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Mathematics
UTME / JAMB 2003
Linear, Simultaneous & Quadratic Equations
Solution
Find the slope of the curve y = 2x\(^2\) + 5x - 3 at (1, 4).
A
4
B
6
C
7
D
9
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
y = 2x\(^2\) + 5x - 3
dy/dx = 4x + 5
= 4 + 5
= 9
Source: Myschool Past Questions Archive (Record #7796)
Mathematics
UTME / JAMB 2003
Integration & Applications
Solution
Evaluate \(\int^{3} _{2}(x^2 - 2x)dx\)
A
4
B
2
C
4/3
D
1/3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\int^{3} _{2}(x^2 - 2x)\)dx
=\(\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{3}_{2}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{3}_{2}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\)
=\(\frac{-8}{3}\) + 4 = \(\frac{4}{3}\)
Source: Myschool Past Questions Archive (Record #7837)
Mathematics
UTME / JAMB 2003
Differentiation & Applications
Solution
If y = 3 sin(-4x), dy/dx is
A
12x cos (4x)
B
-12x cos (-4x)
C
-12 cos (-4x)
D
12 sin (-4x)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = 3 sin(-4x)
dy/dx = 3 x - 4 cos (-4x)
= -12 cos (-4x)
Source: Myschool Past Questions Archive (Record #7838)
Mathematics
UTME / JAMB 2003
Differentiation & Applications
Solution
Determine the maximum value of y = 3x2 - x3
A
zero
B
2
C
4
D
6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = 3x2 - x3
dy/dx = 6x - 3x2
as dy/dx = 0
6x - 3x2 = 0
3x (2 - x) = 0
this implies that 2 -x = 0 and 3x = 0
x = 2 (or) 0
But = dy/dx = 6x - 3x2
d2y/dx2 = 6 - 6x at x = 2
= 6 - 6(2)
= -6
y = 3x2 - x3
= 3(2)2 - 23
= 12 - 8
= 4
Source: Myschool Past Questions Archive (Record #7839)
Mathematics
UTME / JAMB 2003
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
By how much is the mean of 30, 56, 31, 55, 43 and 44 less than the median?
A
0.75
B
0.50
C
0.33
D
0.17
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(Mean = \frac{30+56+31+55+43+44}{6}\)
= \(\frac{259}{6}\) = 43.167
Median = 30,31,43,44,55,56 = \(\frac{43+44}{2}\)
=\( \frac{87}{2}\) = 43.5
Median-mean = 43.5 - 43.17 = 0.33
Source: Myschool Past Questions Archive (Record #7840)
Mathematics
UTME / JAMB 2003
Plane Geometry, Polygons & Circle Theorems
Solution
The range of 4, 3, 11, 9, 6, 15, 19, 23, 27, 24,21 and 16 is
A
16
B
21
C
23
D
24
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Range = (Highest - Lowest) items
= 27 - 3
= 24
Source: Myschool Past Questions Archive (Record #7841)
Mathematics
UTME / JAMB 2003
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the mean of the distribution above
value
0
1
2
3
4
Frequency
1
2
2
1
9
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Mean = \(\frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9}\)
= \(\frac{(0 + 2 + 4 + 3 + 36)}{15}\)
= \(\frac{45}{15}\) =3
Source: Myschool Past Questions Archive (Record #7842)
Mathematics
UTME / JAMB 2003
Fractions, Decimals, Percentages & Approximations
Solution
On a pie chart, there are four sectors of which three angles are 45^\circ, 90^\circ and 135^\circ. If the smallest sector represents N28.00, how much is the largest sector?
A
N96.00
B
N84.00
C
N48.00
D
N42.00
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Let the 4th angle be = x
∴ x + 45 + 90 + 135 = 360
x + 270 = 360
x = 360 - 270
x = 90
∴ smallest angle = 45o
45o = N28.00
1o = 28.00/45
135o = (28.00/45) * (135/1)
= N84.00
Source: Myschool Past Questions Archive (Record #7843)
Mathematics
UTME / JAMB 2003
Permutations, Combinations & Probability
Solution
If \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), find the value of n.
