2016 - WAEC Mathematics Past Questions and Answers - page 3

Median = \(\frac{N}{2}\)th age

= \(\frac{50}{2}\)th age

= 25th age

= 14 years

In the diagram above, x₁ = 50(angles in alternate segment)

x₁ = x₂(base angles of isos. \(\Delta\))

< UXY + x₂ + 50_o = 180_o(sum of angles on a straight line)

< UXY + 50_o + 50_o = 180_p

< UXY + 100_o = 180_o

< UXY = 180_o - 100_o

= 80_o

In In the diagram given, \(\alpha\) = 22^o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)

< PSQ + 133^o + \(\beta\) = 180^o (sum of angles of a \(\Delta\))

< PSQ + 133^o + 22^o = 180^o

< PSQ = 180^o - (133 + 22)^o

45^o

In the diagram above, \(\alpha\) = 2m^o (angle at centre = 2 x angle at circumference)

\(\alpha\) + 10m^o = 360^o (angle at circumference)

\(\alpha\) + 10m^o = 360^o(angles round a point)

2m^o + 10m^o = 360^o

12m^o = 360^o

m^o = \(\frac{360^o}{12}\)

= 30^o

By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\)

\(\frac{8}{3} = \frac{20}{h}\)

h = \(\frac{3 \times 20}{8}\)

= 7.5m

In the diagram above, x₁ = 180^o - 117^o = 63^o(opposite angles of a cyclic quad.)

x₁ = x₂ (base angles of isos. \(\Delta\))

x₁ + x₂ + \(\alpha\) = 180^o (sum of angles of a \(\Delta\)

63^o + 63^o + \(\alpha\) = 180^o

\(\alpha\) = 180^o - (63 + 63)^o

= 54^o

In the diagram, < TUQ + 60^o(corresp. angles)

< QTU = 48^o (alternate angles)

< QU + 60^o + 48^o = 180^o(sum of angles of a \(\Delta\))

< TQU = 180^o - 108^o

= 72^o

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)

\(\alpha\) = tan^-1(\(\frac{1}{\sqrt{3}}\))

= 30^o

In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1

\(\beta\) = tan^-1(1) = 45^o

Hence, < UTW = \(\alpha\) + \(\beta\)

= 30^o + 45^o = 75^o

sectoral angle representing food

= 360^o - (80 + 70 + 90)^o

= 120^o

Amount spent on food

= \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000

= Le 3,600,000

\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,4 & 2,5 & 2,6 \ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \ \hline 4 & 4,1 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)

From the table above, event space, n(E) = 6

sample space, n(S) = 36

Hence, probability sum of scores is at least 10, is;

\(\frac{n(E)}{n(S)}\)

= \(\frac{6}{36}\)

= \(\frac{1}{6}\)