Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

616.

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

617.

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

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

618.

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\(\sqrt{3m}\)

619.

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 \(\frac{1000}{1000}\)

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

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

620.

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