JAMB UTME - Mathematics - 1998
Question 1 Report
In the venn diagram, the shaded region is
Question 2 Report
Answer Details
The cost of overtime = q x 2r = 2qr
The man's total weekly earning = 35r + 2qr
= r(35 + 2q)
Question 3 Report
For what value of \(x\) does \(6\sin(2x - 25)^\circ\) attain its maximum value in the range \(0^\circ \le x \le 180^\circ\)
Answer Details
Question 4 Report
Answer Details
p(1)2 + q(1) + 1 = 2
p + q + 1 = 2
p + q = 1.....(i)
pm2 + qm + 1 = (m - 1)Q(x) + 4
p(-1)2 + q(-1) + 1 = 4
p - q + 1 = 4
p - q = 3....(ii)
p + q = 1, p - q = -3
2p = -2, p = -1
-1 + q = 1
q = 2
Question 5 Report
Answer Details
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3
Question 6 Report
The shaded area represents
Answer Details
= y−y1x−x1
m = y−3x ≥ −32
2(y - 3) ≥ - 3x = 2y - 6 ≥ - 3x
= 2y + 3x ≤ 6 ; x ≤ 0, y ≤ 0
Question 7 Report
Make \( \frac{a}{x} \) the subject of formula \( \frac{x+1}{x-a} \)
Answer Details
x + a = mx - ma
a + ma = mx - x
a(m + 1) = x(m - 1)
ax = m?1m+a
Question 8 Report
The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?
Answer Details
no pupils who failed the test = 1 + 3 + 4 + 7 + 10
= 25
5 of pupils who fail = 2560 x 100%
= 41.70%
Question 9 Report
If \(y = 243(4x + 5)^{-2}\), find \(\frac{dy}{dx}\) when \(x = 1\).
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Question 10 Report
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Question 11 Report
TQ is tangent to circle XYTR, \( \angle YXT = 32^\circ \), \( \angle RTQ = 40^\circ \). Find \( \angle YTR \)
Answer Details
< TWR = < TXR = 40∘ (Angles in the same segments)
< YXR = 40∘ + 32∘ = 72∘
< YXR + < YTR = 180∘ (Supplementary)
72∘ + < YTR = 180∘
< YTR = 180∘ - 72∘
= 108∘
Question 12 Report
Answer Details
a + ar + ar2 = 12 (a1−r )
2(1 + r + r)(r - 1) = 1
= 2r3 = 3
= r3 = 32
r(32 )13 = √32
Question 13 Report
Answer Details
= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -411
Question 14 Report
Given that \( \log_{4}(y - 1) + \log_{4}\left(\frac{1}{2}x\right) = 1 \) and \( \log_{2}(y + 1) + \log_{2}x = 2 \), solve for x and y respectively
Answer Details
log4(y - 1)(12 x) → (y - 1)(12 x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2 → (y + 1)x = 22 = 4.....(ii)
From equation (ii) x = 4y+1 ........(iii)
put equation (iii) in (i) = y (y - 1)[12(4y−1 )] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x = 4−3+1
= 4−2
X = 2
therefore x = -2, y = -3
Question 15 Report
Answer Details
= 3x+33
= 1 + x
X(X−X)(X−X)21−xx2x+1002x+1xx22x2
S.D = √∑(x−7)2∑f
= √(6)2
= 2x23
= 2x2
= 18
x2 = 9
∴ x = ± √9
= ± 3
Question 16 Report
| Average hourly earnings(N) | 5−9 | 10−14 | 15−19 | 20−24 |
| No. of workers | 17 | 32 | 25 | 24 |
Estimate the mode of the above frequency distribution
Answer Details
mode = 9 + D1D2+D1 x C
= 9.5 + 5(32−17)2(32)−17−25
= 9.5 + 7527
= 12.27
≈ 2.3
Question 17 Report
The binary operation \( \oplus \) is defined by \( x \ast y = xy - y - x \) for all real values \( x \) and \( y \). If \( x \ast 3 = 2 \ast \), find \( x \)
Answer Details
2 ∗ x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1
Question 18 Report
If 10112 + x7 = 2510, solve for X.
Answer Details
10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o
= 8 + 0 + 2 + 1
= 1110
x7 = 2510 - 1110
= 1410
71472R00R2
X = 207
Question 19 Report
In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35o, find ∠BAO.
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Question 20 Report
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Question 21 Report
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Question 23 Report
The pie chart shows the monthly expenditure pf a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is ₦7200?
