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Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
Simplify \(\frac{324 - 4x^2}{2x + 18}\)
A
2(x - 9)
B
2(9 + x)
C
81 - x2
D
-2(x - 9)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{324 - 4x^2}{2x + 18}\) = \(\frac{18^2 - (2x)^2}{2x + 18}\)
= \(\frac{(18 - 2x)(18 + 2x)}{(2x + 18)}\)
18 - 2x = 2(9 - x)
or -2(x - 9)
Source: Myschool Past Questions Archive (Record #31524)
Mathematics
UTME / JAMB 1989
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize completely \(y^3 -4xy + xy^3 - 4y\)
A
(y + xy)(y + 2)(y - 2)
B
(y - xy)(y - 2)
C
y(1 + x)(y + 2)(y -2)
D
y(1 - x)(y + 2)(y - 2)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(y^3 -4xy + xy^3 - 4y \)
= \(y^3(1 + x) - 4y(1 + x)
\((y^3 - 4y)(1 + x) = (y^3(1 + x) - 4y(1 + x))\)
∴ = \(y(1 + x)(y + 2)(y - 2)\)
Source: Myschool Past Questions Archive (Record #31529)
Mathematics
UTME / JAMB 1989
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize 4a2 - 12ab - C2 + 9b2
A
4a(a - 3b) + (3b - c)2
B
(2a + 3b - c)(2a + 3b + c)
C
(2a - 3b + c)(2a - 3b - c)
D
4a(a - 3b) + (3b + c)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
4a2 - 12ab - C2 + 9b2
rearranges: (4a2 - 12ab + 9b2) - c2
(2a - 3b)(2a - 3b) - c2 = (2a - 3b)2 - c2
= (2a - 3b + c)(2a - 3b - c)
Source: Myschool Past Questions Archive (Record #31535)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
What are K and L respectively if \(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2)
A
-12, \(\frac{9}{2}\)
B
-6, 9
C
6, 9
D
12, \(\frac{9}{2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{1}{2}\)(3y - 4x)2 = (8x2 + kxy + Ly2)
\(\frac{1}{2}\)(9y2 - 24xy + 16x2) = 8x2 + kxy + Ly2
\(\frac{9}{2}\)y2 - 12xy) = kxy + Ly2
k = -12 ∴ L = \(\frac{9}{2}\)
Source: Myschool Past Questions Archive (Record #31537)
Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
Solve the pair of equation for x and y respectively \(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)
A
-1, 2
B
1, 2
C
2, 1
D
2, -1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)
Let \(x^{-1}\) = a and \(y^{-1}\)= b
2a - 3b = 4 .......(i)
4a + b = 1 .........(ii)
(i) x 3 = 12a + 3b = 3........(iii)
2a - 3b = 4 ...........(i)
(i) + (iii) = 14a = 7
∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\)
From (i), 3b = 2a - 4
3b = 1 - 4
3b = -3
∴ b = -1
From substituting, \(2^{-1} = x^{-1}\)
∴ x = 2
\(y^{-1} = -1, y = -1\)
Source: Myschool Past Questions Archive (Record #31543)
Mathematics
UTME / JAMB 1989
Linear, Simultaneous & Quadratic Equations
Solution
What value of Q will make the expression 4x2 + 5x + Q a complete square?
A
\(\frac{25}{16}\)
B
\(\frac{25}{64}\)
C
\(\frac{5}{8}\)
D
\(\frac{5}{4}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
4x2 + 5x + Q
To make a complete square, the coefficient of x2 must be 1
= x2 + \(\frac{5x}{4}\) + \(\frac{Q}{4}\)
Then (half the coefficient of x2) should be added
i.e. x2 + \(\frac{5x}{4}\) + \(\frac{25}{64}\)
∴ \(\frac{Q}{4}\) = \(\frac{25}{64}\)
Q = \(\frac{4 \times 25}{64}\)
= \(\frac{25}{16}\)
Source: Myschool Past Questions Archive (Record #31548)
Mathematics
UTME / JAMB 1989
Inequalities & Linear Programming
Solution
find the range of values of values of r which satisfies the following inequality, where a, b and c are positive \(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1
A
r > \(\frac{abc}{bc + ac + ab}\)
B
r < abc
C
r > \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\)
D
. \(\frac{1}{abc}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{r}{a}\) + \(\frac{r}{b}\) + \(\frac{r}{c}\) > 1 = \(\frac{bcr + acr + abr}{abc}\) > 1
r(bc + ac + ba > abc) = r > \(\frac{abc}{bc + ac + ab}\)
Source: Myschool Past Questions Archive (Record #31551)
Mathematics
UTME / JAMB 1989
Algebraic Expressions, Polynomials & Factorisation
Solution
Express \(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) as a single algebraic fraction
A
\(\frac{-3}{(x + 1)(2 - x)}\)
B
\(\frac{3}{(x + 1)(2 - x)}\)
C
\(\frac{-1}{(x + 1)}\)
D
\(\frac{1}{(x + 1)(x - 2)}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{1}{x + 1}\) - \(\frac{1}{x - 2}\) = \(\frac{x - 2 - x - 1}{(x + 1)(x - 2)}\)
