2002 - WAEC Mathematics Past Questions and Answers - page 4

31
Simplify \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\)
A
1
B
2
C
4
D
16
correct option: a
\(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)
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32
In the diagram, |SR| = |RQ| and ∠PRQ = 58o ∠VQT = 19o, PQT, SQV and PSR are straight lines. Find ∠QPS
A
42o
B
39o
C
38o
D
30o
correct option: a
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33

The angle of depression of a point Q from a vertical tower PR, 30m high, is 40°. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, |PQ|.

A
35.75m
B
25.00m
C
22.98m
D
19.28
correct option: a

\(\tan 40° = \frac{30}{|PQ|}\)

\(|PQ| = \frac{30}{\tan 40°} = \frac{30}{0.839}\)

= 35.75m

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34
The volume of a cylinder of radius 14cm is 210cm3. What is the curved surface area of the cylinder?
A
15cm2
B
30cm2
C
616cm2
D
1262cm2
correct option: b
\(V = \pi r^2 h\
210 = \frac{22}{7} \times 14^2 \times h \
h = \frac{210}{22 \times 28}\)
Curved surface area \(= 2r\pi h\
= 2 \times \frac{22}{7} \times 14 \times \frac{210}{22 \times 26} = 30cm^2\)
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35

Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

A
\(7\sqrt{3}\)
B
\(10\sqrt{3}\)
C
\(14\sqrt{3}\)
D
\(18\sqrt{3}\)
correct option: c

\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

= \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\)

= \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\)

= \(14\sqrt{3}\)

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36

Factorize m(2a-b)-2n(b-2a)

A
(2a-b)(2n-m)
B
(2a+b)(m-2n)
C
(2a-b)(m+2n)
D
(2a-b)(m-2n)
correct option: c

m(2a - b) - 2n(b - 2a)

= m(2a - b) - (-2n)(2a - b)

= m(2a - b) + 2n(2a - b)

= (m + 2n)(2a - b)

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37

If q oranges are sold for t Naira, how many oranges can be bought for p naira?

A
\(\frac{p}{2}t\)
B
\(\frac{qt}{p}\)
C
\(\frac{q}{pt}\)
D
\(\frac{pq}{t}\)
correct option: d

q oranges = t naira

1 naira = \(\frac{q}{t}\)

p naira = \(p(\frac{q}{t})\)

= \(\frac{pq}{t}\) oranges

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38

In the diagram, QRT is a straight line. If angle PTR = 90°, angle PRT = 60°, angle PQR = 30° and |PQ| = \(6\sqrt{3}cm\), calculate |RT|

A
0.3cm
B
\(\frac{\sqrt{3}}{2}cm\)
C
3cm
D
\(3\sqrt{3}cm\)
correct option: c

In \(\Delta\) QPT,

\(\frac{PT}{6\sqrt{3}} = \sin 30°\)

PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\)

In \(\Delta\) RPT,

\(\frac{PT}{RT} = \tan 60°\)

\(\frac{3\sqrt{3}}{RT} = \tan 60°\)

\(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)

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39

In the diagram, calculate the value of x

A
35o
B
80o
C
100o
D
115o
correct option: c

x - 35° = 65° (corresponding angles)

x = 65° + 35° = 100°

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40

In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ∆PQR. Which of the following is equal to ∠RQT?

A
∠PQT
B
∠SRP
C
∠PQR
D
∠SPR
correct option: d
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