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Mathematics
UTME / JAMB 1984
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize 6x2 - 14x - 12
A
2(x + 3)(3x - 2)
B
6(x - 2)(x + 1)
C
2(x - 3)(3x + 2)
D
6(x + 2)(x - 1)
E
(3x - 4)(2x + 3)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
6x2 - 14x - 12 = 6x2 - 18x + 4x - 12
(3x + 2)(2x - 6)
= 3x(2x - 6) + 2(2x - 6)
= (3x + 2) 2(x - 3)
Source: Myschool Past Questions Archive (Record #32354)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
A straight line y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other
A
(5, 6)
B
(8, 8)
C
(8, 5)
D
(7,7)
E
(7, 5)
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
When y = 5, y = x2 - 12x + 40, becomes
x2 - 12x + 40 = 5
x2 - 12x + 40 - 5 = 0
x2 + 12x + 35 = 0
x2 - 7x - 5x + 35 = 0
x(x - 7) - 5(x - 7) = 0
= (x - 5)(x - 7)
x = 5 or 7
Source: Myschool Past Questions Archive (Record #32355)
Mathematics
UTME / JAMB 1984
Inequalities & Linear Programming
Solution
The table below is drawn for a graph y = x3 - 3x + 1
\(\begin{array}{c|c} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\\hline y = x^3 - 3x + 1 & 1 & -1 & 3 & 1 & -1 & 3 & 1\end{array}\)
From x = -2 to x = 1, the graph crosses the x-axis in the range(s)
A
-1 < x < 0 and 0 < x < 1
B
-2 < x < -1 and 0 < x < 1
C
-2 < x < -1 and -1 < x < 0
D
0 < x < 1
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1
Source: Myschool Past Questions Archive (Record #32360)
Mathematics
UTME / JAMB 1984
Inequalities & Linear Programming
Solution
In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?
A
-1 < x < x < 0
B
-3 < x < 3
C
0 \(\leq\) x < 18
D
15 < x < 18
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total distance covered by Musa in 2 hrs
= x + 10 + 5x
= 6x + 10
Ngozi = 118 km
If they are equal, 6x + 10 = 118
but 6x + 10 < 118
6x < 108
= x < 18
0 < x < 18 = 0 \(\leq\) x < 18
Source: Myschool Past Questions Archive (Record #32384)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
If 2x + 3y = 1 and x - 2y = 11, find (x + y)
A
5
B
-3
C
8
D
2
E
-2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
2x + 3y = 1 x 2.......(i)
x - 2y = 11 x 3.......(ii)
4x + 6y = 2........(iii)
3x - 6Y = 33........(4)
7x = 35
x = 5
Subt. for x
10 + 3y = 1
3y = -9
y = -3
x + y = 5 + -3
5 - 3 = 2
Source: Myschool Past Questions Archive (Record #32401)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?
A
x2 - 5x - 10 = 0
B
x2 - 20x + 360 = 0
C
x2 - 21x - 270 = 0
D
3x2 - 65x + 362 = 0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Tunde and Shola can do the work in 18 days.
Both will do the work in \(\frac{1}{18}\) days.
But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day.
Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day.
\(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\)
\(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\)
\(x^{2} + 15x = 36x + 270\)
\(x^{2} + 15x - 36x - 270 = 0\)
\(x^{2} - 21x - 270 = 0\)
Source: Myschool Past Questions Archive (Record #32421)
Mathematics
UTME / JAMB 1984
Indices, Logarithms & Surds
Solution
If f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4 and g(y) = \(\sqrt{5 + y}\), find g [f(3)] and f[g(4)].
A
3 and 4
B
-3 and 4
C
-3 and -4
D
3 and -4
E
0 and 5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
f(x) = 2(x - 3)\(^2\) + 3(x - 3) + 4
= (2 + 3) (x - 3) + 4
= 5(x - 3) + 4
= 5x - 15 + 4
= 5x - 11
f(3) = 5 x 3 - 11
= 4
g(f(3)) = g(4)
= \(\sqrt{5 + 4}\)
= \(\sqrt{9}\)
= 3
g(4) = 3
f(g(4)) = f(3)
= 4
g[f(3)] and f[g(4)] = 3 and 4 respectively.
