JAMB UTME - Mathematics - 1986
Ajụjụ 1 Ripọtì
Akọwa Nkọwa
= 2601
QY = √2601
= 51
Ajụjụ 3 Ripọtì
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
Akọwa Nkọwa
Number of people in the group is 104360 x 0.9m
= 260000 = 26 x 104
Ajụjụ 4 Ripọtì
Ajụjụ 5 Ripọtì
make U the subject of the formula \(S = \sqrt{\frac{6}{u} - \frac{w}{2}}\)
Akọwa Nkọwa
S = 12−uw2u
2us2 = 12 - uw
u(2s2 + w) = 12
u = 122s2+w
Ajụjụ 6 Ripọtì
Akọwa Nkọwa
x = ky3
k = xy3
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
Ajụjụ 8 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 9 Ripọtì
Find the reciprocal of \( \frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}} \)
Akọwa Nkọwa
= 23 x 65
= 45
reciprocal of 45 = 145
= 54
Ajụjụ 10 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 11 Ripọtì
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)?
Akọwa Nkọwa
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)
Ajụjụ 12 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 13 Ripọtì
Ajụjụ 14 Ripọtì
If P = 18, Q = 21, R = -6 and S = -4, Calculate \( \frac{(P-Q)^3+S^2}{R^3} + S^2 \)
Akọwa Nkọwa
= −27+16R3
= −11−216
= 11216
Ajụjụ 15 Ripọtì
Ajụjụ 16 Ripọtì
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?
Akọwa Nkọwa
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
Ajụjụ 17 Ripọtì
Akọwa Nkọwa
(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
Ajụjụ 18 Ripọtì
Simplify \( \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}} \)
Akọwa Nkọwa
= 323×3233−16×323
= 3563−56
= 1
Ajụjụ 19 Ripọtì
Akọwa Nkọwa
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18
Ajụjụ 20 Ripọtì
In \( \triangle \)XYZ, determine the cosine of angle Z.
Ajụjụ 21 Ripọtì
Akọwa Nkọwa
∴ 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
Ajụjụ 22 Ripọtì
If \( y = \frac{x}{x - 3} + \frac{x}{x + 4} \) find y when \( x = -2 \)
Akọwa Nkọwa
y = −2−5 + (−2)−2+4
= −25 + −22
= 4+−1010
= −1410
= -75
Ajụjụ 23 Ripọtì
If \( \frac{a}{c} = \frac{c}{d} = k \), find the value of \( \frac{3a^2-ac+c^2}{3b^2-bd+d^2} \)
Akọwa Nkọwa
∴ 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
Ajụjụ 24 Ripọtì
Akọwa Nkọwa
R = 100PT
100×50150×6 = 223
= 713 %
Ajụjụ 25 Ripọtì
In the figure, \( \triangle \)PQT is isosceles. PQ = QT, SRQ = \(35^\circ\), TPQ = \(20^\circ\) and PQR is a straight line.Calculate TSR
Akọwa Nkọwa
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∘
Ajụjụ 26 Ripọtì
Simplify \(\left(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\right)x\frac{1}{\sqrt{3}}\)
Akọwa Nkọwa
1√5+√3−1√5−√3 = √5−√3−(√5+√3)(√5+√3)(5√3)
= −2√35−3
= −2√37
Ajụjụ 27 Ripọtì
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
Akọwa Nkọwa
Then the angle marked x i.e. QOP
34∘ x 2 = 68∘
Ajụjụ 28 Ripọtì
Simplify \( \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} \)
Akọwa Nkọwa
= (x+2)+(x−2)+2x(x+2)(x−2)
= 4xx2−4
Ajụjụ 29 Ripọtì
Find all real numbers \(x\) which satisfy the inequality \(\frac{1}{3}(x + 1) - 1 > \frac{1}{5}(x + 4)\)
Akọwa Nkọwa
= x+13 - 1 > x+45
x+13 - x+45 - 1 > 0
= 5x+5−3x−1215
= 2x - 7 > 15
= 2x > 12
= x > 11
Ajụjụ 30 Ripọtì
Akọwa Nkọwa
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = 120+1302
= 125
Ajụjụ 31 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 32 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 33 Ripọtì
Find the smallest number by which 252 can be multiplied to obtain a perfect square
Akọwa Nkọwa
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
Ajụjụ 35 Ripọtì
Akọwa Nkọwa
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log282n
282n = 21
= 2
2n = 28
∴ n = 14
Ajụjụ 36 Ripọtì
Akọwa Nkọwa
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00
Ajụjụ 37 Ripọtì
Find the total surface area of solid cone of radius \(2\sqrt{3}\) cm and slanting side \(4\sqrt{3}\)
Akọwa Nkọwa
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
Ajụjụ 38 Ripọtì
Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ
Akọwa Nkọwa
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Ajụjụ 39 Ripọtì
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.
Akọwa Nkọwa
24 x 8 = TR x 12
TR = 24×812
= = 16cm
Ajụjụ 41 Ripọtì
Ajụjụ 42 Ripọtì
If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)
Akọwa Nkọwa
Tanθ = √b2−a2a2 , Tan 2 = √b2−a2a2
1 + tan2θ = 1 + b2−a2a2
= a2+b2−a2a2
= b2a2
Ajụjụ 43 Ripọtì
The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume
Akọwa Nkọwa
Since the cross section is a trapezium
= 12(6+11)×12×8
= 6 x 17 x 8 = 816m3
Ajụjụ 44 Ripọtì
Akọwa Nkọwa
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)
Ajụjụ 45 Ripọtì
Akọwa Nkọwa
= (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)
Ajụjụ 46 Ripọtì
Akọwa Nkọwa
= 74o
Ajụjụ 47 Ripọtì
Akọwa Nkọwa
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)
Ajụjụ 48 Ripọtì
Simplify \( \frac{1}{5x+5} + \frac{1}{7x+7} \)
Akọwa Nkọwa
= 7+535(x+1)
= 1235(x+1)
