JAMB UTME - Mathematics - 1993
Pregunta 1 Informe
Detalles de la respuesta
∴ Total angle difference = (30 + 13) = 43o
Pregunta 2 Informe
Detalles de la respuesta
Pregunta 3 Informe
The bar chart shows the distribution of marks in a class test. How many students took the test?
Detalles de la respuesta
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 0 + 1 + 0 + 2 = 17
Pregunta 4 Informe
Detalles de la respuesta
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o
Pregunta 5 Informe
Detalles de la respuesta
Pregunta 7 Informe
Detalles de la respuesta
Pregunta 8 Informe
If \(9\left(x - \frac{1}{2}\right)^3x^2\)
Detalles de la respuesta
∴ 2(x - 12 ) = x2
2x - 1 = x2
hence x2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1
Pregunta 9 Informe
The chances of three independent events X, Y, Z occurring are \( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{1}{4} \) respectively. What are the chances of Y and Z only occurring?
Detalles de la respuesta
Change of Y = 23
Chance of Z = 14
Chance of Y and Z only occurring
= Pr (Y ? Z ? Xc)
where Xc = 1 - Pr(X)
1 = 12 = 112
= Pr(Y) x Pr(Z) x Pr(Xc)
= 23 x 14 x 12
= 112
Pregunta 10 Informe
Detalles de la respuesta
= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or 23
Pregunta 11 Informe
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.
Detalles de la respuesta
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o
Pregunta 12 Informe
Detalles de la respuesta
Pregunta 13 Informe
Detalles de la respuesta
Pregunta 14 Informe
Evaluate \( \left(x + \frac{1}{x} + 1\right)^2 - \left(x + \frac{1}{x} + 1\right)^2 \)
Detalles de la respuesta
= (x + 1x + 1 + x + −1x - 1)(x - 1x + 1 - x + 1x + 1)
= (2x) (2 + 2x ) = 2x x 2(1 + 1x )
4x (1 + 1x ) = 4x + 4
= 4(1 + x)
Pregunta 15 Informe
If \( \sin \theta = \cos \theta \), find \( \theta \) between 0o and 360o
Detalles de la respuesta
The acute angle where sin θ = cos θ = 45o
But at the fourth Quadrant Cos θ = +ve
at the 4th quadrant, value with respect to Q is
(360 - Q) where Q = acute angle
(360 - 45) = 315o
The two solution are 45o, 315o
Pregunta 16 Informe
If \( \sqrt{x^2 + 9} = x + 1 \), solve for x
Detalles de la respuesta
x2 + 9 = (x + 1)2 + 1
0 = x2 + 2x + 1 - x2 - 9
= 2x - 8 = 0
2(x - 4) = 0
x = 4
Pregunta 17 Informe
Detalles de la respuesta
Pregunta 18 Informe
Evaluate \( \frac{3524}{0.05} \) correct to 3 significant figures
Detalles de la respuesta
Pregunta 19 Informe
Detalles de la respuesta
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Pregunta 20 Informe
Detalles de la respuesta
but 10y - (5 x 2)y = 5y x 2y
= (Law of indices)
5y x 2y x 5(2y - 2) x 4(y - 1) = 1
but 4(y - 1) = 22(y - 1)
= 2y - 2 (Law of indices)
5y x 5(2y -2) x 2(- 2) = 1
5(3y -2) x 2y x 2(2y -2) = 1
= 5(3y -2) x 2(3y -2) = 1
But ao = 1
10(3y -2) = 10o
3y - 2 = 0
? y = 23
Pregunta 21 Informe
A binary operation \( \ast \) is defined on a set of real numbers by \(x \ast y = x^y\) for all real values of \(x\) and \(y\). If \(x \ast 2 = x\). Find the possible values of \(x\)
Detalles de la respuesta
x ∗ 2 = x2
x ∗ 2 = x
∴ x2 - x = 0
x(x - 1) = 0
x = 0 or 1
Pregunta 22 Informe
Make x the subject of the relation \( \frac{1+ax}{1-ax}=\frac{p}{q} \)
Detalles de la respuesta
Pregunta 23 Informe
The shaded portion in the Venn diagram is
Detalles de la respuesta
Pregunta 24 Informe
In the diagram, O is the centre of the circle and POQ a diameter. If POR = \(96^\circ\), find the value of ORQ.
Detalles de la respuesta
< ROQ = 180 - 86 = 84?
? OQR = Isosceles
R = Q
R + Q + 84 = 180(angle in a ? )
2R = 96 since R = Q
R = 48?
ORQ = 48?
Pregunta 25 Informe
Detalles de la respuesta
Pregunta 26 Informe
PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP
Detalles de la respuesta
Pregunta 27 Informe
| \(Weight(s)\) | \(0-10\) | \(10-20\) | \(20-30\) | \(40-50\) | |
| Number of coconuts | \(10\) | \(27\) | \(19\) | \(6\) | \(2\) |
Estimate the mode of the frequency distribution above.
