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 2941–2955 of 4910 verified questions
Sort by:
Mathematics
UTME / JAMB 2001
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize 4x2 - 9y2 + 20x + 25
A
(2x -3y + 5)(2x - 3y - 5)
B
(2x - 3y)(2x + 3y)
C
(2x - 3y +5)(2x + 3y + 5)
D
(2x + 5)(2x - 9y +5)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Given: 4x2 - 9y2 + 20x + 25
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
Source: Myschool Past Questions Archive (Record #1154)
Mathematics
UTME / JAMB 2001
Mensuration (Areas, Perimeters & Volumes)
Solution
A sector of a circle of radius 7.2cm which subtends an angle of 300^\circ at the centre is used to form a cone. What is the radius of the base of the cone?
A
8cm
B
6cm
C
9cm
D
7cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(r/L) = (\theta/360^\circ)
Given \theta = 300, and L = 7.2cm,
=> r = (300 x 7.2)/360
r = 6cm
Source: Myschool Past Questions Archive (Record #1155)
Mathematics
UTME / JAMB 2001
Trigonometry, Elevation & Bearings
Solution
A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30^\circ
A
4√3cm
B
8cm
C
8√3cm
D
4cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Hint: Make a sketch of an isosceles triangle with two of its sides and angles angles.
: PQ (r) = PR (q) = 8cm
: R^\circ = Q^\circ = 30^\circ
Sum of angles in a triangle = 180^\circ
P^\circ + Q^\circ + R^\circ = 180^\circ
P^\circ + 30^\circ + 30^\circ = 180^\circ
P^\circ = 180^\circ - 60^\circ
p^\circ = 120^\circ.
: PQ = r, PR = q, QR = p
Using sine rule:
\(\frac{p}{sinP}\) = \(\frac{q}{sinq}\)
\(\frac{p}{sin120^\circ}\) = \(\frac{8}{sin30^\circ}\)
cross multiply
p = \(\frac{8 X sin120^\circ}{sin30^\circ}\)
p = \(\frac{8 X \sqrt3/2 }{1/2}\)
p = 8\sqrt3
Source: Myschool Past Questions Archive (Record #1156)
Mathematics
UTME / JAMB 2001
Indices, Logarithms & Surds
Solution
A straight line makes an angle of 30^\circ with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.
A
y = (x/10) + 5
B
y = x + 5
C
√3y = - x + 5√3
D
√3y = x + 5√3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Cos 30 = \(\frac{5}{\text{x}}\)
x cos 30 = 5, => x = 5\(\sqrt{3}\)
Coordinates of P = - 5\(\sqrt{3}\), 0
Coordinates of Q = 0, 5
Gradient of PQ = (y\(_2\) - y\(_1\)) (X\(_2\) - X\(_1\)) = \(\frac{(5 - 0)}{(0 - 5\sqrt{3}}\))
= \(\frac{5}{5\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)
Equation of PQ = y - y\(_1\) = m (x -x\(_1\))
y - 0 = \(\frac{1}{\sqrt{3}}\) (x -( -5\(\sqrt{3}\)))
Thus: \(\sqrt{3}\)y = x + 5\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #1515)
Mathematics
UTME / JAMB 2001
Coordinate Geometry
Solution
Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3).
A
4
B
6
C
3
D
0
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The line joining (P, 4) and (6, -2).
Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\)
The line joining (2, P) and (-1, 3)
Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\)
For perpendicular lines, the product of their gradient = -1.
\((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\)
\(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\)
\(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\)
Source: Myschool Past Questions Archive (Record #1518)
Mathematics
UTME / JAMB 2001
Plane Geometry, Polygons & Circle Theorems
Solution
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
A
6
B
2
C
3
D
8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6
Source: Myschool Past Questions Archive (Record #1526)
Mathematics
UTME / JAMB 2001
Indices, Logarithms & Surds
Solution
P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
A
6.5 units
B
13.0 units
C
3.5 units
D
7.0 units
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
PQ\(^2\) = (x\(_2\) - x\(_1\))\(^2\) + (y\(_2\) - y\(_1\))\(^2\)
= 12\(^2\) + 5\(^2\)
= 144 + 25
= 169
PQ = \(\sqrt{169}\) = 13
But PQ = diameter = 2r, r = \(\frac{\text{PQ}}{2}\) = 6.5 units
Source: Myschool Past Questions Archive (Record #2121)
Mathematics
UTME / JAMB 2001
Trigonometry, Elevation & Bearings
Solution
The bearings of P and Q from a common point N are 020^\circ and 300^\circ respectively. If P and Q are also equidistant from N, find the bearing of P from Q.
