JAMB UTME - Mathematics - 2018
Vraag 1 Verslag
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.
Antwoorddetails
= 34 x 35
= 920
Pr(John and James failed)= 1 - Pr(John and James passed)
1 – 920
= 1120
Vraag 2 Verslag
In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/
Antwoorddetails
∠ 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
Vraag 3 Verslag
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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Vraag 4 Verslag
Vraag 5 Verslag
\[ \frac{0.00256 \times 0.0064}{0.025 \times 0.08} \]
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Explanation:
=0.0000163840.002
=0.008192
Vraag 6 Verslag
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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
Vraag 7 Verslag
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Vraag 8 Verslag
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
Vraag 9 Verslag
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So the answer is 1.0
Vraag 10 Verslag
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= 90100 x 20000 =18000
At the end of second year, its value = 90% of 18000
= 90100 x 18000 = ₦16,200
Vraag 11 Verslag
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
Vraag 12 Verslag
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= 6×5×4×3×2×12×1
= 360
Vraag 13 Verslag
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?
Antwoorddetails
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
Vraag 14 Verslag
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
Vraag 15 Verslag
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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
Vraag 16 Verslag
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HCF of 80, 64, 48 = 16
960 ÷ 16 = 60
Vraag 17 Verslag
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Vraag 18 Verslag
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
Vraag 19 Verslag
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?
Antwoorddetails
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
Vraag 20 Verslag
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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Vraag 21 Verslag
\[ P=\left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \]
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
Vraag 22 Verslag
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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
Vraag 24 Verslag
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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Vraag 25 Verslag
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
Vraag 26 Verslag
Use the cumulative frequency curve to answer the question.
Estimate the median of the date represented on the graph.
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Vraag 27 Verslag
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Vraag 28 Verslag
Use the quadratic equation curve to answer this question.
What is the 80th percentile?
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Vraag 29 Verslag
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Vraag 30 Verslag
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
Vraag 31 Verslag
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Vraag 32 Verslag
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Vraag 33 Verslag
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siny=300 m600 m=12
y=sin−10.5
y=30∘
x=180∘+(90∘−30∘)
x=180∘+60∘
x=240∘
Vraag 34 Verslag
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Vraag 35 Verslag
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Pr(product of x and y NOT > 6) = 816 = 12
Vraag 36 Verslag
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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
Vraag 37 Verslag
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Vraag 38 Verslag
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Vraag 39 Verslag
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
Vraag 40 Verslag
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 \( n((A \cup B)^c) \)?
Antwoorddetails
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
