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.

1,521.

In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110^o and the bearing of Q from P is 050^o. Find angle PQR.

60^o

70^o

120^o

160^o

View answer & explanation

Correct answer is C

Q₁ = 50(alternate angle)

Q₂ = 180^o - 110^o (straight line angle)

Q₂ = 70

PQR = Q₁ + Q₂

= 50^o + 70^o

= 120^o

1,522.

In the diagram P, Q, R, S are points on the circle RQS = 30^o. PRS = 50^o and PSQ = 20^o. What is the value of x^o + y^o?

260^o

130^o

100^o

80^o

View answer & explanation

Correct answer is B

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50^o

Also, < QSR = < QPR(angles in the segment)

< QPR = x^o

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130^o

1,523.

In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58^o. Find < QRS

112^o

116^o

122^o

148^o

View answer & explanation

Correct answer is C

PRQ = 90 - 58 = 32^o(angle in a semi-circle)

Since PRS = 90^o(radius angular to tangent)

QRS = 90 + 32

= 122^o

1,524.

In the diagram, \(\frac{PQ}{RS}\), find x^o + y^o

360^o

300^o

270^o

180^o

View answer & explanation

Correct answer is D

PQR = QRS = Y(alt. amgles)

PQR + x = 180^o(angles on a straight line)

y + x = 180^o

1,525.

In the diagram, |QR| = 5cm, PQR = 60<sup>o</sup> and PSR = 45<sup>o</sup>. Find |PS|, leaving your answe in surd form.

4\(\sqrt{5}\)cm

3\(\sqrt{7}\)cm

4\(\sqrt{6}\)cm

5\(\sqrt{6}\)cm

View answer & explanation

Correct answer is D

tan 6^o = \(\frac{|PR|}{|QR|}\)

\(\sqrt{3} = \frac{|PR|}{5}\)

= |PR| = \(5 \sqrt{3}\)cm

sin 45 = \(\frac{|PR|}{|PS|}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{5 \sqrt{3}}{|PS|}\)

|PS| = \(5 \sqrt{3}\) x \(\sqrt{2}\)

= 5\(\sqrt{6}\)cm