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Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?
(Take pi = 22/7)
A
23 m
B
25 m
C
20 m
D
22 m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Capacity = 3080 m3
Base diameter = 14 m, thus radius = 7 m.
Volume of cylinder = capacity of cylinder = πr2h. = 3080
22/7 x 72 x h = 3080
h = 3080/(22x7)
h = 20 m
Source: Myschool Past Questions Archive (Record #2123)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
Evaluate \(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)
A
3/5(2x-3)5/3 + k
B
6/5(2x-3)5/3 + k
C
2x-3+k
D
2(2x-3)+k
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\int 2(2x - 3)^{\frac{2}{3}} \mathrm d x\)
Let \(u = 2x - 3\)
\(\mathrm d u = 2 \mathrm d x\)
= \(\int u^{\frac{2}{3}} \mathrm d u\)
= \(\frac{u^{\frac{5}{3}}}{\frac{5}{3}} + k\)
= \(\frac{3}{5} (2x - 3)^{\frac{5}{3}} + k\)
Source: Myschool Past Questions Archive (Record #2127)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
Find the area bounded by the curves y = 4 - x2 and y = 2x + 1
A
20(1/3) sq. units
B
20(2/3) sq. units
C
10(2/3) sq. units
D
10(1/3) sq. units
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Hint:
y = 4 - x\(^2\) and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = 1 or x = -3.
Integrating x\(^2\) + 2x - 3
= 3x - x\(^2\) - \(\frac{x^3}{3}\)
from (1, to -3) : 3 (1) - 1\(^2\) - \(\frac{1^3}{3}\) - 3 (-3) - -3\(^2\) - \(\frac{-3^3}{3}\)
= \(\frac{5}{3}\) + 9
will give 32/3 = 10 \(\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #2550)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
Find the rate of change of the volume, V of a sphere with respect to its radius, r when r = 1.
A
12π
B
4π
C
24π
D
8π
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Volume of sphere, V = 4/3 x πr3
Rate of change of V = dv/dr
Thus if V = 4/3 x πr3,
=> dv/dr = 4πr2
At r = 1, Rate = 4 x π x 1 = 4π
Source: Myschool Past Questions Archive (Record #2559)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
If 6Pr = 6, find the value of 6Pr+1
A
30
B
33
C
35
D
15
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
6Pr = 6 Thus r = 1
N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\)
= \(\frac{\text{6!}}{\text{5!}}\)
= \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6
6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\)
= \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\)
= 30
Source: Myschool Past Questions Archive (Record #2869)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
Find the number of ways of selecting 8 subjects from 12 subjects for an examination
A
490
B
495
C
496
D
498
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
To calculate the binomial coefficient \(^{12}C_{8}\):
\(^{12}C_{8} = \frac{12!}{8!4!} = \frac{12 \times 11 \times 10 \times 9}{4!}\)
Calculating \(4!\):
\(4! = 24\)
Thus, \(^{12}C_{8} = \frac{12 \times 11 \times 10 \times 9}{24} = \frac{11880}{24}\) = 495
= \(^{12}C_{8}\) = 495
Source: Myschool Past Questions Archive (Record #7715)
Mathematics
UTME / JAMB 2001
Number and Numeration
Solution
Find the range of \(\frac{1}{6}\), \(\frac{1}{3}\), \(\frac{3}{2}\), \(\frac{2}{3}\), \(\frac{8}{9}\), and \(\frac{4}{3}\)
A
3/4
B
5/6
C
7/6
D
4/3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{1}{6}\), \(\frac{1}{3}\), \(\frac{3}{2}\), \(\frac{2}{3}\), \(\frac{8}{9}\) and \(\frac{4}{3}\).
L.C.M of the denominators = 36
The fraction now becomes
\(\frac{6}{36}\), \(\frac{12}{36}\), \(\frac{54}{36}\), \(\frac{24}{36}\), \(\frac{32}{36}\), and \(\frac{48}{36}\).
Highest fraction = \(\frac{54}{36}\)
Lowest fraction = \(\frac{6}{36}\)
Range = Highest - Lowest
( \(\frac{54}{36}\)) - ( \(\frac{6}{36}\))
= \(\frac{48}{36}\)
= \(\frac{4}{3}\).
