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.
56.77
56.105
28:105
28:49
56:49
Correct answer is B
Area of QRST = 4 x 7 = 28
Area of PQRS = \(\frac{1}{2}\)(4 + 11) x 7
= \(\frac{7}{2}\) x \(\frac{15}{1}\)
= 28:52.5
= 56:105
If PN is perpendicular to QR, find the value of tan x.
\(\frac{5}{9}\)
\(\frac{3}{5}\)
\(\frac{3}{4}\)
\(\frac{4}{3}\)
\(\frac{4}{3}\)
Correct answer is B
By Pythagoras ON = 4
NR = 5
Tan x = \(\frac{3}{5}\)
17.72 meter/sec.
21.67 meters/sec
2.5 meter/sec.
20.45 meters/sec.
13.33 meter/sec.
Correct answer is D
Time(sec) areas of the three sides are \(\frac{1}{2}\) x 2 x 20 = 20 x 2 = 40
\(\frac{1}{2}\) x 4 x 20 = 40
40 + 40 = 80
vel. = \(\frac{80}{4}\)
20 + 0.45 = 20.45 meter/sec.
3\(\sqrt{3}\)
\(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
\(\frac{(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
\(\frac{3 - 1}{\sqrt{3} + 1}\)
Correct answer is B
\(\frac{x + 3}{sin 60^o}\) = \(\frac{x}{sin 306o}\)
3sin 30o = x sin 60o - x sin 30o
= x(sin 60o - sin 30o)
but sin 30o = \(\frac{1}{2}\)
sin 60o = \(\frac{3}{2}\) = \(\frac{3(\sqrt{3} - 1)}{\sqrt{3} + 1}\)
PQ is parallel to RS. Calculate the value of x.
20o
40o
60o
80o
100o
Correct answer is B
< D = 180o - 100v
= 80o (< on a str. line)
< s = 60o - alternate angle
x = 180o - (80o + 60o)
180o - 140o = 40o