A
5
B
6
C
7
D
8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we start by using the formulas for permutations and combinations:
\(^{n}P_{r} = \frac{n!}{(n-r)!}\)
\(^{n}C_{r} = \frac{n!}{r!(n-r)!}\)
\(^{n}P_{3} = \frac{n!}{(n-3)!}\)
\(^{n}C_{4} = \frac{n!}{4!(n-4)!}\)
\(\frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0\)
Factor out \(n!\)
Factoring \(n!\) from both terms gives:
\(n! \left(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!}\right) = 0\)
Since \(n! \neq 0\), we can simplify to:
\(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!} = 0\)
Recall that \(4! = 24\), so:
\(\frac{6}{4!(n-4)!} = \frac{6}{24(n-4)!} = \frac{1}{4(n-4)!}\)
The equation now is:
\(\frac{1}{(n-3)!} - \frac{1}{4(n-4)!} = 0\)
Multiply through by \(4(n-3)!(n-4)!\) gives:
\(4(n-4)! - (n-3)! = 0\)
Recall that \((n-3)! = (n-3)(n-4)!\), so:
\(4(n-4)! - (n-3)(n-4)! = 0\)
\((n-4)!\)
Factoring out \((n-4)!\) gives:
\((n-4)! \left(4 - (n-3)\right) = 0\)
Since \((n-4)! \neq 0\), we have:
\(4 - (n-3) = 0\)
Solving for \(n\):
\(n - 3 = 4 \implies n = 7\)
The value of n is: 7
Source: Myschool Past Questions Archive (Record #7844)
Mathematics
UTME / JAMB 2003
Permutations, Combinations & Probability
Solution
Find the number of committees of three that can be formed consisting of two men and one woman from four men and three women
A
3
B
6
C
18
D
24
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(^{4}C_2 \times ^{3}C_1\)
= \(\left(\frac{4!}{(4-2)!2!}\right)\times\left(\frac{3!}{(3-1)!1!}\right)\)
= \(\frac{4!}{2!2!}\times \frac{3!}{2!}\)
= \(\frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{3 \times 2!}{2!} = 18\)
Source: Myschool Past Questions Archive (Record #7845)
Mathematics
UTME / JAMB 2003
Permutations, Combinations & Probability
Solution
A bag contains 5 blacks balls and 3 red balls. Two balls are picked at random without replacement. What is the probability that a black and red balls are picked?
A
\(\frac{15}{28}\)
B
\(\frac{13}{28}\)
C
\(\frac{5}{14}\)
D
\(\frac{3}{14}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Black balls = 5
Red balls = 3
P(B) = \(\frac{5}{8}\), P(R) = \(\frac{3}{8}\)
P( a Black and a Red) = black ball and red ball or red ball and black ball i.e BR + RB
= (\(\frac{5}{8}\)) * (\(\frac{3}{7}\)) + (\(\frac{3}{8}\)) * (\(\frac{5}{7}\))
= (\(\frac{15}{56}\)) + (\(\frac{15}{56}\))
= \(\frac{30}{56}\)
= \(\frac{15}{28}\)
Source: Myschool Past Questions Archive (Record #7846)
Mathematics
UTME / JAMB 2003
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The result of tossing a fair die 120 times is summarized below. Find the value of x
Number
1
2
3
4
5
6
Frequency
12
20
x
21
x - 1
28
A
19
B
20
C
21
D
22
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
12 + 20 + x + 21 + x-1 +28 = 120
2x + 80 = 120
2x = 120 - 80
2x = 40
x = 20
Source: Myschool Past Questions Archive (Record #7847)
Mathematics
WAEC / WASSCE 2003
Fractions, Decimals, Percentages & Approximations
Solution
Simplify \(\frac{4}{a-3}-\frac{1}{a+2}\)
A
\(\frac{3a+11}{(a-3)(a-2)}\)
B
\(\frac{3a+11}{(a+3)(a-2)}\)
C
\(\frac{3a+11}{(a+3)(a+2)}\)
D
\(\frac{3a+11}{(a-3)(a+2)}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{4}{a-3}-\frac{1}{a+3}\)
= \(\frac{4(a+2)-1(a-3)}{(a-3)(a+2)}\)
\(\frac{4a+8 -a +3}{(a-3)(a+2)}\)
= \(\frac{3a+11}{(a-3)(a+2)}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #26576)
Mathematics
WAEC / WASSCE 2003
Sets, Logic & Venn Diagrams
Diagram
Solution
From the Venn Diagram below, find Q' ∩ R.
Attached Figure / Diagram
A
(e)
B
(c, h)
C
(c, g, h)
D
(c, e, g, h)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Q' ∩ R
Q' = U - Q
Q' = {a, b, c, d, g, h, i}
R = {c, e, h, g}
Q' ∩ R = {c, h, g}
Source: WAEC / JAMB / NECO Classroom Archive (Record #26603)
Mathematics
UTME / JAMB 2003
Trigonometry, Elevation & Bearings
Solution
An aeroplane flies due north from airport P to Q and then flies due east R. If Q is equidistant from P and R, find the bearing of P and R
A
90o
B
135o
C
225o
D
270o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Since PQ = QR, \Delta PQR is a right angle isosceles triangle
∴ ∠PQR = 45o
=> The bearing of P from R = 180 + 45
= 225o
Source: WAEC / JAMB / NECO Classroom Archive (Record #11102)