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 = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)
∠PQT
∠SRP
∠PQR
∠SPR
Correct answer is D
No explanation has been provided for this answer.
In the diagram, calculate the value of x
35o
80o
100o
115o
Correct answer is C
x - 35° = 65° (corresponding angles)
x = 65° + 35° = 100°
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