JAMB UTME - Mathematics - 1985
Question 1 Report
Two points X and Y both on latitude \(60^\circ\text{S}\) have longitude \(147^\circ\text{E}\) and \(153^\circ\text{W}\) respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take \(2\pi R = 4 \times 10^4\text{ km}\), where R is the radius of the earth)
Answer Details
Longitude difference = 147 + 153 = 300oN
distance between xy = θ360 × 2π R cos60o
= 300360 × 4 × 104 × 12
= 1.667 × 104km (1667 km)
Question 2 Report
The number of goals scored by a football team in 20 matches is shown below:
| No. of goals | 0 | 1 | 2 | 3 | 4 | 5 |
| No. of matches | 3 | 5 | 7 | 3 | 1 | 0 |
What are the values of the mean and the mode respectively?
Answer Details
∑f = 20
∑fx = 35
Mean = ∑fx∑f
= 3520
= 74
= 1.75
Mode = 2
= 1.75, 2
Question 3 Report
If \(e^x = 1 + \frac{x}{2} + \frac{x^2}{1.2} + \frac{x^3}{1.2.3} + .....\) Find \(\frac{1}{e^x}\)
Answer Details
Question 4 Report
In the figure, GHIJKLMN is a cube of side a. Find the length of HN.
Answer Details
HJ = √2a2=a√2
HN2 = a2 + (a√2 )2 = a2 + 2a2 = 3a2
HN = √3a2
= a√3 cm
Question 5 Report
Answer Details
Prob. of getting one head + prob. of getting 2 heads
= 14 + 24
= 34
Question 6 Report
In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?
Answer Details
1N = c
= aN+kN
CN = aN + K
30(100) = a(100) + k
3000 = 100a + k.......(i)
60(40) = a(40) + k
2400 = 40a + k.......(ii)
eqn (i) - eqn (ii)
600 = 60a
a = 10
subt. for a in eqn (i) 3000 = 100(10) + K
3000 - 1000 = k
k = 2000
CN = 10N + 2000. when N = 50,
50C = 10(50) + 2000
50C = 500 + 2000
C = 250050
= 50k
Question 7 Report
Make c the subject of formula \(v = 1 - \frac{a}{5}\left(b + \frac{3c}{7}\right)\)
Answer Details
Question 8 Report
Find correct to two decimals places \(100 + \frac{1}{100} + \frac{3}{1000} + \frac{27}{10000}\)
Answer Details
1000,000+100+30+2710000 = 1,000.15710000
= 100.02
Question 9 Report
Find the missing value in the table below
| x |
-4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|
y=4?3x?x^3
|
80 | 18 | 8 | 4 | 0 | -10 | -32 |
Answer Details
= 4 + 9 + 27
= 13 + 27
= 40
Question 10 Report
In the equation below, Solve for x if all the numbers are in base 2: \(\frac{11}{x} = \frac{1000}{x+101}\)
Answer Details
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = 1111101
x = 11
Question 11 Report
22\( \frac{1}{2} \)% of the Nigerian Naira is equal to 17\( \frac{1}{10} \)% of a foreign currency M. What is the conversion rate of the M to the Naira?
Answer Details
M = 17110 %, N = 452
452 x 10171
= 225171
= 1 54171
= 1 1857
Question 12 Report
If \(a\left\{\frac{x+1}{x-2}-\frac{x-1}{x+2}\right\}=6x\). Find a in its simplest form
Answer Details
= 6
6xx2−4 = 6x
a = x2 - 4
Question 13 Report
In the figure, the area of the shaded segment is
Answer Details
Area of triangle = 12×3×3×sin120o
= 92×√32=9√34
Area of shaded portion = 3π−9√34
= 3 π−3√34
Question 14 Report
In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Zo at the centre of the circle, what is Z?
