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.
Find the principal which amounts to N5,500 at a simple interest in 5 years at 2% per annum.
N4,900
N5,000
N4,700
N4,800
Correct answer is B
Principal, P = Amount, A - Interest, I.
A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = N5,000
5/8
5/2
5/32
5/16
Correct answer is B
\(x = \frac{y}{2} \)
\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\)
\(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\)
= \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\)
= \(\frac{1}{8} + \frac{1}{2}\)
= \(\frac{5}{8}\)
\(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\)
= \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\)
= \(\frac{1}{2} - \frac{1}{4}\)
= \(\frac{1}{4}\)
\(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\)
= \(\frac{5}{2}\)
60%
32%
25%
20%
Correct answer is C
Total cost = N(250,000 + 70,000) = N320,000
Selling price = N400,000 (given)
Gain = SP - CP = N(400,000 - 320,000) = N80,000
Gain % = gain/CP x 100 = (80,000/320,000) x 100
Gain % = 25%
Given that \(p = 1 + \sqrt{2}\) and \(q = 1 - \sqrt{2}\), evaluate \(\frac{p^{2} - q^{2}}{2pq}\).
2(2+√2)
-2(2+√2)
2√2
-2√2
Correct answer is D
\(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\)
= \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\)
= \(\frac{(2\sqrt{2})(2)}{-2}\)
= \(-2\sqrt{2}\)
Simplify \((\sqrt[3]{64a^{3}})^{-1}\)
4a
1/8a
8a
1/4a
Correct answer is D
\((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\)
= \((4a)^{-1} \)
= \(\frac{1}{4a}\)