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})

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

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