JAMB UTME - Mathematics - 1984
Pregunta 1 Informe
XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.
Detalles de la respuesta
Also by Pythagoras theorem, XY2 = 62 + 32
XY2 = 36 + 9 = 45
XY = √45=3√3
Pregunta 2 Informe
\( \dfrac{0.0001432}{1940000} = k \times 10^n \) where \( 1 \le k < 10 \) and n is a whole number. The values K and n are
Detalles de la respuesta
0.00014321940000
= k x 10n
where 1 ≤
k ≤
10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11
k = 7.381, n = -11
Pregunta 3 Informe
If \(a = \frac{2x}{1-x}\) and \(b = \frac{1+x}{1-x}\), then \(a^2 - b^2\) in the simplest form is \(\frac{3x+1}{1-x}\)
Detalles de la respuesta
= (2x1−x+1+x1−x )(2x1−x−1+x1−X )
= (3x+11−x )(x−11−x )
= 3x+1x−1
Pregunta 4 Informe
Detalles de la respuesta
x2 - 12x + 40 = 5
x2 - 12x + 40 - 5 = 0
x2 + 12x + 35 = 0
x2 - 7x - 5x + 35 = 0
x(x - 7) - 5(x - 7) = 0
= (x - 5)(x - 7)
x = 5 or 7
Pregunta 5 Informe
Find a factor which is common to all 3 binomial expressions \(4a^2 - 9b^2,\; 8a^3 + 27b^3,\; (4a + 6b)^2\).
Detalles de la respuesta
Pregunta 6 Informe
If \( \sin \theta = \frac{x}{y} \) and \(0^o < 90^o\), then find \( \frac{1}{\tan \theta} \).
Detalles de la respuesta
sinθ = xy
cosθ = √y2−x2y
Pregunta 7 Informe
Using \( \triangle \) XYZ in the figure, find XYZ.
Detalles de la respuesta
sin 120∘ = sin 60∘
5 sin y = 3 sin 60∘
sin y = 3sin60o5
3×0.8665
= 2.5985
y = sin-1 0.5196 = 30∘ 18'
Pregunta 8 Informe
Detalles de la respuesta
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p = −32
q = −32
p + q = 32 - 32
= −62
= -3
Pregunta 9 Informe
What is the volume of this regular three dimensional figure?
Detalles de la respuesta
A = 12 x 4 x 3
= 6cm2
V = 6 x 8
= 48cm2
Pregunta 10 Informe
Detalles de la respuesta
∴ P=KQ2R
36=KQ24
K=36×49
K=16
P=16×522=200
Pregunta 11 Informe
Detalles de la respuesta
672 x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively
Pregunta 12 Informe
The cumulative frequency function of the data below is given below by the equation \(y = cf(x)\). What is \(cf(5)\)?
| Score(\(n\)) | Frequency(\(f\)) |
| 3 | 30 |
| 4 | 32 |
| 5 | 30 |
| 6 | 35 |
| 8 | 20 |
Detalles de la respuesta
cf(5) = 30 + 32 + 30
= 92
Pregunta 13 Informe
Simplify \( \frac{3^n - 3^{n-1}}{3^3 \times 3^n - 27 \times 3^{n-1}} \)
Detalles de la respuesta
= 3n−3n−127(3n−3n−1)
= 127
Pregunta 14 Informe
Detalles de la respuesta
i.e. SP = x
When SP = x at 10% profit
CP = 100100 + 10 of x = 100110 of x
when Q sells to R, SP = ₦209 at loss of 5%
Q's cost price = Q's selling price
CP = 10095 x 209
= 220.00
x = 220
= 220011
= 200
= ₦200.00
Pregunta 15 Informe
If \(pq + 1 = q^2\) and \(t = \frac{1}{p} - \frac{1}{pq}\) express \(t\) in terms of \(q\)
Detalles de la respuesta
t = 1p - 1pq .........(ii)
p = q2−1q
Sub for p in equation (ii)
t = 1q2−1q - 1q2−1q×q
t = qq2−1 - 1q2−1
t = q−1q2−1
= q−1(q+1)(q−1)
= 1q+1
Pregunta 16 Informe
Detalles de la respuesta
Pregunta 17 Informe
Detalles de la respuesta
= 1035 x 145
= 20025