A
040°
B
070°
C
280°
D
320°
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the diagram above,
Source: Myschool Past Questions Archive (Record #2122)
Mathematics
UTME / JAMB 2001
Mensuration (Areas, Perimeters & Volumes)
Solution
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?
(Take pi = 22/7)
A
23 m
B
25 m
C
20 m
D
22 m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Capacity = 3080 m3
Base diameter = 14 m, thus radius = 7 m.
Volume of cylinder = capacity of cylinder = πr2h. = 3080
22/7 x 72 x h = 3080
h = 3080/(22x7)
h = 20 m
Source: Myschool Past Questions Archive (Record #2123)
Mathematics
UTME / JAMB 2001
Plane Geometry, Polygons & Circle Theorems
Solution
Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k.
A
y = 4 \(\pm\) k
B
y = k \(\pm\) 4
C
y = 4 + k
D
y = k - 4
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The line y = 4 is a horizontal line in the Cartesian plane. The distance from any point( x, y) to this line is given by d = | y - 4| and this distance to be equal to k
| y - 4 | = k
case1, y - 4 = k, y = k + 4 ⇒ 4 + k
case2, y - 4 = - k, y = - k + 4 ⇒ 4 - k
combining the two solutions above, we have y = 4 \(\pm\) k
Source: Myschool Past Questions Archive (Record #2124)
Mathematics
UTME / JAMB 2001
Plane Geometry, Polygons & Circle Theorems
Solution
The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST.
A
πr/6
B
πr/2
C
πr/12
D
πr/3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Since ST = the radius r, => SOT is equilateral.
Each angle = 60°
Length of arc ST = θ/360 x 2πr
= 60/360 x 2 x π x r = πr/3
Source: Myschool Past Questions Archive (Record #2125)
Mathematics
UTME / JAMB 2001
Differentiation & Applications
Solution
If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k.
A
4
B
3
C
2
D
1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = 2kx2 + x + 1
Gradient function dy/dx = 4kx + 1
at dy/dx = 9, => 9 = 4kx + 1
=> 8 = 4kx, and k = 8/4x.
At x = 1, k = 8/4 = 2
Source: Myschool Past Questions Archive (Record #2126)
Mathematics
UTME / JAMB 2001
Integration & Applications
Solution
Evaluate \(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)
A
3/5(2x-3)5/3 + k
B
6/5(2x-3)5/3 + k
C
2x-3+k
D
2(2x-3)+k
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)
Let \(u = 2x - 3\)
\(\mathrm d u = 2 \mathrm d x\)
= \(\int u^{\frac{2}{3}} \mathrm d u\)
= \(\frac{u^{\frac{5}{3}}}{\frac{5}{3}} + k\)
= \(\frac{3}{5} (2x - 3)^{\frac{5}{3}} + k\)
Source: Myschool Past Questions Archive (Record #2127)
Mathematics
UTME / JAMB 2001
Differentiation & Applications
Solution
Differentiate \((2x+5)^{2} (x-4)\) with respect to x.
A
4(2x+5)(x-4)
B
4(2x+5)(4x-3)
C
(2x+5)(2x-13)
D
(2x+5)(6x-11)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(y = (2x + 5)^{2} (x - 4)\)
\(\frac{\mathrm d y}{\mathrm d x} = (2x + 5)^{2} (1) + (x - 4)(2)(2)(2x + 5)\)
= \((2x + 5)(2x + 5 + 4x - 16)\)
= \((2x + 5)(6x - 11)\)
Source: Myschool Past Questions Archive (Record #2128)
Mathematics
UTME / JAMB 2001
Integration & Applications
Solution
Find the area bounded by the curves y = 4 - x2 and y = 2x + 1
A
20(1/3) sq. units
B
20(2/3) sq. units
C
10(2/3) sq. units
D
10(1/3) sq. units
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Hint:
y = 4 - x\(^2\) and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = 1 or x = -3.
Integrating x\(^2\) + 2x - 3
= 3x - x\(^2\) - \(\frac{x^3}{3}\)
from (1, to -3) : 3 (1) - 1\(^2\) - \(\frac{1^3}{3}\) - 3 (-3) - -3\(^2\) - \(\frac{-3^3}{3}\)
= \(\frac{5}{3}\) + 9
will give 32/3 = 10 \(\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #2550)