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Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
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
Geometry and Trigonometry
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
Geometry and Trigonometry
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)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
Find x in the diagram below.
Attached Figure / Diagram
A
3\(\sqrt{3}\)
B
\(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
C
\(\frac{(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
D
\(\frac{3 - 1}{\sqrt{3} + 1}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{x + 3}{sin 60^o}\) = \(\frac{x}{sin 306o}\)
3sin 30o = x sin 60o - x sin 30o
= x(sin 60o - sin 30o)
but sin 30o = \(\frac{1}{2}\)
sin 60o = \(\frac{3}{2}\) = \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
Source: Myschool Past Questions Archive (Record #32741)
Mathematics
UTME / JAMB 1979
Geometry and Trigonometry
Diagram
Solution
The area under the speed time graph given the total distance covered by car. What is the average velocity to two places of decimals?
Attached Figure / Diagram
A
17.72 meter/sec.
B
21.67 meters/sec
C
2.5 meter/sec.
D
20.45 meters/sec.
E
13.33 meter/sec.
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Time(sec) areas of the three sides are \(\frac{1}{2}\) x 2 x 20 = 20 x 2 = 40
\(\frac{1}{2}\) x 4 x 20 = 40
40 + 40 = 80
vel. = \(\frac{80}{4}\)
20 + 0.45 = 20.45 meter/sec.
Source: Myschool Past Questions Archive (Record #32742)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
Evaluate without using tables sin(-1290º)
A
\(\frac{3}{2}\)
B
-\(\frac{3}{2}\)
C
\(\frac{2}{2}\)
D
1
E
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
sin(-1290º) = -sin(1290º)
sin([3*360] + 210)
where sin 360 = 0 and sin 210 = 180 + 30 ⇔ -30º
-sin ([3* 0] + [-30])
-sin(-30)
sin30 = \(\frac{1}{2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32724)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Diagram
Solution
In the Figure FD\\AC, the area of AEF = 6sq.cm. AE = 3cm, BC = 3cm, CD = 5cm, < BCD is an obtuse angle. Find the length of BD.
Attached Figure / Diagram
A
\(\sqrt{33}\)
B
6
C
4
D
2\(\sqrt{13}\)
E
4\(\sqrt{5}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of triangle AEF = 6sq. cm
area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines)
6 = \(\frac{1}{2}\) x 3 x h
6 = 3h
3h = \(\frac{12}{3}\)
h = 4 = DX
From D, C x D, CX\(^2\) = 5\(^2\)
- 4\(^2\) = 25 - 16 = 9
Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3)
BD\(^2\) = 4\(^2\) + 6\(^2\)
= 16 + 36 = 52
BD = \(\sqrt{4 \times 13}\)
= 2\(\sqrt{13}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32731)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Diagram
Solution
In the figure, 0 is the centre of the circle ABC, < CED = 30o, < EDA = 40o. What is the size of < ABC?
Attached Figure / Diagram
A
70o
B
110o
C
130o
D
125o
E
145o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
If < CED = 30º, and < EDA = 40º then
Source: WAEC / JAMB / NECO Classroom Archive (Record #32732)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
A rectangular picture 6cm by 8cm is enclosed by a frame \(\frac{1}{2}\)cm wide. Calculate the area of the frame
A
15sq.cm
B
20sq.cm
C
13sq.cm
D
16sq.cm
E
17sq.cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Area of the rectangular picture = L x B = 8 x 6
= 48 sq.cm.
Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm
area of the frame is 63 - 48
= 15 sq. cm
Source: Myschool Past Questions Archive (Record #32685)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
If the four interior angles of a quadrilateral are (p + 10)^\circ, (p - 30)^\circ, (2p - 45)^\circ, and (p + 15)^\circ, then p is
A
125o
B
82o
C
135o
D
105o
E
60o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Sum of interior angles of a polygon = (2n - 4) x 90
where n is the no. of sides
sum of interior angles of the quad. = ({2 x 4} -4) x 90
(8 - 4) x 90 = 4 x 90o
= 360o
(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o
p + 10o + p - 30o + 2p - 45o + p + 15 = 360o
5p = 360o + 50o
= 82o
Source: Myschool Past Questions Archive (Record #32699)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is
A
\(\sqrt{3}\) units
B
3 units
C
\(\sqrt{41 + 20 \sqrt{2}}\) units
D
\(\sqrt{41 - 20 \sqrt{2}}\) units
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Force to the east = 5 units and angle to the east = 0º
force to the Northeast = 4 units and angle to northeast = 45º
Resolve these forces into x and y components
FORCES
x - component
y - component
5
5cos0 = 5
5sin 0 = 0
4
4cos45 = 4 x \(\frac{\sqrt2}{2}\)
4sin45 = \(\frac{\sqrt2}{2}\)
Total force
5 + 2\(\sqrt{2}\)
0 + 2\(\sqrt{2}\)
Resultant force = \(\sqrt{(F_x)^2 + (F_y)^2}\)
= \(\sqrt{ (5 + 2\sqrt{2})^2 + (2\sqrt{2})^2}\)
= \(\sqrt{(25 + 20\sqrt{2} + 8) + 8}\)
= \(\sqrt{33 + 20\sqrt{2} + 8}\) = \(\sqrt{41 + 20\sqrt{2}}\)
Source: Myschool Past Questions Archive (Record #32701)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
The minimum point on the curve y = x2 - 6x + 5 is at
A
(1, 5)
B
(2, 3)
C
(-3, -4)
D
(3, -4)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given the curve \(y = x^{2} - 6x + 5\)
At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)
\(2x - 6 = 0 \implies x = 3\)
Since 3 > 0, it is a minimum point.
When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)
Hence, the turning point has coordinates (3, -4).
Source: Myschool Past Questions Archive (Record #32702)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
A canal has rectangular cross section of width10cm and breadth 1m. If water of uniform density 1 gm cm-3 flows through it at a constant speed of1000mm per minute, the adjacent sea is
A
100000
B
1000000
C
120000
D
30000
E
350000
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The canal's width = 10cm = 100mm (given)
The speed of water = 1000mm
10mm = 1cm
1000mm = 100cm
The adjacent sea must give speed x width
= 1000 x 100 = 100,000
Source: Myschool Past Questions Archive (Record #32704)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
A man runs a distance of 9km at a constant speed for the first 4 km and then 2 km\h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4 km is
A
6 km/h
B
8 km/h
C
9 km\h
D
11 km\h
E
13 km\h
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let the average speed for the first 4 km = x km/h.
Hence, the last 5 km, speed = (x + 2) km/h
Recall: \(Time = \frac{Distance}{Speed}\)
Total time = 1 hour.
\(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\)
\(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\)
\(9x + 8 = x^{2} + 2x\)
\(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\)
\(x^{2} - 8x + x - 8 = 0\)
\(x(x - 8) + 1(x - 8) = 0\)
\(x = -1; x = 8\)
Since speed cannot be negative, x = 8km/h.
Source: Myschool Past Questions Archive (Record #32705)
Mathematics
UTME / JAMB 1978
Geometry and Trigonometry
Solution
A triangle has angles 30o, 15o and 135o. The side opposite to the angle 30o is length 6cm. The side opposite to the angle 135o is equal to
A
12cm
B
6cm
C
6\(\sqrt{2}\)cm
D
12\(\sqrt{2}\)cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 135}\)
\(\frac{6}{\sin 30}\) = \(\frac{x}{\sin 45}\)
x = \(\frac{6 \times \sin 45}{\sin 30}\)
= \(6 \sqrt{2}\)cm
Source: Myschool Past Questions Archive (Record #32709)