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Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
When a dealer sells a bicycle for N81 he makes a profit of 8%. What did he pay for the bicycle?
A
N73
B
N74.52
C
N75
D
N87.48
E
N75.52
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Profit = 8%
\(\therefore S.P = N81 = (100 + 8)% = 108%\)
\(108% = N81\)
\(100% = \frac{81}{108} \times 100\)
= \(\frac{8100}{108} = N75\)
Source: Myschool Past Questions Archive (Record #32610)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
A solid cylinder of radius 3cm has a total surface area of 36\(\pi\)cm2. Find its height
A
2cm
B
3cm
C
4cm
D
5cm
E
6cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Area 2\(\pi\)r2 + 2 \(\pi\)r(r + h)
= 2\(\pi\)r(r + h)
36\(\pi\) = 6\(\pi\)(r + h)
36\(\pi\) = 6\(\pi\)(3 + h)
36\(\pi\) = 18\(\pi\) + 6\(\pi\)h
36\(\pi\) - 18\(\pi\) = 6\(\pi\)h
Divide both side by 6\(\pi\)
h = 3cm
Source: Myschool Past Questions Archive (Record #32614)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
The ratio of the areas of similar triangles is necessarily equal to
A
the ratio of the corresponding sides
B
the ratio of the squares of corresponding sides
C
the ratio of the corresponding heights of the triangles
D
half the ratio of the corresponding heights of the triangles
E
the ratio of the corresponding bases to the heights of the triangles
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.
Source: Myschool Past Questions Archive (Record #32615)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Simplify \(\frac{log_{10}8}{log_{10}4}\)
A
log102
B
log104
C
\(\frac{3}{2}\)
D
2
E
None of the above
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{log_{10}8}{log_{10}4}\) = \(\frac{log_{10}2^3}{log_{10}2^2}\)
= \(\frac{3 log_{10}2}{2log_{10}2}\)
= \(\frac{3}{2}\)
Source: Myschool Past Questions Archive (Record #32618)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Make c the subject of formula v = 1 - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))
A
[\(\frac{7}{3}\) + \(\frac{5}{a}\)(v - 1)] + b
B
[\(\frac{7b}{3}\) + \(\frac{5}{a}\)(v - 1)]
C
\(\frac{-7}{3}\)[b + \(\frac{5}{a}\)(v - 1)]
D
\(\frac{7}{3}\)[b + \(\frac{5v - 1}{a}\)]
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
v = 1 - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))
V - 1 = - \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))
Multiply both sides by -1
1 - V = \(\frac{a}{5}\)(b + \(\frac{3c}{7}\))
Multiply both sides by \(\frac{5}{\text{a}}\)
\(\frac{a}{5}\)(1 - V) = b + \(\frac{3c}{7}\)
\(\frac{a}{5}\)(1 - V) - b = \(\frac{3c}{7}\)
Multiply both sides by \(\frac{7}{3}\)
\(\frac{7}{3}\)( \(\frac{a}{5}\)(1 - V) - b) = c
- \(\frac{7}{3}\)(\(\frac{a}{5}\)(V - 1) + b)
Source: Myschool Past Questions Archive (Record #32619)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
What factor is common to all the expressions x2 - x, 2x2 + x - 1 and x2 - 1?
