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, PS bisects angle QPR. Find the ratio SR:QR.
1:2
1:3
1:4
1:5
1:6
Correct answer is C
From Internal bisection theorem, \(\frac{QP}{PR}\) = \(\frac{QS}{SR}\)
= \(\frac{4}{1}\)
= 1:4
In the figure, If PT is parallel to RS, PQ = PT, and angle SQT = 90o, Find x
35o
50o
55o
70o
80o
Correct answer is D
No explanation has been provided for this answer.
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.