Question Library

(a) A = {m, a, p, e}; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u} (i) \(A \cup B = {a, e, i, m, o, p, u}\) (ii) \(A \cup C = {a, e, l, m, n, o, p, q, r, s, t, u}\) (iii) \(A \cup (B \cap C)\) \(B \cap C = {o, u}\) \(A \cup (B \cap C) = {m, a, p, e, o, u}\) Biology students = n(B) = 190 Chemistry students = n(A) = 300 Let \(n(A \cap B) = a\) (i) From the Venn diagram, \((300 - a) + a + (190 - a) + 70 = 400\) \(560 - a = 400 \implies a = 160\) \(\therefore\) 160 students offer both Biology and Chemistry. (ii) Students offering at least one of Biology and Chemistry = \(n(A) + n(B) + n(A \cap B)\) = \((300 - 160) + (190 - 160) + 160\) = \(140 + 30 + 160\) = 330 students 330 students offer at least one of Biology and Chemistry.

Draw the table of value such that \(-x + 5y \leq 10 \implies -x + 5y = 10\) \(\implies 5y = 10 + x \) \(y = 2 + \frac{x}{5}\) and \(3x - 4y \leq 8 \implies 3x - 4y = 8\) \(y = -2 + \frac{3x}{4}\) Table of value for \(-x + 5y = 10\) x -4 0 4 y 1.2 2 2.8 Table of value for \(3x - 4y = 8\) x -4 0 4 y -5 -2 1

(a) (b) The greatest value of y = 2.25 and this occurs at x = 0.5. (c) Table of values of \(y = 1 - x\) x -2 2 y 3 -1 (d) \(1 + 2x - x^{2} = 0\) \(1 + 1 + 2x - x - x^{2} = 1 - x\) \(2 + x - x^{2} = 1 - x\) The solution of \(1 + 2x - x^{2} = 0\) occurs at the point of intersection of \(y = 2 + x - x^{2}\) and \(y = 1 - x\). This is at points A (x = -0.8) and B (x = 2.8).

(i) The length of the parallel of latitude 46^\circN = \(2 \pi R \cos 46^\circ\) = \(2 \times 3.142 \times 6400 \times 0.6947\) = \(27,939.167 km \approxeq 28000 km\) (to the nearest 100km) (ii) Longitudinal difference between K and Q = 130^\circ - 103^\circ = 27^\circ Distance between K and Q = length of arc KQ = \(\frac{27}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(2,095.31 km \approxeq 2100km\) (to the nearest 100km) (iii) Longitudinal difference between Q and P = 103^\circ + 23^\circ = 126^\circ Distance between Q and P = \(\frac{126}{360} \times 2 \times 3.142 \times 6400 \times \cos 46^\circ\) = \(9778.12 km \approxeq 9780 km\) (to the nearest 10km)

(a) Given any \(\Delta\) ABC, Required: That < A + < B + < C = 180^\circ. Construction : Draw CP // BA and produce BC to Q. Proof: Using the above figure, \(a_{1} = a_{2}\) (alternate angles) \(b_{1} = b_{2}\) (corresponding angles) \(a_{1} + b_{1} + c = 180^\circ\) (angles on a straight line) \(\therefore a_{2} + b_{2} + c = 180^\circ\) \(< A + < B + < C = 180^\circ\) (proved) (b) < LNP = 46^\circ + 80^\circ = 126^\circ (exterior angles of triangle LMN) \(\theta = \frac{126^\circ}{2} = 63^\circ\) (bisector divides into 2 equal halves) \(< MLN = 180^\circ - 80^\circ = 100^\circ\) (angles on a straight line) \(< LQN = 180^\circ - (63^\circ + 100^\circ) = 17^\circ\)

No expanded working has been attached for this question yet. The verified answer is Option .

(a)(i) Marks (%) Freq Cum Freq 0 - 9 10 10 10 - 19 40 50 20 - 29 80 130 30 - 39 140 270 40 - 49 170 440 50 - 59 130 570 60 - 69 100 670 70 - 79 70 740 80 - 89 40 780 90 - 99 20 800 (ii) (iii) From the Ogive, the 50th percentile is 70 marks. (b) Those who scored below 25% (24% and below) = 85 students Those who scored 75% and below = 710 \(\therefore\) Those who scored above 75% = 800 - 710 = 90 students Those to be retained without award = 800 - (85 + 90) = 800 - 175 = 625 students.

\(\frac{x^{2} - y^{2}}{2x^{2} + xy - y^{2}}\) \(2x^{2} + xy - y^{2} = 2x^{2} - xy + 2xy - y^{2}\) = \(x(2x - y) + y(2x - y) \) = \((x + y)(2x - y)\) \(\frac{x^{2} - y^{2}}{2x^{2} + xy - y^{2}} = \frac{(x + y)(x - y)}{(x + y)(2x - y)}\) = \(\frac{x - y}{2x - y}\)

From the diagram, PQRS is a rectangle Area of shaded part = 72sq.cm But 72 = 3h - 4 + 6h - 4 + 4h = 72 - 16h - 8 = 16h - 72 + 8 =16h = 80 h = \(\frac{80}{16}\) = 5cm

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o 50 + 20 + 1 = 7110 Convert 7110 to base 8 \(\begin{array}{c|c} 8 & 71 \\ 8 & 8 R 7\\8 & 1 R 0\\8 & 0 R 1\end{array}\) = 1078

The least length is \(\frac{40}{48}\) = \(\frac{5}{8}\) for the rod to be cut in exactly equal trips Ratio \(\frac{5}{6}\) : \(\frac{48}{40}\) \(\frac{\frac{5}{6}}{\frac{40}{48}}\) = 1 \(\frac{5}{6}\) x \(\frac{48}{40}\) = \(\frac{240}{240}\) = 1 The least length = 240cm

1m = 1.1 yard, length(L)= 50m = (50 x 1.1)yards = 55 yards Area(A) = length(L) y width (W) 1815 = 55y width (W) width (w) = \(\frac{1815}{55}\) = 33 yards But 33 yards = 30 meters

\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.022 \times 1.2}{0.024}\) (to 216) = \(\frac{0.0264}{0.024}\) = 1.1

Let the distance from P to Q = d km (We don't need the actual value of d — it will cancel out.) Time taken from P to Q = \(\frac{\text{d}}{90}\) hours Time taken from Q to P = \(\frac{\text{d}}{45}\) hours Total distance travelled = d + d = 2d km Total time taken = \(\frac{\text{d}}{90}\) + \(\frac{\text{d}}{45}\) = \(\frac{\text{d}}{90}\) + \(\frac{2d}{90}\) = \(\frac{3d}{90}\) = \(\frac{\text{d}}{30}\) hours Average speed = \(\frac{\text{total distance}}{\text{total time}}\) = 2d \div (\(\frac{\text{d}}{30}\)) = 2d \times (\(\frac{30}{\text{d}}\)) = 60 km/h

Let the length of the square be \( L \). The area of the square is: \(A_s = L^2\) After modifications, the new length \( L_r \) and width \( W_r \) of the rectangle are: \(L_r = 1.2L, \quad W_r = 0.8L\) The area of the rectangle \( A_r \) is: \(A_r = L_r \times W_r = (1.2L)(0.8L) = 0.96L^2\) The ratio of the areas is: \(\text{Ratio} = \frac{A_r}{A_s} = \frac{0.96L^2}{L^2} = 0.96\) Thus, the ratio of the area of the rectangle to the area of the square is \( 0.96 \). Or \(\frac{96}{100}\) = 24:25