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 4846–4860 of 4910 verified questions
Sort by:
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
Simplify \(\frac{1 - x^2}{x - x^2}\), where x \(\neq\) 0
A
\(\frac{1}{x}\)
B
\(\frac{1 - x}{x}\)
C
\(\frac{1 + x}{x}\)
D
\(\frac{1}{x - 1}\)
E
\(\frac{-x - 1}{1}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{1 - x^2}{x - x^2}\), where x = \(\neq\) 0
\(\frac{1^2 - x^2}{x - x^2}\)
= \(\frac{(1 + x)(1 - x)}{x(1 - x)}\)
= \(\frac{1 + x}{x}\)
Source: Myschool Past Questions Archive (Record #32674)
Mathematics
UTME / JAMB 1979
Mensuration (Areas, Perimeters & Volumes)
Solution
A cylinder of height h and radius r is open at one end. Its surface area is
A
2\(\pi\)rh
B
\(\pi\)r2h
C
2\(\pi\)rh + \(\pi\)r2
D
2\(\pi\)rh + 2\(\pi\)r2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
A cylinder of height h ans radius r is open at one end, its surface area is 2\(\pi\)rh + \(\pi\)r2
Source: Myschool Past Questions Archive (Record #32675)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
What is the greatest straight line distance between two vertices (corners) of a cube whose sides are 2239cm long?
A
\(\sqrt{2239cm}\)
B
\(\sqrt{2}\) x 2239cm
C
\(\frac{\sqrt{3}}{2}\) 2239cm
D
\(\sqrt{3}\) x 2239cm
E
4478cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x = \(\sqrt{-2239^2 + 2239^2}\)
= -\(\sqrt{10026242}\)
= 3166.42
y = -\(\sqrt{10026242 + 5013121}\)
= -\(\sqrt{15039363}\)
= 3878
= \(\sqrt{3}\) x 2239
Source: Myschool Past Questions Archive (Record #32677)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
What is log\(_7{(49^a)}\) - log\(_{10}^{(0.01)}\)?
A
\(\frac{49^a}{100}\)
B
\(\frac{a}{2}\) + 2
C
72a + 2
D
2a + 2
E
\(\frac{2a}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
log7(49a) - log10(0.01) = log7(72)a - log10100
log772a - log101 - log10 102
= 2a - 2
= 2a + a
Source: Myschool Past Questions Archive (Record #32678)
Mathematics
UTME / JAMB 1979
Fractions, Decimals, Percentages & Approximations
Solution
The size of a quantity first doubles and then increases by a further 16%. After a short time its size decreases by 16%. What is the net increases in size of the quantity?
A
\(\frac{59300}{625}\)
B
\(\frac{50900}{625}\)
C
200%
D
+100%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let x rept. the size of the quantity
2x + \(\frac{116}{100}\) x \(\frac{-84}{100}\)
= 200%
Source: Myschool Past Questions Archive (Record #32679)
Mathematics
UTME / JAMB 1979
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The following table relates the number of objects f corresponding to a certain size x. What is the average size of an object?
\(\begin{array}{c|c} f & 1 & 2 & 3 & 4 & 5 \\ \hline x & 1 & 2 & 4 & 8 & 16\end{array}\)
A
\(\frac{31}{15}\)
B
\(\frac{31}{5}\)
C
\(\frac{129}{5}\)
D
\(\frac{43}{5}\)
E
\(\frac{16}{5}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
F = 1, 2, 3, 4, 5
x = 1, 2, 4, 8, 16
fx = 1, 4, 12, 32, 80, 3f = 15
(average size) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{129}{15}\)
= \(\frac{43}{5}\)
Source: Myschool Past Questions Archive (Record #32680)
Mathematics
UTME / JAMB 1979
Linear, Simultaneous & Quadratic Equations
Solution
If y = x\(^2\) - 2x - 3, find the least value of y and the corresponding value of x
A
x = 3, y = 3
B
x = 1, y = -3
C
x = 4, y = 1
D
x = 1, y = -4
E
x = 2, y = -3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
If the graph of eqn. y = x\(^2\) - 2x - 3 is plot and drawn, the lease value of y = -4 while the corresponding value of x = 1
x = 1, y = -4
OR
\(\frac{dy}{dx}\) = 0 = 2x - 2
2x = 2
x = 1
put x = 1 in y = 2x\(^2\) - 2x - 3
y = - 4
x= 1, y = -4
Source: Myschool Past Questions Archive (Record #32681)
Mathematics
UTME / JAMB 1979
Plane Geometry, Polygons & Circle Theorems
Solution
A father is now three times as old as his son. Twelve years ago he was six times as old as his son. How old are the son and the father?
