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Mathematics
UTME / JAMB 1984
Number and Numeration
Solution
A cone is formed by bending a sector of a circle having an angle of 210o. Find the radius of the base of the cone if the diameter of the circle is 12cm.
A
12cm
B
7.00cm
C
1.75cm
D
21cm
E
3.50cm
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\)
= 6cm
\(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm
\(\frac{210}{360}\) = \(\frac{r}{6}\)
where r = radius of the base of the cone
V = \(\frac{1260}{360}\)
= 3.50cm
Source: Myschool Past Questions Archive (Record #32432)
Mathematics
UTME / JAMB 1984
Number and Numeration
Solution
If a = \(\frac{2x}{1 - x}\) and b = \(\frac{1 + x}{1 - x}\), then a2 - b2 in the simplest form is
A
\(\frac{3x + 1}{x - 1}\)
B
\(\frac{3x^2 - 1}{(x - 1)}\)2
C
x\(\frac{3x - 2}{1 - x}\)
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2
= (\(\frac{2x}{1 - x} + \frac{1 + x}{1 - x}\))(\(\frac{2x}{1 - x} - \frac{1 + x}{1 - X}\))
= (\(\frac{3x + 1}{1 - x}\))(\(\frac{x - 1}{1 - x}\))
= \(\frac{3x + 1}{x - 1}\)
Source: Myschool Past Questions Archive (Record #32434)
Mathematics
UTME / JAMB 1984
Number and Numeration
Solution
Simplify (1 + \(\frac{\frac{x - 1}{1}}{\frac{1}{x + 1}}\))(x + 2)
A
(x2 - 1)(x + 2)
B
x2(x + 2)
C
2 + 34
D
\(\frac{3x^2 - 1}{(x - 1)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\((1 + \frac{x - 1}{\frac{1}{x + 1}})(x + 2)\)
\((x - 1) \div \frac{1}{x + 1} = (x - 1)(x + 1) = x^{2} - 1\)
\((1 + x^{2} - 1)(x + 2) = x^{2}(x + 2)\)
Source: Myschool Past Questions Archive (Record #32435)
Mathematics
UTME / JAMB 1984
Number and Numeration
Solution
A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is
A
r tan yo cm
B
r sin yo cm
C
r cot yo cm
D
r cos yo cm
E
r cosec yo cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{r}{h}\) = tan yo
h = \(\frac{r}{tan y^o}\)
= r cot yo
Source: Myschool Past Questions Archive (Record #32438)
Mathematics
UTME / JAMB 1984
Number and Numeration
Solution
The larger value of y for which (y - 1)2 = 4y - 7 is
A
2
B
4
C
6
D
7
E
8
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4
Source: Myschool Past Questions Archive (Record #32440)
Mathematics
UTME / JAMB 1984
Number and Numeration
Diagram
Solution
What is the volume of this regular three dimensional figure?
Attached Figure / Diagram
A
160cm2
B
48cm2
C
120cm2
D
40cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Volume of the three dimensional figures = v = A x h
A = \(\frac{1}{2}\) x 4 x 3
= 6cm2
V = 6 x 8
= 48cm2
Source: Myschool Past Questions Archive (Record #32776)
Mathematics
UTME / JAMB 1983
Number and Numeration
Solution
Factorize completely 81a\(^4\) - 16b\(^4\)
A
(3a + 2b)(2a - 3b)(9a2 + 4b2)
B
(3a - 2b)(2a - 3b)(4a2 - 9b2)
C
(3a - 2b)(3a + 2b)(9a2 + 4b2)
D
(6a - 2b)(8a - 3b)(4a3 - 9b2)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
81a\(^4\) - 16b\(^4\) = (9a\(^2\))\(^2\) - (4b\(^2\))\(^2\)
= (9a\(^2\) + 4b\(^2\))(9a\(^2\) - 4b\(^2\))
N:B 9a\(^2\) - 4b\(^2\) = (3a - 2b)(3a + 2b)
Source: WAEC / JAMB / NECO Classroom Archive (Record #32471)
Mathematics
UTME / JAMB 1983
Number and Numeration
Solution
If (x + 2) and (x - 1) are factors of the expression \(Lx^{3} + 2kx^{2} + 24\), find the values of L and k.
A
1 = -6, k = -9
B
1 = -2, k = 1
C
k = -1, 1 = -2
D
1 = 0, k = 1
E
k = 0,1 = 6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
L = -6
Source: WAEC / JAMB / NECO Classroom Archive (Record #32452)
Mathematics
UTME / JAMB 1983
Number and Numeration
Solution
A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received N5,000 more than X. What is the total profit of the company for the year?
A
N20, 000
B
N25,000
C
N30,000
D
N15,000
E
N45, 000
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is N5000
Total profit = 5000 x 9
= N45,000
Source: Myschool Past Questions Archive (Record #32445)
Mathematics
UTME / JAMB 1983
Number and Numeration
Solution
If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?
A
4:1 increase
B
2:1 increase
C
1:1 no change
D
1:4 decrease
E
3:4 decrease
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(x \propto y^{3} z^{4}\)
\(x = ky^{3} z^{4}\)
If y is halved and z is doubled, we have
\(x = k (\frac{1}{2} y)^{3} (2z)^{4}\)
\(x = k (\frac{y^{3}}{8}) (16z^{4})\)\)
\(x = 2k y^{3} z^{4}\)
\(\therefore \text{x is increased in the ratio 2 : 1}\)
Source: Myschool Past Questions Archive (Record #32447)
Mathematics
UTME / JAMB 1983
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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)