GRADE A Maths Assessment A 1 Test 6 Answers Number (a) 90 (b) 82/3 (c) 100-1/2 1 4 Answer 2 √80 in form k√5 4√5 3 (a) Max of xy (b) Min of x÷y 6.05x4.15=25.1075 5.95÷4.15=1.43373494 y = kx3 224 = k x 43 (when x = 4 & y = 224 k = 3.5 Formula: y = 3.5x3 4 (a) Find formula Criteria 1 1 1 10 y is proportional to the cube of x. Mark 1 1 1 1 1 2 3 4 1 1 (b) Calculate x when y = 1792 y = 3.5x3 1792 = 3.5x3 X3 = 512 x=8 B 5 Algebra (a) Factorise: x2 - 5x - 24 (b) Solve: x2 - 5x - 24= 0 6 Solve x2 + 6x - 4 = 0 by fomula 1 Answer (x - 8)(x + 3) (x - 8)(x + 3)=0 x = 8 or x = -3 x= Simplify: 7 x 8 - 5 x-4 Show that the equation: 6 b (b 4ac) 2a 5 = x+2 4 – 3x x–1 can be rearranged to make: 3x + 7x – 13 = 0 2 9 Solve: x2 + y2 = 18 y – 2x = 3 (by substitution) 2 2 7 7(x - 4) - 5x x(x – 4) = 7x - 28 - 5x x(x – 4) = 2x - 28 x(x – 4) 1 1 8 4 – 3x x–1 5 = x+2 Criteria 5 2 2 x = -6 ± √(62 – 4x1 x -4) 2 x = -6 + √52 or -6 - √52 2 2 x = 0.61 or -6.61 (2dp) 7 Mark 2 (4 – 2x)(x + 2) 5(x - 1) = 5x – 5 = 8 – 2x – 3x 2 3x2 + 7x – 13 = 0 Substitute y = 2x + 3 x2 + (2x + 3)2 = 40 1 1 1 9 x2 + 4x2 + 12x + 9 = 18 1 5x2 + 12x - 9 = 0 1 (5x - 3)(x + 3) = 0 x = 0.6 or x = -3 When x = 0.6, y = 4.2 1 1 1 When x = -3, y = -3 10 y = pqx (a)find value of p, q & k 10 y = pqx 6 = p x q1 (when x = 1 and y = 6) 3 24 = p x q (when x = 3 and y = 24) 24 = p x q3 6 p x q1 4 = q2 q = 2 and p = 3 (6 = p x q) y = pqx 1 1 1 k = 3 x 24 (when y = k and x = 4) k = 3 x 16 = 48 11 12 13 C 14 Solve: x2 + y2 = 25 y = 2x - 1 (by graphical method) y = sinxº graph Given cos 60º = √3 2 (i) Sin1200 (ii) Sin2400 Make a the subject 2(3a – c) = 5c + 1 Geometry & measures (a) Scale factor (b) Complete image 1 1 Draw line y = 2x - 1 Points of intersection: x = 2.6 , y = 4.2 x = -1.8 , y = -4.6 11 1 1 12 √3 2 -√3 2 1 1 13 2(3a – c) = 5c + 1 6a – 2c = 5c + 1 6a = 7c + 1 a = 7c + 1 6 1 1 1 Answer Mark 1 -2 Criteria 14 C A B E P Q 1 15 16 Show that triangles ABD and BCD are congruent. Find height of larger tank AB = CD (given) Angle ABD = angle BDC (ALTERNATE angles) BD is common Condition of proof: SAS Volume scale factor = 97.2 ÷ 3.6 = 27 Length scale factor = 3√27 = 3 Height of larger tank = 12cm x 3 = 36cm 36cm2 : 100cm2 15 2 16 1 1 17 Calculate length SR QR = √(52 + 72) = √74 Q 5 R 7 SR = √(132 – √742) = 9.7cm S 19 20 Find volume of wood (a) Area of sector (b) Length of arc (a) Size of angle SOB (b) Reason D 21 Handling data Find number of girls from y9 in sample 22 (a)Complete table Area of cross-section = ½ πr2 = ½ x π x 0.62 = 0.5654866776 Volume = area of cross-section x length = 0.5654866776 x 350 = 198 cm3 Length =120 x πx2 0 360 = ⅓ πx2 Length =120 x 2πx 3600 = 2πx 3 460 Angle between tangent and radus = 900 Angle SOP = 680 Angle BOP = 220 (alternate angles) Angle SOB = 68 – 22 = 460 Answer 167 x 50 1385 ≈6 390 400 17 T Q 13 18 1 √74 2 R 18 1 1 1 19 1 1 1 20 1 Mark 1 1 Criteria 21 1 1 22 (b)complete histogram 1 23 P(3 beads are all same colour) 23 p(BBB or RRR or GGG) 2 2 2 3 3 3 5 5 5 = x x + x x + x x 10 10 10 10 10 10 10 10 10 27 125 8 = + + 1000 1000 1000 160 = 1000 1 1 1