2015 - WAEC Mathematics Past Questions and Answers - page 5

41
In the diagram, the shaded part is carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find area of the margin
A
4.7m2
B
4.9m2
C
5.7m2
D
5.9m2
correct option: a
Area of floor = 3.5m x 2.2m = 7.7m2

Area of carpet = (3.5 - 0.5 - 0.5)m x (2.2 - 0.5 - 0.5)m

2.5m x 1.2m = 3m2

hence, area of margine

= area of floor - area of carpet

= (7.7 - 3)m2

= 4.7m2
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42
Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles
A
72o
B
100o
C
120o
D
144o
correct option: d
The diagram given simple illustrates that a pentagon contains three \(\Delta\)s.

Two angles which are in the ratio 2:3 will have actual values 2xo, 3xo respectively. Thus 2xo + 3xo + 3 x 60 = 3x sum of angles of \(\Delta\)a

i.e. 5xo + 180o = 3 x 180o

5xo + 180o = 540o

5xo = 540o - 180o

5xo = 360o

xo = \(\frac{360^o}{5}\)

= 72o

Hence, the smaller of the two angles is

2 x 72o = 144o
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43
A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?
A
\(\frac{9}{13}\)
B
\(\frac{11}{26}\)
C
\(\frac{4}{13}\)
D
\(\frac{1}{26}\)
correct option: c
Two word mathematics is first rewritten as mathematics, since every letter of mathematics appears once in the english alphabet.

Hence the probability that letter selected is from the word mathematics is prob. (matheics) = \(\frac{8}{26}\)

= \(\frac{4}{13}\)
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44
In a circle radius rcm, a chord 16\(\sqrt{3}cm\) long is 10cmfrom the centre of the circle. Find, correct to the nearest cm, the value of r
A
22cm
B
17cm
C
16cm
D
15cm
correct option: b
In the diagram

|AM| = |MB| - \(\frac{|AB|}{2}\)

= \(\frac{16\sqrt{3}}{2}\)cm

= 8\(\sqrt{3}\)cm

in \(\Delta\) AMO, r2 = |AM|Z + |MO|2

r2 = (8\(\sqrt{3}\))2
+ 102

= 64 x 3 + 100

= 192 + 100

= 292

r = \(\sqrt{292}\)

17.088cm

17cm
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45
In the diagram, \(\bar{OX}\) bisects < YXZ and \(\bar{OZ}\) bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ
A
68o
B
72o
C
112o
D
124o
correct option: d
In \(\Delta\) XYZ, 2m + 2n + 68o = 180o

2(m + n) + 68o = 180o...(1)

in \(\Delta\) XOZ, m + n + q = 180o ...(2)

(m + n) = 180o - q...(3)

substituting 180o - q for (m + n) in (1) gives

2(180o - q) + 68o = 180o

360o - 2q = 180o - 68o

360o - 2q = 112o

360o - 112o = 2q

248o = 2q

q = \(\frac{248^o}{2}\)

= 124o

hence, < XOZ = 124o
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