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 9961–9975 of 14893 verified questions
Sort by:
Mathematics
UTME / JAMB 1994
Plane Geometry, Polygons & Circle Theorems
Solution
What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90o?
A
the perpendicular bisector of XY
B
a right-angled triangle
C
a circle
D
a semi circle
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Since XY is a fixed line and
XPY = 90o P is on one side of XY
P1P2P3......Pn are all possible cases where
XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o
Source: Myschool Past Questions Archive (Record #31040)
Mathematics
UTME / JAMB 1994
Plane Geometry, Polygons & Circle Theorems
Solution
If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q
A
P = 2, q = 4
B
p = 3, q = 1
C
p = 5, q = 3
D
p = 6, q = 2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{p + q}{2}\) = 4
p + q = 8 ....(i)
\(\frac{p - 2}{2}\) = q
p - 2q 2 ....(ii)
q = 2, p = 6
3q = 6
Source: Myschool Past Questions Archive (Record #31041)
Mathematics
UTME / JAMB 1994
Trigonometry, Elevation & Bearings
Solution
The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff?
A
\(\frac{5√3m}{3}\)
B
5√3m
C
10√3m
D
\(\frac{10√3m}{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
RQP = 90o
PQS = 30o but
RQS = 90o = \(\frac{RS}{10}\)
RS = 10 tan 60o
RS = 10 tan 60o
RS = 10√3m
Source: Myschool Past Questions Archive (Record #31042)
Mathematics
UTME / JAMB 1994
Trigonometry, Elevation & Bearings
Solution
What is the value of sin(-690)?
A
\(\frac{\sqrt{3}}{2}\)
B
-\(\frac{\sqrt{3}}{2}\)
C
\(\frac{-1}{2}\)
D
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Sin(-690o) = Sin(-360 -3300
sin - 360 = sin 0
∴ sin(-690o) = sin(330o)
Negative angles are measured in clockwise direction
The acute angle equivalent of sin(-330o) = sin(30o)
sin(-330o) = sin(30o)
= \(\frac{1}{2}\)
= 0.5
Source: Myschool Past Questions Archive (Record #31044)
Mathematics
UTME / JAMB 1994
Plane Geometry, Polygons & Circle Theorems
Solution
If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1
A
-1
B
1
C
-2
D
2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = 3t3 + 2t2 - 7t + 3
\(\frac{dy}{dt}\) = 9t2 + 4t - 7
When t = -1
\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
Source: Myschool Past Questions Archive (Record #31047)
Mathematics
UTME / JAMB 1994
Differentiation & Applications
Solution
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient
A
(1, 3)
B
(2, 7)
C
(0, 3)
D
(3, 15)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Equation of curve;
y = 2x2 - 2x + 3
gradient of curve;
\(\frac{dy}{dx}\) = differential coefficient
\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2
∴ \(\frac{dy}{dx}\) = 2
4x - 2 = 2
4x = 4
∴ x = 1
When x = 1, y = 2(1)2 - 2(1) + 3
= 2 - 2 + 3
= 5 - 2
= 3
coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)
Source: Myschool Past Questions Archive (Record #31056)
Mathematics
UTME / JAMB 1994
Integration & Applications
Solution
Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
A
3\(\frac{2}{3}\)
B
4
C
4\(\frac{1}{3}\)
D
4\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #31097)
Mathematics
UTME / JAMB 1994
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y - 1 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\), find y
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x
1
2
3
4
5
Total
f
y + 2
y - 1
2y - 3
y + 4
3y - 4
8y - 2
fx
y + 2
2y - 2
6y - 9
4y + 16
15y - 20
28y - 13
Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)
Source: Myschool Past Questions Archive (Record #31113)
Mathematics
UTME / JAMB 1994
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the mean deviation of the set of numbers 4, 5, 9
A
zero
B
2
C
5
D
6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x = \(\frac{\sum x}{N}\)
= \(\frac{18}{3}\)
= 6
\(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\)
M.D = \(\frac{|x - x|}{N}\)
= \(\frac{6}{3}\)
= 2
Source: Myschool Past Questions Archive (Record #31121)
Mathematics
UTME / JAMB 1994
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Class interval
1 - 5
6 - 10
11 - 15
16 - 20
21 - 25
Frequency
6
15
20
7
2
Estimate the median of the frequency distribution above
A
10\(\frac{1}{2}\)
B
11\(\frac{1}{2}\)
C
12
D
13
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5
= 11 + (\(\frac{25 - 21}{20}\))5
= 11 + (\(\frac{(4)}{20}\))
11 + 1 = 12
Source: Myschool Past Questions Archive (Record #31122)
Mathematics
UTME / JAMB 1994
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above
A
\(\frac{3}{2}\)
B
\(\frac{9}{4}\)
C
\(\frac{5}{2}\)
D
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{24}{8}\)
= 3
Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)
= \(\frac{18}{8}\)
= \(\frac{9}{4}\)
Source: Myschool Past Questions Archive (Record #31123)
Mathematics
UTME / JAMB 1994
Permutations, Combinations & Probability
Solution
In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?
A
\(\frac{1}{2}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{8}\)
D
\(\frac{3}{11}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
40 = 20 - x + x + 35 - x
40 = 55 - x
x = 55 - 40
= 15
∴ P(both) \(\frac{15}{40}\)
= \(\frac{3}{8}\)
Source: Myschool Past Questions Archive (Record #31124)
Mathematics
UTME / JAMB 1994
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
The equation of the graph is
Attached Figure / Diagram
A
y = (x - 3)3
B
y = (x + 3)3
C
y = x3 - 27
D
y = -x3 + 27
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = x3 - 27, y = -27 \(\to\) (0, -27)
when y = 0, x = 3 (3, 0)
Source: Myschool Past Questions Archive (Record #33031)
Mathematics
UTME / JAMB 1994
Inequalities & Linear Programming
Diagram
Solution
Find the inequality which represents the shaded portion in the diagram
Attached Figure / Diagram
A
2x - y - 2 \(\geq\) 0
B
2x - y - 2 \(\leq\) 0
C
2x - y - 2 < 0
D
2x - y - 2 > 0
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
2x - y - 2 \(\geq\) 0 = y \(\leq\) 2x - 2
when x = 0, y = -2, when y = 0, x = 1
Source: Myschool Past Questions Archive (Record #33033)
Mathematics
UTME / JAMB 1994
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the diagram, PQRS is a parallelogram. Find the value of < SQR
Attached Figure / Diagram
A
30o
B
50o
C
80o
D
100o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR
< STP = < RQV = 30o
But SQR + 30o + 50o = 180o
SQR = 180 - 80
= 100o
Source: Myschool Past Questions Archive (Record #33036)