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 the positive number n, such that thrice its square is equal to twelve times the number
1
2
3
4
Correct answer is D
3n2 = 12n
= 3n2 - 12n = 0
= 3n(n - 4) = 0
∴ n = 4
x < \(\frac{12}{13}\)
x < 13
x < 9
\(\frac{13}{12}\)
Correct answer is A
\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1
= \(\frac{6x + 4x + 3x < 12}{12}\)
i.e. 13 x < 12 = x < \(\frac{12}{13}\)
Find the two values of y which satisfy the simultaneous equation 3x + y = 8, x\(^2\) + xy = 6.
-1 and 5
-5 and 1
1 and 5
1 and 1
Correct answer is A
\(3x + y = 8 ... (i)\)
\(x^{2} + xy = 6 ... (ii)\)
From (i), \(y = 8 - 3x\)
From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\)
Equating the two values of y, we have
\(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\)
\(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\)
\(2x^{2} - 8x + 6 = 0\)
\(x^{2} - 4x + 3 = 0\)
\(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\)
\((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\)
\(y = 8 - 3x \)
When x = 1, \(y = 8 - 3(1) = 5\)
When x = 3, \(y = 8 - 3(3) = -1\)
\(\therefore \text{y = -1 or 5}\)
Which of the following is a factor of rs + tr - pt - ps?
(p - s)
(s - p)
(r - p)
(r + p)
Correct answer is C
rs + tr - pt - ps = rs - ps - tr - pt = (r - p)s + (r - p)t = (r - p)(s + t), hence r - p is a factor
(1 - a - b)(1 - a + b)
(1 + a - b)(1 - a + b)
(1 - a + b)(1 - a + b)
(1 + a + b)(1 + a + b)
Correct answer is B
1 - (a - b)2 = [1 + (a - b)][1 - a + b]
= (1 + a - b)(1 - a + b)