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.
In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR
100o
96o
94o
86o
Correct answer is D
< OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58°
< PSR = 36° + 58° = 94°
< PSR + < PQR = 180°
94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
92o
68o
67o
23o
Correct answer is C
Sum of the angles in a triangle = 180°
3x - 180° + 92° + x = 180°
4x - 88° = 180°
4x = 268°
x = 67°
60km/hr
80km/hr
90km/hr
108km/hr
Correct answer is C
\(speed = \frac{distance}{time}\\
72 = \frac{60}{time}\\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)
Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4
2.0
2.5
6.6
6.7
Correct answer is D
Measured value = 0.375
Approximation = 0.4
Error = 0.4 - 0.375 = 0.025
Error% = \(\frac{0.025}{0.375}\) x 100% = 6.67% = 6.7%
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters |QS|
\(10(1+\sqrt{3})\)
\(20\sqrt{3}\)
\(10\sqrt{3}\)
\((10+\sqrt{3})\)
Correct answer is A
In \(\Delta\) PQR,
\(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\)
= 10m
In \(\Delta\) PRS,
\(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\)
= \(\frac{10}\{\frac{1}{\sqrt{3}}\)
= \(10\sqrt{3}\)
PS = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)