Answer Details
Question 24 Report
Differentiate \( \frac{x}{\cos x} \) with respect to x
Answer Details
y = u(x) v (x0
dydx = Udydx + Vdudx
dy x [secx tanx] + secx
x = x secx tanx + secx
Question 25 Report
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Question 26 Report
If x is a positive real number, find the range of values for which \( \frac{1}{3x} + \frac{1}{2} > \frac{1}{4x} \)
Answer Details
= 2+3x6x > 14x
= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0
= x < 0, x < 16 = x < 16 (solution)
Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > 16
Combining solutions in cases(1) and (2)
= x > 0, x < 16
= 0 < x < 16
Question 27 Report
Answer Details
Question 28 Report
Evaluate \[ \frac{1}{0.03} \div \frac{1}{0.024} \]⁻¹ correct to 2 decimal places
Answer Details
= [10.03×0.024 ]-1
= [0.0240.003 ]-1
= 0.030.024
= 3024 = 1.25
Question 29 Report
Evaluate \( \int_{2}^{\pi} (\sec^2 x - \tan^2 x)\,dx \)
Answer Details
∫π2 dx = [X]π2
= π - 2 + c
when c is an arbitrary constant of integration
Question 30 Report
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In the diagram, QTR is a straight line and < PQT = \(30^\circ\). find the sin of < PTR
Answer Details
1520=sinx
sin x = 1520=34
N.B x = < PRQ
Question 31 Report
Let \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) p = \( \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \) Q = \( \begin{pmatrix} u & 4 + u \\ -2v & v \end{pmatrix} \) be 2 x 2 matrices such that PQ = 1. Find (u, v)
Answer Details
= ((2u−6v2(4+u)+3v)4u−10v4(4+u)+5v)
= (1001)
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
4u - 12v = 0-4u - 10v = 0
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -52
∴ (U, V) = (-52 - 1)
Question 32 Report
Answer Details
Question 33 Report
Question 34 Report
A bag contains 16 red balls and 20 blue balls only. How many white balls must be added to the bag so that the probability of randomly picking a red ball is equal to \( \frac{2}{5} \)
Answer Details
Number of blue balls = 20
Let x represent the No of white balls to be added
∴ Total number of balls = 36 + x
2(36 + x) = 80
= 2x + 80 - 72
= 8
x = 82
= 4
Question 35 Report
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P (product of x and y not greater than 6) = 1016
= 58
Question 36 Report
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Question 37 Report
Find the value of k if \( \frac{k}{\sqrt{3}+\sqrt{2}} = k\sqrt{3-2} \)
Answer Details
k√3+√2 x √3−√2√3−√2
= k√3−2
= k(√3−√2 )
= k√3−2
= k√3 - k√2
= k√3−2
k2 = √2
k = 2√2
= √2
Question 38 Report
Answer Details
Z = 150tan60o
= 1503
= 50√3 cm
150XxZ = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50√3
= 50( √3 - √3 )m
Question 39 Report
A chord of a circle radius \( \sqrt{3}\,\mathrm{cm} \) subtends an angle of \(60^\circ\) on the circumference of he circle. Find the length of the chord
Answer Details
Question 40 Report
If \(b^3 = a^{-2}\) and \(c^{\frac{1}{3}} = a^{\frac{1}{2}}b\), express c in terms of a
Answer Details
= a12 b x a-2
= a-32
= (c13 )3
= (a-32 )13
c = a-12
Question 41 Report
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Question 42 Report
In the figure, PQST is a parallelogram and TSR is a straight line. If the area of \( \triangle \)QRS is 20cm2, find the area of the trapezium PQRT.
Answer Details
= 12 x 8 x h = 4h
h = 204
= 5cm
A△ (PQTS) = L x H
A△ PQRT = A△ QSR + A△ PQTS
20 + 50 = 70cm2
ALTERNATIVE METHOD
A△ PQRT = 12 x 5 x 28
= 70cm2
Question 43 Report
Express in partial fractions \( \frac{11+2}{6x^2-x-1} \)
Answer Details
= A3x+1 + B2x−1
11x = 2 = A(2x - 1) + B(3x + 1)
put x = 12
= -−53
= -−53 A → A = 1
∴ 11x+26x2−x−1 = 13x+1 + 32x−1
Question 44 Report
Solve for the equation \( \sqrt{x} - \sqrt{(x - 2)} - 1 = 0 \)
Answer Details
Question 45 Report
The determinant of matrix \(\begin{pmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4 \end{pmatrix}\) in terms of x is
Answer Details
= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1
Question 46 Report
Find the value of x in the diagram.
Answer Details
x = 100 - 30
= 70o
Question 47 Report
The kinetic element with respect to the multiplication shown in the diagram below is
| \(\oplus\) | \(p\) | \(p\) | \(r\) | \(s\) |
| \(p\) | \(r\) | \(p\) | \(r\) | \(p\) |
| \(q\) | \(p\) | \(q\) | \(r\) | \(s\) |
| \(r\) | \(r\) | \(r\) | \(r\) | \(r\) |
| \(s\) | \(q\) | \(s\) | \(r\) | \(q\) |
Answer Details
Question 48 Report
In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities \( \frac{2}{5} \) and \( \frac{1}{3} \) respectively of winning the gold in the football event. What is the probability that either team will win the gold?
Answer Details
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= 25 + 13
= 115
