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.
113\(cm^3\)
131\(cm^3\)
311\(cm^3\)
414\(cm^3\)
Correct answer is A
Surface area of a sphere = \(4 \pi r^2\) \(4 \pi r^2\) = \(\frac{792}{7}cm^2\) 4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\) \(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\) = 9 r = \(\sqrt{9}\) = 3cm Hence, volume of sphere = \(\frac{4}{3} \pi r^3\) = \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) = \(\frac{4 \times 22 \times 9}{7}\) \(\approx\) = 113.143 = 113\(cm^3\) (to the nearest whole number)
14m
10m
7m
5m
Correct answer is C
Volume of a cylinder = \( \pi r^2\)h
385 = \(\frac{22}{7}\) x \(r^2\) x 10
385 x 7 = 22 x \(r^2\) x 10
\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)
= 12.25
r = \(\sqrt{12.25}\)
= 3.5m
Hence, diameter of tank = 2r
= 2 x 3.5 = 7m
A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve.
y = \(x^2 - 5x - 6\)
y = \(x^2 + 5x - 6\)
y = \(x^2 + x - 6\)
y = \(x^2 - x - 6\)
Correct answer is A
Since the curve cuts the x-axis at x = -2 and x = 3,
(x + 2)(x - 3) = 0
\(x^2 - 3x + 2x - 6\) = 0
\(x^2 - x - 6\) = 0
Hence, the equation of the curve is
y = \(x^2 - x - 6\)
Simplify; \(\frac{2 - 18m^2}{1 + 3m}\)
\(2 (1 + 3m)\)
\(2 (1 + 3m^2)\)
\(2(1 - 3m)\)
\(2(1 - 3m^2)\)
Correct answer is C
\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\)
= \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\)
= \(2(1 - 3m)\)
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\)
1\(\frac{1}{2}\)
2
2\(\frac{1}{2}\)
3
Correct answer is A
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)
\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)
Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)
\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)
\(T_1\) = 15 and \(T_2\) = 21
Therefore;
x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9 (again)
z = \(\frac{4}{7}\) x 21 = 12
Hence;
\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\)
\(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)
= 1\(\frac{1}{2}\)