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Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Solution
If a function is defined by f(x + 1) = 3x2 - x + 4, Find f(0).
A
4
B
6
C
9
D
8
E
2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
f(x + 1) = 3x2 - x + 4
f(0) = f(x + 1)
x + 1 = 0 ===> x = -1
f(0) = 3(-1)2 - (-1) + 4
f(0) = 3 + 1 + 4
= 8
Source: Myschool Past Questions Archive (Record #32640)
Mathematics
UTME / JAMB 1980
Mensuration (Areas, Perimeters & Volumes)
Solution
A cylindrical motor of height 12cm has uniform thickness of 2cm. If the diameter of its outer cross section is 10cm, Find the volume of the constituent material. (take \(\pi\) = \(\frac{22}{7}\)
A
\(\frac{6600}{7}\)cm3
B
270cm3
C
660cm3
D
\(\frac{4224}{7}\)cm3
E
\(\frac{1980}{7}\)cm3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
V = \(\pi\)r2h
= \(\frac{22}{7}\) x 52 x 12 - \(\frac{22}{7}\) x 32 x 12
= \(\frac{22}{7}\) x 12(52 - 32)
= \(\frac{4224}{7}\)
Source: Myschool Past Questions Archive (Record #32641)
Mathematics
UTME / JAMB 1980
Variations
If x varies inversely as y, and y varies directly as the square root of z, and z varies directly \(\frac{1}{w^2}\) write down in words how x varies with w
A
x varies inversely as w2
B
x varies directly as w2
C
x varies directly as w
D
x varies inversely as w
E
x varies directly as square root of w
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #32642)
Mathematics
UTME / JAMB 1980
Number Bases & Modular Arithmetic
Solution
Write the decimal number 39 to base 2
A
100111 base 2
B
110111 base 2
C
111001 base 2
D
100101 base 2
E
19.5 base 2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
To convert the decimal number 39 to base 2, we can use the method of successive division by 2 and record the remainders. Here’s the process:
1. Divide 39 by 2:
\(39 \div 2 = 19 \quad \text{remainder: } 1\)
2. Divide 19 by 2:
\(19 \div 2 = 9 \quad \text{remainder: } 1\)
3. Divide 9 by 2:
\(9 \div 2 = 4 \quad \text{remainder: } 1\)
4. Divide 4 by 2:
\(4 \div 2 = 2 \quad \text{remainder: } 0\)
5. Divide 2 by 2:
\(2 \div 2 = 1 \quad \text{remainder: } 0\)
6. Divide 1 by 2:
\(1 \div 2 = 0 \quad \text{remainder: } 1\)
Now, we read the remainders from bottom to top:
Thus, the binary representation of 39 is: 100111\(_2\)
Source: Myschool Past Questions Archive (Record #32643)
Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
If PN is perpendicular to QR, find the value of tan x.
Attached Figure / Diagram
A
\(\frac{5}{9}\)
B
\(\frac{3}{5}\)
C
\(\frac{3}{4}\)
D
\(\frac{4}{3}\)
E
\(\frac{5}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
By Pythagoras QN = 4
Since, QR = 9, and QN = 4, then, NR = 5
Tan x = \(\frac{3}{5}\)
Source: Myschool Past Questions Archive (Record #32743)
Mathematics
UTME / JAMB 1980
Fractions, Decimals, Percentages & Approximations
Diagram
Solution
In the figure, PQ\\SR, ST\\, ST\\RQ, PS = 7cm, PT = 7cm, SR = 4cm. Find the ratio of the area QRST to the area of PQRS.
Attached Figure / Diagram
A
56.77
B
56.105
C
28:105
D
28:49
E
56:49
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area of QRST = 4 x 7 = 28
Area of PQRS = \(\frac{1}{2}\)(4 + 11) x 7
= \(\frac{7}{2}\) x \(\frac{15}{1}\)
= 28:52.5
= 56:105
Source: Myschool Past Questions Archive (Record #32744)
Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, If PT is parallel to RS, PQ = PT, and angle SQT = 90o, Find x
Attached Figure / Diagram
A
35o
B
50o
C
55o
D
70o
E
80o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
from the diagram above, angle b = 180º - 55º - x ( 55º not 56º)
sum of angle at Q = y + 90º + 180º - 55º - x = 180º( sum of angles on a straight line)
y - x = -35 ..................................... eq1
sum of angle in triangle TPQ = 180º - x + 2y = 180º ( sum of angles in a triangle )
2y - x = 0 ............................................eq2
solving eq1 and eq 2 simultaneously
x = 70º
Source: Myschool Past Questions Archive (Record #32745)
Mathematics
UTME / JAMB 1980
Fractions, Decimals, Percentages & Approximations
Diagram
Solution
In the figure, PS bisects angle QPR. Find the ratio SR:QR.
