Question Library

Angle (Y) is (60\(^\circ\)). This construction forms an equilateral triangle, so all its interior angles, including (\(\angle\) Y), are (60\(^\circ\)).

Given the frequency distribution: \(\begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 0-9 & 1 \\ 10-19 & 5 \\ 20-29 & 6 \\ 30-39 & 12 \\ 40-49 & 8 \\ 50-59 & 3 \\ \hline\end{array}\) The modal class is \(30-39\) with frequency \(f_1 = 12\), \(f_0 = 6\) (previous class), \(f_2 = 8\) (next class), and the lower boundary \(L = 29.5\) with class width \(h = 10\). Using the mode formula: \(\text{Mode} = L + \left( \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \right) \times h\) Substituting the values: \(\text{Mode} = 29.5 + \left( \frac{12 - 6}{(12 - 6) + (12 - 8)} \right) \times 10\) \(\text{Mode} = 29.5 + \left( \frac{6}{6 + 4} \right) \times 10 = 29.5 + \left( \frac{6}{10} \right) \times 10 = 29.5 + 6 = 35.5\)

To convert \(54_{10}\) to base 4: \(54 \div 4 = 13\) with remainder \(2\). \(13 \div 4 = 3\) with remainder \(1\). \(3 \div 4 = 0\) with remainder \(3\). Collecting the remainders from bottom to top, we have: \(54_{10} = 312_4\) Thus, \(X = 312.000\).

To solve \(x^2 + 3x - 4 \leq 0\): Factorize: \((x + 4)(x - 1) \leq 0\) Roots: \(x = -4\), \(x = 1\) Solution: \(-4 \leq x \leq 1\) .

Given \(\cos \theta = \frac{x}{y}\), We know \(\sin^2 \theta + \cos^2 \theta = 1\), So \(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{x}{y}\right)^2 = \frac{y^2 - x^2}{y^2}\). Assuming \(\sin \theta > 0\) (acute angle or principal value), \(\sin \theta\) = \(\frac{\sqrt{y^2 - x^2}}{\text{y}}\). Then, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{y^2 - x^2}/y}{x/y} = \frac{\sqrt{y^2 - x^2}}{x}\). Thus, \(\tan \theta = \frac{\sqrt{y^2 - x^2}}{x}\)

Tan \(\theta\) = \(\frac{\text{0pp}}{\text{Adj}}\) Tan \(\theta\) = \(\frac{200}{\text{x}}\) x = \(\frac{200}{\text{tan} 60}\) x = \(\frac{200}{\sqrt{3}}\)

(101\(_2\))\(^2\) ⇒ 101\(_2\) = 1\times2\(^2\) + 0 \times 2\(^1\) + 1 \times 2\(^0\) = 1 \times 4 + 0 + 1 (Recall x⁰ = 1) = 4 + 1 = 5 (101\(_2\))\(^2\) = (5\(_{10}\))\(^2\) = 25. 25 in base 2 = 11001

Total amount = ₦10.50; Ratio = 6: 7: 8; Total ratios = 6 + 7 + 8 = 21 largest share = 8 ∴ Largest amount = \(\frac{8}{21}\) \times ₦10.50 = 8 \times ₦0.50 = ₦4.00

0.000834 = 8.34 x 10\(^{-4}\) Move the decimal point 4 places to your right.

0.027\(^{-\frac{1}{3}}\) = (\(\frac{27}{100}\))\(^{-\frac{1}{3}}\) = (\(\frac{10^3}{3^3}\))\(^{\frac{1}{3}}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)

log\(_2a\) = log\(_84\) Let log\(_84\) = y Then 4 = 8\(^y\) 2\(^2\) = 2\(^{3y}\) 2 = 3y y = \(\frac{2}{3}\) log\(_2a\) = log\(_84\) = \(\frac{2}{3}\) log\(_2a\) = \(\frac{2}{3}\) a = 2\(^{\frac{2}{3}}\) (Note: The logarithm was converted to index form.)

To find the cost price Selling price = ₦600; Profit = 50% Multiply the selling price by 100 and add 100% to the profit new selling price = 600 x 100 = 60,000, new profit = 150% Cost price = = \(\frac{60,000}{150}\) = ₦400

U\(_n\) = a + (n – 1)d ; where a = 11 ; d = –7 U\(_n\) = 11 + (n – 1)(–7) = 11 – 7n + 7 = 18 – 7n

R = {2, 4, 6, 7}, S = {1, 2, 4, 8} R ∪ S ⇒ element in R and S together without replacement. ∴ R ∪ S = {1, 2, 4, 6, 7, 8}

Using common ratios \(\frac{\text{x}}{\frac{16}{9}}\) = \(\frac{1}{\text{x}}\) x\(^2\) = \(\frac{16}{9}\) x = \(\frac{4}{3}\) If x = \(\frac{4}{3}\) Then, the common ratio = \(\frac{3}{4}\) y = ar\(^3\) = \(\frac{16}{9}\) x (\(\frac{3}{4}\))\(^3\) y = \(\frac{16}{9}\) x \(\frac{27}{64}\) = \(\frac{3}{4}\) Therefore, the product of x and y = \(\frac{4}{3}\) x \(\frac{3}{4}\) = 1