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.
12n - 6 = 0
3n - 12 = 0
2n - 35 = 0
5n - 33 = 0
Correct answer is D
12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3
36 + n - 6n = 3
-5n = 3 - 36
-5n = -33
-5n + 33 = 0
5n - 33 = 0
If x + y = 2y - x + 1 = 5, find the value of x
3
2
1
-1
Correct answer is B
x + y = 2y - x + 1 = 5
x + y = 2y - x + 1
x + x + y - 2y = 1
2x - y = 1....(i)
2y - x + 1 = 5
-x + 2y = 5 + 1
-x = 2y = 4
x - 2y = -4 .....(ii)
solve simultaneously (i) x 2x - y = 1
(ii) x x - 2y = -4
2x - y = 1
=2x - 4y = -8
3y = 9
y = \(\frac{9}{3}\)
y = 3
substitute y = 3 into equation (i)
2x - y = 1
2x - 3 = 1
2x = 1 + 3
2x = 4
x = \(\frac{4}{2}\)
= 2
Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)
p = \(\frac{2q - rs}{6}\)
p = 2qr - sr - 3
p = \(\frac{2qr - s}{6}\)
p = \(\frac{2qr - rs}{6}\)
Correct answer is D
q = \(\frac{3p}{r} + \frac{s}{2}\)
q = \(\frac{6p + rs}{2r}\)
6p + rs = 2qr
6p = 2qr - rs
p = \(\frac{2qr - rs}{6}\)
Simplify: \(\frac{54k^2 - 6}{3k + 1}\)
6(1 - 3k2)
6(3k2 - 1)
6(3k - 1)
6(1 - 3k)
Correct answer is C
\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\)
= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\)
= 6(3k - 1)
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)
5
4
3
1
Correct answer is A
\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)
= \(\frac{1}{2}\)
\(\frac{5}{3x} = \frac{1}{3}\)
3x = 15
x = \(\frac{15}{3}\)
= 5