Broadrick EM 2024 Maths Sec Prelim - P2 MS

Broadrick Secondary School4E5N Preliminary Examination 2024Paper 2 Marking Scheme1a B1b = M1A1c = 3750000= M1A12a M1A1b B1c M1 (make linear)M1 (Factorise)A1d M1 (combine)M1 (expand)

= 6.73 or 3.27 (2dp) M1 (quadratic formula using their found eqn)A13a M1 (find m)M1A1b Mid point of BD = (0, ) = (0, 1.5)C is (x, 1.5) and sub into So C is (6, 1.5)M1 (find y value of C)A1cArea = M1A1 (ECF from

=67.5(b))d = 69.6°Or find length of AB, BC and AC and use cosine rule. M1M1A14a20 B1bi When n = 4 and X=2, When n = 5 and X=5, A1A1bii (2)-(1) M1A1A1biii M1

Since number of vertices is an integer, it is possible to have a n-sided polygon with 495 diagonals.M1A15ai (angle at centre = 2x angle at circumference)B1aii (angles in opp seg)B1aiii (base angles of isos triangle) (alt angles, AC//OT) (tan perpendicular to rad) (angle sum of triangle)M1M1A1b They are not angles in opposite segment. A circle cannot be drawn passing through the points A, O, C and D. B1c Minor segment = minor sector – triangle= =12.4 cm2 M1M1A16a-3.33 (2dp)B16bAll points plotted correctlySmooth curveP2C16ciTangent drawn such that it passes through (-1,2) and touches curve at 1 point and gradient is negative.C16cii M1A16d

Draw line on graphx = 2.6 and -1.5(accept 2.5, 2.55) (accept -1.4, -1.45, -1.55, -1.6)M1M1A17aiMedian = 353 gB1aiiUQ: 366 or 367LQ: 342 or 343IQR = = 24g (also accept 23 or 25 g depending on their UQ & LQ)M1A1b20%-->32 applesRead at 128th appleMin mass = 370 gM1A1c = 0.0616M1A1dFalse.The first quartile which represents 25% of the apples are less than 320g.The upper quartile which represents 75% of the apples are less than 360g which implies that 25% of the apples are more than 360gHence there are equal number of apples weighing less than 320g and more than 360g.B1 (states that the whiskers rep 25% of the data)eI disagree. The median mass of apples from tree A is the same as that from tree B. On average, apples from tree A and B weigh the same.IQR (Tree B) = 360 – 320 = 40gHowever, the interquartile range of the masses of the apples from tree B is bigger than that from tree A. The masses of the apples from tree B have a bigger spread, thus are less consistent.B1B1

8a = 17 cmM1A1b = 19.7 cm (3sf)M1A1cLet M be the midpoint of CG and N be the midpoint of BF. Then = 22.97 (4sf) (shown)M1M1M1A1d =19.8° (1dp)If use exact value, angle =19.78165 = 19.8° (1dp)M1A19aCapacity = =69.1150 = 69.1 m3 M1A1bCurved hemisphere = Ring = Slant height of cone, L = Curved area of big cone = Using similar solids, Total area in contact = M1 (Hemisphere or ring)M1 (CSA of big cone using their L)M1 (find ratio of h and then area)M1 (find area of cone in contact)

= 90.0302 = 90.0 m2 A110a B1b M1, A1cItemDescriptionTotal costPrinting of T-shirtsDouble side (Bundle of 500 pcs)$7000 x 2= $14000Goodie BagsNo of packs req = 1000/5 = 200Bulk price (100+ packs)$18 x 200 = $3600Booking of venue6 months