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.
A circle of radius 4 cm and with centre O
The perpendicular bisector of the chords
A straight line passing through centre O
A circle of radius 6 cm and with centre O
Correct answer is A
The locus of the mid-point of all chords of length 6cm within a circle of radius 5cm and with centre O is the perpendicular bisectors of the chords.
6cm
5cm
4cm
3cm
Correct answer is C
Base of pyramid of a square of side 8cm vertex directly above the centre edge = \(4\sqrt{3}\)cm
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC = \(8\sqrt{2}\)
but OC = \(\frac{1}{2}\)AC = 8\(\sqrt{\frac{2}{2}}\) = \(4\sqrt{2}\)cm
OE = h = height
h2 = (\(4\sqrt{3}\))2
16 x 2 - 16 x 2
48 - 32 = 16
h = \(\sqrt{16}\)
= 4
What is the circumference of latitude 0°S if R is the radius of the earth?
R cos\(\theta\)
2\(\pi\) R cos \(\theta\)
R sin \(\theta\)
2\(\pi\) R sin \(\theta\)
Correct answer is B
Circumference of Latitude 0oS where R is the radius of the earth is = 2\(\pi\) R cos \(\theta\)
\(\frac{5}{17}\)
\(\frac{8}{17}\)
\(\frac{8}{15}\)
\(\frac{12}{17}\)
Correct answer is A
ABCD is the floor, by pythagoras
AC2 = 144 + 81 = \(\sqrt{225}\)
AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{15}{17}\)
\(\sqrt{6}\)
4\(\sqrt{3}\)
\(\sqrt{3}\)
\(\frac{12}{\sqrt{3}}\)
Correct answer is B
From the diagram, WYZ = 60o, XYW = 180o - 60o
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o = \(\frac{3}{2}\)
∴ sin 60o = \(\frac{h}{YW}\), sin 60o = \(\frac{h}{8}\)
h = 8 x \(\frac{3}{2}\)
= 4\(\sqrt{3}\)