1. (a) Complete the table for $y=\frac{1}{2}x - 2$. Calculate each value of $y$ for given $x$: For $x = -6$: $y = \frac{1}{2} \times (-6) - 2 = -3 - 2 = -5$ For $x = -4$: $y = \frac{1}{2} \times (-4) - 2 = -2 - 2 = -4$ For $x = 0$: $y = \frac{1}{2} \times 0 - 2 = 0 - 2 = -2$ For $x = 4$: $y = \frac{1}{2} \times 4 - 2 = 2 - 2 = 0$ For $x = 8$: $y = \frac{1}{2} \times 8 - 2 = 4 - 2 = 2$ Table: | x | -6 | -4 | 0 | 4 | 8 | |---|----|----|---|---|---| | y | -5 | -4 | -2 | 0 | 2 | (b) Graph $y=\frac{1}{2}x - 2$: plot points $(-6, -5), (-4, -4), (0, -2), (4, 0), (8, 2)$ and connect with a straight line. It’s a straight line with slope $\frac{1}{2}$ and y-intercept at $-2$. (c) Coordinate of point A is given as $(-6, -5)$ from the graph. (d) A line parallel to y-axis through point A means a vertical line at $x=-6$. 2. (a) Calculate $1 \frac{3}{5} \times \frac{8}{15} \div (3 \frac{2}{3} - 1 \frac{1}{2})$. Convert mixed numbers to improper fractions: $1 \frac{3}{5} = \frac{8}{5}$ $3 \frac{2}{3} = \frac{11}{3}$ $1 \frac{1}{2} = \frac{3}{2}$ Calculate denominator inside parentheses: $\frac{11}{3} - \frac{3}{2} = \frac{22}{6} - \frac{9}{6} = \frac{13}{6}$ Now compute expression: $\frac{8}{5} \times \frac{8}{15} \div \frac{13}{6} = \frac{8}{5} \times \frac{8}{15} \times \frac{6}{13}$ Multiply numerators and denominators: $= \frac{8 \times 8 \times 6}{5 \times 15 \times 13} = \frac{384}{975}$ Simplify by dividing numerator and denominator by 3: $\frac{128}{325}$ Answer (a): $\frac{128}{325}$ (b) Calculate $\frac{4.99 + 3.85}{1.88 \times 0.06}$ round to 3 sig. fig. Sum numerator: $4.99 + 3.85 = 8.84$ Product denominator: $1.88 \times 0.06 = 0.1128$ Divide: $\frac{8.84}{0.1128} \approx 78.36$ Rounded to 3 significant figures: $78.4$ Answer (b): $78.4$ (c) Calculate $\frac{2.7883^2 \times \sqrt{4.15}}{\frac{15}{23} + \frac{3}{7}}$ to 2 decimals. Calculate numerator: $2.7883^2 \approx 7.777$ $\sqrt{4.15} \approx 2.037$ Multiply: $7.777 \times 2.037 \approx 15.84$ Calculate denominator: $\frac{15}{23} + \frac{3}{7} = \frac{15 \times 7}{23 \times 7} + \frac{3 \times 23}{7 \times 23} = \frac{105}{161} + \frac{69}{161} = \frac{174}{161} \approx 1.080$ Divide: $\frac{15.84}{1.080} \approx 14.67$ Answer (c): $14.67$ 3. (a) Expand and simplify $4(x + 2y) - 3(x - 3y)$. Distribute: $4x + 8y - 3x + 9y = (4x - 3x) + (8y + 9y) = x + 17y$ Answer (a): $x + 17y$ (b) Simplify $\frac{5x^5 y}{25 x^2 y^3}$ Divide coefficients: $\frac{5}{25} = \frac{1}{5}$ Subtract powers: $x^{5-2} = x^3$ $y^{1 - 3} = y^{-2} = \frac{1}{y^2}$ Result: $\frac{1}{5} x^3 \times \frac{1}{y^2} = \frac{x^3}{5 y^2}$ Answer (b): $\frac{x^3}{5 y^2}$ 4. Calculate number of blue balls added. Initial ratio red:blue = $8:5$ After adding $b$ blue balls, new ratio $8 : (5 + b) = 2 : 7$ Cross multiply: $8 \times 7 = 2 \times (5 + b)$ $56 = 10 + 2b$ $2b = 46$ $b = 23$ Answer: $23$ 5. Probability that Farah is early is complement of being late. Given $P(late) = 0.4$, so $P(early) = 1 - 0.4 = 0.6$ Answer (a): $0.6$ 6. Tree diagram probabilities: Sum probabilities of weather: $0.3 + P(sunny) = 1$, so $P(sunny) = 1 - 0.3 = 0.7$ Walk probabilities given: $P(walk|rainy) = 0.1$, $P(walk|sunny) = 0.8$ Find $P(by car)$ for each weather: $P(by car|rainy) = 1 - 0.1 = 0.9$ $P(by car|sunny) = 1 - 0.8 = 0.2$ 7. Celsius to Fahrenheit and vice versa (using $F = \frac{9}{5}C + 32$). (a)(i) Convert $10^\circ C$ to Fahrenheit: $F = \frac{9}{5} \times 10 + 32 = 18 + 32 = 50^\circ F$ Answer (a)(i): $50$ (a)(ii) Convert $130^\circ F$ to Celsius: $C = \frac{5}{9} (F - 32) = \frac{5}{9} (130 - 32) = \frac{5}{9} \times 98 = \frac{490}{9} \approx 54.44^\circ C$ Answer (a)(ii): $54.44$ (b) Freezing point $0^\circ C$ in Fahrenheit: $F = \frac{9}{5} \times 0 + 32 = 32^\circ F$ Answer (b): $32$ 8. Bus travel time: Given time with jam is 150 minutes, time increased by 20% means actual time without jam $t$ satisfies: $t + 0.2t = 1.2t = 150 \Rightarrow t = \frac{150}{1.2} = 125$ minutes. (a) Bus reached at 10:25 am, so leave time = arrival time minus travel time: 10:25 am minus 150 minutes = 10:25 - 2 hours 30 min = 7:55 am Answer (a): 7:55 am (b) Travel time without jam: 125 minutes Answer (b): 125 9. Candle volumes: Volume of candle A = length × width × height = $18 \times 8 \times 10 = 1440$ cm³ Volume of candle B = $\frac{2}{5} \times 1440 = 576$ cm³ Given base area of candle B = 48 cm², Height $h = \frac{Volume}{Base area} = \frac{576}{48} = 12$ cm Answer (a): $12$