2006 - WAEC Mathematics Past Questions and Answers - page 4
31
The temperature in a chemical plant was -5oC at 2.00 am. The temperature fell by 6oC and then rose again by 7oC. What was the final temperature?
A
-6oC
B
-4oC
C
4oC
D
8oC
correct option: b
Temperature = -5oC
It fell by 6oC = -5oC - 6oC
= -11oC
Then rose by 7oC = -11 + 7
= -4oC (final temperature)
Users' Answers & CommentsIt fell by 6oC = -5oC - 6oC
= -11oC
Then rose by 7oC = -11 + 7
= -4oC (final temperature)
32
Simplify \(\frac{20}{5\sqrt{28} - 2 \sqrt{63}}\)
A
\(\frac{5\sqrt{7}}{7}\)
B
\(\frac{5\sqrt{5}}{7}\)
C
\(\frac{7\sqrt{5}}{7}\)
D
\(\frac{7\sqrt{7}}{5}\)
correct option: a
\(\frac{20}{5 \sqrt{4 \times 7} - 2\sqrt{9 \times 7}}\)
\(\frac{20}{10 \sqrt{7} - 6\sqrt{7}}\)
= \(\frac{20 \sqrt{7}}{4 \sqrt{7}}\)
= \(\frac{5\sqrt{7}}{7}\)
Users' Answers & Comments\(\frac{20}{10 \sqrt{7} - 6\sqrt{7}}\)
= \(\frac{20 \sqrt{7}}{4 \sqrt{7}}\)
= \(\frac{5\sqrt{7}}{7}\)
33
A ladder 16m long leans against an electric pole. If the ladder makes an angle of 65o with the ground, how far up the electric pole does its top reach
A
6.8m
B
14.5m
C
17.7m
D
34.3m
correct option: b
Sin 65o = \(\frac{x}{16}\)
x = 16x sin 65
= 16 x 0.9063
x = 14.5m
Users' Answers & Commentsx = 16x sin 65
= 16 x 0.9063
x = 14.5m
34
In the diagram, PQUV, PQTU, QRTU and QRST are parallelograms. |UV| = 4.8cm and the perpendicular distance between PR and VS is 5cm. Calculate the area of quadrilateral PRSV
A
96cm2
B
72cm2
C
60cm2
D
24cm2
correct option: c
|VU| = |UT| = |TS|
|VS| = (4.8) x 3 = 14.4cm
|PQ| = |QR| = 4.8
|PR| = (4.8) x 2 = 9.6cm
Since quad. PRSV is a trapezium of height 5cm
Area of quad. PRSV = \(\frac{1}{2}(a + b)h\)
= \(\frac{1}{2}(14.4 + 9.6) \times 5\)
= \(\frac{1}{2}(24) \times 5\)
12 x 5 = 60cm2
Users' Answers & Comments|VS| = (4.8) x 3 = 14.4cm
|PQ| = |QR| = 4.8
|PR| = (4.8) x 2 = 9.6cm
Since quad. PRSV is a trapezium of height 5cm
Area of quad. PRSV = \(\frac{1}{2}(a + b)h\)
= \(\frac{1}{2}(14.4 + 9.6) \times 5\)
= \(\frac{1}{2}(24) \times 5\)
12 x 5 = 60cm2
35
In the diagram, |QR| = 5cm, PQR = 60o and PSR = 45o. Find |PS|, leaving your answe in surd form.
A
4\(\sqrt{5}\)cm
B
3\(\sqrt{7}\)cm
C
4\(\sqrt{6}\)cm
D
5\(\sqrt{6}\)cm
correct option: d
tan 6o = \(\frac{|PR|}{|QR|}\)
\(\sqrt{3} = \frac{|PR|}{5}\)
= |PR| = \(5 \sqrt{3}\)cm
sin 45 = \(\frac{|PR|}{|PS|}\)
\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)
|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)
= 5\(\sqrt{6}\)cm
Users' Answers & Comments\(\sqrt{3} = \frac{|PR|}{5}\)
= |PR| = \(5 \sqrt{3}\)cm
sin 45 = \(\frac{|PR|}{|PS|}\)
\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)
|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)
= 5\(\sqrt{6}\)cm
36
In the diagram, \(\frac{PQ}{RS}\), find xo + yo
A
360o
B
300o
C
270o
D
180o
correct option: d
PQR = QRS = Y(alt. amgles)
PQR + x = 180o(angles on a straight line)
y + x = 180o
Users' Answers & CommentsPQR + x = 180o(angles on a straight line)
y + x = 180o
37
In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58o. Find < QRS
A
112o
B
116o
C
122o
D
148o
correct option: c
PRQ = 90 - 58 = 32o(angle in a semi-circle)
Since PRS = 90o(radius angular to tangent)
QRS = 90 + 32
= 122o
Users' Answers & CommentsSince PRS = 90o(radius angular to tangent)
QRS = 90 + 32
= 122o
38
In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?
A
260o
B
130o
C
100o
D
80o
correct option: b
Draw a line from P to Q
< PQS = < PRS (angle in the sam segment)
< PQS = 50o
Also, < QSR = < QPR(angles in the segment)
< QPR = xo
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
Users' Answers & Comments< PQS = < PRS (angle in the sam segment)
< PQS = 50o
Also, < QSR = < QPR(angles in the segment)
< QPR = xo
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
39
In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find angle PQR.
A
60o
B
70o
C
120o
D
160o
correct option: c
Q1 = 50(alternate angle)
Q2 = 180o - 110o (straight line angle)
Q2 = 70
PQR = Q1 + Q2
= 50o + 70o
= 120o
Users' Answers & CommentsQ2 = 180o - 110o (straight line angle)
Q2 = 70
PQR = Q1 + Q2
= 50o + 70o
= 120o
40
In the diagram, the two circles have a common centre O. If the area of the larger circle is 100\(\pi\) and that of the smaller circle is 49\(\pi\), find x
A
2
B
3
C
4
D
6
correct option: b
area of larger circle = 100\(\pi\)
\(\pi r^2 = 100\pi\)
R2 = 100
R = 10(Radius)
Area of smaller circle = \(49 \pi\)
\(\pi r^2 = 49\pi\)
r2 = 49
r = 7(radius)
Since R = x + r
x = R - r
x = 10 - 7 = 3
Users' Answers & Comments\(\pi r^2 = 100\pi\)
R2 = 100
R = 10(Radius)
Area of smaller circle = \(49 \pi\)
\(\pi r^2 = 49\pi\)
r2 = 49
r = 7(radius)
Since R = x + r
x = R - r
x = 10 - 7 = 3