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WAEC Mathematics Past Questions & Answers - Page 106

526.

The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.

A.

6.0

B.

3.0

C.

2.4

D.

1.8

Correct answer is C

xffx166281638184520

mean x = fxf

= 6025

x = 2.4

528.

The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q

A.

155o

B.

80o

C.

158o

D.

91o

Correct answer is C

< NPQ = < PQB = 50o(alt. < s)

< PQB = 50o

< PQR = < PQR < PQB + < QBR = 72o

< QBR = < PQR - < PQB = 72o - 50o

= 22o

< NQR = 180 - < QBR = 180o - 22 = 158o, the bearing of R from Q = 158o

529.

Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180o iii. x + m + n = 180o iv. x + n = m + z

A.

i and iv

B.

ii

C.

iii

D.

iv

Correct answer is C

m + y + x = 180o(sum of < s on straight line)

y = n(vertically opp. angle)

m + n + x = 180o

530.

In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x

A.

150o

B.

110o

C.

100o

D.

95o

Correct answer is C

L + 30o - 180o(Sum of < s on straight line)

L = 180o - 30o = 150o

L = q = 150o(opposite < s are equal)

y = b = 30o(alt. < s)

b + c = 180o(sum of < s on str. line)

30o + c 180

c = 180 - 30

c = 150o

b = a = 30o (opp < s are equal)

c = d = 150o (opp < s are equal)

a + k + 70o = 180o (sum of < s on )

30o + k + 70o = 180

k + 100o = 180

k = 180 - 100

k = 80o

x + 80o = 180(sum of < s on straight line)

x = 180o - 80o

x = 100o