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Mathematics
UTME / JAMB 1995
Sequences & Series (AP & GP)
Solution
Find the sum to infinity of the following sequence \(1, \frac{9}{10}, (\frac{9}{10})^{2}, (\frac{9}{10})^{3}\)
A
\(\frac{1}{10}\)
B
\(\frac{9}{10}\)
C
\(\frac{10}{9}\)
D
10
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a G.P)
\(a = 1; r = \frac{9}{10}\)
\(S_{\infty} = \frac{1}{1 - \frac{9}{10}}\)
= \(\frac{1}{\frac{1}{10}} = 10\)
Source: Myschool Past Questions Archive (Record #30919)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
\(\begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & k \\1 & 3 & -1 \end{vmatrix}\) = 23
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 23
7k = 14
k = 2
Source: Myschool Past Questions Archive (Record #30921)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
If x = \(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\) and y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\). Find xy.
A
\(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)
B
\(\begin{pmatrix} 2 & 7 \\ 4 & 17 \end{pmatrix}\)
C
\(\begin{pmatrix} 10 & 4 \\ 4 & 6 \end{pmatrix}\)
D
\(\begin{pmatrix} 4 & 3 \\ 10 & 9 \end{pmatrix}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\).
= \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\)
= \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)
Source: Myschool Past Questions Archive (Record #30923)
Mathematics
UTME / JAMB 1995
Trigonometry, Elevation & Bearings
In a triangle XYZ, < YXZ = 44^\circ and < XYZ = 112^\circ. Calculate the acute angle between the internal bisectors of < XYZ and < XZY.
A
12o
B
56o
C
68o
D
78o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #30925)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
Find the distance between two towns P(45^\circN, 30^\circW) and Q(15^\circS, 30^\circW) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]
A
\(\frac{11000}{3}\)km
B
\(\frac{2200}{3}\)km
C
\(\frac{22000}{3}\)km
D
\(\frac{1100}{3}\)km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Angular difference = 45^\circ + 15^\circ = 60^\circ.
\(D = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7000\)
= \(\frac{22000}{3} km\)
Source: Myschool Past Questions Archive (Record #30930)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
Two perpendicular lines PQ and QR intersect at (1, -1). If the equation of PQ is x - 2y + 4 = 0, find the equation of QR
A
x + 2y - 1= -0
B
2x + y - 3 = 0
C
x - 2y - 3 = 0
D
2x + y - 1 = 0
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Line PQ : x - 2y + 4 = 0
2y = x + 4 \(\implies y = \frac{x}{2} + 2 \)
Slope = \(\frac{1}{2}\)
Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\)
Line QR: \(y = mx + b\)
\(y = -2x + b\)
Point of intersection: (1, -1)
\(-1 = -2(1) + b \implies b = -1 + 2 = 1\)
\(y = -2x + 1 \implies y + 2x - 1 = 0\)
\(QR: 2x + y - 1 = 0\)
Source: Myschool Past Questions Archive (Record #30933)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
P is on the locus of points equidiatant from two given points X and Y. UV is a straight line throuh Y parallel to the locus. If < PYU is 40^\circ, find < XPY.
A
100o
B
80 o
C
50 o
D
40 o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< PYU = < YPQ = 40^\circ(Alternative) = PQ\(\parallel\)UV
< XPQ = < YPQ = 40^\circ, Since QX = QY
therefore < XPY = 80^\circ
Source: Myschool Past Questions Archive (Record #30935)
Mathematics
UTME / JAMB 1995
Trigonometry, Elevation & Bearings
Solution
A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank 60o. Calculate the height of the water tank.
A
60m
B
30 \(\sqrt{3}\)m
C
20 \(\sqrt{3}\)m
D
10 \(\sqrt{3}\)m
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
h = 30 tan 60o
= 30\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #30939)
Mathematics
UTME / JAMB 1995
Trigonometry, Elevation & Bearings
Solution
The derivative of cosec x is
A
tan x cosec x
B
-cot x cosec x
C
tan x sec x
D
-cot x sec x
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\csc x = \frac{1}{\sin x}\)
Using the quotient rule,
\(\frac{\mathrm d y}{\mathrm d x} = \frac{vdu - udv}{v^{2}}\)
= \(\frac{\sin x (0) - 1 (\cos x)}{(\sin x)^{2}}\)
= \(\frac{- \cos x}{\sin^{2} x}\)
= \((\frac{- \cos x}{\sin x}) (\frac{1}{\sin x})\)
= \(- \cot x \csc x\)
Source: Myschool Past Questions Archive (Record #30940)
Mathematics
UTME / JAMB 1995
Plane Geometry, Polygons & Circle Theorems
Solution
For what value of x is the tangent to the curve y = x\(^2\) - 4x + 3 parallel to the x-axis?
