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Mathematics
UTME / JAMB 1982
Inequalities & Linear Programming
Solution
What is the least possible value of \(\frac{9}{1 + 2x^2}\) if 0 \(\geq\) x \(\geq\) 2?
A
9
B
5
C
1
D
2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
0 \(\geq\) x \(\geq\) 2 \(\to\) 0, 1, 2
If x = 0, \(\frac{9}{1 + 2x^2}\)
\(\frac{9}{1 + 2(0)^2}\) = \(\frac{9}{1}\)
= 3
If x = 2, \(\frac{9}{1 + 2(1)^2}\)
= \(\frac{9}{3}\)
= 3
If x = 2, \(\frac{9}{1 + 2(2)^2}\)
= \(\frac{9}{9}\)
= 1
The least value of \(\frac{9}{1 + 2x^2}\) is 1 when x = 2
Source: Myschool Past Questions Archive (Record #32491)
Mathematics
UTME / JAMB 1982
Fractions, Decimals, Percentages & Approximations
Solution
A man bought wrist watch for N150 but was only able to sell it for N120. Find the loss per cent on the transaction
A
25%
B
11%
C
20%
D
80%
E
30%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Loss% = \(\frac{\text{actual loss}}{\text{cost price}}\) x 100%
\(\frac{N150 - N120}{N150} \times 100%\)
\(\frac{N30}{N150} \times 100%\)
= 20%
Source: Myschool Past Questions Archive (Record #32492)
Mathematics
UTME / JAMB 1982
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the median of the set set of numbers 110, 116, 113, 119, 118, 127, 118, 117, 113
A
117.5
B
118
C
117
D
116
E
113
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Re-arrange in ascending order: 110, 113, 113, 116, |117|, 118, 119, 127
a= 117
Source: Myschool Past Questions Archive (Record #32493)
Mathematics
UTME / JAMB 1982
Plane Geometry, Polygons & Circle Theorems
Solution
If f(x - 2) = 3\(x^2\) + 4x + 1. Find (1).
A
8
B
40
C
7
D
24
E
32
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
f(x - 2) = 3\(x^2\) + 4x + 1
x - 2 = 1 then x = 3
When x = 3, f(1) = 3(3)2 + 4 x 3 = 1
27 + 12 +1 = 40
Source: Myschool Past Questions Archive (Record #32494)
Mathematics
UTME / JAMB 1982
Mensuration (Areas, Perimeters & Volumes)
Solution
A sector of a circle is bounded by two radii 7cm long and an arc length 6cm. Find the area of the sector.
A
42cm2
B
3cm2
C
21cm2
D
24cm2
E
12cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Length of arc = \(\frac{\theta}{360}\) x 2\(\pi\)r = 6
\(\theta\) x 2\(\pi\)r = 360 x 6
\(\theta\) = \(\frac{360 \times 6}{2\pi r}\)
Area of the sector = \(\frac{\theta}{360}\) x \(\pi\)r2
\(\frac{360 \times 6}{2\pi r}\) x \(\frac{1}{360}\) x \(\pi\)r2 = r
= 3 x 7
= 21cm2
Source: Myschool Past Questions Archive (Record #32495)
Mathematics
UTME / JAMB 1982
Plane Geometry, Polygons & Circle Theorems
Solution
Given that p:q = \(\frac{1}{3}\):\(\frac{1}{2}\) and q:r = \(\frac{2}{5}\): \(\frac{4}{7}\) find p:r
A
4:105
B
7:15
C
20:21
D
2:35
E
3:20
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(p : q = \frac{1}{3} : \frac{1}{2}\)
\(\frac{p}{q} = \frac{2}{3}\)
\(2q = 3p ... (1)\)
\(q : r = \frac{2}{5} : \frac{4}{7}\)
\(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\)
\(10q = 7r ... (2)\)
Eliminating q, we have
(1) : \(2q = 3p \)
\(10q = 15p\)
\(\implies 15p = 7r\)
\(\therefore \frac{p}{r} = \frac{7}{15}\)
\(p : r = 7 : 15\)
Source: Myschool Past Questions Archive (Record #32496)
Mathematics
UTME / JAMB 1982
Trigonometry, Elevation & Bearings
Solution
If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\)
A
\(\frac{(m^2 + n^2)}{m + n}\)
B
\(\frac{(m^2 + n^2 + 2mn)}{4mn}\)
C
\(\frac{2(m^2 + n^2 + mn)}{m + n}\)
D
\(\frac{(m^2 + n^2 + mn)}{m + n}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((m + n)^{2} = (m - n)^{2} + x^{2}\)
\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)
\(x^{2} = 4mn\)
\(x = \sqrt{4mn} = 2\sqrt{mn}\)
1 + tan2\(\theta\) = sec2\(\theta\)
= \(\frac{1}{cos^2\theta}\)
\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)
\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)
\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)
= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)
Source: Myschool Past Questions Archive (Record #32497)
Mathematics
UTME / JAMB 1982
Number Bases & Modular Arithmetic
Solution
The currency used in a country is called 'Matimalik'(M) and is of base seven. A lady in that country bought 4 bags of rice at M56 per bag and and 3 tins of milk at M4 per tin. What is the total cost of the item she bought?
