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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)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
If a rod 10cm in length was measured as 10.5cm, calculate the percentage error
A
5%
B
10%
C
8%
D
7%
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Actual measurement = 10cm
approximated value of measurement = 10.5cm
% error = \(\frac{\text{Actual measurement − Approximated}}{\text{Actual measure}}\) × 100
= \(\frac{10 − 10.5}{10}\) × 100
= \(\frac{-0.5}{10}\) × 100
ignore -sign i.e take absolute value
= \(\frac{0.5}{10}\) × 100
= 5 %
Source: Myschool Past Questions Archive (Record #46348)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Find the principal which amounts to ₦ 5,500 at a simple interest in 5 years at 2% per annum
A
₦ 4,900
B
₦ 5,000
C
₦ 4,700
D
₦ 4,000
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Principal = P, Simple Interest = I, Amount = A
Amount = Principal + Simple Interest
I = \(\frac{PRT}{100}\)
R = rate, T = time
I = \(\frac{P \times 5 \times 2}{100}\)
I = \(\frac{10P}{100}\)
I = \(\frac{P}{10}\)
Amount A = P + I
5500 = P + \(\frac{P}{10}\)
Multiply through by 100
5500 = 10P + P
5500 = 11P
p = \(\frac{5500}{11}\)
p = ₦5000
Source: Myschool Past Questions Archive (Record #46349)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
In how many ways can the word MATHEMATICS be arranged?
A
\(\frac{11!}{9!2!}\)
B
\(\frac{11!}{9!2!2!}\)
C
\(\frac{11!}{2!2!2!}\)
D
\(\frac{11!}{2!2!}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
MATHEMATICS is an eleven-letter word = 11!
There are 2Ms and 2As and 2Ts
Divide the number of repeating letters
= \(\frac{11!}{2!2!2!}\)
Source: Myschool Past Questions Archive (Record #46351)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Factorize completely X\(^2\)+2XY+Y\(^2\)+3X+3Y-18
A
(x + y + 6)(x + y -3)
B
(x - y - 6)(x - y + 3)
C
(x - y + 6)(x - y - 3)
D
(x + y - 6)(x + y + 3)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x^{2} + 2xy + y^{2} + 3x + 3y - 18\)
\(x^{2} + 2xy + 3x + y^{2} + 3y -18\)
\(x^{2} + 2xy - 3x + 6x + y^{2} -3y + 6y -18\)
\(x^{2} + 2xy -3x + y^{2} -3y + 6x + 6y -18\)
\(x^{2} + xy -3x + xy + y^{2} - 3y + 6x + 6y -18\)
x(x + y - 3) + y(x + y - 3) + 6(x + y - 3)
= (x + y - 3)(x + y + 6)
= (x + y + 6)(x + y -3)
Source: Myschool Past Questions Archive (Record #46359)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Make S the subject of the relation
p = s + \(\frac{sm^2}{nr}\)
A
s = \(\frac{nrp}{nr + m^2}\)
B
s = nr + \(\frac{m^2}{mrp}\)
C
s = \(\frac{nrp}{mr}\) + m2
D
s = \(\frac{nrp}{nr}\) + m2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
p = s + \(\frac{sm^2}{nr}\)
p = s + ( 1 + \(\frac{m^2}{nr}\))
p = s (1 + \(\frac{nr + m^2}{nr}\))
nr \times p = s (nr + m2)
s = \(\frac{nrp}{nr + m^2}\)
Source: Myschool Past Questions Archive (Record #46360)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
The operation * on the set R of real number is defined by x * y = 3x + 2y − 1, find 3* − 1
A
9
B
- 9
C
6
D
- 6
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x * y is an operation on 3x + 2y − 1
Find 3A − 1
x = 3, y = −1
3 * − 1 on 3x + 2y − 1
3(3) + 2(−1) −1
= 9 − 2 − 1
= 6
Source: Myschool Past Questions Archive (Record #46361)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
Simplify (\(\sqrt[3]{64a^3})^{-1}\)
A
4a
B
\(\frac{1}{8a}\)
C
8a
D
\(\frac{1}{4a}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
(\(\sqrt[3]{64a^3})^{-1}\)
\(\frac{1}{(\sqrt[3]{64a^3})}\)
= \(\frac{1}{4a}\)
Source: Myschool Past Questions Archive (Record #46363)
Mathematics
UTME / JAMB 2017
Number and Numeration
Solution
If \(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) = m + n √ 6,
find the values of m and n respectively
A
1, − 2
B
− 2, n = 1
C
\(\frac{-2}{5}\), 1
D
\(\frac{2}{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + n√6
\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
\(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\)
\(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\)
= \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\)
= \(\frac{0 - 4 \sqrt{6} - 6}{5}\)
= \(\frac{10 - 5 \sqrt{6}}{5}\)
= − 2 + √6
∴ m + n\(\sqrt{6}\) = − 2 + √6
m = − 2, n = 1
Source: Myschool Past Questions Archive (Record #46364)