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.
Find all median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119
131
125
123
120
Correct answer is B
Arrange in ascending order
89, 100, 108, 119, 120,130, 131, 131, 141, 161
Median = \(\frac{120 + 130}{2}\) = 125
Factorise (4a + 3) \(^2\) - (3a - 2)\(^2\)
(a + 1)(a + 5)
(a - 5)(7a - 1)
(a + 5)(7a + 1)
a(7a + 1)
Correct answer is C
[(4a + 3) \(^2\) - (3a - 2)\(^2\) = a\(^2\) - b\(^2\) = (a + b) (a - b)
= [(4a + 3) + (3a - 2)] [(4a + 3) - (3a - 2)]
= [4a + 3 + 3a - 2] [4a + 3 - 3a + 2]
= (7a + 1)(a + 5)
(a + 5) (7a + 1)
Evaluate (212)\(_3\) - (121)\(_3\) + (222)\(_3\)
(313)\(_3\)
(1000)\(_3\)
(1020)\(_3\)
(1222)\(_3\)
Correct answer is C
Evaluate (212)\(_3\) - (121)\(_3\) + (222)\(_3\)
(212)\(_3\) + (222)\(_3\) = 1211\(_3\)
→ 1211\(_3\) - 121\(_3\)
= 1020\(_3\)
N9,700.00
N6,700.00
N2,000.00
N4,000.00
N2,300.00
Correct answer is E
Degree for food items = 360° - (120° + 80° + 43° + 30° + 18°)
= 360° - 291°
= 69°
∴∴ Amount spent on food items = \(\frac{69}{360}\)×12,000.00
= N2300.00
Evaluate \(\frac{2\sin 30 + 5\tan 60}{\sin 60}\), leaving your answer in surd form.
\(\frac{2\sqrt{3}}{3} + 10\)
\(\frac{3\sqrt{2} - 1}{5}\)
\(\frac{3\sqrt{2} + 1}{5}\)
\(\frac{2\sqrt{3}}{3} - 10\)
Correct answer is A
\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\)
\(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\)
\(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\)
= \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\)
= \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\)
= \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\)
Rationalizing, we get
= \(\frac{2\sqrt{3} + 30}{3}\)
= \(\frac{2}{3} \sqrt{3} + 10\)