JAMB UTME - Mathematics - 1986
Tambaya 1 Rahoto
In \( \triangle \)XYZ, determine the cosine of angle Z.
Tambaya 2 Rahoto
Bayanin Amsa
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = 120+1302
= 125
Tambaya 3 Rahoto
Bayanin Amsa
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)
Tambaya 4 Rahoto
The table below gives the scores of a group of students in a Mathematical test.
| Scores | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 2 | 4 | 7 | 14 | 12 | 6 | 4 | 1 |
If the mode in m and the number of students who scored 4 or less is s. What is (s, m)?
Bayanin Amsa
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)
Tambaya 5 Rahoto
The people in a city with a population of 0.9 million were grouped according to their ages. Use the diagram to determine the number of people in the 15 - 29 years group
Bayanin Amsa
Number of people in the group is 104360 x 0.9m
= 260000 = 26 x 104
Tambaya 6 Rahoto
Bayanin Amsa
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00
Tambaya 7 Rahoto
PQ and PR are tangents from P to a circle centre O as shown in the figure. If QRP = \(34^\circ\), find the angle marked x
Bayanin Amsa
Then the angle marked x i.e. QOP
34∘ x 2 = 68∘
Tambaya 8 Rahoto
Find the reciprocal of \( \frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}} \)
Bayanin Amsa
= 23 x 65
= 45
reciprocal of 45 = 145
= 54
Tambaya 9 Rahoto
In the figure, \( \triangle \)PQT is isosceles. PQ = QT, SRQ = \(35^\circ\), TPQ = \(20^\circ\) and PQR is a straight line.Calculate TSR
Bayanin Amsa
TPQ = 20∘
PQR = is a straight line
Since PQ = QT, angle P = angle T = 20∘
Angle PQR = 180∘ - (20 + 20) = 140∘
TQR = 180∘ - 140∘ = 40∘ < on a straight line
QSR = 180∘ - (40 + 35)∘ = 105∘
TSR = 180∘ - 105∘
= 75∘
Tambaya 11 Rahoto
Find the total surface area of solid cone of radius \(2\sqrt{3}\) cm and slanting side \(4\sqrt{3}\)
Bayanin Amsa
r = 2√3
= πr2 + π rH
H = 4√3 , π r(r + H)
∴ Area = π 2√3 [2√3 + 4√3 ]
= π 2√3 (6√3 )
= 12π x 3
= 36π cm2
Tambaya 13 Rahoto
Bayanin Amsa
∴ x + 2y = 1.....(i)
4(x + 3y) = 16 = 42
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i) = x + 2y = 1
∴ x + 2(1) = 1
x + 2 = 1
∴ x = 1
∴ 3x + y = 3-1 + 1
= 3 = 1
Tambaya 14 Rahoto
make U the subject of the formula \(S = \sqrt{\frac{6}{u} - \frac{w}{2}}\)
Bayanin Amsa
S = 12−uw2u
2us2 = 12 - uw
u(2s2 + w) = 12
u = 122s2+w
Tambaya 15 Rahoto
If P = 18, Q = 21, R = -6 and S = -4, Calculate \( \frac{(P-Q)^3+S^2}{R^3} + S^2 \)
Bayanin Amsa
= −27+16R3
= −11−216
= 11216
Tambaya 16 Rahoto
Simplify \( \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}} \)
Bayanin Amsa
= 323×3233−16×323
= 3563−56
= 1
Tambaya 17 Rahoto
Bayanin Amsa
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)
Tambaya 18 Rahoto
Find all real numbers \(x\) which satisfy the inequality \(\frac{1}{3}(x + 1) - 1 > \frac{1}{5}(x + 4)\)
Bayanin Amsa
= x+13 - 1 > x+45
x+13 - x+45 - 1 > 0
= 5x+5−3x−1215
= 2x - 7 > 15
= 2x > 12
= x > 11
Tambaya 19 Rahoto
Find the smallest number by which 252 can be multiplied to obtain a perfect square
Bayanin Amsa
Let the smallest number be x and the perfect square be y 252x = y.
