JAMB UTME - Mathematics - 2018
Tambaya 1 Rahoto
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?
Bayanin Amsa
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
Tambaya 2 Rahoto
Bayanin Amsa
Tambaya 3 Rahoto
Simplify \(25^{\frac{1}{2}} \times 8^{-\frac{2}{3}}\)
Bayanin Amsa
2512 × 8−23
= √25 x (3√8 ) -2
= 5 x 2-2
= 5 x 122 =
54 = 114
Tambaya 4 Rahoto
\[ \frac{0.00256 \times 0.0064}{0.025 \times 0.08} \]
Bayanin Amsa
Explanation:
=0.0000163840.002
=0.008192
Tambaya 5 Rahoto
Bayanin Amsa
Tambaya 6 Rahoto
Bayanin Amsa
Tambaya 7 Rahoto
Solve for t in the equation \( \frac{3}{4}t + \frac{1}{3}(21 - t) = 11 \)
Bayanin Amsa
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
Tambaya 8 Rahoto
If \[ N=\begin{pmatrix}3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1\end{pmatrix} \] , find \(|N|\).
Bayanin Amsa
3(-3 + 10) - 5(6 - 10) - 4(12 - 6)
3(7) - 5(-4) - 4(6) = 17
Tambaya 9 Rahoto
Bayanin Amsa
Tambaya 10 Rahoto
Find the values of x for which
\( \frac{x+2}{4} - \frac{2x-3}{3} < 4 \)
Bayanin Amsa
3x+6−(8x−12)12<4
−5x+1812<4
−5x+18<48
−5x<30
x>−6
Tambaya 11 Rahoto
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.
Bayanin Amsa
= 34 x 35
= 920
Pr(John and James failed)= 1 - Pr(John and James passed)
1 – 920
= 1120
Tambaya 12 Rahoto
\[ P=\left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \]
Make T the subject of the relation.
Bayanin Amsa
P3=Q(R−T)15
15P3=Q(R−T)
15P3Q=R−T
T=R−15P3Q
Tambaya 13 Rahoto
Use the cumulative frequency curve to answer the question.
Estimate the median of the date represented on the graph.
Bayanin Amsa
Tambaya 14 Rahoto
Bayanin Amsa
a2=122+92
a=√122+92
a=15 m
b2=152+82
b=√152+82
b=17 m
cosθ=ab=1517
Tambaya 15 Rahoto
Bayanin Amsa
Tambaya 16 Rahoto
Bayanin Amsa
HCF of 80, 64, 48 = 16
960 ÷ 16 = 60
Tambaya 17 Rahoto
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 \( n((A \cup B)^c) \)?
Bayanin Amsa
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
Tambaya 18 Rahoto
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?
Bayanin Amsa
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
Tambaya 19 Rahoto
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.
Bayanin Amsa
Tambaya 20 Rahoto
Bayanin Amsa
So the answer is 1.0
Tambaya 21 Rahoto
Bayanin Amsa
Pr(product of x and y NOT > 6) = 816 = 12
Tambaya 22 Rahoto
if \(y = 23_{\mathit{five}} + 101_{\mathit{three}}\) find y leaving your answer in base two
Bayanin Amsa
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
Tambaya 23 Rahoto
Bayanin Amsa
= 90100 x 20000 =18000
At the end of second year, its value = 90% of 18000
= 90100 x 18000 = ₦16,200
Tambaya 24 Rahoto
Bayanin Amsa
Tambaya 25 Rahoto
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?
Bayanin Amsa
Tambaya 26 Rahoto
If \( P = \left[\frac{Q(R-T)}{15}\right]^{\frac{1}{3}} \) make T the subject of the relation
Bayanin Amsa
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
Tambaya 27 Rahoto
Bayanin Amsa
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
Tambaya 28 Rahoto
Bayanin Amsa
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
Tambaya 29 Rahoto
Bayanin Amsa
Tambaya 30 Rahoto
In the figure below, /MX/ = 8cm, /XN/ = 12cm, /NZ/ = 4cm and ∠ XMN = ∠ XZY. Calculate /YM/
Bayanin Amsa
∠ 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
Tambaya 31 Rahoto
Which one of the following gives the members of the set \(A^c \cap B \cap C\)?
Bayanin Amsa
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
Tambaya 32 Rahoto
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?
Bayanin Amsa
Tambaya 33 Rahoto
Bayanin Amsa
Tambaya 34 Rahoto
Bayanin Amsa
siny=300 m600 m=12
y=sin−10.5
y=30∘
x=180∘+(90∘−30∘)
x=180∘+60∘
x=240∘
Tambaya 35 Rahoto
Bayanin Amsa
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
Tambaya 36 Rahoto
Bayanin Amsa
Tambaya 37 Rahoto
Use the quadratic equation curve to answer this question.
What is the 80th percentile?
Bayanin Amsa
Tambaya 38 Rahoto
Bayanin Amsa
= 6×5×4×3×2×12×1
= 360
Tambaya 40 Rahoto
