# 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
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
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
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
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
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
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
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
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
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;

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

= $$\frac{6}{36}$$

= $$\frac{1}{6}$$