Total selling price at ₦245 each
= (103)5 x 245
= 30325
Hence his gain = 30325 - 20025
= 10305
Pregunta 18 Informe
Detalles de la respuesta
= x + 10 + 5x
= 6x + 10
Ngozi = 118 km
If they are equal, 6x + 10 = 188
but 6x + 10 < 118
6x < 108
= x < 18
0 < x < 18 = 0 ≤ x < 18
Pregunta 19 Informe
The sides of a triangle are \( (x + 4)\text{cm} \), \( x\text{cm} \) and \( (x - 4)\text{cm} \), respectively If the cosine of the largest angle is \( \frac{1}{5} \), find the value of \( x \)
Detalles de la respuesta
Cosine B = 15
= 0.2 given
b2 - a2 + c2 - 2a Cos B
Cos B = a2+c2−b22ac
15
= x2+?(x−4)2−(x+4)22x(x−4)
15
= x(x−16)2x(x−4)
15
= x−162x−8
= 5(x - 16)
= 2x - 8
3x = 72
x = 723
= 24
Pregunta 20 Informe
Detalles de la respuesta
h = rtanyo
= r cot yo
Pregunta 21 Informe
Detalles de la respuesta
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4
Pregunta 22 Informe
If the price of oranges was raised by \( \frac{1}{2} \)k per orange. The number of oranges a customer can buy for ?2.40 will be less by 16. What is the present price of an orange?
Detalles de la respuesta
y represent the number of oranges that can be bought
xy = 240k, y = 240x .....(i)
If the price of an oranges is raised by 12 k per orange, number that can be bought for ?240 is reduced by 16
Hence, y - 16 = 240x+12
= 4802x+1
= 4802x+1 .....(ii)
subt. for y in eqn (ii) 240x - 16
= 4802x+1
= 240?16xx
= 4802x+1
= (240 - 16x)(2x + 1)
= 480x
= 480x + 240 - 32x2 - 16
480x = 224 - 32x2
x2 = 7
x = ?7
= 2.5
= 212 k
Pregunta 23 Informe
The quadratic equation whose roots are \(1 - \sqrt{13}\) and \(1 + \sqrt{13}\) is
Detalles de la respuesta
Product of roots = (1 - √13 ) (1 + √13 ) = -12
x2 - (sum of roots) x + (product of roots) = 0
x2 - 2x + 12 = 0
Pregunta 24 Informe
In the diagram, find the size of the angle marked ao
Detalles de la respuesta
S = 280o2
= 140
< O = 360 - 280 = 80o
60 + 80 + 140 + a = 360o
(< in a quad); 280 = a = 360
a = 360 - 280
a = 80o
Pregunta 25 Informe
Find the x co-ordinates of the points of intersection of the two equations in the graph.
Detalles de la respuesta
then x2 - 2x + 1 = 2x + 1
x2 - 4x = 0
x(x - 4) = 0
x = 0 or 4
Pregunta 26 Informe
In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.
Detalles de la respuesta
612=8PS
PS = 12×86
= 16cm
Pregunta 27 Informe
Detalles de la respuesta
Labour Cost = 70120 x 360o1
= 120o
Pregunta 28 Informe
Detalles de la respuesta
Pregunta 29 Informe
In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.
Detalles de la respuesta
Hexagon is a six sided polygon.
Sum of interior angles of polygon = (2n - 4)90∘ = [2 x 6 - 4] x 90 = 8 x 90 = 720∘
each angle = 720o6=120o and TVS = 1202=60o
Pregunta 30 Informe
Simplify \( (1 + \frac{\frac{x - 1}{1}}{\frac{1}{x + 1}})(x + 2) \)
Detalles de la respuesta
Pregunta 31 Informe
Detalles de la respuesta
Total score of girls = A × 8 = 8A
∴ 180 + 8A = 468
8A = 288
A = 36
Pregunta 32 Informe
The table below is drawn for a graph \(y = x^3 - 3x + 1\).