Source: Myschool Past Questions Archive (Record #7718)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
Evaluate (\(\frac{1}{343}\))\(^{\frac{1}{3}}\) + (64)\(^{-\frac{1}{3}}\) - \(\frac{4}{9}\)\(^{-\frac{1}{2}}\)
A
-\(\frac{17}{36}\)
B
-\(\frac{3}{28}\)
C
-\(\frac{31}{28}\)
D
\(\frac{17}{36}\)
E
\(\frac{1}{4}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(\(\frac{1}{343}\))\(^{\frac{1}{3}}\) + (64)\(^{-\frac{1}{3}}\) - \(\frac{4}{9}\)\(^{-\frac{1}{2}}\)
where \(343 = 7^3, 64 = 2^6, 4 = 2^2, 9 = 3^2\)
\(7^{-3}\) \(^{\frac{1}{3}}\) + \(2^6\) \(^{\frac{-1}{3}}\) - \(\frac{2^2}{3^2}\) \(^{\frac{-1}{2}}\)
\(7^{-1} + 2^{-2} - (\frac{2}{3})^{-1}\)
\(\frac{1}{7} + \frac{1}{4} - \frac{3}{2}\)
\(\frac{4+7-42}{28}\)
-\(\frac{31}{28}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #41952)
Mathematics
NECO SSCE 2001
Number and Numeration
Solution
Given that \(\frac{(3 - y)}{2x}\) = \(\frac{(6y + 7)}{(4x - 5)}\), find the value of x when y = 2
A
\(\frac{64}{5}\)
B
6
C
\(\frac{5}{34}\)
D
\(\frac{-5}{34}\)
E
\(\frac{-64}{5}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given:
\(\frac{3 - y}{2x} = \frac{6y + 7}{4x - 5}\)
Substitute y = 2
\(\frac{3 - 2}{2x} = \frac{6(2) + 7}{4x - 5} \implies \frac{1}{2x} = \frac{19}{4x - 5}\)
Cross-multiply: 1 \(\cdot\) (4x - 5) = 19 \(\cdot\) 2x
4x - 5 = 38x
-5 = 38x - 4x
-5 = 34x
x = -\(\frac{5}{34}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #41801)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Diagram
Solution
How many students are there in the group?
Attached Figure / Diagram
A
25
B
30
C
52
D
55
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total number of students = 20 + 16 + 10 + 6
= 52 students
Source: WAEC / JAMB / NECO Classroom Archive (Record #26506)
Mathematics
UTME / JAMB 2001
Number and Numeration
Diagram
Solution
The bar chart above shows different colours of passing a particular point of a certain street in two minutes. What fraction of the total number of cars is yellow?
Attached Figure / Diagram
A
3/25
B
2/25
C
1/5
D
4/15
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Number of yellow cars = 3
Total number of cars = 3 + 4 + 8 + 2 + 6 + 2 = 25
Fraction of yellow cars = 3/25
Source: Myschool Past Questions Archive (Record #10770)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
Which of the following correctly expresses 48 as a product of prime factors?
A
3 x 4 x 4
B
2 x 3 x 8
C
2 x 2 x 2 x 3 x 4
D
2 x 2 x 2 x 2 x 3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
48 = 2 x 2 x 2 x 2 x 3
Source: Myschool Past Questions Archive (Record #26472)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
Evaluate \((20_{three})^2 - (11_{three})^2\) in base three
A
101
B
121
C
202
D
2020
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\((20_{three})^2 - (11_{three})^2\)
= \((20_{3} - 11_{3})(20_{3} + 11_{3})\)
= \((2_{3})(101_{3})\)
= \(202_{3}\)
Source: Myschool Past Questions Archive (Record #26473)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
Evaluate \(log_{10}5 + log_{10}20\)
A
2
B
3
C
4
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(log_{10}5 + log_{10}20\)
\(log_{10}[5\times 20]\)
\(log_{10}100\)
\(log_{10}10^2\)
\(2log_{10}10 = 2\)
Source: Myschool Past Questions Archive (Record #26474)
Mathematics
WAEC / WASSCE 2001
Number and Numeration
Solution
In a class of 80 students, every students studies Economics or Geography or both. If 65 students study Economics and 50 study Geography, how many study both subjects?
A
15
B
30
C
35
D
45
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let c = no of students that offered both subjects
\(\therefore\) No of students offering Economics = 65 - c
No of students offering Geography = 50 - c
65 - c + c + 50 - c = 80
115 - c = 80
c = 35
35 students offer both Economics and Geography.
Source: Myschool Past Questions Archive (Record #26475)