Answer Details
Total students = 120
grade = 18
Zo = 18120
x 3601
= 54o
Question 15 Report
Factorize abx2 + 8y - 4bx - 2axy
Answer Details
abx2 + 8y - 4bx - 2axy = (abx2 - 4bx) + (8y - 2axy)
= bx(ax - 4) 2y(ax - 4) 2y(ax - 4)
= (bx - 2y)(ax - 4)
Question 16 Report
If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of \( \frac{3p-q}{4q+r} \)
Answer Details
P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = 615
, q = 415
, r = 515
= 13
To find 3p−q4q+r
3p - q = 3 x 615
- 415
1815
- 415
= 1415
∴ 4q + r = 4 x 415
+ 515
1615
= 1615
+ 515
= 2115
1415
x 1521
= 1421
= 23
Question 17 Report
Answer Details
180 + 27 = 207o
Note that bearing is always taken from the north
207o = S 27o W
Question 18 Report
Question 19 Report
Question 20 Report
Answer Details
= 32 x √32 =
Area = 16√3 x 6 = 96 √3 cm2
Question 21 Report
Answer Details
35 x = 351
x = 351×53
= 585
Volume of smaller block = 23 x 585
= 234.00
Question 22 Report
Solve the following equation \( \frac{2}{2r-1} - \frac{5}{3} = \frac{1}{r+2} \)
Answer Details
22r−1 - 1r+2 = 53
2r+4−2r+12r−1(r+2) = 53
5(2r+1)(r+2) = 53
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or −52
Question 23 Report
Answer Details
x−2−1012Y=x3+x−10−20210y=x2+152125
The curves intersect at x = 1
Question 24 Report
The factors of 9 - (x2 - 3x - 1)2 are
Answer Details
9 - (x2 - 3x - 1)2 = [3 - (x2 - 3x - 1)] [3 + (x2 - 3x - 1)]
= (3 - x2 + 3x + 1)(3 + x2 - 3x - 1)
= (4 + 3x - x2)(x2 - 3x + 2)
= (4 - x)(1 + x)(x - 1)(x - 2)
= -(x - 4)(x + 1) (x - 1)(x - 2)
Question 25 Report
\((\sqrt{2} + 4\sqrt{2})(\sqrt{3} + \sqrt{2})(\sqrt{3} + \sqrt{2})\) is equal to
Answer Details
(√3+√2 + (√3+√2 ) = 3 + (√6−√6 ) - 2
= 1
Question 26 Report
Answer Details
x = 2 x 1√2
= 2√22
= √2
Question 27 Report
Answer Details
This can be rewritten as (3y)2 + 6(3y) = 27
Let 3y = x
x2 + 6x - 27 = 0
(x + 9)(x - 3) = 0
when x - 3 = 0, x = 3
sub. for x in 3y = x
3y = 3
log33 = y
y = 1
Question 28 Report
Answer Details
x - 2 = 1, x = 3
f(x - 2) = f(1)
= 4(3)2 + 3 + 7
= 36 + 10
= 46
Question 29 Report
Answer Details
a = 1b = - (p - 2), c = 2p + 1
[-(p - 2)]2 - 4 x 1 x (2p + 1) = 0
p2 - 4p + 4 - 4(2p + 1) = 0
p2 - 4p = 4 - 8p - 4 = 0
p2 - 12p = 0
p(p - 12) = 0
p = 0 or 12
Question 30 Report
Answer Details
x2 - 73 = -R3
(x1?76 )2 = -R3 + 4936
?R3 + 4936 = 0
R = 4936 x 31
= 4912
Question 31 Report
Answer Details
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = 610
= 35
Question 32 Report
In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = \( \theta^\circ \) and RQP = \( \phi^\circ \), which of the following relationships between \( \theta^\circ \) and \( \phi^\circ \) is correct
Answer Details
180 = 2ϕ + θ ∘
Question 33 Report
Solve for (x, y) in the equation 2x + y = 4: x^2 + xy = -12
Answer Details
2x + y = 4......(i)
x^2 + xy = -12........(ii)
