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Mathematics
UTME / JAMB 1981
Sequences & Series (AP & GP)
Solution
The first term of an Arithmetic progression is 3 and the fifth term is 9. Find the number of terms in the progression if the sum is 81
A
12
B
27
C
9
D
4
E
36
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
1st term a = 3, 5th term = 9, sum of n = 81
nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9
3 + 4d = 9
4d = 9 - 3
d = \(\frac{6}{4}\)
= \(\frac{3}{2}\)
= 6
Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\))
81 = \(\frac{12n + 3n^2}{4}\) - 3n
= \(\frac{3n^2 + 9n}{4}\)
3n2 + 9n = 324
3n2 + 9n - 324 = 0
By almighty formula positive no. n = 9
= 3
Source: Myschool Past Questions Archive (Record #32595)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\)
A
\(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
B
\(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\)
C
\(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\)
D
\(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\)
= \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\)
= \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
= \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\)
T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
Source: Myschool Past Questions Archive (Record #32596)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
If b = a + cp and r = ab + \(\frac{1}{2}\)cp2, express b\(^2\) in terms of a, c, r.
A
b2 = aV + 2cr
B
b2 = ar + 2c2r
C
b2 = a2 = \(\frac{1}{2}\) cr2
D
b2 = \(\frac{1}{2}\)ar2 + c
E
b2 = 2cr - a2
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
b = a + cp....(i)
r = ab + \(\frac{1}{2}\)cp2.....(ii)
expressing b2 in terms of a, c, r, we shall first eliminate p which should not appear in our answer from eqn, (i)
b - a = cp = \(\frac{b - a}{c}\)
sub. for p in eqn.(ii)
r = ab + \(\frac{1}{2}\)c\(\frac{(b - a)^2}{\frac{ab + b^2 - 2ab + a^2}{2c}}\)
2cr = 2ab + b2 - 2ab + a2
b2 = 2cr - a2
Source: Myschool Past Questions Archive (Record #32597)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
Multiply x2 + x + 1 by x2 - x + 1
A
x4 - x + x2
B
x4 - x2 + x2
C
x4 + x2 + 1
D
x4 + x2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(x2 + x + 1)( x2 - x + 1)
= x2(x2 + x + 1) - x(x2 + x + 1) + (x3 + x + 1)
= x4 - x3 + x2 + x3 - x2 - x + 1
= x4 + x2 + 1
Source: Myschool Past Questions Archive (Record #32598)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
A baking recipe calls for 2.5kg of sugar and 4.5kg of flour. With this recipe some cakes were baked using 24.5kg of a mixture of sugar and flour. How much sugar was used?
A
12.25kg
B
6.75kg
C
8.75kg
D
15.75kg
E
8.25kg
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Sugar : flour = 2.5 : 4.5
Total = 7
sugar used = \(\frac{2.5}{7}\) x 24.5
= \(\frac{61.25}{7}\)
= 8.75
Source: Myschool Past Questions Archive (Record #32599)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
The difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\) is greater than sum of \(\frac{1}{14}\) and 1\(\frac{1}{2}\) by
A
\(\frac{23}{28}\)
B
\(\frac{24}{28}\)
C
\(\frac{50}{56}\)
D
\(\frac{27}{28}\)
E
\(\frac{48}{56}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Difference between 4\(\frac{5}{7}\) and 2\(\frac{1}{4}\)
\(\frac{33}{7}\) - \(\frac{9}{4}\) = \(\frac{69}{24}\)
The sum of \(\frac{1}{14}\) and 1\(\frac{1}{14}\) + \(\frac{3}{2}\)
= \(\frac{11}{7}\)
\(\frac{69}{28}\) - \(\frac{11}{7}\) = \(\frac{25}{28}\)
= \(\frac{50}{56}\)
Source: Myschool Past Questions Archive (Record #32600)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
What is the size of an exterior angle of a regular pentagon?
