Mathematics 2019 Jamb Examination

Perform the calculation: \[ \displaystyle \frac{0.5353659}{5.44} \approx 0.09841 \] Significant figures start at the first non-zero digit (9). The second is 8. Since the next digit is 4 (less than 5), we do not round up. \[ \displaystyle 0.098 \]

From the first equation, \( \displaystyle y = 21 - 3x \). Substitute this into the second equation: \[ \displaystyle x(21 - 3x) = 30 \implies 21x - 3x^2 = 30 \] Rearrange into a quadratic equation: \( \displaystyle 3x^2 - 21x + 30 = 0 \). Divide by 3: \[ \displaystyle x^2 - 7x + 10 = 0 \implies (x - 2)(x - 5) = 0 \] If \( \displaystyle x = 2, y = 30/2 = 15 \). If \( \displaystyle x = 5, y = 30/5 = 6 \).

Point X is at (0, 20) and Y is at (9, 0) relative to origin O. To find the bearing of Y from X, we look at the angle from the North line at X. In the right triangle at X: \[ \displaystyle \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{9}{20} = 0.45 \] \[ \displaystyle \theta = \tan^{-1}(0.45) \approx 24.2^\circ \] The bearing (clockwise from North) is \( \displaystyle 180^\circ - 24.2^\circ = 155.8^\circ \approx 156^\circ \).

Cube both sides: \( \displaystyle P^3 = \frac{Q(R - T)}{15} \)
Multiply by 15: \( \displaystyle 15P^3 = Q(R - T) \)
Divide by Q: \( \displaystyle \frac{15P^3}{Q} = R - T \)
Rearrange for T: \( \displaystyle T = R - \frac{15P^3}{Q} \)

First, find the number of sides (n). The exterior angle is \( \displaystyle 180^\circ - 140^\circ = 40^\circ \). \[ \displaystyle n = \frac{360}{40} = 9 \text{ sides} \] Sum of interior angles = \( \displaystyle (n - 2) \times 180^\circ \): \[ \displaystyle (9 - 2) \times 180 = 7 \times 180 = 1260^\circ \]

Since the selling price is less than the cost price, it is a loss. Loss = \( \displaystyle 1,250,000 - 1,085,000 = 165,000 \). \[ \displaystyle \%\text{Loss} = \frac{165,000}{1,250,000} \times 100 = 13.2\% \]

Differentiate the expression with respect to R, treating \( \displaystyle \pi, h, \) and \( \displaystyle r \) as constants: \[ \displaystyle \frac{d}{dR}(R^2) = 2R, \quad \frac{d}{dR}(Rr) = r, \quad \frac{d}{dR}(r^2) = 0 \] \[ \displaystyle \frac{dV}{dR} = \frac{\pi h}{3}(2R + r) \]

Perform the division: \( \displaystyle 0.0439 / 3.62 \approx 0.01212... \) The closest standard fraction provided in this exam context for this decimal range is \( \displaystyle 12/1000 \), which is \( \displaystyle 0.012 \).

Convert everything to base 5: \[ \displaystyle (5^2)^{1-x} \times 5^{x+2} \div (5^{-3})^x = (5^4)^{-1} \] \[ \displaystyle 5^{2-2x} \times 5^{x+2} \div 5^{-3x} = 5^{-4} \] Using index laws: \( \displaystyle 2 - 2x + x + 2 - (-3x) = -4 \) \[ \displaystyle 2x + 4 = -4 \implies 2x = -8 \implies x = -4 \]

In triangle ABO, OA = OB (radii). Thus, \( \angle OAB = \angle ABO = 26^\circ \). The interior angle \( \angle AOB = 180 - (26 + 26) = 128^\circ \). Angles at a point sum to \( 360^\circ \): \[ \angle AOC = 360 - (\angle AOB + \angle BOC) \] \[ \angle AOC = 360 - (128 + 130) = 360 - 258 = 102^\circ \]