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