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Mathematics
UTME / JAMB 1988
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize completely \((x^2 + x)^2 - (2x + 2)^2\)
A
(x + 1)(x + 2)(x - 2)
B
(x + 1) 2(x + 2) (x - 2)
C
(x + 1)2 (x + 2)2
D
(x + 1)2 (x + 2)(x - 2)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\((x^{2} + x)^{2} - (2x + 2)^{2}\)
= \((x^{2} + x + 2x + 2)(x^{2} + x - (2x + 2))\)
= \((x^{2} + 3x + 2)(x^{2} - x - 2)\)
= \(((x + 1)(x + 2))((x + 1)(x - 2))\)
= \((x + 1)^{2} (x + 2)(x - 2)\)
Source: Myschool Past Questions Archive (Record #31632)
Mathematics
UTME / JAMB 1988
Fractions, Decimals, Percentages & Approximations
Solution
Simplify \(\frac{x - y}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\)
A
x2 + xy + y2
B
x\(\frac{2}{3}\) + x \(\frac{1}{3}\) + y\(\frac{2}{3}\)
C
x\(\frac{2}{3}\) - x\(\frac{1}{3}\)y\(\frac{2}{3}\)
D
y\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{x - y}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\)
( x - y ) = (\(x^{\frac{1}{3}})^3 - (y^{\frac{1}{3}})^3\) = (\(x^{\frac{1}{3}}\) - \(y^{\frac{1}{3}}\))(\(x^{\frac{2}{3}}\) + \(x^{\frac{1}{3}}\)\(y^{\frac{1}{3}}\) + \(y^{\frac{2}{3}}\))
SO, \(\frac{x - y}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\) = \(\frac{(x^{\frac{1}{3}} - y^{\frac{1}{3}})(x^{\frac{2}{3}} + x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}})}{x^{\frac{1}{3}} - y^{\frac{1}{3}}}\)
Common factor in numerator and denominator cancels out.
Final answer = (\(x^{\frac{2}{3}}\) + \(x^{\frac{1}{3}}\)\(y^{\frac{1}{3}}\) + \(y^{\frac{2}{3}}\))
Source: Myschool Past Questions Archive (Record #31633)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
What is the solution of the equation x2 - x - 1 + 0?
A
x = 1.6 and x = -0.6
B
x = -1.6 and x = 0.6
C
x = 1.6 and x = 0.6
D
x = -1.6 and x = -0.6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x^{2} - x - 1 = 0\)
Using the quadratic formula,
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
a = 1, b = -1, c = -1.
\(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\)
\(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\)
\(x = \frac{1 + 2.24}{2} ; x = \frac{1 - 2.24}{2}\)
\(x = \frac{3.24}{2}; x = \frac{-1.24}{2}\)
\(x = 1.62 ; x = -0.61 \)
\(x \approxeq 1.6; -0.6\)
Source: Myschool Past Questions Archive (Record #31637)
Mathematics
UTME / JAMB 1988
Inequalities & Linear Programming
Diagram
For what values of x is the curve y = \(\frac{x^2 + 3}{x + 4}\) decreasing?
Attached Figure / Diagram
A
-3 < x \(\leq\) 0
B
-3 \(\geq\) x < 0
C
0 < x < 3
D
0 \(\leq\) x \(\leq\) 3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #31638)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph
A
y = 1 - x
B
y = 1 + x
C
y = x - 1
D
y = 3x + 3
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The second graph is
\((x^{2} - 2x - 1) + (2 + x - x^{2})\)
= \(1 - x\)
Source: Myschool Past Questions Archive (Record #31639)
Mathematics
UTME / JAMB 1988
Sequences & Series (AP & GP)
Solution
If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference
A
4
B
8
C
6\(\frac{2}{3}\)
D
9\(\frac{1}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let the first term and common difference = a & d respectively.
