JAMB UTME - Mathematics - 2001
Ajụjụ 1 Ripọtì
Akọwa Nkọwa
Ajụjụ 2 Ripọtì
The mean score is
Akọwa Nkọwa
(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7
Ajụjụ 3 Ripọtì
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?
Ajụjụ 4 Ripọtì
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
Akọwa Nkọwa
Cos 30 = x/8
x = 8 cos 30 = 8√3cm
Ajụjụ 6 Ripọtì
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 10 Ripọtì
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
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?
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
The histogram above shows the distribution of passengers in taxis of a certain motor park. How many taxis have more than 4 passengers
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 20 Ripọtì
Akọwa Nkọwa
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)
Ajụjụ 21 Ripọtì
Triangle SPT is the solution of the linear inequalities
Akọwa Nkọwa
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
Ajụjụ 22 Ripọtì
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
| ⊗ | 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
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
Akọwa Nkọwa
=0.14 x 0.275 = 0.0385 Approx. 0.039
Ajụjụ 27 Ripọtì
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Akọwa Nkọwa
So that perimeter of square = 4p
4 x (p/4) = p.
Ajụjụ 29 Ripọtì
P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
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
Akọwa Nkọwa
Thus, the fraction of the total numbers that are yellow is 325
Ajụjụ 31 Ripọtì
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
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
Given that \(p = 1 + \sqrt{2}\) and \(q = 1 - \sqrt{2}\), evaluate \(\frac{(p^2 - q^2)}{2pq}\).
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
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)
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 36 Ripọtì
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
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.
Akọwa Nkọwa
?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?
Ajụjụ 39 Ripọtì
Akọwa Nkọwa
Square of 11 = 121
Ajụjụ 40 Ripọtì
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Akọwa Nkọwa
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
Ajụjụ 43 Ripọtì
Akọwa Nkọwa
To divide the given expression, we can use the long division method.
ax^2 - 26x + 156a - 216b
___________________________________
ax | ax^3 + 3x^2 - 26x^2 + 156ax - 216bx^2 - 24ax + 108
-ax^3
-----
3x^2 - 216bx^2
3x^2 - 3ax^2
--------------
-216bx^2 + 3ax^2
-216bx^2 + 36ax^2
------------------
-33ax^2 + 156a - 216bx^2 - 24ax + 108
-33ax^2 + 33bx^2
------------------
156a - 33ax^2 - 24ax + 108
156a - 33ax^2 - 156ax + 33ax
-------------------------
-24ax + 108
-24ax + 24bx
-------------
108 - 24bx
108 - 2ax
--------
2ax - 24bx + 108
Therefore, the quotient is ax - 2 with a remainder of 2ax - 24bx + 108.
Hence, the answer is ax - 2.
Ajụjụ 44 Ripọtì
Find the value of \( \theta \) in the diagram
Akọwa Nkọwa
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
Ajụjụ 45 Ripọtì
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
Akọwa Nkọwa
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)
Ajụjụ 47 Ripọtì
m2 + n2 = 29
m + n = 7
Akọwa Nkọwa
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)
Ajụjụ 48 Ripọtì
Akọwa Nkọwa
Ajụjụ 49 Ripọtì
If \(x = \frac{y}{2}\), evaluate \(\left(\frac{x^3}{y^3} + \frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^2}{y^2}\right)\)
Akọwa Nkọwa
Ajụjụ 50 Ripọtì
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?
Akọwa Nkọwa
= 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
