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.
\(\frac{1}{18}\)
\(\frac{8}{81}\)
\(\frac{9}{2}\)
8
Correct answer is D
\(x \propto \frac{1}{y}\)
\(x = \frac{k}{y}\)
\(\frac{2}{3} = \frac{k}{9}\)
\(3k = 18 \implies k = 6\)
\(x = \frac{6}{y}\)
When y = \(\frac{3}{4}\),
x = \(\frac{6}{\frac{3}{4}}\)
= \(\frac{6 \times 4}{3}\)
= 8
Given that \(27^{(1+x)}=9\) find x
-3
\(\frac{-1}{3}\)
\(\frac{5}{3}\)
2
Correct answer is B
\(27^{(1+x)}=9\\
3^{3(1+x)}=3^2\\
3(1+x)=2\\
3+3x = 2\\
3x = -1
x = \frac{-1}{3}\)
Given that \(x = -\frac{1}{2}and \hspace{1mm} y = 4 \hspace{1mm} evaluate \hspace{1mm} 3x^2y+xy^2\)
-5
-1
4
11
Correct answer is A
\(x = -\frac{1}{2}, y = 4\\
3x^2y + xy^2\\
3\left[-\frac{1}{2}\right]^2 \times 4 \times + \left(\frac{-1}{2}\right)(4)^2\\
3\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\\
3-8 = -5\)
160o
140o
120o
100o
Correct answer is D
< T = < S = 50° (OS = OT)
< SOT = 180° - 2(50°) = 80°
< ROP = 80° (vertically opposite angle)
\(\therefore\) < OPQ = 180° - 80° = 100° (adjacent angles)
If the interior angles of hexagon are 107°, 2x°, 150°, 95°, (2x-15)° and 123°, find x.
\(57\frac{1}{2}^{\circ}\)
\(65^{\circ}\)
\(106^{\circ}\)
\(120^{\circ}\)
Correct answer is B
Sum of interior angle in a hexagon = (6 - 2) x 180°
= 720°
\(\therefore\) 107° + 2x° + 150° + 95° + (2x - 15)° + 123° = 720°
460 + 4x = 720 \(\implies\) 4x = 720 - 460
4x = 260° \(\implies\) x = 65°