Mathematics 2018 Jamb Examination

Significant figures start from the first non-zero digit. Here, the first significant digit is 4 and the second is 9. The next digit is 4, which is less than 5, so we round down. \[ 0.04945 \approx 0.049 \]

First, find the probability of each failing: \[ P(\text{John fails}) = 1 - \frac{3}{4} = \frac{1}{4} \] \[ P(\text{James fails}) = 1 - \frac{3}{5} = \frac{2}{5} \] Since the events are independent, multiply the probabilities: \[ P(\text{Both fail}) = \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10} \]

The perimeter of a sector is given by the formula: \( P = \text{Arc Length} + 2r \). \[ \text{Arc Length} = \frac{\theta}{360} \times 2\pi r = \frac{300}{360} \times 2 \times \frac{22}{7} \times 14 \] \[ = \frac{5}{6} \times 2 \times 22 \times 2 = \frac{5}{6} \times 88 = 73.33\text{cm} \] \[ \text{Perimeter} = 73.33 + 2(14) = 73.33 + 28 = 101.33\text{cm} \]

Using the laws of indices: Numerator: \( 25^{\frac{2}{3} - \frac{1}{6}} = 25^{\frac{4-1}{6}} = 25^{\frac{3}{6}} = 25^{\frac{1}{2}} = 5 \)
Denominator: \( (\frac{1}{5})^{\frac{7}{6} - \frac{1}{6}} = (\frac{1}{5})^{\frac{6}{6}} = (\frac{1}{5})^1 = \frac{1}{5} \) \[ \text{Expression} = \frac{5}{1/5} = 5 \times 5 = 25 \]

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: \( \displaystyle T = R - \frac{15P^3}{Q} \)

In the number 3.0492, the digits after the decimal point represent tenths, hundredths, thousandths, and ten-thousandths respectively. Since 9 is the third digit after the decimal, its value is: \[ \displaystyle \frac{9}{1000} \]

Using the formula for Simple Interest \( \displaystyle I = \frac{PRT}{100} \), we substitute the known values: \[ \displaystyle 123 = \frac{P \times 3 \times 2.5}{100} \] Cross-multiplying gives: \[ \displaystyle 12300 = 7.5P \implies P = \frac{12300}{7.5} = 1640 \]

Depreciation is calculated using the formula \( \displaystyle V = P(1 - R)^n \): Year 1: \( \displaystyle 20,000 - (10\% \text{ of } 20,000) = 18,000 \) Year 2: \( \displaystyle 18,000 - (10\% \text{ of } 18,000) = 16,200 \) \[ \displaystyle V = 20,000(0.9)^2 = 16,200 \]

First, find the total number of students \( \displaystyle n = \sum f = 22 \). The median is the average of the 11th and 12th positions. Cumulative frequencies are: 1, 3, 10, 15, 19, 22. Since the 11th and 12th students fall into the group that scored 3, the median is 3.

Total number of students \( \displaystyle = 22 \). Number of students who scored 2 or 3 is \( \displaystyle 7 + 5 = 12 \). \[ \displaystyle P(\text{2 or 3}) = \frac{12}{22} = \frac{6}{11} \]