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 diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW
135o
105o
75o
60o
Correct answer is C
In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)
\(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\))
= 30o
In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1
\(\beta\) = tan-1(1) = 45o
Hence, < UTW = \(\alpha\) + \(\beta\)
= 30o + 45o = 75o
In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU.
120o
108o
72o
60o
Correct answer is C
In the diagram, < TUQ + 60o(corresp. angles)
< QTU = 48o (alternate angles)
< QU + 60o + 48o = 180o(sum of angles of a \(\Delta\))
< TQU = 180o - 108o
= 72o
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.
54o
44o
34o
27o
Correct answer is A
In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.)
x1 = x2 (base angles of isos. \(\Delta\))
x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\)
63o + 63o + \(\alpha\) = 180o
\(\alpha\) = 180o - (63 + 63)o
= 54o
4.5m
6.0m
7.5m
9.0m
Correct answer is C
By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\)
\(\frac{8}{3} = \frac{20}{h}\)
h = \(\frac{3 \times 20}{8}\)
= 7.5m
30
36
40
72
Correct answer is A
In the diagram above, \(\alpha\) = 2mo (angle at centre = 2 x angle at circumference)
\(\alpha\) + 10mo = 360o (angle at circumference)
\(\alpha\) + 10mo = 360o(angles round a point)
2mo + 10mo = 360o
12mo = 360o
mo = \(\frac{360^o}{12}\)
= 30o