A
x
B
x + 1
C
x - 1
D
No common factor
E
(2x - 1)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x2 - x = x(x - 1)
= 2x2 + x - 1
= (2x - 1)(x + 10
x2 - 1 = (x + 1)(x - 1)
No factor
Source: Myschool Past Questions Archive (Record #32620)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
In triangle FGH, < G = 90º, < H = 60º, while in triangle XYZ, < X = 60º and < Y = 30º. Form XYZ write down the ratio equal to \(\frac{FG}{FH}\)
A
\(\frac{YX}{ZX}\)
B
\(\frac{YX}{YZ}\)
C
\(\frac{ZX}{YZ}\)
D
\(\frac{YZ}{XY}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
The two triangles are similar
\(\frac{FG}{YZ}\) = \(\frac{FH}{XY}\) = \(\frac{GH}{YZ}\) = \(\frac{FG}{FH}\) = \(\frac{YZ}{XZ}\)
Source: Myschool Past Questions Archive (Record #32625)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Find the product of (2\(\sqrt{y} - 3y\)) and (3y + 2\(\sqrt{y}\))
A
4y = y2
B
4y + 9y2
C
4y - 9y2
D
-4y - 9y2
E
2y + y2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(2\(\sqrt{y} - 3y\)) and (3y + 2\(\sqrt{y}\))
= 6y\(\sqrt{y}\) + 4y - 9y\(^2\) - 6y\(\sqrt{y}\)
= 4y - 9y\(^2\)
Source: Myschool Past Questions Archive (Record #32627)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
If it is given that \(5^{x + 1} + 5^x = 150\), then the value of x is equal to
A
3
B
4
C
1
D
2
E
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(5^{x + 1} + 5^x = 150\)
according to the laws of indices
\(5^1(5^x) + 5^x = 150\)
let \(5^x\) = y
5(y) + y = 150
6y = 150
y = \(\frac{150}{6}\)
= 25
but y = \(5^x\) = 25
\(5^x\) = 25 = \(5^2\)
equating powers since bases are the same
x = 2
Source: Myschool Past Questions Archive (Record #32631)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Find the area of the curved surface of a cone whose base radius is 6cm and whose height is 8cm. (take \(\pi\) = \(\frac{22}{7}\))
A
188.57cm2
B
1320cm2
C
188cm2
D
188.08cm2
E
10cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
S = curved surface area = \(\pi\)rL
= \(\frac{22}{7}\) x 6 x 10
= 188.57cm2
Source: Myschool Past Questions Archive (Record #32634)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Given log 2 = 0.69, log3 = 1, 10 and log7 = 1.90, all to a fixed base, find log 10.5 to the same base without using tables.
A
1.03
B
2.31
C
3.69
D
10.5
E
25
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
log 10.5 = log \(\frac{21}{2}\)
= log 21 - log 2
= log(3 x 7) - log 2
= log 3 + log 7 - log 2
= 1.10 + 1.90 - 0.69
= 3 - 0.69
= 2.31
Source: Myschool Past Questions Archive (Record #32636)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
Simplify 102 + log105
A
500
B
2 log10 5
C
10
D
25
E
log105 x 10100
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
102 + log105 = log10 10100 + log105
= log105 x 10100
Source: Myschool Past Questions Archive (Record #32637)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
The expression x\(^3\) - 4x\(^2\) + cx + d is such that x + 1 is its factor, and its value is 1 when x is -2. Find c and d.
A
c = 4 and d = -9
B
c = -4 and d = 9
C
c = -20 and d = -15
D
c = 20 and d = -15
E
c = -20 and d = 15
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
F(X) = x\(^3\) - 4x\(^2\) + cx + d
= (X + 1) Q(X) + R
x = -1, R = 0,f(-1) = -1\(^3\) - 4(-1)\(^2\) + c(-1) + d = 0
-1 - 4 - c + d = 0
d - c = 5................(i)
f(-2) = -2\(^3\) - 4(-2)\(^2\) + c(-2) + d = 1
= -8 - 16 - 2c + d
= 1
-8 - 16 - 2c + d = 1
-24 - 2c + d = 1
d - 2c = 1 + 24
d - 2c = 25.................(ii)
\(\frac{d - c = 5}{-c = 20}\) d - c = 5
c = -20
d - (-20) = 5
d + 20 = 5
d = 5 - 20
= -15
c = -20, d = -15
Source: Myschool Past Questions Archive (Record #32639)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
If a function is defined by f(x + 1) = 3x2 - x + 4, Find f(0).
A
4
B
6
C
9
D
8
E
2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
f(x + 1) = 3x2 - x + 4
f(0) = f(x + 1)
x + 1 = 0 ===> x = -1
f(0) = 3(-1)2 - (-1) + 4
f(0) = 3 + 1 + 4
= 8
Source: Myschool Past Questions Archive (Record #32640)
Mathematics
UTME / JAMB 1980
Number and Numeration
Solution
A cylindrical motor of height 12cm has uniform thickness of 2cm. If the diameter of its outer cross section is 10cm, Find the volume of the constituent material. (take \(\pi\) = \(\frac{22}{7}\)
A
\(\frac{6600}{7}\)cm3
B
270cm3
C
660cm3
D
\(\frac{4224}{7}\)cm3
E
\(\frac{1980}{7}\)cm3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
V = \(\pi\)r2h
= \(\frac{22}{7}\) x 52 x 12 - \(\frac{22}{7}\) x 32 x 12
= \(\frac{22}{7}\) x 12(52 - 32)
= \(\frac{4224}{7}\)
Source: Myschool Past Questions Archive (Record #32641)