Mathematics 2019 Jamb Examination

The relationship is \( \displaystyle P = \frac{k}{\sqrt{q}} \). First, find k: \( \displaystyle 3 = \frac{k}{\sqrt{16}} \implies 3 = \frac{k}{4} \implies k = 12 \). Now find q when p = 4: \[ \displaystyle 4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4} = 3 \] \[ \displaystyle q = 3^2 = 9 \]

Cost Price (CP): \( \displaystyle \frac{200}{4} \times 2.50 = 50 \times 2.50 = 125.00 \). Remaining mangoes: \( \displaystyle 200 - 30 = 170 \). Selling Price (SP): \( \displaystyle \frac{170}{2} \times 2.40 = 85 \times 2.40 = 204.00 \). Profit: \( \displaystyle 204 - 125 = 79 \). \[ \displaystyle \%\text{Profit} = \frac{79}{125} \times 100 = 63.2\% \]

Using \( \displaystyle I = \frac{PRT}{100} \): \[ \displaystyle 4890 = \frac{8550 \times x \times 3}{100} \] \[ \displaystyle 489000 = 25650x \] \[ \displaystyle x = \frac{489000}{25650} \approx 19.06 \] To the nearest whole number, \( \displaystyle x = 19\% \).

Convert to base form: \( \displaystyle 81^{\frac{-3}{4}} = (3^4)^{\frac{-3}{4}} = 3^{-3} = \frac{1}{27} \)
\( \displaystyle 25^{\frac{1}{2}} = \sqrt{25} = 5 \)
\( \displaystyle 243^{\frac{2}{5}} = (3^5)^{\frac{2}{5}} = 3^2 = 9 \) \[ \displaystyle \text{Expression} = \frac{1}{27} \times 5 \times 9 = \frac{45}{27} = \frac{5}{3} \]

Numerator: \( \displaystyle (0.5436)^3 \approx 0.1606 \). Denominator: \( \displaystyle 0.017 \times 0.219 = 0.003723 \). \[ \displaystyle \text{Result} = \frac{0.1606}{0.003723} \approx 43.137 \] To 3 significant figures, the value is 43.1.

First, expand the expression for \( \displaystyle S \): \[ \displaystyle S = 4t^2 - 8t + 3t - 6 = 4t^2 - 5t - 6 \] Now, differentiate with respect to \( \displaystyle t \): \[ \displaystyle \frac{dS}{dt} = 8t - 5 \] Substitute \( \displaystyle t = 5 \): \[ \displaystyle \frac{dS}{dt} = 8(5) - 5 = 40 - 5 = 35 \text{ units per sec} \]

A polygon with 4 angles is a quadrilateral. The sum of interior angles of a quadrilateral is \( \displaystyle (n-2) \times 180^\circ \), where \( \displaystyle n=4 \): \[ \displaystyle 2x + 5x + x + 4x = (4-2) \times 180 = 360^\circ \] \[ \displaystyle 12x = 360 \implies x = 30^\circ \]

Percentage error is calculated as \( \displaystyle \frac{|\text{Error}|}{\text{Actual value}} \times 100 \): \[ \displaystyle \text{Error} = 0.21 - 0.18 = 0.03 \] \[ \displaystyle \%\text{Error} = \frac{0.03}{0.18} \times 100 = \frac{1}{6} \times 100 \approx 16.66... \approx 16.7\% \]

Solve the expression inside the bracket first: \[ \displaystyle \frac{6}{0.32} \times \frac{0.084}{2} = \frac{0.504}{0.64} = 0.7875 \] Now, find the inverse (to the power of -1): \[ \displaystyle (0.7875)^{-1} = \frac{1}{0.7875} \approx 1.269 \] To 1 decimal place, the value is 1.3.

Convert both equations to base 2:
1. \( \displaystyle 2^{x+y} = 2^4 \implies x + y = 4 \)
2. \( \displaystyle (2^2)^{x-y} = 2^{-5} \implies 2x - 2y = -5 \)
Solving simultaneously: From (1), \( \displaystyle y = 4 - x \). Substitute into (2): \[ \displaystyle 2x - 2(4 - x) = -5 \implies 2x - 8 + 2x = -5 \] \[ \displaystyle 4x = 3 \implies x = \frac{3}{4} \] Substitute \( \displaystyle x \) into (1): \( \displaystyle y = 4 - \frac{3}{4} = \frac{13}{4} \).