JAMB UTME - Mathematics - 1983
Tambaya 1 Rahoto
In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?
Bayanin Amsa
9x = 360∘ + 18∘
x = 3789
= 42∘ , if x = 42∘ , then add maths = 2x−42360 x 60
= 2×42−24360 x 60
= 84−246
= 10
Tambaya 2 Rahoto
A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?
Bayanin Amsa
Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000
Tambaya 3 Rahoto
Bayanin Amsa
n = 6
sum of all interior angles of hexagon polygon will be (2n - 4) x 90o
= [2 x 6 - 4] x 90
= 720o
If one angle is 170o and each of the remaining five angles is 5xo
5xo + 170o = 720o
= 5x + 550
x = 5505
= 110o
Tambaya 4 Rahoto
Bayanin Amsa
Tambaya 5 Rahoto
Bayanin Amsa
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
Tambaya 6 Rahoto
Bayanin Amsa
Tambaya 7 Rahoto
The figure FGHK is a rhombus. What is the value of angle X?
Bayanin Amsa
< KFG = < KHG = x(opposite angles)
x = 120?
Tambaya 8 Rahoto
Given that cos z = L , whrere z is an acute angle, find an expression for \( \frac{\cot z - \cosec z}{\sec z + \tan z} \)
Bayanin Amsa
Given Cos z = L, z is an acute angle
cot z - cosec zsec z + tan z
= cos z
= cos zsin z
cosec z = 1sin z
cot z - cosec z = cos zsin z
- 1sin z
cot z - cosec z = L−1sin z
sec z = 1cos z
tan z = sin zcos z
sec z = 1cos z
+ sin zcos z
= 1l
+ sin zL
the original eqn. becomes
cot z - cosec zsec z + tan z
= L−1sin z1+sinzL
= L(L−1)sin z(1+sin z)
= L(L−1)sin z+1−cos2z
= sin z + 1
= 1 + √1−L2
= L(L−1)1−L+1√1−L2
Tambaya 9 Rahoto
In the figure, find PRQ
Bayanin Amsa
¯PQR = 1252
= 6212 ?
Tambaya 10 Rahoto
Bayanin Amsa
w = k1v
u = k2w3
u = k2(k1v )3
= k2k21v3
k = k2k1k2
u = kv3
k = uv3
= (1)(2)3
= 8
u = 8v3
Tambaya 11 Rahoto
Bayanin Amsa
672 x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively
Tambaya 12 Rahoto
70. What is the sum of the median and modal scores?
Bayanin Amsa
Tambaya 13 Rahoto
Bayanin Amsa
prob. of vowel = 613
Tambaya 14 Rahoto
Bayanin Amsa
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1
Tambaya 15 Rahoto
Bayanin Amsa
let x represent the speed, d represent distance
x = d4
d = 4x
2x = 600−d3
6x = 600 - d
6x = 600 - 4x
10x = 600
x = 60010
= 60km/hr
Tambaya 16 Rahoto
Which of the following equations represents the graph above?
Bayanin Amsa
(x + 2)(4x - 1) = 0
y = 2 - 7x - 4x2
Tambaya 17 Rahoto
Bayanin Amsa
Tambaya 18 Rahoto
Bayanin Amsa
Tambaya 19 Rahoto
PQRS is a desk of dimensions \(2\text{m} \times 0.8\) which is inclined at \(30^\circ\) to the horizontal. Find the inclination of the diagonal PR to the horizontal
Bayanin Amsa
θ = tan-1(0.4)
From the diagram, the inclination of the diagonal PR to the horizontal is 10∘ 42'
Tambaya 20 Rahoto
If \(0.0000152 \times 0.042 = A \times 10^8\), where \(1 \le A < 10\), find A and B
Bayanin Amsa
0.0000152 x 0.042 = A x 108
1 ≤
A < 10, it means values of A includes 1 - 9
0.0000152 = 1.52 x 10-5
0.00042 = 4.2 x 10-4
1.52 x 4.2 = 6.384
10-5 x 10-4
= 10-5-4
= 10-9
= 6.38 x 10-9
A = 6.38, B = -9
Tambaya 21 Rahoto
Without using tables find the numerical value of \( \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) \)
Bayanin Amsa
= log77
= 1
Tambaya 22 Rahoto
Bayanin Amsa
where s = S x k
Area of the paper = (2.524)2
area of the diagram = (0.524)2
area not covered = (2.524)2 - (0.524)2
= 6.370576 - 0.274576
= 6.096
= 6.10cm2 (2 s.f)
Tambaya 23 Rahoto
Simplify \( \frac{3\sqrt{27a^{-9}}}{8} \)
Bayanin Amsa
= 32 x 1a3
= 32a3
Tambaya 24 Rahoto
If O is the centre of the circle in the figure, find the value of x
Bayanin Amsa
= 360 - 260
= 100∘
Tambaya 25 Rahoto
Bayanin Amsa
= 5250 x 100
= 2
Tambaya 26 Rahoto
Bayanin Amsa
PR = 8 sin 60 = 8√32
= 4√3
Cos 45o = PRPS = 4√3PS
PS Cos45o = 4√3
PS = 4√3 x 2
= 4√6
Tambaya 27 Rahoto
Bayanin Amsa
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 = 461r2
= 124
Tambaya 28 Rahoto
In the figure PT is a tangent to the circle with centre at O. If PQT = \(30^\circ\), find the value of PTO
Bayanin Amsa
PTQ = 50∘ (opposite angles are supplementary)
Tambaya 29 Rahoto
Bayanin Amsa
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)
Tambaya 30 Rahoto
In the figure, PQR is a straight line. Find the values of x and y.
