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Mathematics
UTME / JAMB 1984
Linear, Simultaneous & Quadratic Equations
Solution
If (x - 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q
A
9
B
-3
C
3
D
\(\frac{17}{3}\)
E
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x3 + px2 + qx + 1 = (x - 1) Q(x) + R
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p = \(\frac{-3}{2}\)
q = \(\frac{-3}{2}\)
p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\)
= \(\frac{-6}{2}\)
= -3
Source: Myschool Past Questions Archive (Record #32431)
Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Solution
A cone is formed by bending a sector of a circle having an angle of 210o. Find the radius of the base of the cone if the diameter of the circle is 12cm.
A
12cm
B
7.00cm
C
1.75cm
D
21cm
E
3.50cm
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\)
= 6cm
\(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm
\(\frac{210}{360}\) = \(\frac{r}{6}\)
where r = radius of the base of the cone
V = \(\frac{1260}{360}\)
= 3.50cm
Source: Myschool Past Questions Archive (Record #32432)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
The sides of a triangle are(x + 4)cm, xcm and (x - 4)cm, respectively If the cosine of the largest angle is \(\frac{1}{5}\), find the value of x
A
24cm
B
20cm
C
28cm
D
7cm
E
\(\frac{88}{7}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
2 - a2 + c2 - 2a Cos B
Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\)
\(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\)
\(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\)
\(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\)
= 5(x - 16)
= 2x - 8
3x = 72
x = \(\frac{72}{3}\)
= 24
Source: Myschool Past Questions Archive (Record #32433)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
If a = \(\frac{2x}{1 - x}\) and b = \(\frac{1 + x}{1 - x}\), then a2 - b2 in the simplest form is
A
\(\frac{3x + 1}{x - 1}\)
B
\(\frac{3x^2 - 1}{(x - 1)}\)2
C
x\(\frac{3x - 2}{1 - x}\)
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2
= (\(\frac{2x}{1 - x} + \frac{1 + x}{1 - x}\))(\(\frac{2x}{1 - x} - \frac{1 + x}{1 - X}\))
= (\(\frac{3x + 1}{1 - x}\))(\(\frac{x - 1}{1 - x}\))
= \(\frac{3x + 1}{x - 1}\)
Source: Myschool Past Questions Archive (Record #32434)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2)
A
(x2 - 1)(x + 2)
B
x2(x + 2)
C
2 + 34
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\)
\((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\)
\((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)
Source: Myschool Past Questions Archive (Record #32435)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
If pq + 1 = q2 and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q
A
\(\frac{1}{p - q}\)
B
\(\frac{1}{q - 1}\)
C
\(\frac{1}{q + 1}\)
D
1 + 0
E
\(\frac{1}{1 - q}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Pq + 1 = q2......(i)
t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii)
p = \(\frac{q^2 - 1}{q}\)
Sub for p in equation (ii)
t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\)
t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\)
t = \(\frac{q - 1}{q^2 - 1}\)
= \(\frac{q - 1}{(q + 1)(q - 1)}\)
= \(\frac{1}{q + 1}\)
Source: Myschool Past Questions Archive (Record #32436)
Mathematics
UTME / JAMB 1984
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The cumulative frequency function of the data below is given below by the equation y = cf(x). What is cf(5)?
\(\begin{array}{c|c} Score(n) & Frequency(f)\\\hline 3 & 30\\ 4 & 32\\ 5 & 30\\ 6 & 35\\8 & 20 \end{array}\)
A
30
B
32
C
55
D
62
E
92
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If y = cf(x)
cf(5) = 30 + 32 + 30
= 92
Source: Myschool Past Questions Archive (Record #32437)
Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Solution
A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is
A
r tan yo cm
B
r sin yo cm
C
r cot yo cm
D
r cos yo cm
E
r cosec yo cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{r}{h}\) = tan yo
h = \(\frac{r}{tan y^o}\)
= r cot yo
Source: Myschool Past Questions Archive (Record #32438)
Mathematics
UTME / JAMB 1984
Permutations, Combinations & Probability
Solution
Two fair dice are rolled. What is the probability that both show up the same number of points.
A
\(\frac{1}{36}\)
B
\(\frac{7}{36}\)
C
\(\frac{1}{2}\)
D
\(\frac{1}{3}\)
E
\(\frac{1}{6}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #32439)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
The larger value of y for which (y - 1)2 = 4y - 7 is
A
2
B
4
C
6
D
7
E
8
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4
Source: Myschool Past Questions Archive (Record #32440)
Mathematics
UTME / JAMB 1984
Trigonometry, Elevation & Bearings
Solution
If sin \(\theta\) = \(\frac{x}{y}\) and 0o o then find \(\frac{1}{tan\theta}\)
A
\(\frac{x}{\sqrt{y^2 - x^2}}\)
B
\(\frac{y}{x}\)
C
\(\frac{\sqrt{y^2 + x^2}}{y^2 - x^2}\)
D
\(\frac{y^2 - x^2}{x}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\)
sin\(\theta\) = \(\frac{x}{y}\)
cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)
Source: Myschool Past Questions Archive (Record #32441)
Mathematics
UTME / JAMB 1984
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below
\(\begin{array}{c|c} \text{Class boundary in cm} & \text{frequency} & \\\hline 3.35 - 3.45 & 3\\ 3.45 - 3.55 & 6\\ 3.55 - 3.65 & 7\\ 3.65 - 3.75 & 4\end{array}\)
What is the mean diameter of the copper spheres?
A
3.40cm
B
3.58cm
C
3.56cm
D
3.62cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\begin{array}{c|c} \text{x(mid point)} & f & fx\\ \hline 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0\\3.6 & 7 & 25.2\\3.7 & 4 & 14.8\end{array}\)
\(\sum f\) = 20
\(\sum fx\) = 71.2
mean = \(\frac{\sum fx}{\sum f}\)
= \(\frac{71.2}{20}\)
= 3.56
Source: Myschool Past Questions Archive (Record #32442)
Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
What is the volume of this regular three dimensional figure?
Attached Figure / Diagram
A
160cm2
B
48cm2
C
120cm2
D
40cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Volume of the three dimensional figures = v = A x h
A = \(\frac{1}{2}\) x 4 x 3
= 6cm2
V = 6 x 8
= 48cm2
Source: Myschool Past Questions Archive (Record #32776)
Mathematics
UTME / JAMB 1984
Indices, Logarithms & Surds
Diagram
Solution
Find the area of the shaded portion of the semicircular figure.
Attached Figure / Diagram
A
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
B
\(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
C
\(\frac{1}{2}r^2 \pi\)
D
\(\frac{1}{8}r^2 \sqrt{3}\)
E
\(\frac{r^2}{4}(4 \pi - 3 \sqrt{3})\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Asector = \(\frac{60}{360} \times \pi r^2\)
= \(\frac{1}{6} \pi r^2\)
A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\)
\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)
A\(\text{shaded portion}\) = Asector -
A\(\bigtriangleup\)
= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)
= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)
= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
Source: Myschool Past Questions Archive (Record #32778)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS
Attached Figure / Diagram
A
60o
B
90o
C
120o
D
30o
E
80o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
From the diagram, PQRSTW is a regular hexagon.
Hexagon is a six sided polygon.
Sum of interior angles of polygon = (2n - 4)90o = [2 x 6 - 4] x 90 = 8 x 90 = 720o
each angle = \(\frac{720^o}{6} = 120^o\) and TVS = \(\frac{120}{2} = 60^o\)
Source: Myschool Past Questions Archive (Record #32779)