2016 - WAEC Mathematics Past Questions and Answers - page 3
Age(years) & 13 & 14 & 15 & 16 & 17 \
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age
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 + 50o = 180o(sum of angles on a straight line)
< UXY + 50o + 50o = 180p
< UXY + 100o = 180o
< UXY = 180o - 100o
= 80o
In In the diagram given, (\alpha) = 22o (vertically opp. angles), (\alpha) = (\beta) (alternate angles)
< PSQ + 133o + (\beta) = 180o (sum of angles of a (\Delta))
< PSQ + 133o + 22o = 180o
< PSQ = 180o - (133 + 22)o
45o
In the diagram above, (\alpha) = 2mo (angle at centre = 2 x angle at circumference)
(\alpha) + 10mo = 360o (angle at circumference)
(\alpha) + 10mo = 360o(angles round a point)
2mo + 10mo = 360o
12mo = 360o
mo = (\frac{360^o}{12})
= 30o
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 = 180o - 117o = 63o(opposite angles of a cyclic quad.)
x1 = x2 (base angles of isos. (\Delta))
x1 + x2 + (\alpha) = 180o (sum of angles of a (\Delta)
63o + 63o + (\alpha) = 180o
(\alpha) = 180o - (63 + 63)o
= 54o
In the diagram, < TUQ + 60o(corresp. angles)
< QTU = 48o (alternate angles)
< QU + 60o + 48o = 180o(sum of angles of a (\Delta))
< TQU = 180o - 108o
= 72o
In (\Delta) UXT, tan(\alpha) = (\frac{1}{\sqrt{3}})
(\alpha) = tan-1((\frac{1}{\sqrt{3}}))
= 30o
In (\Delta)WXT, tan(\beta) (\frac{\sqrt{3}}{\sqrt{3}}) = 1
(\beta) = tan-1(1) = 45o
Hence, < UTW = (\alpha) + (\beta)
= 30o + 45o = 75o
sectoral angle representing food
= 360o - (80 + 70 + 90)o
= 120o
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})