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
\((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\)
= \((4a)^{-1} \)
= \(\frac{1}{4a}\)