A
20 and 45
B
100 and 150
C
45 and 65
D
35 and 75
E
20 and 60
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Son = x, Father = 3x, 12 yrs ago
son = x - 12, father = 3x - 12
3x - 12 = 6(x - 12)
3x - 12 = 6x - 72
3x = 60
x = 20
Source: Myschool Past Questions Archive (Record #32682)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
If \(\sqrt{3^{\frac{1}{x}}}\) = \(\sqrt{9}\) then the value of x is:
A
\(\frac{3}{4}\)
B
\(\frac{4}{3}\)
C
\(\frac{1}{3}\)
D
\(\frac{2}{3}\)
E
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(\sqrt{3^{\frac{1}{x}}}\) = \(\sqrt{9}\)
3\(^{\frac{1}{2x}}\) = \(9^{\frac{1}{2}}\)
3\(^{\frac{1}{2x}}\) = 3\(^{2 \times \frac{1}{2}}\)
3\(\frac{1}{2x}\) = 3\(\frac{2}{2}\) = 3
\(3^{\frac{1}{2x}}\) = \(3^{1}\)
\(\frac{1}{2x}\) = \(\frac{1}{1}\)
x = \(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #32683)
Mathematics
UTME / JAMB 1979
Trigonometry, Elevation & Bearings
Solution
A man is standing in the corridor of a 10-storey building and looking down at a tall tree in front of the building. He sees the top of the tree at angle of depression of 30o. If the tree is 200m tall and the man's eyes are 300m above the ground, calculate the angle of depression of the foot tree as seen by the man
A
30o
B
60o
C
45o
D
25o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let x rep. the angle of depression of the foot of the tree.
tan 30o = \(\frac{y}{100}\)
y = 100 tan 30o
= 57.8
By Pythagoras, AC2 = 3002 + 582
= 900 + 3354
tan x = \(\frac{opp}{adj}\)
= \(\frac{58}{300}\)
= 0.19
tan x = 0.19
x = tan 0.19
= 60o
Source: Myschool Past Questions Archive (Record #32684)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Diagram
Solution
In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and
Attached Figure / Diagram
A
3\(\sqrt{2}\)
B
2\(\sqrt{3}\)
C
\(\frac{\sqrt{3}}{2}\)
D
\(\frac{2}{\sqrt{3}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
BC = 6 : DC = \(\frac{6}{2}\) = 3cm
By construction < EDE = 180o(90o + 60o) = 180o - 150o
= 30o(angle on a strt. line)
From rt < triangle ADC, AD2 = 52 - 32
= 25 - 9 = 6
AD = 4
From < AEC, let AS = x
\(\frac{x}{sin 60^o}\) - \(\frac{4}{sin 90^o}\)
sin 90o = 1
sin 60o = \(\frac{\sqrt{3}}{2}\)
x = 4sin 60o
x = 3 x \(\frac{\sqrt{3}}{2}\)
= 2\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #32733)
Mathematics
UTME / JAMB 1979
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of
Attached Figure / Diagram
A
125o
B
120o
C
115o
D
110o
E
10o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Both pairs of opp. angles are equal
< STP = 55\(^o\) - isosceles angle
< TSR = 55\(^o\) - alternate angle to < STP
Hence, < PSR = 55\(^o\) + 55\(^o\) = 110\(^o\)
\(\therefore\) < PQR = 110\(^o\)
Source: Myschool Past Questions Archive (Record #32734)
Mathematics
UTME / JAMB 1979
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
If O is the centre of the circle, < POS equls
Attached Figure / Diagram
A
70o
B
75o
C
105o
D
140o
E
150o
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Since O is the centre of the circle o
i.e. < substended at the centre is twice that substended at any part of the circumference
Source: Myschool Past Questions Archive (Record #32735)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Diagram
Solution
(Numbers indicate the lengths of the sides of the triangles) If the area of \(\bigtriangleup\) PQR is k2sq. units what is the area of the shades portion?
Attached Figure / Diagram
A
\(\frac{5}{9}\)k2 sq. units
B
\(\frac{1}{3}\)k2 sq. units
C
\(\frac{8}{9}\)k2 sq. units
D
\(\frac{7}{9}\)k2 sq. units
E
\(\frac{2}{3}\)k2 sq. u
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of shaded portion = Area of triangle PQR - Area of inner triangle
Area of triangle given 3 sides a, b, c = \(\sqrt{s(s - a)(s - b)(s - c)}\)
where \(s = \frac{a + b + c}{2} \)
Area of PQR :
\(s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7\)
Area = \(\sqrt{7(7 - 3)(7 - 5)(7 - 6)}\)
= \(\sqrt{7(4)(2)(1)} = \sqrt{56}\)
\(\implies K^{2} = \sqrt{56}\)
Area of inner triangle :
\(s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}\)
Area = \(\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}\)
= \(\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}\)
= \(\sqrt{\frac{896}{81}}\)
= \(\sqrt{\frac{16}{81}} \times \sqrt{56}\)
= \(\frac{4}{9} K^{2}\)
\(\therefore \text{The area of the shaded portion} = K^{2} - \frac{4}{9}K^{2} = \frac{5}{9}K^{2}\)
Source: Myschool Past Questions Archive (Record #32738)
Mathematics
UTME / JAMB 1979
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
PQ is parallel to RS. Calculate the value of x.
Attached Figure / Diagram
A
20o
B
40o
C
60o
D
80o
E
100o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< D = 180º - 100º
= 80º (< on a str. line)
< S = 60º - alternate angle
x = 180º - (80º + 60º)
180º - 140º = 40º
Source: Myschool Past Questions Archive (Record #32740)