JAMB UTME - Mathematics - 1986
Question 1 Report
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.
Answer Details
24 x 8 = TR x 12
TR = 24×812
= = 16cm
Question 2 Report
Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ
Answer Details
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Question 3 Report
Answer Details
(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
Question 4 Report
If \( y = \frac{x}{x - 3} + \frac{x}{x + 4} \) find y when \( x = -2 \)
Answer Details
y = −2−5 + (−2)−2+4
= −25 + −22
= 4+−1010
= −1410
= -75
Question 5 Report
Simplify \(\left(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\right)x\frac{1}{\sqrt{3}}\)
Answer Details
1√5+√3−1√5−√3 = √5−√3−(√5+√3)(√5+√3)(5√3)
= −2√35−3
= −2√37
Question 6 Report
Question 7 Report
Answer Details
= 74o
Question 8 Report
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?
Answer Details
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
Question 9 Report
Answer Details
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = 120+1302
= 125
Question 10 Report
Simplify \( \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} \)
Answer Details
= (x+2)+(x−2)+2x(x+2)(x−2)
= 4xx2−4
Question 11 Report
Find all real numbers \(x\) which satisfy the inequality \(\frac{1}{3}(x + 1) - 1 > \frac{1}{5}(x + 4)\)
Answer Details
= x+13 - 1 > x+45
x+13 - x+45 - 1 > 0
= 5x+5−3x−1215
= 2x - 7 > 15
= 2x > 12
= x > 11
Question 12 Report
Answer Details
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
Question 13 Report
Find the total surface area of solid cone of radius \(2\sqrt{3}\) cm and slanting side \(4\sqrt{3}\)
Answer Details
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
Question 14 Report
Answer Details
x = ky3
k = xy3
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
Question 15 Report
Simplify \( \frac{0.0324 \times 0.00064}{0.48 \times 0.012} \)
Answer Details
324×64×10−948×12×10−5 = 36 x 10-4
= 3.6 x 10-3
Question 16 Report
Answer Details
24 - y = 12 (45 - y)
45 - y = 2(24 - y)
45 - y = 48 - 2y
2y - y = 48 - 45
∴ y = 3
The exact year = 1984
1984 - 3 = 1981
Question 17 Report
Answer Details
prob. of selecting a prime No. is 311
Question 18 Report
Answer Details
R = 100PT
100×50150×6 = 223
= 713 %
Question 19 Report
Simplify \( \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}} \)
Answer Details
= 323×3233−16×323
= 3563−56
= 1
Question 22 Report
Find the smallest number by which 252 can be multiplied to obtain a perfect square
Answer Details
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
Question 23 Report
Question 24 Report
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
Answer Details
Then the angle marked x i.e. QOP
34∘ x 2 = 68∘
Question 25 Report
If P = 18, Q = 21, R = -6 and S = -4, Calculate \( \frac{(P-Q)^3+S^2}{R^3} + S^2 \)
Answer Details
= −27+16R3
= −11−216
= 11216
Question 26 Report
Answer Details
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
Question 27 Report
In \( \triangle \)XYZ, determine the cosine of angle Z.
Question 28 Report
Answer Details
= 2601
QY = √2601
= 51
Question 29 Report
The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume
Answer Details
Since the cross section is a trapezium
= 12(6+11)×12×8
= 6 x 17 x 8 = 816m3
Question 30 Report
Answer Details
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00
Question 31 Report
make U the subject of the formula \(S = \sqrt{\frac{6}{u} - \frac{w}{2}}\)
Answer Details
S = 12−uw2u
2us2 = 12 - uw
u(2s2 + w) = 12
u = 122s2+w
Question 32 Report
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)?
Answer Details
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)
Question 33 Report
Answer Details
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
Question 34 Report
Answer Details
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)
Question 35 Report
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
Answer Details
Number of people in the group is 104360 x 0.9m
= 260000 = 26 x 104
Question 36 Report
Answer Details
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log282n
282n = 21
= 2
2n = 28
∴ n = 14
Question 37 Report
Find the reciprocal of \( \frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}} \)
Answer Details
= 23 x 65
= 45
reciprocal of 45 = 145
= 54
Question 38 Report
If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)
Answer Details
Tanθ = √b2−a2a2 , Tan 2 = √b2−a2a2
1 + tan2θ = 1 + b2−a2a2
= a2+b2−a2a2
= b2a2
Question 39 Report
Question 40 Report
In the figure, \( \triangle \)PQT is isosceles. PQ = QT, SRQ = \(35^\circ\), TPQ = \(20^\circ\) and PQR is a straight line.Calculate TSR
Answer Details
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∘
Question 41 Report
Simplify \( \frac{1}{5x+5} + \frac{1}{7x+7} \)
Answer Details
= 7+535(x+1)
= 1235(x+1)
Question 42 Report
Answer Details
= (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)
Question 43 Report
Answer Details
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)
Question 44 Report
Answer Details
∴ 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
Question 45 Report
Answer Details
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
Question 46 Report
Answer Details
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
Question 47 Report
If \( \frac{a}{c} = \frac{c}{d} = k \), find the value of \( \frac{3a^2-ac+c^2}{3b^2-bd+d^2} \)
Answer Details
∴ 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
Question 48 Report
Answer Details
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18
