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.
What is the number whose logarithm to base 10 is 2.3482?
223
2228
2.235
22.37
0.2229
Correct answer is A
\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\)
N : b Look for the antilog of .3482 which gives 222.9
4
-4
2
-2
None of the above
Correct answer is E
x\(^2\) + 4 = 0
x\(^2\) = -4
You can't apply a square root to a negative number. So the answer is None of the above.
Find the square root of 170 - 20\(\sqrt{30}\)
2 \(\sqrt{10}\) - 5\(\sqrt{3}\)
2 \(\sqrt{5}\) - 5\(\sqrt{6}\)
5 \(\sqrt{10}\) - 2\(\sqrt{3}\)
3 \(\sqrt{5}\) - 8\(\sqrt{6}\)
Correct answer is B
\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\)
Squaring both sides,
\(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\)
Equating the rational and irrational parts, we have
\(a + b = 170 ... (1)\)
\(2 \sqrt{ab} = 20 \sqrt{30}\)
\(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \)
\(ab = 3000 ... (2)\)
From (2), \(b = \frac{3000}{a}\)
\(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\)
\(a^{2} - 170a + 3000 = 0\)
\(a^{2} - 20a - 150a + 3000 = 0\)
\(a(a - 20) - 150(a - 20) = 0\)
\(\text{a = 20 or a = 150}\)
\(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\)
\(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\)
= \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)
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
\(S_{n} = \frac{1}{1 - x}\)