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Mathematics
UTME / JAMB 2018
Algebra
Solution
Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm
A
1920cm\(^2\)
B
592cm\(^2\)
C
296cm\(^2\)
D
148cm\(^2\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Total surface area of a cupboard is given by the equation A = 2(lb + bh + lh) L = 12, b = 10, h = 8
A = 2((12 x 10) + (10 x 8) + (12 x 8))
A = 2(120 + 80 + 96)
A = 2 x 296
A = 592cm\(^2\)
Answer is B
Source: Myschool Past Questions Archive (Record #46988)
Mathematics
UTME / JAMB 2018
Number and Numeration
Solution
Convert 0.04945 to two significant figures
A
0.040
B
0.049
C
0.050
D
0.49
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
0.04945 to 2s.f is 0.049
Source: Myschool Past Questions Archive (Record #46989)
Mathematics
UTME / JAMB 2018
Statistics and Probability
Solution
The probabilities that John and James pass an examination are \(\frac {3}{4}\)and \(\frac {3}{5}\) respectively. Find the probability of both boys failing the examination.
A
\(\frac {1}{10}\)
B
\(\frac {2}{10}\)
C
\(\frac {9}{20}\)
D
\(\frac {11}{20}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
To find the probability of both John and James failing the examination:
Probabilities of failing:
\(P(\text{John fails}) = 1 - P(\text{John passes}) = 1 - \frac{3}{4} = \frac{1}{4}\)
\(P(\text{James fails}) = 1 - P(\text{James passes}) = 1 - \frac{3}{5} = \frac{2}{5}\)
Probability of both failing:
\(P(\text{Both fail}) = P(\text{John fails}) \times P(\text{James fails}) = \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10}\)
Thus, the probability of both boys failing the examination is:
\(\frac{1}{10}\)
Source: Myschool Past Questions Archive (Record #46991)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
An arc of a circle of radius 14cm subtends angle 300^\circ at the centre. Find the perimeter of the sector formed by the arc (take \pi = \(\frac{22}{7}\))
A
14.67cm
B
73.33 cm
C
101.33cm
D
513.33cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Length of the arc = \(\frac{\theta}{360} \times 2\pi r\)
= \(\frac{300}{360} \times 2 \times \frac{22}{7} \times 14\)
= \(\frac{220}{3}\)
= 73.33 cm
Perimeter of the sector = Length of the arc + 2r
= 73.33 + 2(14)
= 101.33 cm
Source: Myschool Past Questions Archive (Record #46993)
Mathematics
UTME / JAMB 2018
Number and Numeration
Solution
Simplify
\(\frac {25^{\frac{2}{3}} \div 25^{\frac{1}{6}}} {(\frac{1}{5})^{\frac{7}{6}} \div (\frac{1}{5})^{\frac{1}{6}}}\)
A
25
B
1
C
\(\frac{1}{5}\)
D
\(\frac{1}{25}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac {25^{\frac{2}{3}} \div 25^{\frac{1}{6}}} {(\frac{1}{5})^{\frac{7}{6}} \div (\frac{1}{5})^{\frac{1}{6}}}\)
= \(\frac{25^{\frac{2}{3} - \frac{1}{6}}}{(\frac{1}{5})^{\frac{7}{6} - \frac{1}{6}}}\)
= \(\frac{25^{\frac{1}{2}}}{(\frac{1}{5})}\)
= \(5 \div \frac{1}{5}\)
= 25
Source: Myschool Past Questions Archive (Record #46994)
Mathematics
UTME / JAMB 2018
Number and Numeration
Diagram
Solution
Make T the subject of the relation.