\(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P)
Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\)
\(T_{8} + T_{9} = 72\)
\(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\)
From (i), \(a = -6 - 3d\)
\(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\)
\(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\)
\(d = \frac{84}{9} = 9\frac{1}{3}\)
Source: Myschool Past Questions Archive (Record #31640)
Mathematics
UTME / JAMB 1988
Sequences & Series (AP & GP)
Solution
If 7 and 189 are the first and fourth terms of geometric progression respectively, find the sum of the first three terms of the progression
A
182
B
91
C
63
D
28
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(T_{n} = ar^{n - 1}\) (nth term of a G.P)
\(T_{4} = ar^{3} = 189\)
\(7 \times r^{3} = 189 \implies r^{3} = 27\)
\(r = \sqrt[3]{27} = 3\)
\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\)
\(S_{3} = \frac{7(3^{3} - 1)}{3 - 1} \)
= \(\frac{7 \times 26}{2} = 91\)
Source: Myschool Past Questions Archive (Record #31642)
Mathematics
UTME / JAMB 1988
Fractions, Decimals, Percentages & Approximations
Solution
Find correct to one decimal place, 0.24633 \(\div\) 0.0306
A
0.8
B
1.8
C
8.0
D
8.1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{0.24633}{0.03060}\) multiplying throughout by 100,000
= \(\frac{24633}{3060}\)
= 8.05
= 8.1
Source: Myschool Past Questions Archive (Record #31644)
Mathematics
UTME / JAMB 1988
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12o. How many students are offering Physics?
A
18
B
15
C
10
D
5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No of students offering Physics are \(\frac{12}{360}\) x 150
= 5
Source: Myschool Past Questions Archive (Record #31646)
Mathematics
UTME / JAMB 1988
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
\(\begin{array}{c|c} Scores(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency(f) & 7 & 11 & 6 & 7 & 7 & 5 & 3\end{array}\)
In the distribution above, the mode and median respectively are
A
1,3
B
1, 2
C
3, 3
D
0, 2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the distribution, Mode = 1 and
Median = \(\frac{2 + 2}{2}\)
= 2
= 1, 2
Source: Myschool Past Questions Archive (Record #31647)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
if x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y
A
27
B
30
C
33
D
90
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Prime numbers between 1 and 6 are 2, 3, and 5
x = 2 + 3 + 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x Y = 10 x 3 = 30
Source: Myschool Past Questions Archive (Record #31649)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle
A
110o
B
120o
C
130o
D
140o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Let the fifth interior angle be y: sum of interior angle of a pentagon
= (2 x 5 - 4) x 90o
= 6 x 90o
= 540o
(90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540o
400o + y = 540o
y = 540 - 400o
y = 140o
Source: Myschool Past Questions Archive (Record #31651)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
For which of the following exterior angles is a regular polygon possible? i. 35^\circ ii. 18^\circ iii. 15^\circ
A
i and ii
B
ii only
C
ii and iii
D
iii only
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
for a regular polygon to be possible, it must have all sides angles equal. \(\frac{360}{18}\) = 20 sides and \(\frac{360}{15}\) = 24 sides
(ii) and (iii) are right
Source: Myschool Past Questions Archive (Record #31655)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify \(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{of } 32}\)
A
\(\frac{3}{256}\)
B
\(\frac{3}{32}\)
C
6
D
85
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{1\frac{1}{2}}{2 \div \frac{1}{4} \text{ of }32}\)
= \(\frac{\frac{3}{2}}{2 \div \frac{1}{4} \times 32}\)
\(\frac{3}{2} \times 4 = 6\)
Source: Myschool Past Questions Archive (Record #31661)
Mathematics
UTME / JAMB 1988
Fractions, Decimals, Percentages & Approximations
Solution
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?
A
20
B
2
C
1.1
D
0.2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
% error = \(\frac{\text{actual error}}{\text{true value}}\) x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = \(\frac{0.1}{5.0}\) x 100
= \(\frac{10}{5}\)
= 2%
Source: Myschool Past Questions Archive (Record #31662)