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.
Solve the simultaneous linear equations: 2x + 5y = 11, 7x + 4y = 2
x = -8, y = 1
x = -2, y = 4
x = \(\frac{-34}{27}\), y = \(\frac{73}{27}\)
x = 7, y = -9
Correct answer is C
2x + 5y = 11.......(i)
7x + 4y = 2.......(ii)
(i) x 7 \(\to\) 14x + 35y = 77.........(iii)
(ii) x 2 \(\to\) 14x + 8y = 4........(iv)
(iii) - (iv)
27y = 73
y = \(\frac{73}{27}\)
1\(\frac{2}{3}\)
2
5
6
10
Correct answer is C
Let x rept. the usual speed = \(\frac{Distance}{Time}\)
= \(\frac{20}{1}\) x \(\frac{1}{2}\)
= 5
(a), (b), (c)
(c)
None of the choices
All of the above
Correct answer is B
0 < \(\frac{x + 3}{x - 1}\) < 2
Put x = 0, -3 and 9
0 < \(\frac{9 + 3}{9 - 1}\) \(\leq\) 2
i.e. 0 < 1.5 \(\leq\) 2 (true)
but 0 < \(\frac{0 + 3}{0 - 1}\) \(\leq\) 2
i.e. 0 < -3 \(\leq\) 2 (not true)
-3 \(\leq\) 2
-3 is not greater than 0
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0
c = \(\frac{2d - 5 - b}{ad}\)
c = \(\frac{2d - 5 - abd}{ad}\)
c = \(\frac{2d - 5 - ab}{ad}\)
c = \(\frac{5 - 2d - b}{ad}\)
Correct answer is B
a(b + c) + \(\frac{5}{d}\) - 2 = 0
ab + ac + \(\frac{5}{d}\) - 2 = 0
abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d
acd = 2d - 5 - abd
c = \(\frac{2d - 5 - abd}{ad}\)
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If < PQS = 15o find < QRS
75o
37\(\frac{1}{2}\)o
127\(\frac{1}{2}\)o
105o
Correct answer is D
PSO = 90o(angle in a semicircle)
SPO = 180o - (90o + 15o) = 75o
< QRS = 108o - 75o
= 105o(opposite angles in cyclic quadrilateral are supplementary)