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Mathematics
UTME / JAMB 1992
Indices, Logarithms & Surds
Solution
Simplify 5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\)
A
17\(\sqrt{4}\)
B
4\(\sqrt{17}\)
C
17\(\sqrt{2}\)
D
12\(\sqrt{4}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
5\(\sqrt{18}\) - 3\(\sqrt{72}\) + 4\(\sqrt{50}\) = 5(3\(\sqrt{2}\)) - 3(6\(\sqrt{2}\)) + 4(5\(\sqrt{2}\))
15\(\sqrt{2}\) - 18\(\sqrt{2}\) + 20\(\sqrt{2}\) = 35\(\sqrt{2}\) - 18\(\sqrt{2}\)
= 17\(\sqrt{2}\)
Source: Myschool Past Questions Archive (Record #31296)
Mathematics
UTME / JAMB 1992
Indices, Logarithms & Surds
Solution
If x = 3 - \(\sqrt{3}\), find x2 + \(\frac{36}{x^2}\)
A
9
B
18
C
24
D
27
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x = 3 - \(\sqrt{3}\)
x2 = (3 - \(\sqrt{3}\))2
= 9 + 3 - 6\(\sqrt{34}\)
= 12 - 6\(\sqrt{3}\)
= 6(2 - \(\sqrt{3}\))
∴ x2 + \(\frac{36}{x^2}\) = 6(2 - \(\sqrt{3}\)) + \(\frac{36}{6(2 - \sqrt{3})}\)
6(2 - \(\sqrt{3}\)) + \(\frac{6}{2 - \sqrt{3}}\) = 6(- \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}\)
= 6(2 - \(\sqrt{3}\)) + \(\frac{6(2 + \sqrt{3})}{4 - 3}\)
6(2 - \(\sqrt{3}\)) + 6(2 + \(\sqrt{3}\)) = 12 + 12
= 24
Source: Myschool Past Questions Archive (Record #31297)
Mathematics
UTME / JAMB 1992
Sets, Logic & Venn Diagrams
Solution
If x = (all prime factors of 44) and y = (all prime factors of 60), the elements of X ∪ Y and X ∩ Y respectively are
A
(2, 4, 3, 5, 11) and (4)
B
(4, 3, 5, 11) and (3, 4)
C
(2, 5, 11) and (2)
D
(2, 3, 5, 11) and (2)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x = (all prime factors of 44) and y = (all prime factors of 60)
∴ x = (2, 11), y = (2, 3, 5)
X ∪ Y = (2, 3, 5, 11),
X ∩ Y = (2)
Source: Myschool Past Questions Archive (Record #31304)
Mathematics
UTME / JAMB 1992
Sets, Logic & Venn Diagrams
Solution
If U = (1, 2, 3, 6, 7, 8, 9, 10) is the universal set. E = (10, 4, 6, 8, 10) and F = {x: 1x\(^2\) = 2\(^6\), x is odd}. Find (E ∩ F)', where ' means the complement of a set.
A
(0)
B
U
C
(8)
D
\(\phi\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
U = (1, 2, 3, 6, 7, 8, 9, 10)
E = (10, 4, 6, 8, 10)
F = (x : x\(^2\) = 2\(^6\), x is odd)
∴ F = \(\phi\) Since x\(^2\) = 2\(^6\) = 64
x = \(\pm 8\) which is even
∴ E ∩ F = \(\phi\) Since there are no common elements
Since E ∩ F = \(\phi\), then ( E ∩ F)' = U = (1, 2, 3, 6, 7, 8, 9, 10)
Source: Myschool Past Questions Archive (Record #31306)
Mathematics
UTME / JAMB 1992
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize \(9p^2 - q^2 + 6qr - 9r^2\)
A
(3p - 3q + r)(3p - q - 3r)
B
(6p - 3q - 3r)(3p - q - 4r)
C
(3p - q + 3r)(3p + q - 3r)
D
(3q - p + 3r)(3q - p + 3r)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(9p^{2} - q^{2} + 6qr - 9r^{2}\)
= \(9p^{2} - (q^{2} - 6qr + 9r^{2})\)
= \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\)
= \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\)
= \(9p^{2} - (q - 3r)^{2}\)
= \((3p + (q - 3r))(3p - (q - 3r))\)
= \((3p + q - 3r)(3p - q + 3r)\)
Source: Myschool Past Questions Archive (Record #31307)
Mathematics
UTME / JAMB 1992
Linear, Simultaneous & Quadratic Equations
Solution
Solve the equation: \(y - 11\sqrt{y} + 24 = 0\)
A
8, 3
B
64, 9
C
6, 4
D
9, -8
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(y - 11\sqrt{y} + 24 = 0 \implies y + 24 = 11\sqrt{y}\)
Squaring both sides,
\(y^{2} + 48y + 576 = 121y\)
\(y^{2} + 48y - 121y + 576 = 0 \implies y^{2} - 73y + 576 = 0\)
\(y^{2} - 64y - 9y + 576 = 0\)
\(y(y - 64) - 9(y - 64) = 0\)
\((y - 9)(y - 64) = 0\)
\(\therefore \text{y = 64 or y = 9}\)
Source: Myschool Past Questions Archive (Record #31308)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
Make t the subject of formula S = ut + \(\frac{1}{2} at^2\)
A
\(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
B
\(\frac{1}{a}\) {u \(\pm\) (U2 - 2as)}
C
\(\frac{1}{a}\) {u \(\pm\) \(\sqrt{2as}\)}
D
\(\frac{1}{a}\) {-u + \(\sqrt{( 2as)}\)}
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given S = ut + \(\frac{1}{2} at^2\)
S = ut + \(\frac{1}{2} at^2\)
∴ 2S = 2ut + at2
= at2 + 2ut - 2s = 0
t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\)
= -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\)
= \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))
Source: Myschool Past Questions Archive (Record #31311)
Mathematics
UTME / JAMB 1992
Fractions, Decimals, Percentages & Approximations
Solution
A man invested a sum of N280.00 partly at 5% and partly at 4%. if the total interest is N12.80 per annum, find the amount invested at 5%
A
14.00
B
120.00
C
140.00
D
160.00
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let the amounts invested at 4% and 5% respectively be x and y.
