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.
Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR
11o
73o
107o
138o
149o
Correct answer is C
No explanation has been provided for this answer.
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.
15o
25o
40o
50o
55o
Correct answer is B
Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o
3.3cm
4.4cm
5cm
5.5cm
6.6cm
Correct answer is B
\(\frac{PQ}{RQ} = \frac{RQ}{SQ}\) (Similar triangles)
\(\therefore \frac{PS + 10}{12} = \frac{12}{10}\)
\(10 (PS + 10) = 12 \times 12\)
\(10 PS + 100 = 144 \implies 10 PS = 44\)
\(PS = 4.4 cm\)
34 sides
30 sides
26 sides
17 sides
15 sides
Correct answer is D
Sum of interior angles in a polygon = \((2n - 4) \times 90°\)
\(\therefore (2n - 4) \times 90° = 30 \times 90°\)
\(\implies 2n - 4 = 30 \)
\(2n = 34 \implies n = 17\)
The polygon has 17 sides.
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68°, find ∠PQT.
34o
48o
56o
68o
73o
Correct answer is C
< STQ = < SQR = 68° (alternate segment)
\(\therefore\) < STQ = 68°
< TQS = \(\frac{180° - 68°}{2}\)
= \(\frac{112}{2} = 56°\)
\(\therefore\) < PQT = 180° - (68° + 56°)
= 180° - 124°
= 56°