JAMB UTME - Mathematics - 1983
Ajụjụ 1 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
Circumference of PRS = π2
= 227
x 71
x 12
= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm
Ajụjụ 2 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 3 Ripọtì
Simplify \( \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} \)
Akọwa Nkọwa
log10 a13
+ 14
log10 a - 112
log10a7 = log10 a13
+ log1014
- log10 a712
= log10 a712
- log10 a712
= log10 1 = 0
Ajụjụ 4 Ripọtì
If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?
Ajụjụ 5 Ripọtì
If O is the centre of the circle in the figure, find the value of x
Akọwa Nkọwa
From the diagram; The value of x = 360∘ - 2(130∘ )
= 360 - 260
= 100∘
Ajụjụ 6 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 7 Ripọtì
If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2
Akọwa Nkọwa
W α
1v
u α
w3
w = k1v
u = k2w3
u = k2(k1v
)3
= k2k21v3
k = k2k1k2
u = kv3
k = uv3
= (1)(2)3
= 8
u = 8v3
Ajụjụ 8 Ripọtì
A ship H leaves a port P and sails 30 km due south. Then it sails 50km due west. What is the bearing of H from P
Akọwa Nkọwa
Ajụjụ 9 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
(23
)m (34
)n = 252729
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 10 Ripọtì
The scores of 16 students in a mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65,
70. What is the sum of the median and modal scores?
Akọwa Nkọwa
Ajụjụ 11 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 12 Ripọtì
In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS
Akọwa Nkọwa
From the diagram, sin 60o = PR8
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 13 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 = 80?
< SPQ + < SRQ = 180(Supplementary)
80 + < QRS = 180? - 80?
= 100?
Ajụjụ 14 Ripọtì
The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x
Akọwa Nkọwa
(3x + 1)2 = (3x - 1)2 + 2 (Pythagoras's theorem)
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12
Ajụjụ 15 Ripọtì
PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?
Akọwa Nkọwa
Ajụjụ 16 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
FROM the diagram, PQT = 50∘
PTQ = 50∘ (opposite angles are supplementary)
Ajụjụ 17 Ripọtì
One interior angle of a convex hexagon is 170o and each of the remaining interior angles is equal to xo. Find x
Akọwa Nkọwa
An hexagon polygon is a six sided polygon
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 18 Ripọtì
If x + 2 and x - 1 are factors of the expression 1x3 + 2kx2 + 24, find the values of 1 and K
Akọwa Nkọwa
f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6
Ajụjụ 19 Ripọtì
In the figure, PQR is a straight line. Find the values of x and y.
Akọwa Nkọwa
23 + 3y + 45∘ = 180∘
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 20 Ripọtì
Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77
Akọwa Nkọwa
24.57+25.63+24.32+26.01+25.775
mean = 126.35
= 25.26
Ajụjụ 21 Ripọtì
Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12
Akọwa Nkọwa
4x - 3 = 3x + y = x - y = 3.......(i)
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 22 Ripọtì
If a rod of length 250cm is measured as 255cm long in error, what is the percentage error in the measurement?
Akọwa Nkọwa
% error = Actual errorreal value
x 100
= 5250
x 100
= 2
Ajụjụ 23 Ripọtì
Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26
Akọwa Nkọwa
6 + 10 + 14 + 16 + 26 = 72
672
x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively
Ajụjụ 24 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
2x - 24)∘ + (3x - 18)∘ + (2x + 12)∘ + (x + 12)∘ + x∘ = 360∘
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 25 Ripọtì
Simplify \( \frac{3\sqrt{27a^{-9}}}{8} \)
Akọwa Nkọwa
3√27a−98
= 3a−32
= 32
x 1a3
= 32a3
Ajụjụ 26 Ripọtì
If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is
Ajụjụ 28 Ripọtì
Correct each of the numbers 59.81798 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures
Akọwa Nkọwa
59.81798 = 59.8 (3 s.f)
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)
Ajụjụ 29 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 30 Ripọtì
Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0
Akọwa Nkọwa
x2 + y - 8 = 0, y + 5x - 2 = 0
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
Ajụjụ 31 Ripọtì
In the figure, find PRQ
Akọwa Nkọwa
Angle subtended at any part of the circumference of the circle 125o2 at centre = 360? - 235? = 125?
¯PQR = 1252
= 6212 ?
Ajụjụ 32 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 33 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
Adding the values of all the items together, it gives 70
Okro sector = 145270 x 360o1
= 19.33∘
= 1913 ∘
Ajụjụ 34 Ripọtì
Simplify \( \frac{3}{2x-1} + \frac{2-x}{x-2} \)
Akọwa Nkọwa
32x−1
+ 2−xx−2
= 32x−1
- x−2x−2
= 32x−1
- 1
= 3−(2x−1)2x−1
= 3−2x+12x−1
= 4−2x2x−1
Ajụjụ 35 Ripọtì
The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.
Akọwa Nkọwa
By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median = 47+502
972
= 4812
Ajụjụ 36 Ripọtì
The letters of the word MATRICULATION are cut and put into a box. One letter is drawn at random from the box. Find the probability of drawing a vowel
Akọwa Nkọwa
Vowels of letters are 6 in numbers
prob. of vowel = 613
Ajụjụ 37 Ripọtì
Given a regular hexagon, calculate each interior angle of the hexagon
Akọwa Nkọwa
Sum of interior angles of polygon = (2n - 4)rt < s
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 38 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
tanθ = 0.82 = (0.4)
θ = tan-1(0.4)
From the diagram, the inclination of the diagonal PR to the horizontal is 10∘ 42'
Ajụjụ 40 Ripọtì
The figure FGHK is a rhombus. What is the value of angle X?
Akọwa Nkọwa
< HKF = 60? , < KFG = 120?
< KFG = < KHG = x(opposite angles)
x = 120?
Ajụjụ 41 Ripọtì
Find x if (xbase4)2 = 100100base2
Akọwa Nkọwa
(x2)4 to base10 gives (1 x 25) + (1 x 25) + (1 x 2)
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 = 461r2
= 124
Ajụjụ 42 Ripọtì
A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?
Akọwa Nkọwa
Speed = distancetime
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 43 Ripọtì
On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.
Akọwa Nkọwa
Area of the paper = area of square = L x B or S2
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 44 Ripọtì
Factorize completely 81a4 - 16b4
Akọwa Nkọwa
81a4 - 16b4 = (9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)
Ajụjụ 45 Ripọtì
Without using tables find the numerical value of \( \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) \)
Akọwa Nkọwa
log749 + log717
= log77
= 1
Ajụjụ 46 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
By using cosine formula, p2 = Q2 + R2 - 2QR cos p
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 47 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
xf132133154285216107784
Median is half of total frequency = 50th term
4 falls in the range = 4
Ajụjụ 48 Ripọtì
Which of the following equations represents the graph above?
Akọwa Nkọwa
x = -2 or x = 14
(x + 2)(4x - 1) = 0
y = 2 - 7x - 4x2
