Mathematics 2018 Jamb Examination

First, convert the distance to kilometers: \( \displaystyle 800\text{m} = \frac{800}{1000} = 0.8\text{km} \).
Next, convert the time to hours: \( \displaystyle 20\text{ mins} = \frac{20}{60} = \frac{1}{3}\text{ hour} \).
Average speed \( \displaystyle = \frac{\text{Distance}}{\text{Time}} \): \[ \text{Speed} = \frac{0.8}{1/3} = 0.8 \times 3 = 2.4\text{km/h} \]

First, find the total number of litres N900.00 can buy: \[ \text{Litres} = \frac{900}{63} = \frac{100}{7}\text{ litres} \] Since the car goes 14km per litre, the total distance is: \[ \text{Distance} = \frac{100}{7} \times 14 = 100 \times 2 = 200\text{km} \]

< EHG = < EFH = 54º ( alternate segment)

54º + 54º + < EHF = 180º ( sum of angles in a triangle)

< EHF = 180º - 108º = 72º

 One lap equals the circumference of the track: \[ C = \pi d = \frac{22}{7} \times 100 = 314.28\text{m} \] Number of times \( \displaystyle = \frac{\text{Total Distance}}{\text{Circumference}} \): \[ \text{Laps} = \frac{1000}{314.28} \approx 3.18 \] Correct to the nearest whole number, he will run round 3 times.

Convert to standard form: Numerator: \( (2.56 \times 10^{-3}) \times (6.4 \times 10^{-3}) = 16.384 \times 10^{-6} \) Denominator: \( (2.5 \times 10^{-2}) \times (8 \times 10^{-2}) = 20 \times 10^{-4} = 2 \times 10^{-3} \) Divide: \[ \frac{16.384 \times 10^{-6}}{2 \times 10^{-3}} = 8.192 \times 10^{-6 - (-3)} = 8.192 \times 10^{-3} \]

Let the total shares be \( \displaystyle 11x + 9x = 20x \).
Kehinde's share is \( \displaystyle 9x \). She sells \( \displaystyle \frac{2}{3} \) of her share: \[ \displaystyle \frac{2}{3} \times 9x = 6x \] We are told this \( \displaystyle 6x \) portion is worth N720.00: \[ \displaystyle 6x = 720 \implies x = 120 \] The total value of the store is \( \displaystyle 20x \): \[ \displaystyle 20 \times 120 = N2,400.00 \]

First, find the diagonal of the floor (base diagonal, \( \displaystyle d_f \)): \[ \displaystyle d_f = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{m} \] Next, find the space diagonal (room diagonal, \( \displaystyle D \)): \[ \displaystyle D = \sqrt{d_f^2 + \text{height}^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17\text{m} \] The angle \( \displaystyle \theta \) is between the room diagonal (hypotenuse) and the floor diagonal (adjacent side): \[ \displaystyle \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{15}{17} \]

Find the prime factors: 
\( \displaystyle 48 = 2^4 \times 3 \) 
\( \displaystyle 64 = 2^6 \) 
\( \displaystyle 80 = 2^4 \times 5 \)
HCF: Common factors with lowest power = \( \displaystyle 2^4 = 16 \).
LCM: All factors with highest power = \( \displaystyle 2^6 \times 3 \times 5 = 64 \times 15 = 960 \).
Divide LCM by HCF: \[ \displaystyle \frac{960}{16} = 60 \]

Parallel lines have the same gradient. The line \( \displaystyle 7x + 5y = 12 \) can be written as \( \displaystyle 5y = -7x + 12 \), so the gradient \( \displaystyle m = -\frac{7}{5} \). The required line also has \( \displaystyle m = -\frac{7}{5} \) and passes through (5, 7): \[ \displaystyle y - 7 = -\frac{7}{5}(x - 5) \] \[ \displaystyle 5(y - 7) = -7(x - 5) \] \[ \displaystyle 5y - 35 = -7x + 35 \] \[ \displaystyle 7x + 5y = 70 \]

The salary increases every 6 months. We need to find how many 6-month periods have passed by the 8th month of the 3rd year.
Year 1: 12 months (two 6-month periods)
Year 2: 12 months (two 6-month periods)
Year 3: 8 months (one full 6-month period, plus 2 months into the next)
Total completed 6-month periods of increase = \( \displaystyle 2 + 2 + 1 = 5 \) increases. \[ \displaystyle \text{Salary} = 540 + (5 \times 36) \] \[ \displaystyle \text{Salary} = 540 + 180 = N720.00 \]