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Mathematics
UTME / JAMB 1995
Indices, Logarithms & Surds
Solution
Express the product of 0.0014 and 0.011 in standard form
A
1.54 x 10-2
B
1.54 x 10-3
C
1.54 x 10-4
D
1.54 x 10-5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(0.0014 = 1.4 \times 10^{-3}\)
\(0.011 = 1.1 \times 10^{-2}\)
\(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\)
= \(1.54 \times 10^{-5}\)
Source: Myschool Past Questions Archive (Record #30877)
Mathematics
UTME / JAMB 1995
Indices, Logarithms & Surds
Solution
Evaluate \(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)
A
27
B
1
C
\(\frac{1}{3}\)
D
\(\frac{1}{8}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\)
= \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\)
= \(\frac{3^{3} - 3^{1}}{24}\)
= \(\frac{27 - 3}{24} = 1\)
Source: Myschool Past Questions Archive (Record #30878)
Mathematics
UTME / JAMB 1995
Indices, Logarithms & Surds
Solution
Simplify \(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)
A
\(\frac{1}{3}\)
B
9
C
16cm
D
3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
\(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\)
= \(\frac{\sqrt{3}}{3\sqrt{3}}\)
= \(\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #30881)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
Four members of a school first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?
A
25
B
21
C
16
D
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Number of people playing both rugby and cricket = 4
Number that play cricket only = 11 - 4 = 7
Number that play rugby only = 14 - 4 = 10.
Number that play for at least one of the teams = 4 + 7 + 10 = 21.
Source: Myschool Past Questions Archive (Record #30882)
Mathematics
UTME / JAMB 1995
Sets, Logic & Venn Diagrams
If S = (x : x\(^2\) = 9, x > 4), then S is equal to
A
4
B
{0}
C
\(\emptyset\)
D
{\(\emptyset\)}
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #30884)
Mathematics
UTME / JAMB 1995
Indices, Logarithms & Surds
Solution
If x - 1 and x + 1 are both factors of the equation x\(^3\) + px\(^2\) + qx + 6 = 0, evaluate p and q
A
-6, -1
B
6, 1
C
1, -1
D
6, -6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x^{3} + px^{2} + qx + 6 = 0\)
f(x - 1) = 0; f(1) = 0.
\(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\)
f(x + 1) = 0; f(-1) = 0.
\((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\)
Subtract (2) from (1).
\(2q = -2 \implies q = -1\)
\(p - (-1) = -5 \implies p = -5 - 1 = -6\)
\((p, q) = (-6, -1)\)
Source: Myschool Past Questions Archive (Record #30885)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
Finds a positive value of p if the expression 2x2 - px + p leaves a remainder 6 when divided by x - p and q
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x - p,x = p
2p2 - p2 + p = 6
= p2 + p - 6
= 0
p = 3, 2
Source: Myschool Past Questions Archive (Record #30894)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
Find T in terms of K, Q, and S if S = 2r\(\sqrt{\pi(QT + K)}\)
A
\(\frac{S^2}{2 \pi r^2Q} - \frac{k}{Q}\)
B
\(\frac{S^2}{2 \pi r^2Q}\) - k
C
\(\frac{S^2}{4 \pi r^2Q} - \frac{k}{Q}\)
D
\(\frac{s^2}{4 \pi r^2Q}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{s^2}{4r^2}\) = \(\pi\)(QT + K)
\(\frac{s^2}{4r^2}\) - k\(\pi\) = QT\(\pi\)
T = \(\frac{s^2}{4Q\pi r^2}\) - \(\frac{k}{Q}\)
Source: Myschool Past Questions Archive (Record #30903)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points
A
(-6, 0), (-1, 0)
