frac{6}{7}\)
C
\(\frac{-2}{7}\)
D
\(\frac{-6}{7}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The given line is 2y + 8x - 17 = 0.
Rewrite in slope form: 2y = -8x + 17
y = -4x + \(\frac{17}{2}\)
Slope = −4.
The line through points \((-1, -p)\) and \((-2p, 2)\) must have the same slope (parallel lines).
Slope = \(\frac{2 - (-p)}{-2p - (-1)} = \frac{2 + p}{1 - 2p}\)
\(\frac{2 + p}{1 - 2p} = -4\)
Cross-multiply:
2 + p = -4(1 - 2p)
2 + p = -4 + 8p
2 + 4 = 8p - p
6 = 7p
p = \(\frac{6}{7}\)
Source: Myschool Past Questions Archive (Record #76327)
Mathematics
UTME / JAMB 2003
Inequalities & Linear Programming
Solution
If \(\frac{\pi}{2}\) \(\leq\) \(\theta\) \(\leq\) 2\(\pi\), find the maximum value of f(\(\theta\)) = \(\frac{4}{6 + 2cos \theta}\)
A
\(\frac{1}{2}\)
B
4
C
1
D
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
To find the maximum of \(f(\theta) = \frac{4}{6 + 2\cos \theta}\) for \(\frac{\pi}{2} \leq \theta \leq 2\pi\), simplify to \(f(\theta) = \frac{2}{3 + \cos \theta}\).
The maximum occurs when \(3 + \cos \theta\) is minimized. In the interval, \(\cos \theta\) ranges from \(-1\) to \(1\), with minimum \(-1\) at \(\theta = \pi\).
Thus, minimum denominator is \(3 - 1 = 2\), so maximum \(f(\theta) = \frac{2}{2} = 1\).
Source: Myschool Past Questions Archive (Record #76328)