By trial and error method, 252 x 9 = 1764
Check if y = 1764
y2 = 42
x = 7
Tambaya 20 Rahoto
If \( y = \frac{x}{x - 3} + \frac{x}{x + 4} \) find y when \( x = -2 \)
Bayanin Amsa
y = −2−5 + (−2)−2+4
= −25 + −22
= 4+−1010
= −1410
= -75
Tambaya 21 Rahoto
Tambaya 23 Rahoto
Bayanin Amsa
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log282n
282n = 21
= 2
2n = 28
∴ n = 14
Tambaya 24 Rahoto
Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ
Bayanin Amsa
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Tambaya 25 Rahoto
Bayanin Amsa
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18
Tambaya 26 Rahoto
In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.
Bayanin Amsa
24 x 8 = TR x 12
TR = 24×812
= = 16cm
Tambaya 27 Rahoto
Tambaya 28 Rahoto
Bayanin Amsa
Area of sector = θ360 x 2π r2........(i)
Length of arc = θ360 x 2π r........(ii)
from eqn. (ii) θ = 240π , subt. for θ in eqn (i)
Area x 2401 x 1360 x π61
= 24cm2
Tambaya 29 Rahoto
If \( \frac{a}{c} = \frac{c}{d} = k \), find the value of \( \frac{3a^2-ac+c^2}{3b^2-bd+d^2} \)
Bayanin Amsa
∴ ab = bk
cd = k
∴ c = dk
= 3a2−ac+c23b2−bd+d2
= 3(bk)2−(bk)(dk)+dk23b2−bd+a2
= 3b2k2−bk2d+dk23b2−bd+d2
k = 3b2k2−bk2d+dk23b2−bd+d2
Tambaya 30 Rahoto
Bayanin Amsa
= 2601
QY = √2601
= 51
Tambaya 31 Rahoto
Simplify \( \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} \)
Bayanin Amsa
= (x+2)+(x−2)+2x(x+2)(x−2)
= 4xx2−4
Tambaya 32 Rahoto
Bayanin Amsa
x = ky3
k = xy3
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
Tambaya 33 Rahoto
Bayanin Amsa
R = 100PT
100×50150×6 = 223
= 713 %
Tambaya 34 Rahoto
Bayanin Amsa
To get his total mark when he scored 57, subtract 57 from 75 to give 18, then subtract this 18 from the new total mark(ie. 240)
= 240 - 18
= 222
Tambaya 35 Rahoto
Bayanin Amsa
L.C.M = 26 x 3 x 5
H.C.F = 24
26×3×524
= 22 x 3 x 5
= 4 x 3 x 5
= 12 x 5
= 60
Tambaya 36 Rahoto
Bayanin Amsa
= 74o
Tambaya 37 Rahoto
Bayanin Amsa
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
∴1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges
Tambaya 38 Rahoto
Simplify \( \frac{1}{5x+5} + \frac{1}{7x+7} \)
Bayanin Amsa
= 7+535(x+1)
= 1235(x+1)
Tambaya 39 Rahoto
If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)
Bayanin Amsa
Tanθ = √b2−a2a2 , Tan 2 = √b2−a2a2
1 + tan2θ = 1 + b2−a2a2
= a2+b2−a2a2
= b2a2
Tambaya 40 Rahoto
Tambaya 41 Rahoto
Simplify \(\left(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\right)x\frac{1}{\sqrt{3}}\)
Bayanin Amsa
1√5+√3−1√5−√3 = √5−√3−(√5+√3)(√5+√3)(5√3)
= −2√35−3
= −2√37
Tambaya 42 Rahoto
Tambaya 43 Rahoto
Bayanin Amsa
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0
Tambaya 44 Rahoto
Bayanin Amsa
Using almighty formula i.e. x = b±√b2−4ac2a
a = 3, b = 6, c = -2
x = −6±√62−4(3)(−2)2(3)
x = 6±√36−246
x = 6±√606
x = −6±√4×156
x = −1±√153
Tambaya 45 Rahoto
Bayanin Amsa
= (a + b)(a - b)
= [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)]
= [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)]
= (7a + 1)(a + 5)
∴ (a + 5)(7a + 1)
Tambaya 46 Rahoto
The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume
Bayanin Amsa
Since the cross section is a trapezium
= 12(6+11)×12×8
= 6 x 17 x 8 = 816m3
Tambaya 48 Rahoto
A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2 : 3 : 5 respectively. If Bisi got 5, how many were share out?
Bayanin Amsa
Let x r3epresent total number of pencils shared
B : S : T = 2 + 3 + 5 = 10
2 : 3 : 5
= 210
x y
= 5
2y =5
2y = 50
∴ y = 502
= 25
