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.
\(2x^2 + 13xy - 15y^2\)
\(2x^2 - 13xy - 15y^2\)
\(2x^2 + 13xy + 15y^2\)
\(2x^2 - 13xy + 15y^2\)
Correct answer is D
\((2x-3y)(x-5y)=2x^2 - 10xy - 3xy + 15y^2\\
=2x^2 - 13xy + 15y^2\)
Solve the equation \(2^7 = 8^{5-x}\)
\(\frac{5}{8}\)
\(\frac{8}{3}\)
\(\frac{3}{2}\)
\(\frac{15}{4}\)
Correct answer is B
\(2^7 = 2^{3(5-x)}\Rightarrow 7 = 3^{5-x} \Rightarrow 7 15 - 3x\\
\Rightarrow -8 = -3x \Rightarrow x = \frac{8}{3}\)
Convert 101101two to a number in base ten
61
46
45
44
Correct answer is C
1011012 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = 32 + 0 + 8 + 4 + 0 + 1 = 4510
In the diagram, PR is a diameter, ∠PRQ = (3x-8)° and ∠RPQ = (2y-7)°. Find x in terms of y
\(x=\frac{75-2y}{3}\)
\(x=\frac{105-3y}{2}\)
\(x=\frac{105-2y}{3}\)
\(x=\frac{75-3y}{2}\)
Correct answer is C
180 = ∠RPQ + ∠PRQ + ∠PQR Since PQR = 90 (theorem: angle in a semi circle)
180 = ∠RPQ + ∠PRQ + 90 => 180° = (3x-8)°+(2y-7)°+90°; 90+8+7 = 3x+2y =>\(\frac{105-2y}{3}=x\)
In the diagram, KL//MN, ∠LKP = 30o and ∠NMP = 45o. Find the size of the reflex ∠KPM.
285o
255o
225o
210o
Correct answer is A
∠KPM = 360o - (30 + 45)o
= 360o - 75o = 285o