Detalles de la respuesta
= L1+Δ1xcΔ1+Δ2
= 10+(20−10)(27−10)2(27)−10−19
= 10 + 17025
= 10 + 6.8
= 16.8
Pregunta 28 Informe
Detalles de la respuesta
let the polygon has n-sides
therefore, Total interior angle 150 x n = 150n
hence 150n = (2n - 4)90
150n = 180n - 360
360 = (180 - 150)n
30n = 360
n = 12
Pregunta 29 Informe
Detalles de la respuesta
y - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1
Pregunta 30 Informe
Detalles de la respuesta
(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -32
Since 5 is a root, then (x - 5) is a factor
Pregunta 31 Informe
In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = \(32^\circ\). Find PSQ
Detalles de la respuesta
< RTS = < PQS = 32∘ (Alternative angle)
< PSQ = 90 - < PSQ = 90∘ - 32∘
= 58∘
Pregunta 32 Informe
Detalles de la respuesta
Pregunta 33 Informe
Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of \( \frac{1}{2} \)o at the centre of the circle
Detalles de la respuesta
∴ Radius = 4cm
Length of arc = θ360 x 2 π r but Q = 2212
∴ Length 2212360 x 2 x π x 4
= 2212×8π360
= 180360
= π2
Pregunta 34 Informe
Detalles de la respuesta
100 = 242 - 2x2 - 2x2 x -12
100 = 3x2 + x2
= 3x2
x = √1003
= 10√3 x 3√3
x = 1033 cm
Pregunta 35 Informe
Detalles de la respuesta
Pregunta 36 Informe
| Class | Frequency |
| 1−5 | 2 |
| 6−10 | 4 |
| 11−15 | 5 |
| 16−20 | 2 |
| 21−25 | 3 |
| 26−30 | 2 |
| 31−35 | 1 |
| 36−40 | 1 |
Find the median of the observation in the table given.
Detalles de la respuesta
∑f2
= 202
= 10, L1 = 10.5, fo = 6, fm = 5
Median = 10.5 + (10−6)5 5
= 10.5 + 4
= 14.5
Pregunta 37 Informe
simplify \( \frac{1}{\sqrt{3}-2} - \frac{1}{\sqrt{3}+2} \)
Detalles de la respuesta
L.C.M = (3- 2) (3 + 2)
∴ 1√3−2 - 1√3−2 = √3+2−√3−2√3−2+√3−2
√3+2−√3+23−2√3+2√3−4
= 43−2
= 4−1
= -4
Pregunta 38 Informe
Detalles de la respuesta
increment = ₦36 (every 6 months)
Period of increment = 2 yrs and 6 months
amount(increment) = ₦36 x 5 = ₦180
The man's new salary = ₦540 = ₦180
= ₦720.00
Pregunta 39 Informe
Detalles de la respuesta
Pregunta 40 Informe
find the radius of a sphere whose surface area is 154cm2 \( \left(\pi=\frac{22}{7}\right) \)
Detalles de la respuesta
4π r2 = 154
r√1544π
= 3.50cm
Pregunta 41 Informe
Solve without using tables \( \log_{5}(62.5) - \log_{5}\left(\frac{1}{2}\right) \)
Detalles de la respuesta
= log5(62.5)12 - log5(2 x 62.5)
= log5(125)
= log553 - 3log55
= 3
Pregunta 42 Informe
In the diagram, \(QP // ST\): \(PQR = 34^\circ\) \(QRS = 73^\circ\) and \(RS = RT\). Find \(SRT\)
Detalles de la respuesta
R = 180∘ - 107∘
< p = 180∘ - (107∘ - 34∘ )
108 - 141∘ = 39∘
Angle < S = 39∘ (corr. Ang.) But in △ SRT
< S = < T = 39∘
SRT = 180 - (39∘ + 39∘ )
= 180∘ - 78∘
= 102∘
Pregunta 43 Informe
Simplify \( \frac{1}{p} - \frac{1}{q} + \frac{p}{q} - \frac{q}{p} \)
Detalles de la respuesta
q?ppq x pqp2q2 = q?pp2q2
?(p?q)(p+q)(p?q)
= ?1p+q
Pregunta 44 Informe
In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST
Detalles de la respuesta
SQ2 = 142 - 52
196 - 25 = 171
ST2 + TQ2 = SQ2
ST2 + 22 = 171
ST2 = 171 - 4
= 167
ST = √167
= 12.92 = 12.9cm
Pregunta 45 Informe
Detalles de la respuesta
Pregunta 46 Informe
Integrate 1−xx3 with respect to x
Detalles de la respuesta
∫1−xx3
= ∫1x3−∫xx3
= x-3 dx - x-2dx
= 12x2+1x
Pregunta 47 Informe
If x is negative, what is the range of values of x within which \( \frac{x+1}{3} > \frac{1}{X+3} \)
Detalles de la respuesta
= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0
= x > 0, x < -3, x < -4 = x < -3(solution only)
Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)
Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)
combining the solutions -4 < x < -3
Pregunta 48 Informe
Detalles de la respuesta
∴ for the △ to be isosceles, either
2x - 3 = x + 3 or 2x + 3 = x + 3
solve the two equations we arrive at
x = 6 or x = 0
When x = 6, the sides are 9, 15, 9
When x = 0, the sides are 3, 4, -3 since lengths of a △ can never be negative then the value of x = 6
Pregunta 49 Informe
From the figure, calculate TH in centimeters
Detalles de la respuesta
TH5+QH = tan 30∘
TH = (b + QH) tan 30∘
QH = 56 (5 + QH) 1√3
QH(1 - 1√3 ) = 5√3
QH = 5√3√3−1√3
= 5√3−1
Pregunta 50 Informe
In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.
Detalles de la respuesta
QS/QP = SR/PR
3/4 = SR/g
4SR = 27
SR = 274
= 634 cm
