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.
N = \(\frac{kp_A}{D^2} + {cp_B}{D^2}\)
N = \(\frac{k P_{A} P_{B} }{D^2}\)
N = \(\frac{kD_AP_D}{B^2}\)
N = \(\frac{kD^2_AP_D}{B}\)
Correct answer is B
\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\)
\(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\)
\(N = \frac{k P_{A} P_{B}}{D^{2}}\)
The sum of the progression is 1 + x + x2 + x3 + ......
\(\frac{1}{1 - x}\)
\(\frac{1}{1 + x}\)
\(\frac{1}{x - 1}\)
\(\frac{1}{x}\)
Correct answer is A
Sum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\)
In the given series, a (the first term) = 1 and r (the common ratio) = x.
\(S_{n} = \frac{1(1 - x^{n})}{1 - x}\)
a = 1, and as n tends to infinity
\(S_{n} = \frac{1}{1 - x}\)
\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{71}{97}\)
\(\frac{9}{16}\), \(\frac{34}{59}\), \(\frac{3}{5}\) , \(\frac{71}{97}\)
\(\frac{3}{5}\), \(\frac{9}{16}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
\(\frac{9}{16}\), \(\frac{3}{5}\), \(\frac{71}{97}\), \(\frac{34}{59}\)
Correct answer is B
\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73
Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\)
Multiply (x + 3y + 5) by (2x2 + 5y + 2)
2x2 + 3yx2 + 10xy + 15y2 + 13y + 10x2 + 2x + 10
2x3 + 6yx2 + 5xy + 15y2 + 31y + 5x2 + 2x + 10
2x3 + 6xy2 + 5xy + 15y2 + 12y + 10x2 + 2x = 10
2x2 + 6xy2 + 5xy + 15y2 + 13y + 10x2 + 2x + 10
2x3 + 2yx2 + 10xy + 10y2 + 31y + 5x2 + 10
Correct answer is B
\((x + 3y + 5)(2x^{2} + 5y + 2)\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\)
= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)
3\(\frac{2}{3}\)
5\(\frac{1}{4}\)
6\(\frac{1}{2}\)
8
8\(\frac{1}{8}\)
Correct answer is A
\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)
= \(4\frac{29}{24}\)
\(\equiv 5\frac{5}{24}\)
\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)
= \(\frac{40 - 3}{24}\)
= \(\frac{37}{24}\)
\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)
= \(\frac{88}{24}\)
= 3\(\frac{2}{3}\)