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Mathematics
UTME / JAMB 2017
Mensuration (Areas, Perimeters & Volumes)
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
Indices, Logarithms & Surds
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
Trigonometry, Elevation & Bearings
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
Differentiation & Applications
Solution
If y = x Sin x, find \(\frac{dy}{dx}\) when x = \(\frac{\pi}{2}\)
A
\(\frac{- \pi}{2}\)
B
-1
C
1
D
\(\frac{ \pi}{2}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
y = xsinx
\(\frac{dy}{dx}\) = \(1 \sin x + x \cos x\)
= \(\sin x + x \cos x\)
At x = \(\frac{\pi}{2}\)
= sin\(\frac{\pi}{2}\) + \(\frac{\pi}{2} \cos {\frac{\pi}{2}}\)
= 1 + \(\frac{\pi}{2}\) \times 0
= 1
Source: Myschool Past Questions Archive (Record #46327)
Mathematics
UTME / JAMB 2017
Variations
Solution
If temperature t is directly proportional to heat h, and when t = 20oC, h = 50 J, find t when h = 60J
A
24oC
B
20oC
C
34oC
D
30oC
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
t ∝ h, t = 20, h
t = ? h = 60
t = kh where k is constant
20 = 50k
k = \(\frac{20}{50}\)
k = \(\frac{2}{5}\)
when h = 60, t = ?
t = \(\frac{2}{5}\) × 60
t = 24oC
Source: Myschool Past Questions Archive (Record #46328)
Mathematics
UTME / JAMB 2017
Plane Geometry, Polygons & Circle Theorems
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
Algebraic Expressions, Polynomials & Factorisation
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
Indices, Logarithms & Surds
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
Plane Geometry, Polygons & Circle Theorems
Solution
The locus of a point which is equidistant from the line PQ forms a
A
circle centre P
B
pair of parallel lines each opposite to PQ
C
circle centre Q
D
perpendicular line to PQ
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The locus of points equidistant from a given straight line PQ consists of two parallel lines, one on each side of PQ, each at the same perpendicular distance from it.
Source: Myschool Past Questions Archive (Record #46332)
Mathematics
UTME / JAMB 2017
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
Given the quadrilateral RSTO inscribed in the circle with O as centre. Find the size angle x and given RST = 60o
Attached Figure / Diagram
A
100o
B
140o
C
120o
D
10o
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
If RST = 60o
RXT = 2 × RST
(angle at the centre twice angle at the circumference)
RXT = 2 × 60
= 120o
Source: Myschool Past Questions Archive (Record #46334)
Mathematics
UTME / JAMB 2017
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5
A
16
B
14
C
12
D
10
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Range = Highest Number - Lowest Number
Mode is the number with highest occurrence
10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5
Range = 10 − 4 = 6
Mode = 10
Sum of range and mode = range + mode = 6 + 10
= 16
Source: Myschool Past Questions Archive (Record #46335)
Mathematics
UTME / JAMB 2017
Sequences & Series (AP & GP)
Solution
Find the sum to infinity of the series
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........
A
\(\frac{1}{2}\)
B
\(\frac{3}{5}\)
C
\(\frac{-1}{5}\)
D
\(\frac{73}{12}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Sum to infinity
∑ = arn − 1
= \(\frac{a}{1}\) − r
a = \(\frac{1}{4}\)
r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)
r = \(\frac{1}{s}\) × \(\frac{4}{1}\)
= \(\frac{1}{2}\)
S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)
= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{2}{1}\)
= \(\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #46336)
Mathematics
UTME / JAMB 2017
Number Bases & Modular Arithmetic
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
Indices, Logarithms & Surds
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
Plane Geometry, Polygons & Circle Theorems
Solution
In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon
A
8
B
6
C
4
D
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2x + x = 180o
3x = 180o
x = 60o (exterior angle of the polygon)
angle = \(\frac{\text{total angle}}{\text{number of sides}}\)
60 = \(\frac{360}{n}\)
n = \(\frac{360}{60}\)
n = 6 sides
Source: Myschool Past Questions Archive (Record #46339)