Bayanin Amsa
3x + 6y + 90∘ = 360∘
3x + 67 = 270.......(i) x 2, 5x + y + y = 180∘
5x + 2y = 180∘ .......(ii) x 6
6 x 12y = 540..........(iii)
30x + 12y = 1080.........(iv)
egn(iv) - eqn(iii)
24x = 540
x = 22.5 and y = 33.75∘
Tambaya 31 Rahoto
In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per household:
| No. of members per household | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Total |
| No. of households | 3 | 12 | 15 | 28 | 21 | 10 | 7 | 4 | 100 |
What is the median?
Bayanin Amsa
Median is half of total frequency = 50th term
4 falls in the range = 4
Tambaya 33 Rahoto
Bayanin Amsa
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6
Tambaya 34 Rahoto
If \( \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} \), find the values of m and n
Bayanin Amsa
2m4n x 3n3m = 283-6
2m - 2n x 3n - m = 28 x 3-6
m - 2n = 8........(i)
-m + n = -6........(ii)
Solving the equations simultaneously
m = 4, n = -2
Tambaya 35 Rahoto
Bayanin Amsa
The median = 47+502
972
= 4812
Tambaya 36 Rahoto
Bayanin Amsa
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i)
3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2
Tambaya 37 Rahoto
PQRS is a cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of \(50^\circ\) with the tangent as shown in the figure. What is the size of QRS?
Bayanin Amsa
< SPQ + < SRQ = 180(Supplementary)
80 + < QRS = 180? - 80?
= 100?
Tambaya 38 Rahoto
Bayanin Amsa
mean = 126.35
= 25.26
Tambaya 39 Rahoto
Bayanin Amsa
sum of interior angles of an hexagon
(2 x 6 - 4) x 90o = (12 - 4) x 90o
= 8 x 90o
= 720o
each interior angle will have 720o6
= 120o
Tambaya 40 Rahoto
Simplify \( \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} \)
Bayanin Amsa
= log10 a712 - log10 a712
= log10 1 = 0
Tambaya 41 Rahoto
Make T the subject of the equation \( \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} \)
Bayanin Amsa
av1?v
= ?2v+Ta+2T
(av)3(1?v)3
= 2v+Ta+2T
a3v3(13?v)3
= 2v+Ta+2T
= 2v(1?v)3?a4v32a3v3?(1?v)3
Tambaya 43 Rahoto
The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?
Bayanin Amsa
Okro sector = 145270 x 360o1
= 19.33∘
= 1913 ∘
Tambaya 44 Rahoto
y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.
Bayanin Amsa
y = kx2 + c√x
y = 2 when x = 1
2 = k + c1
k + c = 2
y = 6 when x = 4
6 = 16k + c2
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = 1031
c = 2 - 1031
= 62−1031
= 5231
y = 10x231+5231√x
Tambaya 45 Rahoto
Simplify \( \frac{3}{2x-1} + \frac{2-x}{x-2} \)
Bayanin Amsa
= 32x−1 - x−2x−2
= 32x−1 - 1
= 3−(2x−1)2x−1
= 3−2x+12x−1
= 4−2x2x−1
Tambaya 46 Rahoto
PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60o. Find the perimeter of the figure PTRS. \( \pi = \frac{22}{7} \)
Bayanin Amsa
= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm
Tambaya 47 Rahoto
Bayanin Amsa
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)
Tambaya 48 Rahoto
In a triangle PQT, QR = \( \sqrt{3}\text{ cm} \), PR = 3cm, PQ = \( 2\sqrt{3} \)cm and PQR = 30o. Find angles P and R
Bayanin Amsa
Cos P = Q2+R2−p22QR
= (3)2+2(√3)2−322√3
= 3+12−912
= 612
= 12
= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60∘
= < P = 60∘
If < P = 60∘ and < Q = 30
< R = 180∘
- 90∘
angle P = 60∘
and angle R is 90∘
