1997 - WAEC Mathematics Past Questions and Answers - page 3
21
Which of the following statements is/are true when two straight lines intersect?
I. Adjacent angles are equal II. Vertically opposite angles are equal III. Adjacent angles are supplementary
A
II only
B
I and III only
C
II and III only
D
I, II and III
correct option: c
Users' Answers & Comments22
PQRS and GHRS are parallelograms on the same base SR and between the same parallel straight line PH and SR. Which of the following is true?
A
Δ PGS = Δ QHR
B
Parallelogram PQRS = Δ QHR
C
Trapezium PQXS = Δ QHR
D
Δ PGS = Parallelogram GHRS
correct option: e
Users' Answers & Comments23
From the diagram above, which of the following is true?
A
a +b + x =180
B
a = b + x
C
a – b = 180 - x
D
a + b = x + 180
correct option: d
From the diagram, the angles in the triangle are:
(180 - a)°, (180 - b)° and x°.
\(\therefore\) (180 - a) + (180 - b) + x = 180°
360 + x - a - b = 180°
360 - 180 + x = a + b
180 + x = a + b.
24
The interior angles of a pentagon are 126o, 114o, y, 92o and 83o. Find the value of y.
A
85o
B
95o
C
105o
D
115o
correct option: e
Sum of the angle of a pentagon
= (2 x 5 - 4) x 90
= 540o
126 + 114 + y + 92 + 83 = 540
y = 540 - 415 = 125o
Users' Answers & Comments= (2 x 5 - 4) x 90
= 540o
126 + 114 + y + 92 + 83 = 540
y = 540 - 415 = 125o
25
Solve the inequality 3x - 8 ≥ 5x
A
x ≤ 4
B
x ≥ 1
C
x ≤ -4
D
x ≤ -1
26
Divide 3.6721 by 4
A
0.9180
B
1.4180
C
1.1680
D
1.9180
27
The eleventh term of an AP, is 25 and its first term is -3. Find its common difference.
A
19/10
B
21/5
C
23/11
D
21/2
28
Find the common ratio in the GP: log 3, log 9, log 81 ..........
A
1
B
2
C
3
D
6
29
Find the value of t for which \(\frac{64}{27} = (\frac{3}{4})^{t - 1}\)
A
-4
B
-2
C
11/3
D
2
correct option: b
\(\frac{64}{27} = (\frac{3}{4})^{t-1}\)
\((\frac{3}{4})^t = \frac{64}{27} \times \frac{3}{4} = \frac{16}{9}\)
\((\frac{3}{4})^t = (\frac{9}{16})^{-1}\)
\((\frac{3}{4})^t = (\frac{3}{4})^{-2}\)
t = -2
30
O is the centre of the circle PQRS. PR and QS intersect at T POR is a diameter, ∠PQT = 42o and ∠QTR = 64o; Find ∠QRT
A
68o
B
64o
C
42o
D
32o
correct option: a
TQR = 90 - 42 = 48o
QRT = 180 - (64 - 48)
= 180 - 112
= 68o
Users' Answers & CommentsQRT = 180 - (64 - 48)
= 180 - 112
= 68o