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Mathematics
UTME / JAMB 1997
Plane Geometry, Polygons & Circle Theorems
Solution
Each of the base angles of a isosceles triangle is 58^\circ and the verticles of the triangle lie on a circle. Determine the angle which the base of the triangle subtends at the centre of the circle.
A
128o
B
16o
C
64o
D
58o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Base angle of the triangle = 58^\circ.
The third angle = 180^\circ - (58^\circ + 58^\circ)
= 64^\circ
Angle subtended at the centre = \(2 \times 64^\circ = 128^\circ\) (angle at the centre is twice the angle at the circumference).
Source: Myschool Past Questions Archive (Record #30822)
Mathematics
UTME / JAMB 1997
Plane Geometry, Polygons & Circle Theorems
Solution
A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc
A
22cm
B
44cm
C
110cm
D
220cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Diameter = 42cm
Length of the arc = \(\frac{\theta}{360^\circ} \times \pi d\)
= \(\frac{60}{360} \times \frac{22}{7} \times 42cm\)
= \(22cm\)
Source: Myschool Past Questions Archive (Record #30823)
Mathematics
UTME / JAMB 1997
Mensuration (Areas, Perimeters & Volumes)
Solution
An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(\(\pi\) = \(\frac{22}{7}\))
A
22cm2
B
44cm2
C
66cm2
D
88cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of a sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\)
r = 6cm; \(\theta\) = 70^\circ.
Area of the sector = \(\frac{70}{360} \times \frac{22}{7} \times 6^{2}\)
= \(22 cm^{2}\)
Source: Myschool Past Questions Archive (Record #30824)
Mathematics
UTME / JAMB 1997
Mensuration (Areas, Perimeters & Volumes)
Solution
A cone with a sector angle of 45^\circ is cut out of a circle of radius r cm. Find the base radius of the cone.
A
\(\frac{r}{16}\) cm
B
\(\frac{r}{6}\) cm
C
\(\frac{r}{8}\) cm
D
\(\frac{r}{2}\) cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360^\circ}\)
= \(\frac{r \times 45^\circ}{360^\circ}\)
= \(\frac{r}{8} cm\)
Source: Myschool Past Questions Archive (Record #30825)
Mathematics
UTME / JAMB 1997
Coordinate Geometry
Solution
The angle between the positive horizontal axis and a given line is 135o. Find the equation of the line if it passes through the point (2,3)
A
x - y = 1
B
x + y = 1
C
x + y = 5
D
x - y = 5
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
m = tan 135o = -tan 45o = -1
\(\frac{y - y_1}{x - x_1}\) = m
\(\frac{y - 3}{x - 2}\) = -1
= y - 3 = -(x - 2)
= -x + 2
x + y = 5
Source: Myschool Past Questions Archive (Record #30833)
Mathematics
UTME / JAMB 1997
Indices, Logarithms & Surds
Solution
A point P moves so that its equidistant from point L and M. If LM is16cm, find the distance of P from LM when P is 10cm from L
A
12cm
B
10cm
C
8cm
D
6cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
p from LM = \(\sqrt{10^2 - 8^2}\)
= \(\sqrt{36}\) = 6cm
Source: Myschool Past Questions Archive (Record #30836)
Mathematics
UTME / JAMB 1997
Indices, Logarithms & Surds
Solution
Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2
A
3√10
B
3√5
C
√26
D
√13
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
Source: Myschool Past Questions Archive (Record #30840)
Mathematics
UTME / JAMB 1997
Trigonometry, Elevation & Bearings
Solution
The angle of elevation of a building from a measuring instrument placed on the ground is 30o. If the building is 40m high, how far is the instrument from the foot of the building?
A
\(\frac{20}{√3}\)m
B
\(\frac{40}{√3}\)m
C
20√3m
D
40√3m
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{40}{x}\) = tan 30o
x = \(\frac{40}{tan 30}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40\sqrt3m
Source: Myschool Past Questions Archive (Record #30850)
Mathematics
UTME / JAMB 1997
Indices, Logarithms & Surds
Solution
In a triangle XYZ, if < XYZ is 60º, XY = 3cm, and YZ = 4cm, calculate the length of the side XZ.