A
36o
B
60o
C
72o
D
120o
E
360o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o
= 5
= (10 - 4) x 90o
= 540o
each interior angle = \(\frac{540}{5}\)
= 108o
= 180o let x represent ext. angle
x + 108o = 180v
x = 180o - 108o
= 72o
Source: Myschool Past Questions Archive (Record #32601)
Mathematics
UTME / JAMB 1981
Linear, Simultaneous & Quadratic Equations
Solution
The sum of the root of a quadratic equation is \(\frac{5}{2}\) and the product of its root is 4. The quadratic equation is
A
x + 8x2 - 5 = 0
B
2x2 - 5x + 8 = 0
C
2x2 - 8x + 5 = 0
D
x2 + 8x - 5 = 0
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
x\(^2\) - (sum of roots) x + (product of roots) = 0
x\(^2\) - \(\frac{5x}{2}\) + 4 = 0
2x\(^2\) - 5x + 8 = 0
Source: Myschool Past Questions Archive (Record #32602)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Solve the given equation \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)
A
27
B
9
C
\(\frac{1}{27}\)
D
18
E
81
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)
Let \(\log_{3} x = a\).
\(a^{2} - 6a + 9 = 0\)
\(a^{2} - 3a - 3a + 9 = 0\)
\(a(a - 3) - 3(a - 3) = 0\)
\((a - 3)(a - 3) = 0\)
\(\implies a = 3 (twice)\)
\(\log_{3} x = 3 \implies x = 3^{3} = 27\)
Source: Myschool Past Questions Archive (Record #32603)
Mathematics
UTME / JAMB 1981
Mensuration (Areas, Perimeters & Volumes)
Solution
What is the length of an arc of a circle that subtends 2\(\frac{1}{2}\) radians at the centre when the radius of the circle is 8cm?
A
8cm
B
10cm
C
40cm
D
20cm
E
16cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given: \(\theta\) = 2\(\frac{1}{2}\) radians, r = 8cm
but 360º = 2 \(\pi\) radian.
Using: L = length of an arc = \(\frac{\theta}{2 \pi}\) x 2\(\pi\) r
= \(\frac{2.5}{2\pi}\) x 2 x \(\pi\) x 8
= 2.5 x 8 = 20cm.
Source: Myschool Past Questions Archive (Record #32604)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, the chords XY and ZW are produced to meet at T such that YT = WT, ZYW = 40o and YTW = 30o. What is YXW?
Attached Figure / Diagram
A
50o
B
75o
C
35o
D
115o
E
30o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
< WYT = \(\frac{180 - 30}{2}\)
= \(\frac{150}{2}\)
= 75o(base angles of isosceles D)
YXW = 75 - 40 = 35o(Exterior angle is equal to sum of interior angles)
Source: Myschool Past Questions Archive (Record #32753)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Diagram
Solution
In the figure, find the area of XYZW
Attached Figure / Diagram
A
60cm2
B
54cm2
C
27cm2
D
54\(\sqrt{2}\)cm2
E
27\(\sqrt{2}\)cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of trapezium XYZW = \(\frac{1}{2}\)(a + b)h
h = 6 sin 30º = 3
area = \(\frac{1}{2}\)(10 + 8) x 3
\(\frac{1}{2}\)(18) x 4
= 27cm\(^2\)
Source: Myschool Past Questions Archive (Record #32754)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, find x in terms of a, b and c.
Attached Figure / Diagram
A
a + b + c
B
180o - (a + b + c)
C
a - b - c
D
a + b
E
a + c
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
180 - x + a + b + c
= 180(sum of interior angle in triangle)
a + b + c = x.
Source: Myschool Past Questions Archive (Record #32755)
Mathematics
UTME / JAMB 1981
Mensuration (Areas, Perimeters & Volumes)
Diagram
Solution
In the diagram, XZ is the diameter of a circle's radius \(\frac{5}{2}\). If XY is 4cm, then the area of the triangle XYZ is
Attached Figure / Diagram
A
12cm2
B
28\(\frac{1}{2}\)cm2
C
16cm2
D
10cm2
E
6cm2
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Area of the triangle XYZ = \(\frac{1}{2}\)bh = \(\frac{1}{2}\)ZY x XY
= \(\frac{1}{2}\) x 3 x 4
= 6cm2.
Source: Myschool Past Questions Archive (Record #32756)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
In the figure, WU//YZ, WY//VU = 12cm, VZ = 6cm, XU = 8cm. Determine the length of WU.
Attached Figure / Diagram
A
1cm
B
6cm
C
2cm
D
4cm
E
3cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
From similar triangle, \(\frac{x}{6}\) = \(\frac{8}{12}\)
12x = 48
x = \(\frac{48}{12}\)
= 4
Source: Myschool Past Questions Archive (Record #32757)