JAMB UTME - Mathematics - 2001
Question 1 Report
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Question 2 Report
The graph shows the cumulative frequency of the distribution of masses of fertilizer for 48 workers in one institution. Which of the following gives the inter-quartile range?
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Question 3 Report
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Question 4 Report
The distribution of colours of beads in a bowl is given above. What is the probability that a bead selected at random will be blue or white?
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= 1+2+4+5+3
=15
Number of blue beads = 1
P(Blue beads) = 1/15
Numbers of white beads = 4
P(white beads) = 4/15
∴P(Blue of white beads)
= P(Blue) + P(White)
= 1/15 + 4/15
= 5/15
= 1/3
Question 5 Report
Given that \(p = 1 + \sqrt{2}\) and \(q = 1 - \sqrt{2}\), evaluate \(\frac{(p^2 - q^2)}{2pq}\).
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Question 6 Report
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So that perimeter of square = 4p
4 x (p/4) = p.
Question 7 Report
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Question 8 Report
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Question 9 Report
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Question 10 Report
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This implies that Team P and Q wins
∴ P(P wins) = 1/2
P(Q wins) = 1/2
∴ P(game ends in a draw) = 1/2*1/2 = 1/4
Question 11 Report
| ⊗ | k | l | m |
| k | l | m | k |
| l | m | k | l |
| m | k | l | m |
The identity element with respect to the multiplication shown in the table above is
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Question 12 Report
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Question 13 Report
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y = 4 - x2 and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = -3 or x = 1.
Integrating x2 + 2x - 3 from (-3, to 1) w.r.t x will give 31/3 = 10(1/3)
Question 14 Report
The histogram above shows the distribution of passengers in taxis of a certain motor park. How many taxis have more than 4 passengers
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Question 15 Report
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Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6
Question 16 Report
In the figure above, PQR is a straight line segment, PQ = QT. Triangle PQT is an isosceles triangle, ?SQR is 75\(^{\circ}\) and ?QPT is 25\(^{\circ}\). Calculate the value of ?RST.
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?PTQ = 25? (base ?s of isosceles ?)
In ? QSR,
?RQS = ?QPT + ?QTP
(Extr = sum of interior opposite ?s)
?RQS = 25 + 25
= 50?
Also in ? QSR,
75 + ?RQS + ?QSR = 180?
(sum of ?s of ?)
?75 + 50 + ?QSR = 180
125 + ?QSR = 180
?QSR = 180 - 125
?QSR = 55?
But ?QSR and ?RST are the same
?RST = 55?
Question 17 Report
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Question 18 Report
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Question 19 Report
The bar chart above shows different colours of passing a particular point of a certain street in two minutes. What fraction of the total number of cars
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Question 20 Report
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Question 21 Report
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A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = ₦5,000
Question 22 Report
The bar chart shows different colours of cars passing a particular point of a certain street in two minutes. What fraction of the cars is yellow
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Thus, the fraction of the total numbers that are yellow is 325
Question 24 Report
m2 + n2 = 29
m + n = 7
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m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)
Question 25 Report
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Question 26 Report
If \(x = \frac{y}{2}\), evaluate \(\left(\frac{x^3}{y^3} + \frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^2}{y^2}\right)\)
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Question 27 Report
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=0.14 x 0.275 = 0.0385 Approx. 0.039
Question 28 Report
Triangle SPT is the solution of the linear inequalities
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2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1
Question 29 Report
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Question 30 Report
P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
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Question 31 Report
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Square of 11 = 121
Question 32 Report
Find the value of \( \theta \) in the diagram
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3t2 = 2t2 - 2t2 cos? = 2t2(1 - cos? )
1 - cos? = 3t22t2 = 32
cos = 1 - 32=?12
? = cos-1(-12 ) = 120? and 240?
N.B 0 ? ? 360
Question 33 Report
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Question 34 Report
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Cos 30 = x/8
x = 8 cos 30 = 8√3cm
Question 35 Report
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To divide the given expression, we can use the long division method.
ax^2 - 26x + 156a - 216b
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ax | ax^3 + 3x^2 - 26x^2 + 156ax - 216bx^2 - 24ax + 108
-ax^3
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3x^2 - 216bx^2
3x^2 - 3ax^2
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-216bx^2 + 3ax^2
-216bx^2 + 36ax^2
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-33ax^2 + 156a - 216bx^2 - 24ax + 108
-33ax^2 + 33bx^2
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156a - 33ax^2 - 24ax + 108
156a - 33ax^2 - 156ax + 33ax
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-24ax + 108
-24ax + 24bx
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108 - 24bx
108 - 2ax
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2ax - 24bx + 108
Therefore, the quotient is ax - 2 with a remainder of 2ax - 24bx + 108.
Hence, the answer is ax - 2.
Question 36 Report
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Question 37 Report
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Question 38 Report
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Question 39 Report
A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?
(Take pi = 22/7)
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Question 40 Report
The mean score is
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(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7
Question 42 Report
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Question 44 Report
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Question 45 Report
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Question 46 Report
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x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3
Question 47 Report
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Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
Question 48 Report
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Question 49 Report
A sector of a circle of radius 7.2cm which subtends an angle of \(300^\circ\) at the centre is used to form a cone. What is the radius of the base of the cone?
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Question 50 Report