Attached Figure / Diagram
A
1:2
B
1:3
C
1:4
D
1:5
E
1:6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From Internal bisection theorem, \(\frac{QP}{PR}\) = \(\frac{QS}{SR}\)
= \(\frac{4}{1}\)
= 1:4
Source: Myschool Past Questions Archive (Record #32746)
Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Diagram
fG is a given straight line and H is a fixed point. The construction marks shown in the diagram indicate the
Attached Figure / Diagram
A
perpendicular bisector of the line
B
perpendicular bisector of th eline JK
C
perpendicular bisector of the line HI
D
perpendicular from H to the line FG
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #32747)
Mathematics
UTME / JAMB 1980
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
In the figure, FGHJ is a circle of radius 3cm centre O. FOH, GOJ are perpendicular diameters. With G as centre of an arc of a circle is drawn to pass through F and H. Find the length of the perimeter of the lunar portion shaded.
Attached Figure / Diagram
A
3\(\pi\)\(\sqrt{2 - 1}\)cm
B
\(\frac{9}{2}\)\(\pi\)cm
C
3 \(\pi\)(1 + \(\frac{2}{2}\))
D
3(1 + \(\frac{2}{2}\))
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Perimeter of lunar portion = 3\(\pi\) + \(\frac{3\pi \sqrt{2}}{2}\)
= 3 \(\pi\)(1 + \(\frac{2}{2}\))
Source: Myschool Past Questions Archive (Record #32748)
Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
The angles marked in the figure are measured in degrees. Find x.
Attached Figure / Diagram
A
45o
B
90o
C
60o
D
15o
E
30o
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Sum of external angles = 360o
3x + 5x + x + 2x + x = 360o
12x = 360o
x = 30o
Source: Myschool Past Questions Archive (Record #32749)
Mathematics
UTME / JAMB 1980
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, PS = SR, and Rx is a tangent to the circle at R. < QRX is equal to 72º. Find angle SPR
Attached Figure / Diagram
A
20o
B
36o
C
72o
D
30o
E
15o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Source: Myschool Past Questions Archive (Record #32750)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90º, calculate \(\frac{tan(90 - \theta)}{cos^2\theta}\)
A
\(\frac{1}{2}\)
B
\(\frac{4}{\sqrt{3}}\)
C
\(\sqrt{3}\)
D
4\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\), then \(\theta\) = 60º
∴ \(\frac{tan(90 - \theta)}{cos^2\theta}\) = \(\frac{tan(90 - 60)}{cos^2 60}\) = \(\frac{tan30}{cos^2 60}\)
= \(\frac{1}{\sqrt3} \times (\frac{2}{1})^2 = \frac{4}{\sqrt{3}}\)
Source: Myschool Past Questions Archive (Record #32751)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Diagram
Solution
Find the area of the shades segments in the figure
Attached Figure / Diagram
A
\(\sqrt{3}
B
4 \(\pi - \sqrt{3}\)
C
-\(\frac{2}{3} \pi\)
D
\(\frac{2\pi}{3}\) -3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of section = \(\frac{60^o}{360^o}\) x 11r2
= \(\frac{60}{360} \times \pi \times 2^2\)
= \(\frac{1}{6}\) x 4
= \(\frac{4\pi}{6}\)
= \(\frac{2\pi}{3}\)
Area of triangle = \(\frac{1}{2x}\)
= 2 x 28.......60
Segment Area = Area of section - Area of triangle
= \(\frac{2\pi}{3}\) -3
Source: Myschool Past Questions Archive (Record #32752)
English Language
UTME / JAMB 1980
Tenses, Aspect & Conditionals
Solution
Fill in the blank spaces in the following sentences making use of the best of the five option:
The telephone .... in the nineteenth century and is now used in most countries in the world
A
had been invented
B
was invented
C
would have been invented
D
has been invented
E
was being invented
medium Difficulty
Verified Correct Answer: Option B
Grammar & Contextual Explanation
The sentence refers to a specific past event—the invention of the telephone in the nineteenth century. In such cases, the simple past tense ("was invented") is the most appropriate choice.
Source: Myschool Past Questions Archive (Record #1369)