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 this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of <PQR is
125o
120o
115o
110o
10o
Correct answer is D
Both pairs of opp. angles are equal
< STP = 55\(^o\) - isosceles angle
< TSR = 55\(^o\) - alternate angle to < STP
Hence, < PSR = 55\(^o\) + 55\(^o\) = 110\(^o\)
\(\therefore\) < PQR = 110\(^o\)
3\(\sqrt{2}\)
2\(\sqrt{3}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{2}{\sqrt{3}}\)
Correct answer is B
BC = 6 : DC = \(\frac{6}{2}\) = 3cm
By construction < EDE = 180o(90o + 60o) = 180o - 150o
= 30o(angle on a strt. line)
From rt < triangle ADC, AD2 = 52 - 32
= 25 - 9 = 6
AD = 4
From < AEC, let AS = x
\(\frac{x}{sin 60^o}\) - \(\frac{4}{sin 90^o}\)
sin 90o = 1
sin 60o = \(\frac{\sqrt{3}}{2}\)
x = 4sin 60o
x = 3 x \(\frac{\sqrt{3}}{2}\)
= 2\(\sqrt{3}\)
70o
110o
130o
125o
145o
Correct answer is D
If < CED = 30º, and < EDA = 40º then
<EOD = 180-(30-40) (angles in a triangle sum to 180) → 110º
<AOC = <EOD = 110º
At centre O: 360 - 110 = 250º
The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.
250 * \(\frac{1}{2}\) → 125º
<ABC = 125º
\(\sqrt{33}\)
6
4
2\(\sqrt{13}\)
4\(\sqrt{5}\)
Correct answer is D
Area of triangle AEF = 6sq. cm
area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines)
6 = \(\frac{1}{2}\) x 3 x h
6 = 3h
3h = \(\frac{12}{3}\)
h = 4 = DX
From D, C x D, CX2 = 52
- 42
= 25 - 16 = 9
Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3)
BD2 = 42 + 62
= 16 + 36 = 52
BD = \(\sqrt{4 \times 13}\)
= 2\(\sqrt{13}\)
In the figure, AB is parallel to CD then x + y + z is
185o
200o
270o
360o
195o
Correct answer is D
x + y + z = 360o