2010 - JAMB Mathematics Past Questions and Answers - page 4

In the diagram above 88^o + (3x + 20)^o = 180^o

(oppsite < s of circlic quad. are supplementary)

88^o + 20^o + 3x = 180^o

108^o + 3x = 180^o

3x = 180^o - 108^o

= 72^o

Let A denote the area of \(\bigtriangleup\)PQR, then A = \(\frac{1}{2} bh\)

Using Sin 60^o = \(\frac{h}{q}\)

h = q sin 60^o

So A = \(\frac{1}{2}b(q \sin 60^o)\)

12\(\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}\)

12\(\sqrt{3}\) - 2q\(\sqrt{3}\)

q = \(\frac{12}{2} = 6\)cm

\(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}=\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\\ =\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\\ =\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\\ =\frac{2}{3}\times \frac{1}{2}\\ =\frac{1}{3}\)

2√3+√5

√5-√3

=	2√+√5	x √5+√3 √5+√3
√5-√3

=	2√3(√5+√3)+ √5(√5+√3)
____(√5)2-(√3)2

=	2√15+2(3) +5+√15
_____5 - 3

=	2√15+6+5+√15
______2

=	11+3√15
___2

0.21 x 0.34 = 0.0714
= 7.14 x 10^-2

Let C n G = X
n C only = 40 - x
n G only = 30 - x
40 - X + 30 - X = 50
70 - X = 50
X = 70 - 50
X = 20

x2 + 3x2

2x - 8

=	x(x2+3x10)
2(x2-4)

=	x(x+5)(x-2)
2(x2-2 2)

=	x(x+5)(x-2)
2(x-2)(x+2)

=	x(x+5)
2(x+2)

y ∝ √x
y = K√x
K = ^y/_√x = ³/_√16
= ³/₄
y = ³/₄√x
= ³/₄√64 when x = 64
= ³/₄ x ⁸/₁
= 6

x * y = x + y²
2 * 3 = 2 + 3²
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 5²
= 11 + 25
= 36

Using substitution If 18(p+18) = (18 + p)q
let q = 18
imply 18(p+18) = (18+p)18
i.e 18p + 18 x 18 = 18 x 18 + 18p
Since the left hand side = the right hand side
imply that q = 18