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.
Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)
121
144
169
196
Correct answer is C
\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)
\([8 + 5]^2\) = \([13]^2\)
= 169
Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\)
\(\sqrt{3} + 5\sqrt{5}\)
\(6 \sqrt{3} - 5 \sqrt{5}\)
\(6 \sqrt{3} + \sqrt{2}\)
\(6\sqrt{3} - \sqrt{2}\)
Correct answer is A
\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)
= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\)
= \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\)
= \(\sqrt{3} + 5\sqrt{5}\)
Add 54 \(_{eight}\) and 67\(_{eight}\) giving your answers in base eight
111
121
123
143
Correct answer is D
54 \(_{eight}\) and 67\(_{eight}\) = 1438
Starting with normal addition, 4 + 7 gives 11
(it is more than the base, 8) 8 goes in 11 just 1 time, remaining 3, the remainder will be written, and the 1 will be added to the sum of 5 and 6 which gives 12 altogether, 8 goes in 12 one time remaining 4, the remainder 4 was written and then the 1 that was the quotient was then written since nothing to add the 1 to.
So answer is 143 in base eight
026\(^o\)
045\(^o\)
210\(^o\)
240\(^o\)
Correct answer is D
Cos θ = \(\frac{adj}{hyp}\)
= \(\frac{300}{600}\)
= 0.5
θ = Cos - 10.5
= 60
∠ RPQ = ∠ PQs
So the bearing of P from Q is 180 + 60 = 240\(^o\)
Answer is D
Simplify 25\(\frac{1}{2}\) × 8\(\frac{-2}{3}\)
1\(\frac{1}{4}\)
2\(\frac{1}{4}\)
6
10
Correct answer is A
Using law of indices
25\(\frac{1}{2}\) × 8\(\frac{-2}{3}\)
= √25 x (\(\sqrt[3]{8}\)) -2
= 5 x 2-2
= 5 x \(\frac{1}{2^2}\) =
\(\frac{5}{4}\) = \(\frac{11}{4}\)
Answer is A