B
(-3, 0), (-2,0)
C
(-6, 0),(1, 0)
D
(2, 0), (3, 0)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
When X = 3, Y = 0(3, 0),
When x = 2, y = 0(2, 0)
Source: Myschool Past Questions Archive (Record #30904)
Mathematics
UTME / JAMB 1995
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize completely the expression \(abx^2 + 6y - 3ax - 2byx\)
A
(ax - 2y)(bx - 3)
B
(bx + 3)(2y - ax)
C
(bx + 3)(ax - 2y)
D
(ax - 2y)(ax - b)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(abx^{2} + 6y - 3ax - 2byx\)
Collecting like terms, we have
\(abx^{2} - 3ax + 6y - 2byx\)
= \(ax(bx - 3) + 2y(3 - bx)\)
= \(ax(bx - 3) - 2y(bx - 3)\)
= \((ax - 2y)(bx - 3)\)
Source: Myschool Past Questions Archive (Record #30905)
Mathematics
UTME / JAMB 1995
Inequalities & Linear Programming
Solution
Solve the inequality (x - 3)(x - 4) \(\leq\) 0
A
3 \(\leq\) x \(\leq\) 4
B
3 < x < 4
C
3 \(\leq\) x < 4
D
3 < x \(\leq\) 4
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(x - 3)(x - 4) \(\leq\) 0
Case 1 (+, -) = x - 3 \(\geq\) 0, X - 4 \(\geq\) 0
= X \(\leq\) 3, x \(\geq\) 4
= 3 < x \(\geq\) 4 (solution)
Case 2 = (-, +) = x - 3 \(\leq\) 0, x - 4 \(\geq\) 0
= x \(\leq\) 3, x \(\geq\) 4
therefore = 3 \(\leq\) x \(\leq\) 4
Source: Myschool Past Questions Archive (Record #30906)
Mathematics
UTME / JAMB 1995
Sequences & Series (AP & GP)
Solution
The 4th term of an A.P. is 13 while the 10th term is 31. Find the 21st term.
A
175
B
85
C
64
D
45
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
a + 3d = 13
a + 9d = 31
6d = 18
= d = 3
a = 13 - 9
= 4
a + 20d = 4 + (20 x 3)
= 64
Source: Myschool Past Questions Archive (Record #30910)
Mathematics
UTME / JAMB 1995
Linear, Simultaneous & Quadratic Equations
Solution
Simplify \(\frac{x^2 - 1}{x^3 + 2x^2 - x - 2}\)
A
\(\frac{1}{x + 2}\)
B
\(\frac{x - 1}{x + 1}\)
C
\(\frac{x - 1}{x + 2}\)
D
\(\frac{1}{x - 2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{x^2 - 1}{(x - 1)(x + 2)(x + 1)}\) = \(\frac{(x -
1)(x + 1)}{(x - 1)(x + 2)(x + 1)}\)
= \(\frac{1}{x + 2}\)
Source: Myschool Past Questions Archive (Record #30912)
Mathematics
UTME / JAMB 1995
Algebraic Expressions, Polynomials & Factorisation
Solution
Express \(\frac{5x - 12}{(x - 2)(x - 3)}\) in partial fractions
A
\(\frac{2}{x + 2} - \frac{3}{x - 3}\)
B
\(\frac{2}{x - 2} + \frac{3}{x - 3}\)
C
\(\frac{2}{x - 3} - \frac{3}{x - 2}\)
D
\(\frac{5}{x - 3} - \frac{4}{x - 2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{5x - 12}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}\)
= \(\frac{A(x - 3) + B(x - 2)}{(x - 2)(x - 3)}\)
\(\implies 5x - 12 = Ax - 3A + Bx - 2B\)
\(A + B = 5 ... (i)\)
\(-(3A + 2B) = -12 \implies 3A + 2B = 12 ... (ii)\)
From (i), \(A = 5 - B\)
\(3(5 - B) + 2B = 12\)
\(15 - 3B + 2B = 12 \implies B = 3\)
\(A + 3 = 5 \implies A = 2\)
\(\frac{5x - 12}{(x - 2)(x - 3)} = \frac{2}{x - 2} + \frac{3}{x - 3}\)
Source: Myschool Past Questions Archive (Record #30913)
Mathematics
UTME / JAMB 1995
Binary Operations, Matrices & Determinants
Solution
Which of the following binary operations is cumulative on the set of integers?
A
a \(\ast\) b = a + 2b
B
a \(\ast\) b = a + b - ab
C
a \(\ast\) b = a2 + b
D
a \(\ast\) b = \(\frac{a(b + 1)}{2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(a \ast b = a + b - ab\)
\(b \ast a = b + a - ba\)
On the set of integers, the two above are cumulative as multiplication and addition are cumulative on the set of integers.
Source: Myschool Past Questions Archive (Record #30916)