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: \(\frac{2x^2 - 5x - 12}{4x^2 - 9}\)
\(\frac{x + 4}{2x + 3}\)
\(\frac{x + 4}{2x - 3}\)
\(\frac{x - 4}{2x + 3}\)
\(\frac{x - 4}{2x - 3}\)
Correct answer is D
\(\frac{2x^2 - 5x - 12}{4x^2 - 9}\) = \(\frac{3x^2 - 8x + 3x - 12}{(2x)^2 - 3^2}\)
= \(\frac{32(x - 4) + 3(x - 4)}{(2x - 3)(2x + 3)} - \frac{(x - 4) + (2x + 3)}{(2x - 3) (2x + 3)}\)
= \(\frac{x - 4}{2x - 3}\)
If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r
\(\frac{9}{25} pr^2\)
\(\frac{9}{25} p^2r\)
\(\frac{25}{9} p^2r\)
\(\frac{25}{9} pr^2\)
Correct answer is C
p = \(\frac{3}{5} \sqrt{\frac{q}{r}}; \frac{5p}{3} = \sqrt{\frac{q}{r}}\)
= (\(\frac{5}{3}p\))2
= \(\frac{q}{r}\)
= \(\frac{25p^2}{9} = \frac{q}{r}\)
q = \(\frac{25}{9} p^2 r\)
If 4y is 9 greater than the sum of y and 3x, by how much is y greater than x?
3
6
9
12
Correct answer is A
4y - 9 > y + 3x; 4y - y > 3x + 9
3y > 3(x + 3); y = > \(\frac{3(x + 3)}{3}\)
y > x + 3; y - 3 > x
y is greater than x by 3
(a - y)(5y - 3a)
(y - a)(5y - 3a)
(y - a)(5y + 3a)
(y + a)(5y - 3a)
Correct answer is D
5y2 + 2ay - 3a2 = 5y2 + 5ay - 3a2
= 5y(y + a) - 3a(y + a)
= (y + a)(5y - 3a)
Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\)
zero
\(\frac{1}{5}\)
1
2
Correct answer is A
Given; x = 2; y = \(\frac{-1}{4}\)
= \(\frac{x^2y - 2xy}{5}\)
= \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\)
= \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\)
= \(\frac{1 + 1}{5}\)
= \(\frac{0}{5}\)
= 0