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.
The figure is an example of the construction of a
perpendicular bisector of a given straight line QR
perpendicular from a given point to a given line QR
perpendicular to a line from a given point P on that line
given angle
Correct answer is B
QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR
In the figure, find the value of x
110o
100o
90o
80o
Correct answer is C
Z = X = Y = 180o .........(i)
2Z + Y + Y = 190 = 2Z + 2Y = 180
Z + Y = 90o........(ii)
hence x + (z + y) = 180
x + 90o = 180o
x = 180o - 90o
= 90o
210o
150o
105o
50o
Correct answer is D
Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the nonagon = 9
where 3 are equal and 6 other angles = 1110o
( 2 x 9 - 4)90o = (18 - 4)90o
= 14 x 90o = 1260o
9 angles = 12600, 6 angles = 1110o
Remaining 3 angles = 1260o - 1110o = 150o
size of one of the3 angles \(\frac{150}{3}\) = 50o
4:9
4:5
1:3
2:3
Correct answer is B
h1 = \(\frac{h_1 + h_2}{1.5}\) = \(\frac{h_1}{h_2} = 2\)
\(\frac{A_{\bigtriangleup PQR}}{A_{TP(QRST)}} = \frac{\frac{1}{2} \times 2 \times h_1}{\frac{1}{2} \times h_2 (2 + 3)}\)
\(\frac{2}{5} \times \frac{h_1}{h_2} = \frac{2}{5} \times 2\)
= \(\frac{4}{5}\)
30o
40o
45o
50o
Correct answer is A
From the diagram, POQ is a diameter; o is the centre of the circle and TP is a tangent where RTP = 30o
RTP = RQP = x(at circumference = made by tangent outside the circle) i.e x = 30o