\(\therefore x + y = 280 ... (i)\)
Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\)
Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\)
\(\therefore 0.04x + 0.05y = 12.80\)
\(\implies 4x + 5y = 1280 ... (ii)\)
From (i), \(x = 280 - y\).
Put into (ii), \(4(280 - y) + 5y = 1280\)
\(1120 - 4y + 5y = 1280\)
\(1120 + y = 1280 \implies y = 1280 - 1120 = N160\)
\(\therefore\) N160 was invested at the rate of 5% per annum.
Source: Myschool Past Questions Archive (Record #31312)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
If x + 1 is a factor of x3 + 3x2 + kx + 4, find the value of k
A
6
B
-6
C
8
D
-8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x + 1 is a factor of x3 + 3x2 + kx + 4
Let f(x) = x3 + 3x2 + kx + 4
∴ f(-1) = (-1)3 + 3(-1)2 + k(-1) + 4 = 0
-1 + 3 - k + 4 = 0
∴ k = 6
Source: Myschool Past Questions Archive (Record #31316)
Mathematics
UTME / JAMB 1992
Linear, Simultaneous & Quadratic Equations
Solution
Resolve \(\frac{3}{x^2 + x - 2}\) into partial fractions
A
\(\frac{1}{x - 1} - \frac{1}{x + 2}\)
B
\(\frac{1}{x + 1} + \frac{1}{x - 2}\)
C
\(\frac{1}{x + 1} - \frac{1}{x - 2}\)
D
\(\frac{1}{x - 2} + \frac{1}{x + 2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\)
\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)
A(x + 2) + B(x - 1) = 3
when x = 1, 3A = 3 \(\to\) a = 1
when x = -2, -3B = 3 \(\to\) B = -1
= \(\frac{1}{x - 1} - \frac{1}{x + 2}\)
Source: Myschool Past Questions Archive (Record #31319)
Mathematics
UTME / JAMB 1992
Inequalities & Linear Programming
Solution
Find all values of x satisfying the inequality -11 \(\leq\) 4 - 3x \(\leq\) 28
A
-5 \(\leq\) x \(\geq\) 8
B
5 \(\leq\) x \(\leq\) 8
C
-8 \(\leq\) x \(\leq\) 5
D
-5 < x \(\leq\) 8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
To solve -11 \(\leq\) 4 - 3x \(\leq\) 28
-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28
- 11 - 4\(\leq\) - 3x and -3x \(\leq\) 28 - 4
-15\(\leq\) -3x and -3x \(\leq\) 24
i.e. x \(\leq\) 5 and x \(\geq\) -8
∴ -8 \(\leq\) x \(\leq\) 5
Source: Myschool Past Questions Archive (Record #31320)
Mathematics
UTME / JAMB 1992
Sequences & Series (AP & GP)
Solution
Find the sum to infinity to the following series 3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
A
1270
B
190
C
18
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....
a = 3
r = \(\frac{2}{3}\)
s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)
= \(\frac{3}{\frac{1}{3}}\)
= 3 x 3
= 9
Source: Myschool Past Questions Archive (Record #31321)
Mathematics
UTME / JAMB 1992
Sequences & Series (AP & GP)
Solution
For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms
A
157
B
187
C
197
D
200
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
a = 2, d = 3 and n = 11
To find Sn/sub> = \(\frac{n}{2}\) [2a + (n - 1) \(\delta\)]
= \(\frac{11}{2}\) [2(2) + (11 - 1) 3]
= \(\frac{11}{2}\)n [4 + 10(3)]
= \(\frac{11}{2}\)(34)
= 11 x 17
= 187
Source: Myschool Past Questions Archive (Record #31328)
Mathematics
UTME / JAMB 1992
Binary Operations, Matrices & Determinants
Solution
p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)
Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\) Where pT is the transpose P calculate /pT/ when x = 0, y = 1 and z = 2
A
48
B
24
C
-24
D
-48
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
PT = \(\begin{vmatrix}0 & 2 & 4 \\ 2 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
/pT/ = \(\begin{vmatrix}0 & 2 & 4 \\ 3 & 1 & 3\\ 0 & 3 & 2 \end{vmatrix}\)
= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]
= 0 - 12 + 36 = 24
Source: Myschool Past Questions Archive (Record #31332)
Mathematics
UTME / JAMB 1992
Plane Geometry, Polygons & Circle Theorems
Solution
p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)
Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)
PQ is equivalent to
A
PPT
B
pp-1
C
qp
D
pp
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = pT
pq = ppT
Source: Myschool Past Questions Archive (Record #31333)