JAMB UTME - Mathematics - 2000
Ajụjụ 2 Ripọtì
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
Akọwa Nkọwa
A = ₦150,000, r = 25%, n = 3.
150,000 = P(1+0.25)3 = P(1.25)3
P = 150,000/1.253 =₦76,800.00
Ajụjụ 4 Ripọtì
Akọwa Nkọwa
1. Integrate the given first derivative of f(x) at the boundaries, (0,0)
Then solve accordingly to get f(x) = y = (3x2/)2 + 2x
Ajụjụ 5 Ripọtì
The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
Find the value of \( \displaystyle \int_{0}^{\pi}\frac{\cos^{2}\theta-1}{\sin^{2}\theta}\,d\theta \)
Ajụjụ 7 Ripọtì
Ajụjụ 8 Ripọtì
Akọwa Nkọwa
dy/dx = 2 cos2x +(-2x sin2x) - 2 cos2x
= 2 cos2x - 2x sin2x - cos2x
= -2x sin2x
= -2 x (π/4) sin2 x (π/4)
= -(π/2) x 1 = -(π/2)
Ajụjụ 9 Ripọtì
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
Akọwa Nkọwa
= π(9/2 + 3) = 15π/2
Ajụjụ 13 Ripọtì
Akọwa Nkọwa
Convert 2567 and 340x to base 10, such that:
314 - 139 = 3x2 + 4x
=> 3x2 + 4x - 175 = 0 (quadratic)
Factorising, (x - 7) (3x + 25) = 0,
either x = 7 or x = -25/3 ( but x cannot be negative)
Therefore, x = 7.
Ajụjụ 14 Ripọtì
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
Given that the various faces of a fair dice 1, 2, 3, 4, 5, 6 appeared 30, 43, 54, 40, 41, 32 times respectively in a single toss. Picture the figures as being represented in a simple table with number (X) against frequency (f).
If a pie chart is used to depict the data, the angle corresponding to 4 is?
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
If \(U = \{x : x\text{ is an integer and }1 \le x \le 20\}\)
\(E_1 = \{x : x\text{ is a multiple of }3\}\)
\(E_2 = \{x : x\text{ is a multiple of }4\}\)
and an integer is picked at random from \(U\), find the probability that it is not in \(E_2\)
Akọwa Nkọwa
E1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
P(E1) = 5/20
P(not E1) = 1 - (5/20) = 15/20 = 3/4
Ajụjụ 18 Ripọtì
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
Akọwa Nkọwa
y = 4x/3 - 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3
Ajụjụ 20 Ripọtì
Ajụjụ 21 Ripọtì
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
Akọwa Nkọwa
Number of letter M = 2
Number of letter E = 2
Number of letter A = 2
Arrangement = 11!/(2! 2! 2!) ways
Ajụjụ 24 Ripọtì
Ajụjụ 25 Ripọtì
Akọwa Nkọwa
210P - 42P = 434 + 406
168P = 840
P = 840/168 = 5
Ajụjụ 26 Ripọtì
Akọwa Nkọwa
3 x 2n x 21 - 22 x 2n x 2-1 2n x 2 -2n
Solving the equation above gives;
2n x 21 x 2 2n = 2 x 2 = 4
Ajụjụ 27 Ripọtì
A predator moves in a circle of radius \( \sqrt{2} \) centre \((0,0)\), while a prey moves along the line \(y = x\). If \(0 \le x \le 2\), at which point(s) will they meet?
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
Akọwa Nkọwa
f(1) => a + b + c = 5 .....(1)
Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2)
From (2),=> b = 1, and 2ax = 2x, => a = 1.
substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3
Thus a = 1, b = 1 and c = 3
Ajụjụ 30 Ripọtì
Simplify \( \frac{1}{1+\sqrt{5}} - \frac{1}{1-\sqrt{5}} \)
Akọwa Nkọwa
= 3−√5−3−√5(3+√5)(3−√5
= −2√59−5
= −2√54
= - 12√5
Ajụjụ 32 Ripọtì
Akọwa Nkọwa
i.e. P-1 + P - P-1.P = 0
P-1 - P-1.P = -P
P-1(1 - P) = -P
P-1 = -P/(P-1)
= P/(P-1)
Ajụjụ 33 Ripọtì
Akọwa Nkọwa
Sum of (x) = 15x, algebraic sum of (deviation) = zero (0)
Sum of (squared-deviation) = 10x2
Variance = ∑(squared-deviation)/n = 10x2/5 = 2x2
Ajụjụ 35 Ripọtì
Let P = {1, 2, u, v, w, x}; Q = {2, 3, u, v, w, 5, 6, y} and R = {2, 3, 4, v, x, y}.
Determine (P-Q) ? R
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
Akọwa Nkọwa
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides
Ajụjụ 37 Ripọtì
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
Akọwa Nkọwa
Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2
Ajụjụ 39 Ripọtì
Akọwa Nkọwa
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o
Ajụjụ 40 Ripọtì
Akọwa Nkọwa
a * 2 = a2 = 2 - a
a2 + 2a - a - 2 = 0
a(a+2) - 1(a+2) = 0
(a-1)(a+2) = 0
a = 1, -2
Ajụjụ 41 Ripọtì
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
Akọwa Nkọwa
0.7 = 0.4 + P(Y)
P(Y) = 0.7 - 0.4 = 0.30
Ajụjụ 43 Ripọtì
Akọwa Nkọwa
60-3x+3x+50-3x = 94-x.
110-3x+x = 94
-2x = 94-110
=>-2x = -16, this x = -8.
Members that like only one game:
= 60 - 3x + 50 - 3x
= 60 - 3x8 + 50 - 3x8
= 60 - 24 + 50 - 24
= 36 + 26 = 62
Ajụjụ 44 Ripọtì
Akọwa Nkọwa
∴2-3 = 5p
∴5-3log52 = 5log52-3 = 5pii 22log23 = q => log532 = q
∴32 = 2q
∴222log23 = 2q
= 32 = 9
i x ii = 2-3 x 9 = 1/23 x 9
= 1/8 x 9
= 9/8 = 11/8
Ajụjụ 45 Ripọtì
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
Akọwa Nkọwa
Since ST is parallel to UV => angle USR = 50°(alternate to VUS.
Angle UST = 180° - 50° = 130° (angle on a straight line)
Ajụjụ 47 Ripọtì
In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.
Akọwa Nkọwa
