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.
Correct 241.34(3 x 10-\(^3\))\(^2\) to 4 significant figures
0.0014
0.001448
0.0022
0.002172
Correct answer is D
first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)
(3 x 10-\(^3\))\(^2\)............(B)
= 3\(^2\)x\(^2\)
= \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\)
(Note that x\(^2\) = \(\frac{1}{x^3}\))
= 24.34 x 3\(^2\) x \(\frac{1}{10^6}\)
= \(\frac{2172.06}{10^6}\)
= 0.00217206
= 0.002172(4 s.f)
The H.C.F. of a2bx + ab2x and a2b - b2 is
b
a + b
b(a \(\div\) b)
abx(a2 - b2)
Correct answer is B
a2bx + ab2x; a2b - b2
abx(a + b); b(a2 - b2)
b(a + b)(a + b)
∴ H.C.F. = (a + b)
Simplify \(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\)
-7\(\frac{7}{8}\)
\(\frac{-2}{7}\)
\(\frac{-10}{21}\)
\(\frac{10}{21}\)
Correct answer is B
\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\)
\(\frac{19}{4}\) - \(\frac{25}{4}\)............(A)
\(\frac{21}{5}\) x \(\frac{5}{4}\).............(B)
Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\)
A = \(\frac{19}{4}\) - \(\frac{25}{4}\)
= \(\frac{-6}{4}\)
B = \(\frac{21}{5}\) x \(\frac{5}{4}\)
= \(\frac{105}{20}\)
= \(\frac{21}{4}\)
But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\)
= \(\frac{-6}{4}\) x \(\frac{4}{21}\)
= \(\frac{-24}{84}\)
= \(\frac{-2}{7}\)
\(\frac{5}{12}\)
\(\frac{1}{3}\)
\(\frac{3}{4}\)
\(\frac{7}{12}\)
Correct answer is D
Coca-cola = 10 bottles, Fanta = 8 bottles
Sprite = 6 bottles, Total = 24
P(cola-cola) = \(\frac{10}{24}\)
P(not coca-cola) = 1 - \(\frac{10}{24}\)
\(\frac{24 - 10}{24}\) = \(\frac{14}{24}\)
= \(\frac{7}{12}\)
96
94
92
90
Correct answer is A
% of boxes containing at least 5 defective bolts each
= \(\frac{7 + 17 + 10 + 8 + 6}{50}\)
= \(\frac{48}{50}\) x \(\frac{100}{1}\)
= 96%