2018 - WAEC Mathematics Past Questions and Answers - page 4
Find the median of 2, 1, 0, 3, 1, 1, 4, 0, 1 and 2
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
Find the mean deviation of 20, 30, 25, 40, 35, 50, 45, 40, 20 and 45
Mean = \(\frac{\sum f x}{\sum f}\)
= \(\frac{350}{10}\)
= 35
= \(\frac{\sum f |d|}{\sum f}\)
= \(\frac{90}{10}\)
= 9
Find the value of t in the diagram
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, PR is a tangent to the circle at Q, QT//RS,
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\(^o\)
In the diagram, WXYZ is a rectangle with dimension 8cm by 6cm. P, Q, R and S are the midpoints of the sides of the rectangle as shown. Using this information, what type of quadrilateral is the shaded region?
The solution of x + 2 \(\geq\) 2x + 1 is illustrated
x + 2 \(\geq\) 2x + 1
x - 2x \(\geq\) 1 - 2
-x \(\geq\) -1
\(\frac{-x}{-1}\) \(\geq\) \(\frac{-1}{-1}\)
x \(\leq\) 1
The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m
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, PQ//RS. Find x in terms of y and z
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
In the diagram, PQ is a straight line, (m + n) = 110\(^o\) and (n + r) = 130\(^o\) and (m + r) = 120\(^o\). Find the ratio of m : n : r
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
Donations during the launching of a church project were sent in sealed envolopes. The table shows the distribution of the amount of money in the envelope. How much was the donation?
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