JAMB UTME - Mathematics - 2018
Question 1 Report
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Second quadrant: Sin is positive, Cos is negative and Tan is negative
Third quadrant: Tan is positive, Sin is negative and Cos is negative
Fourth quadrant: Cos is positive, Sin is negative and Tan is negative
The correct option is the third quadrant only where Tanθ is positive and Sinθ is negative
Question 3 Report
Simplify \(25^{\frac{1}{2}} \times 8^{-\frac{2}{3}}\)
Answer Details
2512 × 8−23
= √25 x (3√8 ) -2
= 5 x 2-2
= 5 x 122 =
54 = 114
Question 4 Report
Solve for t in the equation \( \frac{3}{4}t + \frac{1}{3}(21 - t) = 11 \)
Answer Details
Multiply through by the LCM of 4 and 3 which is 12
12 x(34 t) + 12 x (13 (21 - t)) = (11 x 12)
9t + 4(21 - t) = 132
9t + 84 - 4t = 132
5t + 84 = 132
5t = 132 - 84 = 48
t = 485
t = 9 35
Question 5 Report
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Using 90 as the angle for transport
So, x + 2x + 120 + 90 = 360
3x + 210 = 3605
3x = 360 - 210
= 150
x = 1503 = 50
So the angle for housing is 2x = 2 × 50
= 100
Amount spent on housing = 100360 × 7200
= ₦2000
Question 6 Report
In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/
Answer Details
∠ XMN = ∠ XZY
Angle X is common
So, ∠ XNM = ∠ XYZ
Then from the angle relationship
XMXZ = XNXY = MNZY
XM = 8, XZ = 12 + 4 = 16,
XN = 12, XY = 8 + YM
816 = 12(8+YM)
Cross multiply
8(8 + YM) = 192
64 + 8YM = 192
8YM = 128
YM = 1288
= 16cm
Question 7 Report
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?
Answer Details
Question 8 Report
Question 9 Report
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So the answer is 1.0
Question 10 Report
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Question 11 Report
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a2=122+92
a=√122+92
a=15 m
b2=152+82
b=√152+82
b=17 m
cosθ=ab=1517
Question 12 Report
\[ \frac{0.00256 \times 0.0064}{0.025 \times 0.08} \]
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Explanation:
=0.0000163840.002
=0.008192
Question 13 Report
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Question 14 Report
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Question 15 Report
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Question 16 Report
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Question 17 Report
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Every 6 months that can be counted on the eight month of the third year is 5
(i.e. 2 times in the first year, 2 times in the second year and once in the third year)
His salary then = initial salary + increment
= 540 + 5(36)
= 540 + 180
= ₦720.00
It can also be solves using a sequence in form of an AP
Question 18 Report
A trader realises 10x - x\(^{2}\) Naira profit from the sale of x bags of corn. How many bags will give him the maximum profit?
Answer Details
Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0
i.e. δpδx = 0
δpδx = 10 - 2x = 0
10 = 2x
Then x = 102 = 5
Question 19 Report
if \(y = 23_{\mathit{five}} + 101_{\mathit{three}}\) find y leaving your answer in base two
Answer Details
23five = 2 x 51 + 3 x 50
= 10 + 3 = 13
101five = (1 x 32) + (0 x 31) + (1 x 30)
= 9 + 0 + 1 = 10
So, y = 13 + 10 = 23
To convert 23 to base 2 (as in the diagram above)
y = 23
= 10111five
Question 20 Report
Which one of the following gives the members of the set \(A^c \cap B \cap C\)?
Answer Details
B = {r, s. t, u}
C = {t, u, v, w, x}
A1 n B n C = elements common to the three sets = none = empty set = Φ
Question 21 Report
Use the cumulative frequency curve to answer the question.
Estimate the median of the date represented on the graph.
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Question 22 Report
Use the quadratic equation curve to answer this question.
What is the 80th percentile?
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Question 23 Report
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= 6×5×4×3×2×12×1
= 360
Question 24 Report
Find the values of x for which
\( \frac{x+2}{4} - \frac{2x-3}{3} < 4 \)
Answer Details
3x+6−(8x−12)12<4
−5x+1812<4
−5x+18<48
−5x<30
x>−6
Question 25 Report
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= 90100 x 20000 =18000
At the end of second year, its value = 90% of 18000
= 90100 x 18000 = ₦16,200
Question 26 Report
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.
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= 34 x 35
= 920
Pr(John and James failed)= 1 - Pr(John and James passed)
1 – 920
= 1120
Question 27 Report
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HCF of 80, 64, 48 = 16
960 ÷ 16 = 60
Question 28 Report
\[ P=\left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \]
Make T the subject of the relation.
Answer Details
P3=Q(R−T)15
15P3=Q(R−T)
15P3Q=R−T
T=R−15P3Q
Question 29 Report
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Question 30 Report
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?
Answer Details
Median is the middle number
= Nth2
Term = 22th2 = 11th term
Going in ascending order, 11th term is 3, going in descending order 11th term is 3
Median = 3 + 32 = 62 = 3
Question 31 Report
The figure below is a Venn diagram showing the elements arranged within sets A, B, C, \( \epsilon \).
Use the figure to answer this question
What is \( n((A \cup B)^c) \)?
Answer Details
B = (r, s, t, u)
A U B = Elements in both A and B = (p, q, r, s, t, u, v)
(A U B)1 = elements in the universal set E but not in (A U B)= (w, x, y, z)
n(A U B) 1 = number of the elements in (A U B)1 = 4
Question 32 Report
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Question 33 Report
The table below shows the frequency of children of age x years in a hospital:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| f | 3 | 4 | 5 | 6 | 7 | 6 | 5 | 4 |
Use the table to answer the question below:
What is the modal age?
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Question 34 Report
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Question 35 Report
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siny=300 m600 m=12
y=sin−10.5
y=30∘
x=180∘+(90∘−30∘)
x=180∘+60∘
x=240∘
Question 36 Report
If \( P = \left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \) make T the subject of the relation
Answer Details
P3 = Q(R−T)15
Cross multiplying
15P3 = Q(R - T)
Divide both sides by Q
15P3Q = R - T
Rearranging gives
T = R - 15P3Q
= RQ−15P3Q
Question 37 Report
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Question 38 Report
If \[ N=\begin{pmatrix}3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1\end{pmatrix} \] , find \(|N|\).
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3(-3 + 10) - 5(6 - 10) - 4(12 - 6)
3(7) - 5(-4) - 4(6) = 17
Question 39 Report
If the simple interest on a sum of money invested at 3% per annum for \(2\frac{1}{2}\) years is ₦123, find the principal.
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Question 40 Report
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Pr(product of x and y NOT > 6) = 816 = 12
