WAEC Past Questions and Answers - Page 3788

18,936.

A man made a loss of 15% by selling an article for N595. Find the cost price of the article

A.

N600.00

B.

N684.25

C.

N700.00

D.

N892.50

Correct answer is C

15% loss = N595

\(\implies\) (100 - 15)% of CP = N595

85% of CP = N595

CP = \(\frac{595 \times 100}{85}\)

= N700.00

18,937.

A right circular cone is such that its radius r is twice its height h. Find its volume in terms of h

A.

\(\frac{2}{3}\pi h^2\)

B.

\(\frac{1}{12}\pi h^3\)

C.

\(\frac{4}{3}\pi h^2\)

D.

\(\frac{4}{3}\pi h^3\)

Correct answer is D

Volume of a cone = \(\frac{\pi r^2 h}{3}\)

r = 2h.

V = \(\frac{\pi \times (2h)^2 \times h}{3}\)

= \(\frac{4}{3} \pi h^3\)

18,938.

In the diagram /PS/ = 9 cm, /OR/ = 5cm, \(P\hat{S}R = 63^o\) and \(S\hat{P}Q = P\hat{Q}R = 90^o\). Find, correct to the nearest whole number, the area of the trapezium,

A.

55cm2

B.

25cm2

C.

22cm2

D.

13cm2

Correct answer is A

Considering the triangle in the diagram
\(tan\theta = \frac{opp}{hyp}\\
Tan 63^{\circ} = \frac{h}{4}\\
h = 4 tan 63^{\circ}\\
Area of trapezium = \frac{1}{2}h(a+b)\\
\left(\frac{1}{2} \times 4 tan 63^{\circ}[5+9]\right)\\
=28\times 1.963 = 54.96 = 55cm^2\)

18,940.

Evaluate Cos 45o Cos 30o - Sin 45o Sin 30o leaving the answer in surd form

A.

\(\frac{\sqrt{2}-1}{2}\)

B.

\(\frac{\sqrt{3}-\sqrt{2}}{4}\)

C.

\(\frac{\sqrt{6}-\sqrt{2}}{2}\)

D.

\(\frac{\sqrt{6}-\sqrt{2}}{4}\)

Correct answer is D

\(cos45^o \times cos30^o - sin45^o \times sin30^o\\
\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\times \frac{1}{2}\\
\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}; = \frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\)