JAMB UTME - Mathematics - 1983
Ajụjụ 1 Ripọtì
Akọwa Nkọwa
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)
Ajụjụ 2 Ripọtì
Akọwa Nkọwa
mean = 126.35
= 25.26
Ajụjụ 4 Ripọtì
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. \( \pi = \frac{22}{7} \)
Akọwa Nkọwa
= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm
Ajụjụ 5 Ripọtì
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?
Akọwa Nkọwa
9x = 360∘ + 18∘
x = 3789
= 42∘ , if x = 42∘ , then add maths = 2x−42360 x 60
= 2×42−24360 x 60
= 84−246
= 10
Ajụjụ 8 Ripọtì
Ajụjụ 9 Ripọtì
If \(0.0000152 \times 0.042 = A \times 10^8\), where \(1 \le A < 10\), find A and B
Akọwa Nkọwa
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
Ajụjụ 10 Ripọtì
Akọwa Nkọwa
w = k1v
u = k2w3
u = k2(k1v )3
= k2k21v3
k = k2k1k2
u = kv3
k = uv3
= (1)(2)3
= 8
u = 8v3
Ajụjụ 11 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 12 Ripọtì
In the figure PT is a tangent to the circle with centre at O. If PQT = \(30^\circ\), find the value of PTO
Akọwa Nkọwa
PTQ = 50∘ (opposite angles are supplementary)
Ajụjụ 13 Ripọtì
Which of the following equations represents the graph above?
Akọwa Nkọwa
(x + 2)(4x - 1) = 0
y = 2 - 7x - 4x2
Ajụjụ 14 Ripọtì
Simplify \( \frac{3}{2x-1} + \frac{2-x}{x-2} \)
Akọwa Nkọwa
= 32x−1 - x−2x−2
= 32x−1 - 1
= 3−(2x−1)2x−1
= 3−2x+12x−1
= 4−2x2x−1
Ajụjụ 15 Ripọtì
Make T the subject of the equation \( \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} \)
Akọwa Nkọwa
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
Ajụjụ 16 Ripọtì
Akọwa Nkọwa
672 x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively
Ajụjụ 17 Ripọtì
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.
Akọwa Nkọwa
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
Ajụjụ 18 Ripọtì
If O is the centre of the circle in the figure, find the value of x
Akọwa Nkọwa
= 360 - 260
= 100∘
Ajụjụ 19 Ripọtì
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 21 Ripọtì
In a triangle PQT, QR = \( \sqrt{3}\text{ cm} \), PR = 3cm, PQ = \( 2\sqrt{3} \)cm and PQR = 30o. Find angles P and R
Akọwa Nkọwa
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∘
Ajụjụ 22 Ripọtì
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?
Akọwa Nkọwa
< SPQ + < SRQ = 180(Supplementary)
80 + < QRS = 180? - 80?
= 100?
Ajụjụ 23 Ripọtì
The figure FGHK is a rhombus. What is the value of angle X?
Akọwa Nkọwa
< KFG = < KHG = x(opposite angles)
x = 120?
Ajụjụ 24 Ripọtì
Without using tables find the numerical value of \( \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) \)
Akọwa Nkọwa
= log77
= 1
Ajụjụ 25 Ripọtì
Ajụjụ 26 Ripọtì
Simplify \( \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} \)
Akọwa Nkọwa
= log10 a712 - log10 a712
= log10 1 = 0
Ajụjụ 27 Ripọtì
Akọwa Nkọwa
PR = 8 sin 60 = 8√32
= 4√3
Cos 45o = PRPS = 4√3PS
PS Cos45o = 4√3
PS = 4√3 x 2
= 4√6
Ajụjụ 28 Ripọtì
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
Akọwa Nkọwa
θ = tan-1(0.4)
From the diagram, the inclination of the diagonal PR to the horizontal is 10∘ 42'
Ajụjụ 29 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 30 Ripọtì
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?
Akọwa Nkọwa
Okro sector = 145270 x 360o1
= 19.33∘
= 1913 ∘
Ajụjụ 31 Ripọtì
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?
Akọwa Nkọwa
Median is half of total frequency = 50th term
4 falls in the range = 4
Ajụjụ 32 Ripọtì
Akọwa Nkọwa
The median = 47+502
972
= 4812
Ajụjụ 33 Ripọtì
Akọwa Nkọwa
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 = 461r2
= 124
Ajụjụ 34 Ripọtì
Akọwa Nkọwa
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)
Ajụjụ 35 Ripọtì
Akọwa Nkọwa
prob. of vowel = 613
Ajụjụ 36 Ripọtì
Akọwa Nkọwa
= 5250 x 100
= 2
Ajụjụ 37 Ripọtì
Given that cos z = L , whrere z is an acute angle, find an expression for \( \frac{\cot z - \cosec z}{\sec z + \tan z} \)
Akọwa Nkọwa
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
Ajụjụ 38 Ripọtì
Simplify \( \frac{3\sqrt{27a^{-9}}}{8} \)
Akọwa Nkọwa
= 32 x 1a3
= 32a3
Ajụjụ 39 Ripọtì
Akọwa Nkọwa
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)
Ajụjụ 40 Ripọtì
In the figure, find PRQ
Akọwa Nkọwa
¯PQR = 1252
= 6212 ?
Ajụjụ 41 Ripọtì
Akọwa Nkọwa
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6
Ajụjụ 42 Ripọtì
Akọwa Nkọwa
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
Ajụjụ 43 Ripọtì
In the figure, PQR is a straight line. Find the values of x and y.
Akọwa Nkọwa
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∘
Ajụjụ 44 Ripọtì
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?
Akọwa Nkọwa
Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000
Ajụjụ 45 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 46 Ripọtì
70. What is the sum of the median and modal scores?
Akọwa Nkọwa
Ajụjụ 47 Ripọtì
If \( \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} \), find the values of m and n
Akọwa Nkọwa
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
Ajụjụ 48 Ripọtì
Akọwa Nkọwa
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
