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Mathematics
UTME / JAMB 1990
Algebra
Solution
Divide x3 - 2x2 - 5x + 6 by (x - 1)
A
x2 - x - 6
B
x2 - 5x + 6
C
x2 - 7x + 6
D
x2 - 5x - 6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Step 1: Just multiply each of the options by ( x - 1 )
Step 2: Then collect like terms to derive the same equation found in the question.
Hope this helps!
Source: Myschool Past Questions Archive (Record #31455)
Mathematics
UTME / JAMB 1990
Algebra
Solution
Find the solution of the equation x - 8\(\sqrt{x}\) + 15 = 0
A
3, 5
B
-3, -5
C
9, 25
D
-9, 25
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x - 8\(\sqrt{x}\) + 15 = 0
x + 15 = 8\(\sqrt{x}\)
square both sides = (x + 15)2 = (8 \(\sqrt{x}\)2
x2 + 225 + 30x = 64x
x2 + 225 + 30x - 64x = 0
x2 - 34x + 225 = 0
(x - 9)(x - 25) = 0
x = 9 or 25
Source: Myschool Past Questions Archive (Record #31460)
Mathematics
UTME / JAMB 1990
Algebra
Find the sum of the first twenty terms of the progression log a, log a2, log a3.....
A
log a20
B
log a21
C
log a200
D
log a210
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 #31471)
Mathematics
UTME / JAMB 1990
Algebra
Diagram
Solution
What is the equation of the quadratic function represented by the graph?
Attached Figure / Diagram
A
y = x2 + x - 2
B
y = -x2 - x + 2
C
y = x2 - x - 2
D
y = -x2 - x + 2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The required equation is y = x2 - x - 2
i.e. B where the graph touches the graph touches the x-axis y = 0
x2 - x - 2 = 0 = (x + 1)(x - 2) = 0
Hence roots of the equation are -1 and 2 as shown in the graph
Source: Myschool Past Questions Archive (Record #32874)
Mathematics
WAEC / WASSCE 1990
Algebra
Solution
(a) Simplify \(\frac{0.016 \times 0.084}{0.48}\) [Leave your answer in standard form].
(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{0.016 \times 0.084}{0.48}\)
= \(\frac{16 \times 10^{-3} \times 84 \times 10^{-3}}{48 \times 10^{-2}}\)
= \(28 \times 10^{-3 - 3 - (-2)}\)
= \(28 \times 10^{-4} = 2.8 \times 10^{-3}\)
(b) \(T_{n} = a + (n - 1)d\) (Terms of an A.P)
\(T_{2} = a + d = 2 ... (1)\)
\(T_{6} = a + 5d = 5 ... (2)\)
\((2) - (1) : 4d = 3 \implies d = \frac{3}{4} = 0.75\)
Put the value of d in (1), so we have
\(a + 0.75 = 2 \implies a = 1.25\)
Therefore, the length of the poles in metres = \(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\)
Source: Myschool Past Questions Archive (Record #54630)
Mathematics
WAEC / WASSCE 1990
Algebra
Show on a graph, the area which gives the solution set of the inequalities: \(y - 2x \leq 4 ; 3y + x \geq 6 ; y \geq 7x - 9\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option .
Source: Myschool Past Questions Archive (Record #54632)
Mathematics
WAEC / WASSCE 1990
Algebra
Solution
The following is an incomplete table for the relation \(y = 2x^{2} - 5x + 1\)
x
-3
-2
-1
0
1
2
3
4
5
y
8
1
-1
26
(a) Copy and complete the table.
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \(y = 2x^{2} - 5x + 1\) for \(-3 \leq x \leq 5\).
(c) Using the same scale and axes, draw the graph of \(y = x + 6\).
(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \(2x^{2} - 5x + 1 = x + 6\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(y = 2x^{2} - 5x + 1\)
x
-3
-2
-1
0
1
2
3
4
5
y
34
19
8
1
-2
-1
4
13
26
Table of value for \(y = x + 6\)
x
-2
0
2
y
4
6
8
(b)
(d) (i) Least value of y = -2 occurs at x = 1
(ii) Solution of \(2x^{2} - 5x + 1 = x + 6\) are x = -0.7 and x = 3.7.