A
M2467
B
M2427
C
M2367
D
M3417
E
M3387
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
4 bags of rice - M 56 each
3 tins of milk - M 4 each
\(M 56 \times 4 = M 323\)
\(M 4 \times 3 = M 15\)
\(M (323 + 15) = M 341\)
Source: Myschool Past Questions Archive (Record #32498)
Mathematics
UTME / JAMB 1982
Plane Geometry, Polygons & Circle Theorems
Solution
Solve for x, If \(\frac{\frac{2}{x}}{\frac{1}{p^2} + \frac{1}{q^2}}\) = m
A
\(\frac{4pq}{m(p + q)}\)
B
\(\frac{2p^2q^2}{m(q^2 + p^2)}\)
C
\(\frac{2pq}{m(q^2 + p^2)}\)
D
\(\frac{2p^2q^2}{m(p^2)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{1}{p^2}\) + \(\frac{1}{q^2}\) = \(\frac{q^2 + p^2}{p^2 + q^2}\)
\(\frac{\frac{2}{x}}{\frac{p^2 + q^2}{p^2 q^2}}\)
m = \(\frac{2p^2q^2}{x(p^2 + q^2)}\)
= 2p2q2 = mx(p2 + q2)
x = \(\frac{2p^2q^2}{m(q^2 + p^2)}\)
Source: Myschool Past Questions Archive (Record #32505)
Mathematics
UTME / JAMB 1982
Fractions, Decimals, Percentages & Approximations
Solution
Express 37.05 x 0.0042 in standard form
A
15.561 x 102
B
1.5561 x 10-4
C
1.556 x 10-1
D
1.5561 x 101
E
1.55 x 101
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
37.05 x 0.0042 in standard form
\(\begin{array}{c|c}No. & log \\\hline 37.05 & 1.5688\\ 0.0042 & 3.6232 \\ \hline & 1.1920\end{array}\)
= 0.1556
= 1.556 x 10-1
Source: Myschool Past Questions Archive (Record #32510)
Mathematics
UTME / JAMB 1982
Algebraic Expressions, Polynomials & Factorisation
Solution
Find the values of x for which the expression \(\frac{(x - 3)(x - 2)}{x^2 + x - 2}\) is undefined?
A
1, -2
B
-1, 2
C
2, 3
D
-1. -2
E
-2, -3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
to find the values of x for which the expression is undefined, let x\(^2\) + x - 2 = 0
By factorizing, we have (x + 2)(x - 1) = 0
when x + 2 = 0, when x - 1 = 0, x = -2 or x = 1
The two values are -2 and 1
Source: Myschool Past Questions Archive (Record #32513)
Mathematics
UTME / JAMB 1982
Trigonometry, Elevation & Bearings
Solution
A stone Q is tied to a point P vertically above Q by an inelastic string of length 2 meters. How high does the stone rise when the string is inclined at an angle 60o to the vertical?
A
It does not rise
B
2\(\sqrt{3}\) meters
C
3 meters
D
1 meters
E
3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Cos 60o = \(\frac{PD}{2}\)
Cos 60o x 2 = 0.5 x 2
= 1m
Source: Myschool Past Questions Archive (Record #32517)
Mathematics
UTME / JAMB 1982
Plane Geometry, Polygons & Circle Theorems
Solution
Express 130 kilometers per second in meters per hour
A
7.8 x 10-5
B
4.68 x 106
C
7,800,000
D
4.68 x 108
E
7.80 x 105
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
1km = 1000m
60sec. = 1 mins
60 mins. = 1 hr
130000m per sec = 130000 x 3600
= 468000000m/hr
= 468 x 108 m/hr
Source: Myschool Past Questions Archive (Record #32520)
Mathematics
UTME / JAMB 1982
Variations
Solution
The quantity y is partly constant and partly varies inversely as the square of x. With P and Q as constants, a possible relationship between x and y is
A
y = Q + \(\frac{P}{x^2}\)
B
y = Q + px
C
y = \(\frac{PQ}{x^2}\)
D
y = Q - \(\frac{P}{x^2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given the above statement,
\(y = Q + \frac{P}{x^{2}}\)
Source: Myschool Past Questions Archive (Record #32524)
Mathematics
UTME / JAMB 1982
Plane Geometry, Polygons & Circle Theorems
Solution
If \(\frac{3e + f}{3f - e}\) = \(\frac{2}{5}\), find the value of \(\frac{e + 3f}{f - 3e}\)
A
\(\frac{5}{2}\)
B
1
C
\(\frac{26}{7}\)
D
\(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{3e + f}{3f - e}\) = \(\frac{2}{5}\)
= 3e + f
= 2 x 1
\(\frac{-e + 3f}{3e - f}\) = \(\frac{5 \times 3}{2}\)
= \(\frac{3e + 9f = 15}{10f = 17}\)
f = \(\frac{17}{10}\)
Sub. for equ. (1)
3e + \(\frac{17}{10}\) = 2
3e = 2 - \(\frac{17}{10}\)
\(\frac{3}{10}\)
e = \(\frac{3}{10}\) x \(\frac{1}{3}\)
= \(\frac{1}{10}\)
= e + 3f = \(\frac{1}{10}\) + \(\frac{3 \times}{10}\) = \(\frac{52}{10}\)
f - 3e = \(\frac{17}{10}\) - 3 x \(\frac{1}{10}\)
= \(\frac{14}{10}\)
= \(\frac{52}{10}\) x \(\frac{10}{14}\)
= \(\frac{26}{7}\)
Source: Myschool Past Questions Archive (Record #32529)