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.
N42,000
N18,000
N16,000
N12,000
Correct answer is C
Angle of sector subtended by yam
= 360o - (70 + 80 + 50)o
= 360o - 200o
= 160o
But \(\frac{80^o}{360^o}\) x T = 8000
T = \(\frac{8000 \times 360^o}{80^o}\)
= N36,000
Hence the amount spent on yam = \(\frac{160^o}{260} \times N36,000\)
= N16,000
y = x + 5
y = -x + 5
y = x - 5
y = -x - 5
Correct answer is B
(x1, y1) = (0,5)
(x2, y2) = (5, 0)
Using \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)
\(\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0}\)
\(\frac{y - 5}{-5} = \frac{x}{5}\)
5(y - 5) = -5x
y - 5 = -x
x + y = 5
y = -x + 5
Find the value of x in the figure above
20\(\sqrt{3}\)cm
10\(\sqrt{3}\)cm
5\(\sqrt{3}\)cm
4\(\sqrt{3}\)cm
Correct answer is B
In the figure above, \(\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}\) (Sine rule)
x = \(\frac{10 \sin 60^o}{\sin 30^o}\)
= 10 x \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\)
= 10 x \(\frac{\sqrt{3}}{2} \times \frac{2}{1}\)
= 10\(\sqrt{3}\)cm
From the figure above, what is the value of p?
135o
90o
60o
45o
Correct answer is B
In the figure above, qo = 30o (vertically opposite angles)
(P + 2q)o + 30o = 180o(angles on a straight line)
p + 2 x 30o + 30o = 180o
p + 60o + 30o = 180o
p + 90o = 180o
p = 180o - 90o
= 90o
91o
89o
37o
19o
Correct answer is C
In the diagram above, \(\alpha\) = 54o(alternate angles; KL||MN) < KNM = 2\(\alpha\) (LN is bisector of < KNM) = 108o
35o + < KMN + 108o = 180o(sum of angles of \(\bigtriangleup\))
< KMN + 143o = 180o
< KMN = 180o - 143o
= 37o