= \(\frac{-3}{(x + 1)(2 - x)}\)
Source: Myschool Past Questions Archive (Record #31553)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{x(x + 1)^{-\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\)
A
\(\frac{-1}{x + 1}\)
B
\(\frac{1}{x + 1}\)
C
\(\frac{1}{x}\)
D
\(\frac{1}{x - 1}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\[\frac{x(x + 1)^{-\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\]
The numerator can be rewritten as: \[x(x + 1)^{-\frac{1}{2}} - (x + 1)^{\frac{1}{2}} = \frac{x}{(x + 1)^{\frac{1}{2}}} - (x + 1)^{\frac{1}{2}}\]
Combining the terms in the numerator over a common denominator: \[= \frac{x - (x + 1)}{(x + 1)^{\frac{1}{2}}}\]
\[= \frac{x - x - 1}{(x + 1)^{\frac{1}{2}}} = \frac{-1}{(x + 1)^{\frac{1}{2}}}\]
Substituting this back into the original expression gives:
\[\frac{\frac{-1}{(x + 1)^{\frac{1}{2}}}}{(x + 1)^{\frac{1}{2}}}\]
This simplifies to: \[\frac{-1}{(x + 1)^{\frac{1}{2}} \cdot (x + 1)^{\frac{1}{2}}} = \frac{-1}{(x + 1)}\]
= \[{\frac{-1}{x + 1}}\]
Source: Myschool Past Questions Archive (Record #31558)
Mathematics
UTME / JAMB 1989
Sequences & Series (AP & GP)
Solution
The sum of the first two terms of a geometric progression is x and sum of the last terms is y. If there are n terms in all, then the common ratio is
A
\(\frac{x}{y}\)
B
\(\frac{y}{x}\)
C
(\(\frac{x}{y}\))\(\frac{1}{n - 2}\)
D
(\(\frac{y}{x}\))\(\frac{1}{n - 2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Sum of nth term of a G.P = Sn = \(\frac{ar^n - 1}{r - 1}\)
sum of the first two terms = \(\frac{ar^2 - 1}{r - 1}\)
x = a(r + 1)
sum of the last two terms = Sn - Sn - 2
= \(\frac{ar^n - 1}{r - 1}\) - \(\frac{(ar^{n - 1})}{r - 1}\)
= \(\frac{a(r^n - 1 - r^{n - 2} + 1)}{r - 1}\) (r2 - 1)
∴ \(\frac{ar^{n - 2}(r + 1)(r - 1)}{1}\)= arn - 2(r + 1) = y
= a(r + 1)r^n - 2
y = xrn - 2
= yrn - 2
\(\frac{y}{x}\) = r = (\(\frac{y}{x}\))\(\frac{1}{n - 2}\)
Source: Myschool Past Questions Archive (Record #31570)
Mathematics
UTME / JAMB 1989
Sequences & Series (AP & GP)
Solution
If -8, m, n, 19 are in arithmetic progression, find (m, n)
A
1, 10
B
2, 10
C
3, 13
D
4, 16
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
-8, m, n, 19 = m + 8
= 19 - n
m + n = 11
i.e. 1, 10
Source: Myschool Past Questions Archive (Record #31571)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
A regular polygon of (2k + 1) sides has 140^\circ as the size of each interior angle. Find k
A
4
B
4\(\frac{1}{2}\)
C
8
D
8\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
A regular has all sides and all angles equal. If each interior angle is 140^\circ each exterior angle must be
180^\circ - 140^\circ = 40^\circ
The number of sides must be \(\frac{360^o}{40^o}\) = 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k = \(\frac{8}{2}\)
= 4
Source: Myschool Past Questions Archive (Record #31576)
Mathematics
UTME / JAMB 1989
Plane Geometry, Polygons & Circle Theorems
Solution
PQRS is a rhombus. If PR\(^2\) + QS\(^2\) = kPQ\(^2\), determine k.
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
PR\(^2\) + QS\(^2\) = kPQ\(^2\)
SQ\(^2\) = SR\(^2\) + RQ\(^2\)
PR\(^2\) + SQ\(^2\) = PQ\(^2\) + SR\(^2\) + 2RQ\(^2\)
= 2PQ\(^2\) + 2RQ\(^2\)
= 4PQ\(^2\)
∴ K = 4
Source: Myschool Past Questions Archive (Record #31581)
Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Solution
In XYZ, y = z = 30^\circ and XZ = 3cm. Find YZ
A
\(\frac{\sqrt{3}}{2}\) cm
B
3\(\frac{\sqrt{3}}{2}\) cm
C
3\(\sqrt{3}\) cm
D
2\(\sqrt{3}\) cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Y \(\to\) 30^\circ
X \(\to\) 120^\circ
Z \(\to\) 30^\circ
\(\frac{3}{\text{sin 30}}\) = \(\frac{YZ}{\text{sin 120}}\)
YZ = \(\frac{3\times \sin 120}{\sin 30}\)
\(3 \times \frac{\sqrt{3}}{2} \times 2 = 3\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #31585)
Mathematics
UTME / JAMB 1989
Mensuration (Areas, Perimeters & Volumes)
Solution
if x : y = 5 : 12 and z = 52cm, find the perimeter of the triangle.
A
68cm
B
84cm
C
100cm
D
120cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
13 = 52
1 = \(\frac{52}{13}\)
= 4
5 + 12 + 13 = 30
Total perimeter = 30 x 4
= 120cm
Source: Myschool Past Questions Archive (Record #31594)