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
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
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 & Comments
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
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
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
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 & Comments
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
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
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
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
Please share this, thanks: