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Mathematics
UTME / JAMB 1983
Plane Geometry, Polygons & Circle Theorems
Solution
Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12
A
x = 5, y = 2
B
x = 2, y = 5
C
x = 5, y = -2
D
x = -2, y = -5
E
x = -5, y = -2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
4x - 3 = 3x + y = x - y = 3.......(i)
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i) 3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2
Source: Myschool Past Questions Archive (Record #32448)
Mathematics
UTME / JAMB 1983
Indices, Logarithms & Surds
Diagram
Solution
In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS
Attached Figure / Diagram
A
2\(\sqrt{3}\)cm
B
4\(\sqrt{6}\)cm
C
2\(\sqrt{6}\)cm
D
8\(\sqrt{6}\)cm
E
8cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
From the diagram, sin 60o = \(\frac{PR}{8}\)
PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\)
= 4\(\sqrt{3}\)
Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\)
PS Cos45o = 4\(\sqrt{3}\)
PS = 4\(\sqrt{3}\) x 2
= 4\(\sqrt{6}\)
Source: Myschool Past Questions Archive (Record #32449)
Mathematics
UTME / JAMB 1983
Trigonometry, Elevation & Bearings
Solution
Given that cos z = L , whrere z is an acute angle, find an expression for \(\frac{\cot z - \csc z}{\sec z + \tan z}\)
A
\(\frac{1 - L}{1 + L}\)
B
\(\frac{L^2 \sqrt{3}}{1 + L}\)
C
\(\frac{1 + L^3}{L^2}\)
D
\(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given Cos z = L, z is an acute angle
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z
= \(\frac{\text{cos z}}{\text{sin z}}\)
cosec z = \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\)
sec z = \(\frac{1}{\text{cos z}}\)
tan z = \(\frac{\text{sin z}}{\text{cos z}}\)
sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\)
= \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)
the original eqn. becomes
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\)
= \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\)
= \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\)
= sin z + 1
= 1 + \(\sqrt{1 - L^2}\)
= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
Source: Myschool Past Questions Archive (Record #32450)
Mathematics
UTME / JAMB 1983
Indices, Logarithms & Surds
Solution
If 0.0000152 x 0.00042 = A x 10\(^{B}\), where 1 \(\leq\) A < 10, find A and B
A
A = 9, B = 6.38
B
A = 6.38, B = -9
C
A = -6.38, B = -9
D
A = -9, B = -6.38
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
0.0000152 x 0.00042 = A x 10\(^{B}\)
1 \(\leq\) A < 10, it means values of A include 1 - 9
0.0000152 = 1.52 x 10\(^{-5}\)
0.00042 = 4.2 x 10\(^{-4}\)
1.52 x 4.2 = 6.384
10\(^{-5}\times 10^{-4}\)
= 10\(^{-5-4}\)
= 10\(^{-9}\)
= 6.38 x 10\(^{-9}\)
A = 6.38, B = -9
Source: Myschool Past Questions Archive (Record #32451)
Mathematics
UTME / JAMB 1983
Indices, Logarithms & Surds
Solution
Make T the subject of the equation \(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\)
A
T = \(\frac{3av}{1 - v}\)
B
T = \(\frac{1 + v}{2a^2v^3}\)
C
T = \(\frac{2v(1 - v)^3 - a^4v^3}{2a^3v^3 + (1 - v)^2}\)
D
\(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\)
\(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
\(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\)
= \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)
Source: Myschool Past Questions Archive (Record #32453)
Mathematics
UTME / JAMB 1983
Plane Geometry, Polygons & Circle Theorems
Solution
The value of (0.03)\(^3\) - (0.02)\(^3\) is
A
0.019
B
0.0019
C
0.00019
D
0.000019
E
0.000035
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Using the method of difference of two cubes,
\(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\)
\((0.03)^{3} - (0.02)^{3} = (0.03 - 0.02)((0.03)^{2} + (0.03)(0.02) + (0.02)^{2}\)
= \((0.01)(0.0009 + 0.0006 + 0.0004)\)
= \(0.01 \times 0.0019\)
= \(0.000019\)
Source: Myschool Past Questions Archive (Record #32454)
Mathematics
UTME / JAMB 1983
Plane Geometry, Polygons & Circle Theorems
Solution
y varies partly as the square of x and partly as the inverse of the square root of x.Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4
