JAMB UTME - Mathematics - 2018
Question 1 Report
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Question 2 Report
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Question 3 Report
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HCF of 80, 64, 48 = 16
960 ÷ 16 = 60
Question 4 Report
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Question 5 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 6 Report
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Question 7 Report
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So the answer is 1.0
Question 8 Report
Use the quadratic equation curve to answer this question.
What is the 80th percentile?
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Question 9 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 10 Report
In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/
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∠ 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 11 Report
Which one of the following gives the members of the set \(A^c \cap B \cap C\)?
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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 12 Report
Solve for t in the equation \( \frac{3}{4}t + \frac{1}{3}(21 - t) = 11 \)
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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 13 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 14 Report
Question 15 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?
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Question 16 Report
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Question 17 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 18 Report
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Question 19 Report
\[ \frac{0.00256 \times 0.0064}{0.025 \times 0.08} \]
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Explanation:
=0.0000163840.002
=0.008192
Question 20 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 21 Report
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Question 22 Report
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Question 23 Report
Find the values of x for which
\( \frac{x+2}{4} - \frac{2x-3}{3} < 4 \)
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3x+6−(8x−12)12<4
−5x+1812<4
−5x+18<48
−5x<30
x>−6
Question 24 Report
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Question 25 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 26 Report
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Pr(product of x and y NOT > 6) = 816 = 12
Question 28 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 29 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?
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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 30 Report
if \(y = 23_{\mathit{five}} + 101_{\mathit{three}}\) find y leaving your answer in base two
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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 31 Report
If \( P = \left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \) make T the subject of the relation
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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 32 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 33 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?
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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 34 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 35 Report
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= 6×5×4×3×2×12×1
= 360
Question 36 Report
\[ P=\left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \]
Make T the subject of the relation.
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P3=Q(R−T)15
15P3=Q(R−T)
15P3Q=R−T
T=R−15P3Q
Question 37 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 38 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 39 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 \( n((A \cup B)^c) \)?
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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 40 Report
Simplify \(25^{\frac{1}{2}} \times 8^{-\frac{2}{3}}\)
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2512 × 8−23
= √25 x (3√8 ) -2
= 5 x 2-2
= 5 x 122 =
54 = 114
