2016 - WAEC Mathematics Past Questions and Answers - page 3
21
\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age
Age(years) & 13 & 14 & 15 & 16 & 17 \
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age
A
13
B
14
C
15
D
16
correct option: b
Median = \(\frac{N}{2}\)th age
= \(\frac{50}{2}\)th age
= 25th age
= 14 years
Users' Answers & Comments= \(\frac{50}{2}\)th age
= 25th age
= 14 years
22
In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < VXY.
A
70o
B
80o
C
105o
D
110o
correct option: b
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
Users' Answers & Commentsx1 = 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
23
In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113o and < RSt = 220, find < PSQ
A
22o
B
45o
C
67o
D
89o
correct option: b
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
Users' Answers & Comments< PSQ + 133o + \(\beta\) = 180o (sum of angles of a \(\Delta\))
< PSQ + 133o + 22o = 180o
< PSQ = 180o - (133 + 22)o
45o
24
In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m.
A
30
B
36
C
40
D
72
correct option: a
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
Users' Answers & Comments\(\alpha\) + 10mo = 360o (angle at circumference)
\(\alpha\) + 10mo = 360o(angles round a point)
2mo + 10mo = 360o
12mo = 360o
mo = \(\frac{360^o}{12}\)
= 30o
25
Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?
A
4.5m
B
6.0m
C
7.5m
D
9.0m
correct option: c
By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\)
\(\frac{8}{3} = \frac{20}{h}\)
h = \(\frac{3 \times 20}{8}\)
= 7.5m
Users' Answers & Comments\(\frac{8}{3} = \frac{20}{h}\)
h = \(\frac{3 \times 20}{8}\)
= 7.5m
26
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.
A
54o
B
44o
C
34o
D
27o
correct option: a
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
Users' Answers & Commentsx1 = 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
27
In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU.
A
120o
B
108o
C
72o
D
60o
correct option: c
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
Users' Answers & Comments< QTU = 48o (alternate angles)
< QU + 60o + 48o = 180o(sum of angles of a \(\Delta\))
< TQU = 180o - 108o
= 72o
28
In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW
A
135o
B
105o
C
75o
D
60o
correct option: c
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
Users' Answers & Comments\(\alpha\) = 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
29
The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?
A
Le 2,250,000.00
B
Le 22,700,000.00
C
Le 3,600,000.00
D
Le 4,500,000.00
correct option: c
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
Users' Answers & Comments= 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
30
A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?
A
\(\frac{5}{36}\)
B
\(\frac{1}{6}\)
C
\(\frac{5}{18}\)
D
\(\frac{2}{3}\)
correct option: b
\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,4 & 2,5 & 2,6 \ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \ \hline 4 & 4,1 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)
From the table above, event space, n(E) = 6
sample space, n(S) = 36
Hence, probability sum of scores is at least 10, is;
\(\frac{n(E)}{n(S)}\)
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)
Users' Answers & Comments& 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;
\(\frac{n(E)}{n(S)}\)
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)