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.
Simplify \(\frac{x^2 + y^2 + xy}{x + y}\) - \(\frac{x^2 + y^2}{x - y}\)
\(\frac{x^2 + y^2} {(x - y)^2}\)
\(\frac{2y^3} {y^2 - x^2}\)
\(\frac{3x^2 + y^2} {(2x - y)^2}\)
\(\frac{x^2 + y^2} {(x^2 - y)}\)
Correct answer is B
\(\frac{x^2 + y^2 + xy}{x + y}\) - \(\frac{x^2 + y^2}{x -y}\)
= \(\frac{-y^3 - y^3}{x^2 - y^2}\)
= \(\frac{2y^3}{x^2 - y^2}\)
= \(\frac{2y^3}{y^2 - x^2}\)
\(\frac{44}{3}\)\(\pi\)cm3
12\(\pi\)cm3
\(\frac{38}{3}\)\(\pi\)cm3
\(\frac{40}{3}\)\(\pi\)cm3
\(\frac{32}{33}\)\(\pi\)cm3
Correct answer is A
Volume of steel ball = \(\frac{4\pi r^2}{3}\)
= \(\frac{4}{3}\) \(\pi\) x 1
= \(\frac{4 \pi}{3}\)cm3
Vol. of cylinder = \(\pi\)r2h
= \(\pi\) x 22 x 3
Vol. of water = 16\(\pi\) - \(\frac{4 \pi}{3}\)
= \(\frac{48 - 4 \pi}{3}\)
= \(\frac{44 \pi}{3}\)cm3
Simplify \(\frac{5^x \times 25^{x - 1}}{125^{x + 1}}\)
52x + 1
5x + 1
5-5
52
53
Correct answer is C
\(\frac{5^x \times 25^{x - 1}}{125^{x + 1}}\) = \(\frac{5^x \times 5^{2x - 2}}{5^{3x + 3}}\)
= \(\frac{5^{x + 2x - 2}}{5^{3x + 3}}\)
= \(\frac{5^{3x - 2}}{5^{3x + 3}}\)
= 5\(^{3x - 2 - 3x - 3}\)
= 5\(^{-5}\)
Rationalize the denominator of the expression \(\frac{6 + 2\sqrt{5}}{4 - 3\sqrt{6}}\)
\(\frac{12+ 4\sqrt{5 + 7} 5 + 6\sqrt{3}}{39}\)
\(\frac{-(24 + 18\sqrt{6} + 8\sqrt{5} + 6\sqrt{30})}{38}\)
\(\frac{24 + 3\sqrt{6 + 8} 5 + 6\sqrt{30}}{19}\)
\(\frac{-15 + 3\sqrt{5 + 18} 5 + 6\sqrt{30}}{36}\)
\(\frac{-(12 + 4\sqrt{5} +9\sqrt{6} + 3\sqrt{30})}{19}\)
Correct answer is B
Rationalize using the reciprocal of the denominator to multiply through
(i.e. Multiply both numerator and denominator using \(4 + 3\sqrt{6}\) )
Watch your signs in the course of this.
20sq.cm
100sq.cm
25sq.cm
16sq.cm
36sq.cm
Correct answer is C
Area of a square = 4(5) where S is each sides of the square
Perimeter = 20(given)
4S = 20
S = \(\frac{20}{4}\)
= 5
Area s2 = 52
= 25