Source: Myschool Past Questions Archive (Record #32424)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
Find a factor which is common to all three binomial expressions 4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2
A
4a + 6b
B
4a - 6b
C
2a + 3b
D
2a - 3b
E
none
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b)
8a3 + 27b3 = (2a)3 + (3b)3
= (2a + 3b)(4a - 6ab = 9a2)
(4a + 6b)2 = 2(2a + 3b)2
Source: Myschool Past Questions Archive (Record #32430)
Mathematics
UTME / JAMB 1984
Linear, Simultaneous & Quadratic Equations
Solution
If (x - 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q
A
9
B
-3
C
3
D
\(\frac{17}{3}\)
E
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x3 + px2 + qx + 1 = (x - 1) Q(x) + R
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p = \(\frac{-3}{2}\)
q = \(\frac{-3}{2}\)
p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\)
= \(\frac{-6}{2}\)
= -3
Source: Myschool Past Questions Archive (Record #32431)
Mathematics
UTME / JAMB 1984
Mensuration (Areas, Perimeters & Volumes)
Solution
A cone is formed by bending a sector of a circle having an angle of 210o. Find the radius of the base of the cone if the diameter of the circle is 12cm.
A
12cm
B
7.00cm
C
1.75cm
D
21cm
E
3.50cm
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\)
= 6cm
\(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm
\(\frac{210}{360}\) = \(\frac{r}{6}\)
where r = radius of the base of the cone
V = \(\frac{1260}{360}\)
= 3.50cm
Source: Myschool Past Questions Archive (Record #32432)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
The sides of a triangle are(x + 4)cm, xcm and (x - 4)cm, respectively If the cosine of the largest angle is \(\frac{1}{5}\), find the value of x
A
24cm
B
20cm
C
28cm
D
7cm
E
\(\frac{88}{7}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
2 - a2 + c2 - 2a Cos B
Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\)
\(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\)
\(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\)
\(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\)
= 5(x - 16)
= 2x - 8
3x = 72
x = \(\frac{72}{3}\)
= 24
Source: Myschool Past Questions Archive (Record #32433)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
If a = \(\frac{2x}{1 - x}\) and b = \(\frac{1 + x}{1 - x}\), then a2 - b2 in the simplest form is
A
\(\frac{3x + 1}{x - 1}\)
B
\(\frac{3x^2 - 1}{(x - 1)}\)2
C
x\(\frac{3x - 2}{1 - x}\)
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2
= (\(\frac{2x}{1 - x} + \frac{1 + x}{1 - x}\))(\(\frac{2x}{1 - x} - \frac{1 + x}{1 - X}\))
= (\(\frac{3x + 1}{1 - x}\))(\(\frac{x - 1}{1 - x}\))
= \(\frac{3x + 1}{x - 1}\)
Source: Myschool Past Questions Archive (Record #32434)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2)
A
(x2 - 1)(x + 2)
B
x2(x + 2)
C
2 + 34
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\)
\((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\)
\((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)
Source: Myschool Past Questions Archive (Record #32435)
Mathematics
UTME / JAMB 1984
Plane Geometry, Polygons & Circle Theorems
Solution
If pq + 1 = q2 and t = \(\frac{1}{p}\) - \(\frac{1}{pq}\) express t in terms of q
A
\(\frac{1}{p - q}\)
B
\(\frac{1}{q - 1}\)
C
\(\frac{1}{q + 1}\)
D
1 + 0
E
\(\frac{1}{1 - q}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Pq + 1 = q2......(i)
t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii)
p = \(\frac{q^2 - 1}{q}\)
Sub for p in equation (ii)
t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\)
t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\)
t = \(\frac{q - 1}{q^2 - 1}\)
= \(\frac{q - 1}{(q + 1)(q - 1)}\)
= \(\frac{1}{q + 1}\)
Source: Myschool Past Questions Archive (Record #32436)
Mathematics
UTME / JAMB 1984
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The cumulative frequency function of the data below is given below by the equation y = cf(x). What is cf(5)?
\(\begin{array}{c|c} Score(n) & Frequency(f)\\\hline 3 & 30\\ 4 & 32\\ 5 & 30\\ 6 & 35\\8 & 20 \end{array}\)
A
30
B
32
C
55
D
62
E
92
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If y = cf(x)
cf(5) = 30 + 32 + 30
= 92
Source: Myschool Past Questions Archive (Record #32437)