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.

1,656.

In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.

A.

16cm

B.

14cm

C.

12cm

D.

8cm

Correct answer is A

PT x QT = TR x TS

24 x 8 = TR x 12

TR = \(\frac{24 \times 8}{12}\)

= = 16cm

1,657.

In the figure, find angle x

A.

100o

B.

120o

C.

60o

D.

110o

E.

140o

Correct answer is E

In the figure, angle x = 20o + 80o + 40o

= 140o

1,658.

In the figure, the area of the shaded segment is

A.

3\(\pi\)

B.

9\(\frac{\sqrt{3}}{4}\)

C.

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

D.

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

E.

\(\pi + \frac{9 \sqrt{3}}{4}\)

Correct answer is C

Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\)

Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\)

= \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\)

Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\)

= 3 \(\pi - 3 \frac{\sqrt{3}}{4}\)

1,659.

The figure is an example of the construction of a

A.

perpendicular bisector of a given straight line QR

B.

perpendicular from a given point to a given line QR

C.

perpendicular to a line from a given point P on that line

D.

given angle

Correct answer is B

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

1,660.

In the figure, find the value of x

A.

110o

B.

100o

C.

90o

D.

80o

Correct answer is C

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o