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Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
An (n - 2)2 sided figure has n diagonals. Find the number n diagonals for 25-sided figure
A
7
B
8
C
9
D
10
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(n - 2)2 = 25
n - 2 = 25 = 5
n = 5 + 2
= 7
Source: Myschool Past Questions Archive (Record #31903)
Mathematics
UTME / JAMB 1987
Inequalities & Linear Programming
Solution
Solve the inequality x - 1 > 4(x + 2)
A
x > -3
B
x < -3
C
2 < x < 3
D
-3 < x < -2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8
= 3x < -9
∴ x < -3
Source: Myschool Past Questions Archive (Record #31909)
Mathematics
UTME / JAMB 1987
Differentiation & Applications
Solution
The minimum value of y in the equation y = x\(^2\) - 6x + 8 is
A
8
B
3
C
7
D
-1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = x\(^2\) - 6x + 8
\(\frac{dy}{dx}\) = 2x - 6
\(\frac{dy}{dx}\) = 0
2x - 6 = 0
x = 3
\(\therefore\) y = 3\(^2\) - 6(3) + 8
= 9 - 18 + 8
= -1
Source: Myschool Past Questions Archive (Record #31914)
Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
Find the eleventh term of the progression 4, 8, 16.....
A
213
B
212
C
211
D
210
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
a = 4, r = \(\frac{4}{2}\)
\(\frac{8}{4}\) = 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212
Source: Myschool Past Questions Archive (Record #31921)
Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles
A
210o
B
150o
C
105o
D
50o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Nonagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 1260^\circ; 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles \(\frac{150}{3}\) = 50o
Source: Myschool Past Questions Archive (Record #31924)
Mathematics
UTME / JAMB 1987
Mensuration (Areas, Perimeters & Volumes)
Solution
Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ
A
35o
B
37o
C
49o
D
59o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59^\circ
PRQ = QNP = 59^\circ(angles in the same segment of a circle are equal)
Source: Myschool Past Questions Archive (Record #31927)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
The sine, cosine and tangent of 210o are respectively
A
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
B
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{3}\)
C
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
D
\(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
210o = 180o - 210o = - 30o
From ratio of sides, sin -30o = -\(\frac{1}{2}\)
Cos 210o = 180o - 210o = -30o
= cos -30o = \(\frac{-3}{2}\)
But tan 30o = \(\frac{1}{\sqrt{3}}\), rationalizing this
= \(\frac{1}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{3}\)
∴ = \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
Source: Myschool Past Questions Archive (Record #31933)
Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
If tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) find sec\(\theta\)
A
\(\frac{m^2 + n^2}{(m^2 - n^2)}\)
B
\(\frac{m^2 + n^2}{2mn}\)
C
\(\frac{mn}{2(m^2 + n^2)}\)
D
\(\frac{m^2n^2}{2(m^2 - n^2)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\)
\(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 = \(\frac{m^2 + n^2}{2mn}\)
Source: Myschool Past Questions Archive (Record #31942)
Mathematics
UTME / JAMB 1987
Trigonometry, Elevation & Bearings
Solution
From two points x and y, 8m apart, and in line with a pole, the angles of elevation of the top of the pole are 30ºand 60º respectively. Find the height of the pole assuming that x, y, and the foot of the pole are on the same horizontal plane and x and y are on the same side of the pole.
A
\(\sqrt{6}\)
B
4\(\sqrt{3}\)
C
\(\sqrt{3}\)
D
\(\frac{12}{\sqrt{3}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From < ZWY
\(\frac{h}{a}\) = Tan60º
h = a tan 60º .............................1
From < XWZ
\(\frac{h}{ 8 + a }\) = tan 30º
h = ( 8 + a ) tan 30º ...................2
equate 1 and 2
(8 + a ) tan30º = a tan60º
\(\frac{( 8 + a )}{\sqrt3}\) = a\(\sqrt{3}\)
8 + a = 3a
2a = 8
a = 4
but h = a tan 60º
= 4\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #31950)
Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
A room is 12m long, 9m wide, and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room
A
\(\frac{15}{17}\)
B
\(\frac{8}{17}\)
C
\(\frac{8}{15}\)
D
\(\frac{12}{17}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
ABCD is the floor, by pythagoras
AC2 = 144 + 81 = \(\sqrt{225}\)
AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{15}{17}\)
Source: Myschool Past Questions Archive (Record #31962)
Mathematics
UTME / JAMB 1987
Mensuration (Areas, Perimeters & Volumes)
Solution
What is the circumference of latitude 0^\circS if R is the radius of the earth?
A
R cos\(\theta\)
B
2\(\pi\) R cos \(\theta\)
C
R sin \(\theta\)
D
2\(\pi\) R sin \(\theta\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Circumference of Latitude 0oS where R is the radius of the earth is = 2\(\pi\) R cos \(\theta\)
Source: Myschool Past Questions Archive (Record #31967)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4\(\sqrt3\)cm
A
6cm
B
5cm
C
4cm
D
3cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Base of pyramid of a square of side 8cm vertex directly above the centre edge = \(4\sqrt{3}\)cm
From the diagram, the diagonal of one base is AC\(^2\) = 8\(^2\) + 8\(^2\)
AC\(^2\) = 64 + 64 = 128
AC = \(8\sqrt{2}\)
but OC = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) x 8\(\sqrt{2}\) = \(4\sqrt{2}\)cm
OE = h = height
h\(^2\) = EC\(^2\) - OC\(^2\)
h\(^2\) = (\(4\sqrt{3}\))\(^2\) - (\(4\sqrt{2}\))\(^2\)
h\(^2\) = 16 x 3 - 16 x 2
48 - 32 = 16
h = \(\sqrt{16}\)
= 4cm
Source: Myschool Past Questions Archive (Record #31975)
Mathematics
UTME / JAMB 1987
Plane Geometry, Polygons & Circle Theorems
Solution
What is the locus of the mid-points of all chords of length 6cm within a circle of radius 5cm and with centre 0
A
A circle of radius 4 cm and with centre O
B
The perpendicular bisector of the chords
C
a straight line passing through centre O
D
A circle of radius 6 cm and with centre O
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The locus of the mid-point of all chords of length 6cm within a circle of radius 5cm and with centre O is the perpendicular bisectors of the chords.
Source: Myschool Past Questions Archive (Record #31978)
Mathematics
UTME / JAMB 1987
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day
A
187o30'
B
172o30'
C
135o225o
D
202o30' 157o 30'
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight = \(\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}\)
= \(\frac{810}{1440}\) x 360o
= \(\frac{291600}{1440}\)
= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'
Source: Myschool Past Questions Archive (Record #31988)
Mathematics
UTME / JAMB 1987
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The goals scored by 40 football teams from three league divisions are recorded below
No of goals
0
1
2
3
4
5
6
Frequency
4
3
15
16
1
0
1
What is the total number of goals scored by all the teams?
A
21
B
40
C
91
D
96
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let x rep. number of goals
x
f
fx
0
4
0
1
3
3
2
15
30
3
16
48
4
1
4
5
0
0
6
1
6
\(\sum fx\) = 91
Total number of goals scored is \(\sum fx\)= 91
Source: Myschool Past Questions Archive (Record #31990)