A
3
B
2
C
1
D
6
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
At the point where the tangent is parallel to the x- axis, the slope of the curve = 0.
Given: \(y = x^{2} - 4x + 3\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x - 4\)
\(2x - 4 = 0 \implies 2x = 4 \)
\(x = 2\)
When x = 2, the tangent of the curve is parallel to the x- axis.
Source: Myschool Past Questions Archive (Record #30941)
Mathematics
UTME / JAMB 1995
Linear, Simultaneous & Quadratic Equations
Solution
Two variables x and y are such that \(\frac{dy}{dx}\) = 4x - 3 and y = 5 when x = 2. Find y in terms of x
A
2x2 - 3x + 5
B
2x2 - 3x + 3
C
2x2 - 3x
D
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
∫dy = ∫(4x - 3)dx, y = 2x2 - 3x + C
when y = 5, x = 2, C = 3
y = 2x2 - 3x + 3
Source: Myschool Past Questions Archive (Record #30942)
Mathematics
UTME / JAMB 1995
Integration & Applications
Solution
Find the area bounded by the curve y = 3x\(^2\) - 2x + 1, the ordinates x = 1 and x = 3 and the x-axis.
A
24
B
22
C
21
D
20
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\)
\(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\)
\(y = [x^{3} - x^{2} + x]_{1} ^{3}\)
= \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\)
= \(21 - 1 = 20\)
Source: Myschool Past Questions Archive (Record #30943)
Mathematics
UTME / JAMB 1995
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
\(\begin{array}{c|c} \text{Age in years} & 13 & 14 & 15 & 16 & 17 \\ \hline \text{No. of students} & 3 & 10 & 30 & 42 & 15\end{array}\)
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
A
27o
B
30o
C
54o
D
108o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Number of 15- year old pupils = 30.
Total number of students: 3 + 10 + 30 + 42 + 15 = 100.
Angle = \(\frac{30}{100} \times 360^\circ\)
= \(108^\circ\)
Source: Myschool Past Questions Archive (Record #30944)
Mathematics
UTME / JAMB 1995
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The variance of the scores 1, 2, 3, 4, 5 is
A
1, 2
B
1, 4
C
2.0
D
3.0
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(x\)
1
2
3
4
5
sum =15
\(x - \bar{x}\)
-2
-1
0
1
2
\((x - \bar{x})^{2}\)
4
1
0
1
4
10
Mean = \(\frac{15}{5} = 3\)
\(Variance = \frac{\sum (x - \bar{x})^{2}}{n}\)
= \(\frac{10}{5}\)
= \(2.0\)
Source: Myschool Past Questions Archive (Record #30947)
Mathematics
UTME / JAMB 1995
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
\(\begin{array}{c|c} \text{Class Interval} & Frequency & \text{Class boundaries} & Class Mid-point \\ \hline 1.5 - 1.9 & 2 & 1.45 - 1.95 & 1.7\\ 2.0 - 2.4 & 1 & 1.95 - 2.45 & 2.2\\ 2.5 - 2.9 & 4 & 2.45 - 2.95 & 2.7 \\ 3.0 - 2.9 & 15 & 2.95 - 3.45 & 3.2\\ 3.5 - 3.9 & 10 & 3.45 - 3.95 & 3.7\\ 4.0 - 4.4 & 5 & 3.95 - 4.45 & 4.2\\ 4.5 - 4.9 & 3 & 4.45 - 4.95 & 4.7\end{array}\)
Find the mode of the distribution.
A
3.2
B
3.3
C
3.7
D
4.2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Mode = a + \(\frac{ (b - a)(fm - Fb)}{2Fm - Fa - Fb}\)
= 3.0 + \(\frac{(3.45 - 2.95)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(0.5)(11)}{30 - 14}\)
= 3 + \(\frac{5.5}{16}\)
= 3 + 0.34375
= 3.34375
= 3.3cm
Source: Myschool Past Questions Archive (Record #30948)