2005 - JAMB Mathematics Past Questions and Answers - page 2

x = -2, x = -1 and x = 3
∴x+2 = 0, x+1 = 0 and x-3 = 0
Product of the factors
(x+2)(x+1)(x-3) = 0
(x² + 3x + 2)(x-3)
x(x² + 3x + 2) -3(x² + 3x + 2) = 0
x³ + 3x² + 2x - 3x² - 9x - 6 = 0
x³ - 7x - 6 = 0

t = time taken and N = number of men
t ∝ ¹/_N
t = ^K/_N
K = Nt
K = 30 * 6
K = 180
∴t = ¹⁸⁰/_N
4 = ¹⁸⁰/_N
4N = 180
N = ¹⁸⁰/₄
45 men

\(W\infty LD^2\W=KLd^2\K=\frac{W}{Ld^2}\=\frac{140}{54}\times\left(4\frac{2}{3}\right)^2 \=\frac{140}{54}\times\left(\frac{14}{3}\right)^2\=\frac{140\times 9}{54\times 14\times 14}\=\frac{5}{42}\∴W=\frac{5}{42Ld^2}\42W=5Ld^2\\frac{42W}{5L}=d^2\d=\sqrt{\frac{42W}{5L}}\)

7x - 3 > 25 + 3x
7x - 3x > 25 + 3
4x > 28
x > ²⁸/₄
x > 7

U₇ = a + (7 - 1)d
= a + 6d
U₃ = a + (3 - 1)d
= a + 2d
But U₇ = 2(U₃)
∴a + 6d = 2(a + 2d)
a + 6d = 2a + 4d
2a - a + 4d - 6d = 0
a - 2d = 0 → eqn1
Sn = ⁿ/₂ (2a + (n - 1)d)
42 = ⁴/₂ (2a + (4 - 1)d)
42 = 2(2a + 3d)
21 = 2a + 3d → eqn2
eqn1 * eqn2 0 = 2a - 4d
21 = 7d
∴d = ²¹/₇
d = 3

8, 12, 16, .....96
a = 8, d = 4, l = 96, n = 20
S₂₀ = ⁿ/₂(a + l)
= ²⁰/₂(8 + 96)
= 10 * 104
= 1040

a * b = ab + 2(a + b + 1)
let e be the identity element
∴ a * e = e * b = a
∴ a * e
ae + 2(a + e + 1) = a
ae + 2a + 2e + 2 = a
ae + 2e = a - 2a = 2
(a + 2)e = -a - 2
e = -(a-2) / (a+2)
e = -(a+2) / (a+2)
e = -1

p + 40^o = 100^o (exterior ∠ = sum of two interior opp ∠s)
p = 100^o - 40^o
P = 60^o
But q + p = 180^o (∠s on a straight line)
q + 60^o
q = 180^o - 60^o
q = 120^o
x = q (corresponding ∠)
∴x = 120^o

∠GOK = 85^o (vertical opposite angle ∠s)
∠EOG + ∠GOK + ∠KOF = 180 (∠s on a straight line)
∠EOg + 85^o + 52^o = 180^o
∠EOG + 137^o = 180^o
∠EOG = 43^o