Attached Figure / Diagram
A
T = \(\frac{R + P3}{15Q}\)
B
T = \(\frac{R - 15P^3}{Q}\)
C
T =R - \(\frac{15P^3}{Q}\)
D
T = \(\frac{15R + Q}{P^3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
P = (\(\frac{Q( R - T )}{ 15})^\frac{1}{3}\)
take cube of both sides
\(P^3 =\frac{Q( R - T)}{ 15}\)
cross multiply
\(15P^3 = Q( R - T)\)
\(\frac{15P^3}{Q}\) = R - T
T = R - \(\frac{15P^3}{Q}\)
Source: Myschool Past Questions Archive (Record #46995)
Mathematics
UTME / JAMB 2018
Number and Numeration
What is the place value of 9 in the number 3.0492?
A
\(\frac{9}{10000}\)
B
\(\frac{9}{1000}\)
C
\(\frac{9}{100}\)
D
\(\frac{9}{10}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #46996)
Mathematics
UTME / JAMB 2018
Number and Numeration
If the simple interest on a sum of money invested at 3% per annum for 2\(\frac{1}{2}\) years is N123, find the principal.
A
N676.50
B
N820
C
N1,640
D
N4,920
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #46997)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?
A
N16,200
B
N14,200
C
N12,000
D
N8,000
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Since it depreciates by 10% At the end of first year, its value = 90% of 20000
= \(\frac{90}{100}\) x 20000 =18000
At the end of second year, its value = 90% of 18000
= \(\frac{90}{100}\) x 18000 = ₦16,200
Answer is A
Source: Myschool Past Questions Archive (Record #46998)
Mathematics
UTME / JAMB 2018
Statistics and Probability
Solution
The table shown gives the marks scored by a group of student in a test. Use the table to answer the question given.
Mark
0
1
2
3
4
5
Frequency
1
2
7
5
4
3
What is the median mark?
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total frequency = 1 + 2 + 7 + 5 + 4 + 3 = 22
Median is the middle number
= \(\frac{Nth}{2}\)
term = \(\frac{22th}{2}\) = 11th term
Going in ascending order, 11th term is 3, going in descending order 11th term is 3
Median = 3 + \(\frac{3}{2}\) = \(\frac{6}{2}\) = 3
Answer is C
Source: Myschool Past Questions Archive (Record #46999)
Mathematics
UTME / JAMB 2018
Statistics and Probability
The table shown gives the marks scored by a group of student in a test. Use the table to answer the question given.
Mark
0
1
2
3
4
5
Frequency
1
2
7
5
4
3
What is the probability of selecting a student from the group that scored 2 or 3
A
\(\frac{1}{11}\)
B
\(\frac{5}{22}\)
C
\(\frac{7}{22}\)
D
\(\frac{6}{11}\)
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 #47000)
Mathematics
UTME / JAMB 2018
Number and Numeration
A boy walks 800m in 20 minutes. Calculate his average speed in Km/H
A
2.4
B
4
C
24
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #47001)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
A car uses one litre of petrol for every 14km. If one litre of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?
A
420Km
B
405Km
C
210Km
D
200Km
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
since 1litre cost ₦63
So ₦900 is the cost of \(\frac{900}{63}\) litres
= 14.2857litres of petrol
nbsp; The car uses 1 litre for 14km
So 14.2857 litres will be used for (14.2857 x 14)
= 200km
Answer is D
Source: Myschool Past Questions Archive (Record #47002)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Diagram
Solution
In the diagram, GI is a tangent to the circle at H. If EF||GI, calculate the size of ∠EHF
Attached Figure / Diagram
A
126°
B
72°
C
54°
D
28°
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Source: Myschool Past Questions Archive (Record #47003)
Mathematics
UTME / JAMB 2018
Geometry and Trigonometry
Solution
How many times, correct to the nearest whole number, will a man run round a circular track of diameter 100m to cover a distance of 1000m?
A
3
B
4
C
5
D
6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In terms of distance, a circle has a total distance or perimeter of 2\pir or \pid
Where r is radius and d is the diameter
So perimeter = \(\frac{22}{7}\) x 100
= 314.2857m
To cover a distance of 1000m, he is going to round the circular track for \(\frac{1000}{314.2857}\) = 3.18
\(\approxeq\) 3 (to the nearest whole number)
Source: Myschool Past Questions Archive (Record #47004)