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.
In the figure, GHIJKLMN is a cube of side a. Find the length of HN.
\(\sqrt{3a}\)
3a
3a2
a\(\sqrt{2}\)
a\(\sqrt{3}\)
Correct answer is E
HJ2 = a2 + a2 = 2a2
HJ = \(\sqrt{2a^2} = a \sqrt{2}\)
HN2 = a2 + (a\(\sqrt{2}\))2 = a2 + 2a2 = 3a2
HN = \(\sqrt{3a^2}\)
= a\(\sqrt{3}\)cm
In the figure POQ is the diameter of the circle PQR. If PSR = 145o, find xo
25o
35o
45o
125o
55o
Correct answer is E
< PRQ = \(\frac{1}{2}\) < POQ = 90o
< PSR + < PQR = 180o
< PQR = 180o - 145o = 35o
\(\bigtriangleup\)PQR is a right angled triangle
x = 90 - < PQR
= 90o - 35o
= 55o
28o, 36o
36o, 28o
43o, 61o
61o, 43o
36o, 43o
Correct answer is A
yo = 180o - (86o + 58o)
180 - 144 = 36o
xo = 180 - (94 + 58)
180 -152 = 28
(xo, yo) = (28o, 36o)
190 sq.cm
20 sq.cm
210 sq.cm
160sq.cm
320sq.cm
Correct answer is C
By Pythagoras, KZ2 = 252 - 52
KZ2 = (25 + 15)(25 - 15) = 400
KZ = \(\sqrt{400}\) = 20
area of XYZ = \(\frac{1}{2} \times 28 \times 15\)
= 210 sq. cm
XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.
5\(\sqrt{3}\)cm
3\(\sqrt{5}\)cm
3\(\sqrt{3}\)cm
5cm
6cm
Correct answer is B
by Pythagoras theorem, XW = 3cm
Also by Pythagoras theorem, XY2 = 62 + 32
XY2 = 36 + 9 = 45
XY = \(\sqrt{45} = 3 \sqrt{3}\)