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.
An (\(n - 2)^2\) sided figure has n diagonals. Find the number n diagonals for a 25-sided figure
7
8
9
10
Correct answer is A
(n-2)\(^2\)=25
n - 2 = 5
n = 5 + 2 = 7
\(\theta\)o + \(\phi\)o = 902
\(\phi\)o = 902 - 2\(\theta\)o
\(\theta\)o = \(\phi\)o
\(\phi\)o = 2\(\theta\)o
\(\theta\)o + 2\(\phi\)o
Correct answer is E
180 - \(\phi\)o = \(\theta\)o + \(\phi\)o (Sum of opposite interior angle equal to its exterior angle)
180 = 2\(\phi\) + \(\theta\)o
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)