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