A
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
B
y = x2 + \(\frac{1}{\sqrt{x}}\)
C
y = x2 + \(\frac{1}{x}\)
D
y = \(\frac{x^2}{31} + \frac{1}{31\sqrt{x}}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
y = kx2 + \(\frac{c}{\sqrt{x}}\)
y = 2when x = 1
2 = k + \(\frac{c}{1}\)
k + c = 2
y = 6 when x = 4
6 = 16k + \(\frac{c}{2}\)
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = \(\frac{10}{31}\)
c = 2 - \(\frac{10}{31}\)
= \(\frac{62 - 10}{31}\)
= \(\frac{52}{31}\)
y = \(\frac{10x^2}{31} + \frac{52}{31\sqrt{x}}\)
Source: Myschool Past Questions Archive (Record #32455)
Mathematics
UTME / JAMB 1983
Linear, Simultaneous & Quadratic Equations
Solution
Simplify \(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
A
\(\frac{x}{(x - 3)(x - 7)}\)
B
\(\frac{x}{(x + 3)(x - 7)}\)
C
\(\frac{x}{(x + 3)(x + 7)}\)
D
\(\frac{x}{(x - 3)(x + 7)}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{x - 7}{x^2 - 9}\) x \(\frac{x^2 - 3x}{x^2 - 49}\)
= \(\frac{x - 7}{(x - 3)(x + 3)}\) x \(\frac{x(x - 3)}{(x - 7)(x + 7)}\)
= \(\frac{x}{(x + 3)(x + 7)}\)
Source: Myschool Past Questions Archive (Record #32456)
Mathematics
UTME / JAMB 1983
Plane Geometry, Polygons & Circle Theorems
Solution
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x
A
2
B
6
C
18
D
12
E
5
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem)
\(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\)
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
The sides cannot be 0 hence x = 12.
Source: Myschool Past Questions Archive (Record #32457)
Mathematics
UTME / JAMB 1983
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.
A
47
B
50
C
48\(\frac{1}{2}\)
D
48
E
49
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median = \(\frac{47 + 50}{2}\)
\(\frac{97}{2}\)
= 48\(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #32458)
Mathematics
UTME / JAMB 1983
Number Bases & Modular Arithmetic
Solution
Find x if (x\(_4\))\(^2\) = 100100\(_2\)
A
6
B
12
C
100
D
210
E
110
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x\(_4\) = x \(\times\) 4\(^0\) = x in base 10.
100100\(_2\) = 1 x 2\(^5\) + 1 x 2\(^2\)
= 32 + 4
= 36 in base 10
\(\implies\) x\(^2\) = 36
x = 6 in base 10.
Convert 6 to a number in base 4.
4
6
4
1 r 2
0 r 1
= 12\(_4\)
Source: Myschool Past Questions Archive (Record #32459)
Mathematics
UTME / JAMB 1983
Indices, Logarithms & Surds
Solution
Simplify log\(_{10}\) a\(\frac{1}{3}\) + \(\frac{1}{4}\)log\(_{10}\) a - \(\frac{1}{12}\)log\(_{10}\)a7
A
1
B
\(\frac{7}{6}\)log10 a
C
zero
D
10
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
log10 a\(\frac{1}{3}\) + \(\frac{1}{4}\)log10 a - \(\frac{1}{12}\)log10a7 = log10 a\(\frac{1}{3}\) + log10\(\frac{1}{4}\) - log10 a\(\frac{7}{12}\)
= log\(_{10}\) a\(\frac{7}{12}\) - log\(_{10}\) a\(\frac{7}{12}\)
= log\(_{10}\) 1 = 0
Source: Myschool Past Questions Archive (Record #32460)
Mathematics
UTME / JAMB 1983
Variations
Solution
If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2
A
u = \(\frac{8}{v^3}\)
B
v = \(\frac{8}{u^2v^3}\)
C
u = 8v3
D
v = 8u2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
W \(\alpha\) \(\frac{1}{v}\)u \(\alpha\) w3
w = \(\frac{k1}{v}\)
u = k2w3
u = k2(\(\frac{k1}{v}\))3
= \(\frac{k_2k_1^2}{v^3}\)
k = k2k1k2
u = \(\frac{k}{v^3}\)
k = uv3
= (1)(2)3
= 8
u = \(\frac{8}{v^3}\)
Source: Myschool Past Questions Archive (Record #32461)
Mathematics
UTME / JAMB 1983
Linear, Simultaneous & Quadratic Equations
Solution
Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0
A
-28, 7
B
6, -28
C
6, -1
D
-1, 7
E
3, 2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1
Source: Myschool Past Questions Archive (Record #32462)
Mathematics
UTME / JAMB 1983
Plane Geometry, Polygons & Circle Theorems
Solution
Find the missing value in the following table \(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline Y = x^3 - x + 3 & & 3 & 3 & 3 & 9 & 27\end{array}\)
A
-3
B
3
C
-9
D
13
E
9
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
When x = -2, y = x3 - x + 3
= -23 - (-2) + 3
= -8 + 2 + 3
= -3
Source: Myschool Past Questions Archive (Record #32463)