2018 - WAEC Mathematics Past Questions and Answers - page 4

First arrange the numbers in ascending order of magnitude; 0, 0, 1, 1, 1, 1, 2, 2, 3, 4

Counting from the right, the fifth number is 1

Counting from the left, the fifth number is 1

Hence, median = \(frac{1 + 1}{2}\)

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

= 1

Mean = \(\frac{\sum f x}{\sum f}\)

= \(\frac{350}{10}\)

= 35

= \(\frac{\sum f |d|}{\sum f}\)

= \(\frac{90}{10}\)

= 9

In the diagram,

\(\alpha\) = 180\(^o\) - 177\(^o\) (angles on a straight line)

\(\alpha\) = 63\(^o\)

t = 2 x 63\(^o\) (angle at centre = 2 x angle at circum)

= 126\(^o\)

In the diagram,

a = 50\(^o\) (alternate angles)

b\(_1\) + a 35\(^o\) = 180\(^o\) (sum of angles on a straight line)

i.e; b\(_1\) + 50\(^o\) + 35\(^o\)

= 180v

b\(_1\) + 180\(^o\) - 85\(^o\) = 90\(^o\)

But b\(_2\) = \(b_1\) = 95\(^o\) (angles in alternate segement)

<QST = b\(_2\) = 95\(^o\)

x + 2 \(\geq\) 2x + 1

x - 2x \(\geq\) 1 - 2

-x \(\geq\) -1

\(\frac{-x}{-1}\) \(\geq\) \(\frac{-1}{-1}\)

x \(\leq\) 1

In the diagram, < WOZ = 180\(^o\) (angle on a straight line)

< WOX = < XOY = < YOZ

(|WX| = |XY| = |YZ|)

\(\frac{180^o}{3}\) = 60\(^o\)

= 60\(^o\)

M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)

< YOZ + 2m (base angles of a \(\bigtriangleup\))

60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))

60\(^o\) + 2m = 180\(^o\)

2m = 180\(^o\) - 60\(^o\)

2m = 120\(^o\)

m = \(\frac{120^o}{2}\)

= 60\(^o\)

In the diagram,

a = z (alternate angles)

b = 180\(^o\) - a (angles on a straight line)

b = 180\(^o\) - z

c = 180\(^o\) - x (angles on a straight line)

y = b + c (sum of oposite interior angles)

y = 180\(^o\) - z + 180\(^o\) - x

y = 360\(^o\) - z - x

x = 360\(^o\) - y - z

m + n = 110\(^o\), (n + r) = 130\(^o\)

(m + n) = 120\(^o\)

then, r = 130\(^o\) - n

and;

m + (130^o - n) = 120\(^o\)

m - n = -10\(^o\)

2m + (n + r) = 110 + 120 = 230

2m + 130 = 230

2m = 230 - 130

m = \(\frac{100}{2}\) = 50\(^o\)

n = 110\(^o\) - 50\(^o\)

= 60\(^o\)

r = 130\(^o\) - 60\(^o\) = 70\(^o\)

Hence, the ratio m : n : r

= 50 : 60 : 70

= 5 : 6 : 7

Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10

= 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20

= N62,972