A
√23cm
B
√13cm
C
2√5cm
D
2√3cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(XZ)\(^2\) = 3\(^2\) + 4\(^2\) - 2 x 3 x 4 cos60º
= 25 - 24\(\frac{1}{2}\)
XZ = \(\sqrt{13}\)cm
Source: Myschool Past Questions Archive (Record #30851)
Mathematics
UTME / JAMB 1997
Differentiation & Applications
Solution
Differentiate \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x
A
\(\frac{2 + 2}{3x^3}\)
B
2 + \(\frac{1}{6x}\)
C
2 - \(\frac{2}{3x^3}\)
D
\(\frac{1}{5}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{6x^3 - 5x^2 + 1}{3x^2}\)
let y = 3x2
y = \(\frac{6x^3}{3x^2}\) - \(\frac{6x^2}{3x^2}\) + \(\frac{1}{3x^2}\)
Y = 2x - \(\frac{5}{3}\) + \(\frac{1}{3x^2}\)
\(\frac{dy}{dx}\) = 2 + \(\frac{1}{3}\)(-2)x-3
= 2 - \(\frac{2}{3x^3}\)
Source: Myschool Past Questions Archive (Record #30852)
Mathematics
UTME / JAMB 1997
Linear, Simultaneous & Quadratic Equations
Solution
\(\frac{d}{dx}\) cos(3x\(^2\) - 2x) is equal to
A
-sin(6x - 2)
B
-sin(3x2 - 2x)dx
C
(6x - 2) sin(3x2 - 2x)
D
-(6x - 2)sin(3x2 - 2x)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let \(3x^{2} - 2x = u\)
\(y = \cos u \implies \frac{\mathrm d y}{\mathrm d u} = - \sin u\)
\(\frac{\mathrm d u}{\mathrm d x} = 6x - 2\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (6x - 2) . - \sin u\)
= \(- (6x - 2) \sin (3x^{2} - 2x)\)
Source: Myschool Past Questions Archive (Record #30854)
Mathematics
UTME / JAMB 1997
Integration & Applications
Solution
Integrate \(\frac{1}{x}\) + cos x with respect to x
A
-\(\frac{1}{x^2}\) + sin x + k
B
x + sin x - k
C
x - sin x + k
D
-\(\frac{1}{x^2}\) - sin x + k
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int \frac{1}{x} + \cos x = ln x - \sin x + k\)
Source: Myschool Past Questions Archive (Record #30856)
Mathematics
UTME / JAMB 1997
Integration & Applications
Solution
If \(y = x(x^4 + x + 1)\), evaluate \(\int \limits_{0} ^{1} y \mathrm d x\).
A
\(\frac{11}{12}\)
B
1
C
\(\frac{5}{6}\)
D
zero
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(y = x(x^{4} + x + 1) = x^{5} + x^{2} + x\)
\(\int \limits_{0} ^{1} (x^{5} + x^{2} + x) \mathrm d x = \frac{x^{6}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2}\)
= \([\frac{x^{6}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2}]_{0} ^{1}\)
= \(\frac{1}{6} + \frac{1}{3} + \frac{1}{2}\)
= \(1\)
Source: Myschool Past Questions Archive (Record #30857)
Mathematics
UTME / JAMB 1997
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Ages
20
25
30
35
40
45
Number of people
3
5
1
1
2
3
Find the median age of the frequency distribution in the table above.
A
20
B
25
C
30
D
35
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Total freq = 15
Median position = \(\frac{n + 1}{2}\) for odd numbers
Median position = \(\frac{15 + 1}{2}\) = \(\frac{16}{2}\) = 8th position.
From the table, the 8th position corresponds to age 25.
Source: Myschool Past Questions Archive (Record #30859)
Mathematics
UTME / JAMB 1997
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the difference between the range and the variance of the following set of numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7 where \(\sum d^2\) = 60
A
2
B
3
C
4
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Range : 10 - 2 = 8
Variance = \(\frac{\sum d^{2}}{n}\)
= \(\frac{60}{10}\)
= 6
8 - 6 = 2.
Source: Myschool Past Questions Archive (Record #30863)