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
| \(y = x^3 - 3x + 1\) | \(1\) | \(-1\) | \(3\) | \(1\) | \(-1\) | \(3\) | \(1\) |
From x = -2 to x = 1, the graph crosses the x-axis in the range(s)
Detalles de la respuesta
Pregunta 33 Informe
Detalles de la respuesta
x1 = y4
y = 4x
Hence the sign of the graph is y + 3 = 4x
Pregunta 34 Informe
In the figure, 0 is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is
Detalles de la respuesta
< R = < P = 45∘ (corresponding angles)
< PSO = < P = 45∘ (△ PSO is a right angle)
Pregunta 35 Informe
Detalles de la respuesta
Prob. of both showing the same number of points
= 636
= 16
Pregunta 36 Informe
Detalles de la respuesta
= (2 + 3) + (x - 3) + 4
5(x - 3) + 4
5x - 15 + 4
= 5x - 11
f(3) = 5 x 3 - 11
= 4
Pregunta 37 Informe
Rationalize \( \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} \)
Detalles de la respuesta
= (5×7)+(5√7×5)−(7×√5×7)(−7×5)(√7)2
= 5√35−7√352
= −2√352
= - √35
Pregunta 39 Informe
Find, without using logarithm tables, the value of \( \frac{\log_{3} 27-\log_{\frac{1}{4}} 64}{\log_{3} \frac{1}{81}} \)
Detalles de la respuesta
=3 log3181
= -4 log33 = -4
let log14 64 = (14 )x
= 64
4-x = 43
log327−log1464log3181 = 3−(−3)−4
= −64
= −32
Pregunta 41 Informe
Simplify \( \frac{\left(\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{3}\text{ of }\frac{2}{5}}{3-\frac{1}{1\frac{1}{2}}} \)
Detalles de la respuesta
23−15
= 10−315
= 715
13
Of 25
= 13
x 25
= 25
(23−15
) - 13
of 25
= 715−215
= 13
3 - 1112
= 3 - 23
= 73
23−15of2153−1112
= 1373
= 13
x 37
= 17
Pregunta 42 Informe
Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below:
| Class boundary in cm | frequency |
| 3.35 − 3.45 | 3 |
| 3.45 − 3.55 | 6 |
| 3.55 − 3.65 | 7 |
| 3.65 − 3.75 | 4 |
What is the mean diameter of the copper spheres?
Detalles de la respuesta
∑f = 20
∑fx = 71.2
mean = ∑fx∑f
= 71.220
= 3.56
Pregunta 43 Informe
Find the area of the shaded portion of the semicircular figure.
Detalles de la respuesta
= 16πr2
A△ = 12r2sin60o
12r2×√32=r2√34
Ashaded portion
= Asector -
A△
= (16πr2−r2√34)3
= πr22−3r2√34
= r24(2π−3√3)
Pregunta 44 Informe
Detalles de la respuesta
Total amount invested = ₦50,000
Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest
then yield on x = 6100 x and yield on y = 8100 y
Hence, 6100 x + 8100 y
= 3,700.........(i)
x + y = 50,000........(ii)
6x + 8y = 370,000 x 1
x + y = 50,000 x 6
6x + 8y = 370,000.........(iii)
6x + 6y = 3000,000........(iv)
Eqn (iii) - Eqn (ii)
2y = 70,000
y = 70,0002
= 35,000
Money invested at 8% is ₦35,000
Pregunta 45 Informe
Detalles de la respuesta
Let x represent numbers divisible by 4, nth term = a + (n - 1)d
a = 4, d = 4
Last term = 4 + (n - 1)4
288 = 4 + 4n - n
= 2884
= 72
rn(Note: 288 is the last Number divisible by 4 between 1 and 300)
Prob. of x = 72288
= 14
