Simplify 3.72 x 0.025 and express your answer in the standard form
\(9.3\times 10^3\)
\(9.3\times 10^2\)
\(9.3\times 10^{-2}\)
\(9.3\times 10^{-3}\)
Correct answer is C
\(3.72 \times 0.025 = 0.093 = 9.3\times 10^{-2}\)
3.3cm
5.3cm
4.0cm
8.0cm
Correct answer is C
Volume = \(\frac{1}{3} b^2 h\)
\(\therefore b = \sqrt{\frac{3v}{h}}\)
\(b = \sqrt{\frac{3(80)}{15}} \)
\(b = \sqrt{16} = 4.0 cm\)
In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.
60cm2
50cm2
40cm2
30cm2
Correct answer is D
PR = 4 + 4 = 8 cm
Area of trapezium = \(\frac{1}{2} (a + b) h\)
= \(\frac{1}{2} (4 + 8) \times 5\)
= 30 cm\(^2\)
\(16\sqrt{3}m\)
\(4\sqrt{3}m\)
\(\frac{\sqrt{3}}{16}\)
\(\frac{16\sqrt{3}}{3}\)
Correct answer is D
\(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\)
When we rationalize gives \(\frac{16\sqrt{3}}{3}\)
040o
070o
110o
290o
Correct answer is D
< ABC = 40° (alternate angles)
\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)
= 70°
\(\therefore\) Bearing of A from C = 360° - 70°
= 290°
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