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.
36
27
24.5
13.5
Correct answer is B
\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)
Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\)
-8
-13
13
19
Correct answer is B
\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 8y - 4 - 9y + 3 = 12 \\
\Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)
Solve the equation 2x - 3y = 22; 3x + 2y = 7
-5
-4
4
5
Correct answer is B
2x - 3y = 22 ---- eqn I
3x + 2y = 7 ---- eqn II
multiply eqn I by 2
4x - 6y = 44 ---- eqn III
multiply eqn II by 3
9x + 6y = 21 ---- eqn IV
Adding eqn III and IV
=> 13x = 65 => x = 5
Substituting 5 for x in eqn II
3x5 + 2y = 7 => 2y = -8
y = -4
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)]
\(\frac{\sqrt{2}}{7}cm\)
\(3\frac{1}{2}cm\)
7cm
\(7\sqrt{2}cm\)
Correct answer is D
\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\
\Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)
026o
045o
210o
240o
Correct answer is D
\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\\
\theta = cos^{-1}(0.5000)=60^{\circ}\)
The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)