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Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group
A
10.1 years
B
9.3 years
C
8.7 years
D
8 . 3 years
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
x = 10 ; 10 = \(\frac{x}{25}\)
x = 250
x = \(\frac{250 + 12.4}{26}\)
x = 10.09
x = 10.1 years
Source: WAEC / JAMB / NECO Classroom Archive (Record #50023)
Mathematics
WAEC / WASSCE 2017
Number and Numeration
Solution
Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
A
x = \(\frac{6 + 12}{d^2 + y}\)
B
x = \(\frac{12}{d^2 - y}\)
C
x = \(\frac{12}{y} - 2d^2\)
D
x = \(\frac{12}{2d^2 + y}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
\(d^2 = \frac{6}{x} - \frac{y}{2}\)
\(2xd^2 = 12 - xy\)
\(2xd^2 + xy = 12\)
x = \(\frac{12}{2d^2 + y}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #50125)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Given T = { even numbers from 1 to 12 }
N = {common factors of 6, 8 and 12}
Find T ∩ N
A
{2, 3}
B
{2, 3, 4}
C
{3, 4, 6}
D
{2}
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
T = {evenn numbers from 1 to 12}
N = {common factors of 6,8 and 12}
Find T ∩ N
T = {2, 4, 6, 8, 10, 12}
N = {2}
T ∩ N = {2} i.e value common to T & N
Source: Myschool Past Questions Archive (Record #46321)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
The curved surface area of a cylinder 5cm high is 110cm2. Find the radius of its base
π = \(\frac{22}{7}\)
A
2.6cm
B
3.5cm
C
3.6cm
D
7.0cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Curved surface area of cylinder = 2πrh
110 = 2 × \(\frac{22}{7}\) × r × 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
Source: Myschool Past Questions Archive (Record #46324)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
If two graphs Y = px2 + q and y = 2x2 − 1 intersect at x =2, find the value of p in terms of q
A
q − \(\frac{8}{7}\)
B
7 − \(\frac{q}{4}\)
C
8 − \(\frac{q}{2}\)
D
7 + \(\frac{q}{8}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Y = Px2 + q
Y = 2x2 - 1
Px2 + q = 2x2 - 1
Px2 = 2x2 - 1 - q
p = \(\frac{2x^2 - 1 - q}{x^2}\)
at x = 2
P = \(\frac{2(2)^2 - 1 - q}{2^2}\)
= \(\frac{2(4) - 1 -q}{4}\)
= \(\frac{8 - 1 - q}{4}\)
P = \(\frac{7 - q}{4}\)
Source: Myschool Past Questions Archive (Record #46325)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Evaluate (\(\sin\)45º + \(\sin\)30º ) in surd form
A
\(\frac{\sqrt{3}}{2\sqrt{2}}\)
B
√3 − \(\frac{1}{2}\)
C
\(\frac{1}{2}\)√2
D
\(\frac{1 +\sqrt{2}}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
hypotenuse
sin = \(\frac{1}{2}\)
\(\sin45 = \frac{1}{\sqrt{2}}\)
= \(\frac{2}{2}\)
∴ (sin45 + sin30)
= \(\frac{1}{\sqrt{2}} + \frac{1}{2}\)
= \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\)
= \(\frac{\sqrt{2} + 1}{2}\)
= \(\frac{1 + \sqrt{2}}{2}\)
Source: Myschool Past Questions Archive (Record #46326)
Mathematics
UTME / JAMB 2017
Number and Numeration
Evaluate 1 - (\(\frac{1}{5}\) x \(\frac{2}{3}\)) + ( 5 + \(\frac{2}{3}\))
A
4
B
3
C
2\(\frac{2}{3}\)
D
\(\frac{98}{15}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #46329)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Given m = N\(\sqrt{\frac{SL}{T}}\) make T the subject of the formula
A
\(\frac{\text{NSL}}{M}\)
B
\(\frac{N^2SL}{M^2}\)
C
\(\frac{N^2SL}{M}\)
D
\(\frac{NSL}{M^2}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
M = N \(\sqrt{\frac{SL}{T}}\),
make T subject of formula square both sides
M\(^{2}\) = \(\frac{N^2SL}{T}\)
TM\(^{2}\) = N\(^{2}\)SL
T = \(\frac{N^2SL}{M^2}\)
Source: Myschool Past Questions Archive (Record #46330)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Simplify 3 \(^{n − 1}\) \times \(\frac{27^{n + 1}}{81^n}\)
A
3\(^{2n}\)
B
9
C
3n
D
3 \(^{n + 1}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
3\(^{n - 1}\) \times \(\frac{27^{n + 1}}{81^n}\)
= 3\(^{n - 1}\) \times \(\frac{3^{3(n + 1)}}{3^{4n}}\)
= 3\(^{n - 1 + 3n + 3 − 4n}\)
= 3\(^{4n − 4n − 1 + 3}\)
= \(3^{2}\)
= 9
Source: Myschool Past Questions Archive (Record #46331)
Mathematics
UTME / JAMB 2017
Number and Numeration
Diagram
Solution
The base in which the operation was performed was
Attached Figure / Diagram
A
6
B
2
C
4
D
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The operation was performed in base 2 because when counting in base 2, there are only two possible digits: 0 and 1.
Source: Myschool Past Questions Archive (Record #46337)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
The value of x + x ( xx) when x = 2 is
A
16
B
10
C
18
D
24
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
when x=2, we have
\(2 + 2(2^{2})\) = \(2 + 8 = 10\)
Source: Myschool Past Questions Archive (Record #46338)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
A cylindrical tank has a capacity of 3080m3. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.
A
23m
B
25m
C
20m
D
22m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Capacity = Volume = 3080m3
base diameter = 14m
radius = \(\frac{\text{diameter}}{2}\)
= 7m
Volume of Cylidner = Capacity of cylinder
πr2h = 3080
\(\frac{22}{7}\) × 7 × 7 × h = 3080
h = \(\frac{3080}{22 \times 7}\)
h = 20m
Source: Myschool Past Questions Archive (Record #46340)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Simplify 4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)
A
7
B
− 7
C
− 7\(\sqrt{3}\)
D
7\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\)
= 4\(\sqrt{3}\) × 9 + 5\(\sqrt{3}\) × 4 − 3\(\sqrt{3}\) × 25
= 4 × 3\(\sqrt{3}\) + 5 × 2\(\sqrt{3}\) − 3 × 5\(\sqrt{3}\)
= 12\(\sqrt{3}\) + 10\(\sqrt{3}\) − 15\(\sqrt{3}\)
= (12 + 10 − 15)\(\sqrt{3}\)
= 7\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #46341)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Divide 4x\(^3\) - 3x + 1 by 2x - 1
A
2x2 -x + 1
B
2x2 - x -1
C
2x2 + x + 1
D
2x2 + x -1
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
by method of long division, we get the answer.
Source: Myschool Past Questions Archive (Record #46344)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
A car dealer bought a second-hand car for 250,000 and spent N 70,000 refurbishing it. He then sold the car for N400,000. What is the percentage gain?
A
60%
B
32%
C
25%
D
20%
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total Cost Price = N(250,000 + 70,000)
= N 320,000
Selling Price = N 400,000(Given)
Gain = Selling Price - Cost Price
= 400,000 - 320,000
= 80,000
% gain = \(\frac{\text{Gain}}{\text{Cost Price}}\) \times 100
= \(\frac{80,000}{320,000}\) \times 100
Gain % = 25%
Source: Myschool Past Questions Archive (Record #46345)