from eqn (i), y = 4 - 2x
= x2 + x(4 - 2x)
= -12
x2 + 4x - 2x2 = -12
4x - x2 = -12
x2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
sub. for x = 6, in eqn (i) y = -8, 8
=(6,-8); (-2, 8)
Question 34 Report
List all integers satisfying the inequality \( -2 \le 2x - 6 < 4 \)
Answer Details
-2 ≤
2x - 6 < 4 = 2x - 6 < 4
= 2x < 10
= x < 5
2x ≥
-2 + 6 ≥
= x ≥
2
∴ 2 ≤
x < 5 [2, 3, 4]
Question 35 Report
Answer Details
= 23
P(R1 n R11) = P(both red)
35 x 59
= 13
Question 36 Report
Answer Details
= 2√3
(2)23
= 43
1 + sec2 30o = sec 30o
= 1 + 4√3
= 213
Question 37 Report
In \( \triangle \) XYZ, XKZ = 90O, XK = 15cm, XY = 35cm and YK = 8cm. Find the area of \( \triangle \) XYZ
Answer Details
KZ2 = (25 + 15)(25 - 15) = 400
KZ = √400 = 20
area of XYZ = 12×28×15
= 210 sq. cm
Question 38 Report
Answer Details
Janet now has 132 x + 105 = x+3153
John's money left = 23 x
= 14(x+315)3
= 23
24x = 3x + 945
∴ x = 45
Question 39 Report
A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?
Answer Details
1√3(12)2
= 4√3
= √3√3
= 4√3√3
m = 64kg, V = 4πr33
= 4π(4)33
= 256π3
x 10-6m3
density(P) = MassVolume
= 64256π3×10−6
= 64×3×10−6256
= 34×10−6
m = PV = 34π×10−6
x 43
π
[32 - 22] x 10-6
34×10−6
x 43
x 5 x 10-6
= 5kg
Question 40 Report
In \( \triangle \) XYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = \( \theta \). Find cos\( \theta \)o
Answer Details
= 169+81−2126×9
cosθ = 12926×9
= 4378
Question 41 Report
Write h in terms of a, b, c, d if \(a = \frac{b(1-ch)}{a-dh}\)
Answer Details
a = b(1?ch)a?dh
a = b?bch1?dh
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = a?bad?bc
Question 42 Report
In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = \(86^\circ\) and NQP = \(122^\circ\) find (\(x^\circ\), \(y^\circ\))
Answer Details
180 - 144 = 36∘
x∘ = 180 - (94 + 58)
180 -152 = 28
(x∘ , y∘ ) = (28∘ , 36∘ )
Question 43 Report
Answer Details
2x - 3y = -10 x 2
10x - 6y = 5
4x - 6y = -20 .......(i)
10x - 6y = 5.......(ii)
eqn(ii) - eqn(1)
6x = 15
x = 156
= 52
x = 212
Sub. for x in equ.(ii) 10(52 ) - 6y = 5
y = 312
Question 44 Report
Answer Details
T = 4yrs, R = 8%, a = N330.00
330 - P = PTR100 , A = P + I
i.e. A = P + PTR100
330 = P + P(4)(8)100
33000 = 32P + 100p
132P = 33000
P = N250.00
Question 45 Report
Arrange the following numbers in ascending order of magnitude \( \frac{6}{7} \), \( \frac{13}{15} \), 0.8650
Answer Details
67
, 1315
, 0.8650
In ascending order, we have 0.8571, 0.8650, 0.8666
i.e. 67
< 0.8650 < 1315
Question 46 Report
Answer Details
Area of trapezium = 12 (a + b)h
14 = 12 (4 + 3)h
14 = 72 h
h = 14×27
= 4
Area of SQR = 12 (3 x 4)
= 122
= 6.0
Question 47 Report
If \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \theta \) is less than 90o. Calculate \( \cos \frac{90-\theta}{\sin^2\theta} \)
Answer Details
= tanθsin2θ
Sin2θ =14
cos(90−θ)sin2θ
= 1√3
= 4√3
