WAEC Mathematics Past Questions & Answers - Page 60

296.

If P and Q are two statements, under what condition would p|q be false?

A.

If p is true and q is true

B.

If p is true and q is false

C.

If p is false and q is false

D.

If p is false and q is true

Correct answer is B

From the truth table above, p \(\to\) q  would be false if p is true and q is false

297.

In the diagram, which of the following ratios is equal to \(\frac{|PN|}{|PQ|}\)?

A.

\(\frac{|PN|}{|PR|}\)

B.

\(\frac{|PM|}{|PQ|}\)

C.

\(\frac{|PM|}{|PR|}\)

D.

\(\frac{|PR|}{|PQ|}\)

Correct answer is C

\(\frac{|PN|}{|PQ|}\) = \(\frac{|PM|}{|PR|}\)

298.

The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.

A.

27m

B.

13.5 \(\sqrt{3m}\)

C.

13.5 \(\sqrt{2m}\)

D.

9\(\sqrt{3m}\)

Correct answer is D

From the diagram,

tan 30\(^o\) = \(\frac{h}{27}\)

h = 27 tan 30\(^o\)

= 27 x \(\frac{1}{\sqrt{3}}\)

= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{27 \sqrt{3}}{3}\)

= 9\(\sqrt{3m}\)

299.

Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0

A.

\(\frac{153}{250}\)

B.

\(\frac{68}{111}\)

C.

\(\frac{61}{100}\)

D.

\(\frac{21}{33}\)

Correct answer is A

0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)

= \(\frac{612}{1000}\)

= \(\frac{153}{250}\)

300.

If  x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z

A.

2:3

B.

3:4

C.

3:8

D.

1:2

Correct answer is D

\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)

\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)

But,

x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\)

y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\)

Using y = y

\(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\)

 \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\)

\(T_1 = 15\) and \(T_2 = 21\)

Thus , x = \(\frac{2}{5}\) x 15 = 6

y = \(\frac{3}{5}\) x 15 = 9

y = \(\frac{3}{7}\) x 21 = 9

z = \(\frac{4}{7}\) x 21 = 12

Hence; x : z = 6 : 12

= 1 : 2