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.
3k2
2k - k2
\(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
k2
Correct answer is C
\(\frac{a}{c}\) = \(\frac{c}{d}\) = k
∴ \(\frac{a}{b}\) = bk
\(\frac{c}{d}\) = k
∴ c = dk
= \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)
= \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\)
= \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0
1 and 4
-2 and 2
0 and 1
-1 and 0
Correct answer is C
16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0
Make U the subject of the formula S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)
u = \(\frac{12}{2s^2}\)
u = \(\frac{12}{2s+ w}\)
u = \(\frac{12}{2s^2 + w}\)
u = \(\frac{12}{2s^2}\) + w
Correct answer is C
S = \(\sqrt{\frac{6}{u} - \frac{w}{2}}\)
S = \(\frac{12 - uw}{2u}\)
2us2 = 12 - uw
u(2s2 + w) = 12
u = \(\frac{12}{2s^2 + w}\)
Simplify \(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\)
\(\frac{2x}{(x - 2)(x + 2)(x^2 - 4)}\)
\(\frac{2x}{x^2 - 4}\)
\(\frac{x}{x^2 - 4}\)
\(\frac{4x}{x^2 - 4}\)
Correct answer is D
\(\frac{1}{x - 2}\) + \(\frac{1}{x + 2}\) + \(\frac{2x}{x^2 - 4}\)
= \(\frac{(x + 2) + (x - 2) + 2x}{(x + 2)(x - 2)}\)
= \(\frac{4x}{x^2 - 4}\)
If \(5^{(x + 2y)} = 5\) and \(4^{(x + 3y)} = 16\), find \(3^{(x + y)}\).
7
1
3
27
Correct answer is B
\(5^{(x + 2y)} = 5\)
∴ x + 2y = 1.....(i)
\(4^{(x + 3y)} = 16 = 4^2\)
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i)
\(x + 2y = 1 \implies x + 2(1) = 1\)
\(x + 2 = 1 \implies x = -1\)
\(\therefore 3^{(x + y)} = 3^{(-1 + 1)}\)
\(3^{0} = 1\)