WAEC Mathematics Past Questions & Answers - Page 51

251.

Three exterior angles of a polygon are 30\(^o\), 40\(^o\) and 60\(^o\). If the remaining exterior angles are 46\(^o\) each, name the polygon.

A.

decagon

B.

nonagon

C.

octagon

D.

hexagon

Correct answer is C

Sum of all exterior angles is 360\(^o\)

360\(^o\) (30\(^o\) - 40\(^o\))

360 - (130\(^o\))

230\(^o\)

remaining is 46\(^o\) = \(\frac{230}{46}\) = 5

5 + 3 = 8 sides; Octagon

252.

Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)

A.

4

B.

\(\sqrt{3} + 4\)

C.

4 \(\sqrt{2}\)

D.

4\(\sqrt{3} + 4\)

Correct answer is A

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

\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)

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

= 4

253.

In what number base was the addition 1 + nn = 100, where n > 0, done?

A.

n - 1

B.

n + 1

C.

n

D.

n + 2

Correct answer is C

No explanation has been provided for this answer.

254.

The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group

A.

10.1 years

B.

9.3 years

C.

8.7 years

D.

8 . 3 years

Correct answer is A

x = 10 ; 10 = \(\frac{x}{25}\)

x = 250

x = \(\frac{250 + 12.4}{26}\)

x = 10.09

x = 10.1 years

255.

Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o - q}{1 - tan 30^o}\)

A.

\(\sqrt{3 - 2}\)

B.

2 - \(\sqrt{3}\)

C.

\(\sqrt{3}\)

D.

-2

Correct answer is C

Tan 60 = 3; Tan 30 = 1

\(\frac{\tan 60^o - 1}{1 - tan 30^o}\)  = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)

= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= \(\sqrt{3}\)