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.
5.0km
7.5km
7.6km
8.7km
Correct answer is D
\((PR)^2 = (PQ)^2 + (QR)^2\)
\(10^2 = 5^2 + (QR)^2\)
\((QR)^2 = 100 - 25\)
\(QR = \sqrt{75}\)
= \(8.660\)
\(\approxeq\) 8.7 km
Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
\(\frac{79}{156}\)
\(\frac{85}{156}\)
\(\frac{7}{13}\)
\(\frac{8}{1}\)
Correct answer is A
If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\\
=169-25\\
BC = \sqrt{144} = 12\\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\\
=\frac{79}{156}\)
The bearing S40°E is the same as
040o
050o
130o
140o
Correct answer is D
= 90° + 50°
= 140°
If 2x : (x +1) = 3:2, what is the value of x?
\(\frac{1}{2}\)
1
\(1\frac{1}{2}\)
3
Correct answer is D
\(2x : (x + 1) = 3:2\\
\frac{2x}{x+1}=\frac{3}{2}\\
∴ 4x = 3x + 3 x =3\)
If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24
\(\frac{1}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
Correct answer is A
\(y \propto \frac{1}{\sqrt{x}}\)
\(y = \frac{k}{\sqrt{x}}\)
When x = 16, y = 2.
\(2 = \frac{k}{\sqrt{16}} \implies 2 = \frac{k}{4}\)
\(k = 8\)
\(y = \frac{8}{\sqrt{x}}\)
When y = 24,
\(24 = \frac{8}{\sqrt{x}}\)
\(\sqrt{x} = \frac{8}{24} = \frac{1}{3}\)
\(\therefore x = (\frac{1}{3})^2\)
\(x = \frac{1}{9}\)