Source: Myschool Past Questions Archive (Record #54650)
Mathematics
WAEC / WASSCE 1989
Algebra
Solution
If the second and fourth term of a G.P are 8 and 32 respectively,what is the sum of the first four terms?
A
28
B
40
C
48
D
60
E
68
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
ar = 8 .....(1)
ar3 = 32 .....(2)
divide (2) by (1)
r2 = 4 r = 2 ∴ a = 4
S = a(rn - 1)/r - 1
S = 4(24 - 1)/2 - 1 = 60
Source: Myschool Past Questions Archive (Record #22342)
Mathematics
WAEC / WASSCE 1989
Algebra
Diagram
Solution
The graph above is that of a quadratic function y = 2x\(^2\) - x - 6
The roots of the equation are
Attached Figure / Diagram
A
1.5, 2
B
-1.5, 2
C
1, -6
D
1, 6
E
-1, 6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The roots are; x = -1.5 and 2
Source: Myschool Past Questions Archive (Record #22348)
Mathematics
WAEC / WASSCE 1989
Algebra
Solution
Solve the following equation: 6x\(^2\) - 7x - 5 = 0
A
x = 1/2 or x = -21/2
B
x = 1/3 or x = 21/2
C
x = 1 2/3 or x = -1/2
D
x = 12/3 or x = 1/2
E
x = 5/6 or x = -1
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
6x\(^2\) - 7x - 5 = 0
6x\(^2\) - 10x + 3x - 5 = 0
2x(3x - 5) + 1(3x - 5) = 0
(2x + 1)(3x - 5) = 0
x = \(\frac{5}{3}\) or x = \(-\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #22362)
Mathematics
WAEC / WASSCE 1989
Algebra
Solution
What is the probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank
A
1/6
B
1/3
C
1/2
D
2/3
E
5/6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Probability that 3 customers are attended to in the sequence they arrive
= prob of attending to the 1st arrival x prob of attending to the second last x prob of attending to the last
= \(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\)
= \(\frac{1}{6}\)
Source: Myschool Past Questions Archive (Record #22419)
Mathematics
WAEC / WASSCE 1989
Algebra
Solution
In an A.P. the first term is 2, and the sum of the 1st and the 6th term is 16\(\frac{1}{2}\). What is the 4th term
A
12
B
91/2
C
8
D
7
E
51/2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
1st term a = 2; 6th term = a + (6 - 1)d = 2 + 5d
2 + (2 + 5d) = 16\(\frac{1}{2}\); 4 + 5d = 16\(\frac{1}{2}\)
5d = 12.5 ∴ = 2.5
the 4th term is 2 + (4 - 1)2.5= 9\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #22421)
Mathematics
UTME / JAMB 1989
Algebra
Solution
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ?
A
-1 and 0
B
-1 and 1
C
1 and 3
D
0 and 1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(9^{y} - 4 \times 3^{y} + 3 = 0\)
\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\)
\((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\)
Let \(3^{y}\) be r. Then,
\(r^{2} - 4r + 3 = 0\)
Solving the equation,
\(r^{2} - 3r - r + 3 = 0\)
\(r(r - 3) - 1(r - 3) = 0\)
\((r - 3)(r - 1) = 0\)
\(\therefore \text{r = 3 or 1}\)
Recall, \(3^{y} = r\)
\(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\)
\(\implies \text{y = 1 or 0}\)
Source: Myschool Past Questions Archive (Record #31507)
Mathematics
UTME / JAMB 1989
Algebra
Solution
The cost of dinner for a group of students is partly constant and partly varies directly as the number of students. If the cost is N74.00 when the number of students is 20 and N96.00 when the number of students is 30, find the cost when there are 15 students
A
N68.50
B
N63.00
C
N60.00
D
N52.00
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
where C = cost, S = number of students, a and K are constants.
C = a + kS. If C = 74, S = 20
C = 96, S = 30, C= ? S = 15
74 = a + 20k......(1)
96 = a + 30k......(2)
subtract equation (1) from (2)
96 = a + 30k
-
74 = a + 20k
--------------
22 = 10k
k = 2.2
find a
74 = a + 44
a = 30
C = 30 + 2.2S
when S = 15, C = 30 + 2.2 x 15
= 30 + 33
= N63
Source: Myschool Past Questions Archive (Record #31516)
Mathematics
UTME / JAMB 1989
Algebra
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)