2011 - WAEC Mathematics Past Questions and Answers - page 4

volume of rectangular container = L x B x H

1 litre = 1000cm³

96 litres = (\frac{1000cm^3}{1 litres}) x 96 litres

volume = 96000cm³

96000 = 220 x 40 x 11

96--- = 8800H

H = (\frac{96000}{8800})

= 10.97cm

= 11cm

volume of rectangular container = L x B x H

1 litre = 1000cm³

96 litres = (\frac{1000cm^3}{1 litres}) x 96 litres

volume = 96000cm³

96000 = 220 x 40 x 11

96--- = 8800H

H = (\frac{96000}{8800})

= 10.97cm

= 11cm

let the number of question = n

n(n + 50) = 600

n² + 50n - 600 - 0 by quadratic formular

x = (\frac{b \pm \sqrt{b^2 - 4ac}}{2})

n = (\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)})

= (\frac{-50 \pm \sqrt{4900}}{2})

= (\frac{-50 + 70}{2})

= (\frac{20}{2})

= 10

27^x = 9^y

3^3x = 3^2y

3x = 2y

tan (\theta) = (\frac{9}{50}) = 0.18

(\theta = tan^{-1} 0.18)

(\theta) = 10.20

< Q = (\frac{240}{2}) (angle at centre twice that at the circumference)

< Q = 120^o

Also < POR = 360 - 240

= 120^o

( < s at centre) since /PQ/ = /QR/, < x = < R

Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)

x + R + 120 = 360^o

x + R = 360 - 240 = 120; Since x = R

x + x = 120

2x = 120

Since x = R

x + x = 120

2x = 120

x = (\frac{120}{2})

= 60^o

x + 52^o = 90

x = 90 - 52

x = 38^o

k = opposite angle Z

k = 38^o

y + k = 90^o

y + 38^o = 90^o

y = 90^o - 38^o

y = 52^o

y = x = 180^o(sum of angles on straight line)

52 + x = 180^o

x = 180 - 52

x = 128

2 + 2 + 5 + 8 + 12 + H + 7 + 3 + 2 = 52