solutions 2023 Beatty Sec 4NA Math P1

1 Beatty 4NA Math Prelim 2023 P1 Marking scheme Qn Solutions Marks allocation 1 3 5 √14 6 ߨ 5 17 25 B2 – all correct B1 – 1mistake B0 - >1 mistake 2 250 189 100%250 24.4%    M1 A1 3a 250 : 3000 1 : 12 B1 3b ݔ+ ݕ ݏݐ݅݊ݑ →݇ 1 ݐ݅݊ݑ→ ݇ ݔ+ ݕ ݔ ݏݐ݅݊ݑ→ ݔ݇ ݔ+ ݕ B1 4a sinݔ= 0.813 ݔ= ݊݅ݏିଵ(0.813) ݔ= 54.39° ݎ݋ ݔ= 180° − 54.39° = 125.6° (1 dp) Answer : 125.6° B1 4b 124.8° B1 5 ݕ= ݇ √ݔ 1.5 =݇ √16 ݇= 1.5 × √16 ݇= 6 ݕ= 6 √ݔ M1 A1 6 ට൫3 − (−2)൯ ଶ + (−1 − 5)ଶ =7.81 units (3sf) M1 A1 63%

2 7 B2 – correct figure B1 – one pair of sides drawn correctly 8a 6 15 6 15 2.5 y y y     Largest integer value = −3 M1 A1 8b 2 2 (3 5)(7 2) 21 6 35 10 21 29 10 x x x x x x x          M1 A1 9a 180° − 18° − 18° (angle sum of triangle) = 144° B2 9b (݊− 2) × 180 ݊= 144 180݊− 360 = 144݊ 180݊− 144݊= 360 36݊= 360 ݊= 10 M1 A1

3 Or One exterior angle 180° − 144° = 36° Number of sides = ଷ଺଴° ଷ଺° =10 M1 A1 10a   16 3 1 16 3 2 8 3 11 11 n n n n a a a a a a a a a a a n         M1 A1 10b 0 1 0 3 3 3 1 3(3 ) 3 3 3 1 0 1 x x x x x x x          M1 A1 11b 15 8 B1 12a 2 26 15 3 (2 5 ) x y xy xy x y    B1 12b 2 1 4 9 3 1 4 ( 3)( 3) 3 1 4( 3) ( 3)( 3) 1 4 12 ( 3)( 3) 4 11 ( 3)( 3) x x x x x x x x x x x x x x                   M1(for correct numerator) A1 13a −2 B1 13b −4n + 26 B1

4 13c 8 2 278 8 278 2 8 280 35 n n n n       Since n is a whole number, 278 is a term in the sequence. M1 A1 14 Reflex angle AOC 360 150 210      Perimeter of major sector = 210 2 (6)360  +6+6 =7π + 12 M1 – addition of two radius M1 – Arc length A1 15 120 180 120 60 180 60 50 70 EDB BDC BCD                 Since angle BDC ≠ angle BCD, triangle BCD is not an isosceles triangle. M1 M1 A1 16a 3 1 6 2 B1 16b 2 1 6 3 B1 16c 5 6 B1 17a     2 2 2 2 8 6 4 4 6 ( 4) 10 x x x x           4, 10a b  B2 B1 for one correct or (x 4)2 + k (opposite angles of //gram (Adjacent angles on a straight line) (angle sum of triangle)

5 17b 2( 4) 10 4 10 10 4 7.16(2 ) x x x x dp        M1 (allow ft, but b must be < 0) B1 (c.a.o) 18a ( 4,0) B1 18b (0 4)(0 2) 8 y y     Answer : (0,8) B1 18c 4 2 2 1 x x    B1 18d When x = −1, ( 1 4)( 1 2) 9 y y       Answer : ( 1,9) B1 19a Gradient 4 0 2 2 1     Equation of PR is 2y x  . (shown) M1 (finding gradient) A1 (conclusion must be seen. working to find y-intercept is not necessary). 19b 4 10 10 4 0.84(2 ) x x x dp      or R P

6 Draw the line 2 1y x  : Plotting of at least 2 points within the graph : B1 Drawing a straight line passing through the points : B1 Answer : x = 1, y = 1. (B1) -> answer must be seen on graph. x 0 1 2 y −1 0 3 20a Number of minutes (t) Frequency 0 ≤ m < 10 5 10 ≤ m < 20 13 20 ≤ m < 30 20 30 ≤ m < 40 30 40 ≤ m < 50 12 Total : 80 students Answer : 30 ≤ m < 40 20b Mean = 5 × 5 + 15 × 13 + 25 × 20 + 35 × 30 + 45 × 12 80 = 28.875 M1 A1 20c 30 + 12 80 × 100% = 42 80 × 100% =52.5% M1 A1 20d No, I do not agree. Histogram shows the comparison of data more clearly than pie chart. B1 Accept other equivalent answers 21a Area of square = x2 Perimeter of equilateral triangle = 117 – 4x Therefore, ݔଶ = 117 − 4ݔ ݔଶ + 4ݔ− 117 = 0(ݏℎ݊ݓ݋) M1 - Attempt to find area of square and equate it to perimeter of equilateral triangle must be seen A1

7 21b 2 4 117 0 ( 9)( 13) 0 9 0 9 13 0 13 x x x x x x or x x             M1 (factorisation must be correct, with eqn = 0) A2 (both answers are correct) 21c Area of square = perimter of equilateral triangle Therefore, length of one side of equilateral triangle 9 9 3 27cm   Or 117 4(9) 3 27   B1 22 Construction 23a 2 2 2 1 ( )(12) 1003 12 300 300 12 300 12 5 r r r r r        M1 (forming correct eqn, with h = 12 ) M1 (square root) A1 23b 2 212 5 13 AB AB cm    M1 (allow ecf) A1

8 23c 2 2 (5) (5)(13) 282.74 283 (3 )cm sf      M1 (1 mark awarded for finding curved surface area, allow ecf) A1