PHSS 2022 P1 SA1 Marking Scheme

2022 MYE Math S4E5N P1 Marking Scheme Question Answer Marks Guidance 1a 5 12 3 xx=+ 5 12 3 xx−= 14 123 x= 18 4 or 277x= A1 A1 for correct working and answer. Answer corrected to 3 significant figures is not accepted. 1b ( )7 2 3 2m n n m− − − = 7 2 3 6m n n m− − + = 13 5mn− M1 A1 M1 for expanding ( )32nm−− . Question Answer Marks Guidance 2 ( )533 26 xx −− ( )9 5 3 6 xx−−= 9 5 15 6 xx−+= 4 15 6 x+= M1 A1 For combining fractions into single fraction with common denominator correctly Question Answer Marks Guidance 3 210 5 2x y xy x− + − = 210 5 2x xy x y+ − − = ( ) ( )5 2 2x x y x y+ − + = ( )( )2 5 1x y x+− M1 A1 Other accepted answer: ( )( )2 2 1 5x y x− − −

Question Answer Marks Guidance 4 26 11 10 0xx− − = ( )( )2 5 3 2 0xx− + = x = 2 3− or 5 2 M2 A1 M1 for each factor. A1 for both answers. Answer corrected to 3 significant figures is not accepted. No mark if student did not use factorisation method. Question Answer Marks Guidance 5a 300 children – 360° x = 360 60300 = 72 B1 5b Sector angle for mangoes = 360 84 102 72− − −  = 102° The number of children who like rambutans to those who like mangoes = 72 : 102 = 12 : 17 M1 A1 5c Percentage of children who like durians = 102 100%360 = 28.3% A1 A1 for correct working with percentage symbol and answer in decimal. Question Answer Marks Guidance 6a 2198 2 3 11=   B1 6b HCF = 22 3 18= B1 Question Answer Marks Guidance 7 p = 1 q = 400 B2 Must correct for both p and q to get B2. Question Answer Marks Guidance 8 Let the monthly payment be $x. M1

20% 1455 12 1587x + = 291 12 1587x+= 108x= The monthly payment is $108. A1 M1 for finding the deposit 20% 1455 $291= Question Answer Marks Guidance 9a k = 3 m = 5 B1 B1 9b 1 B1 Question Answer Marks Guidance 10 Price of cap without use of voucher discount = 15+7 = $22 Original price = 100 1222 100 − = 22 0.88 = $25 M1 A1 No A1 if student show this working: 22 100% 2588= Question Answer Marks Guidance 11a F ma= 5 20= 100N= A1 A1 for working and answer. 5.00 20.00 is not accepted 11b Evidence: 5.27 and 24.92 are rounded to 1 significant figure. Concept: Both numbers are rounded down to smaller values for estimation. Conclusion: Hence, the estimated force is underestimated. B1 B1 for concept and conclusion. Question Answer Marks Guidance

12 3 23 54 x x   = 3 5 2 34x x    = 533 322 3 2 4 x x  = 5 2 3 2 8x x = 8x M1 M1 A1 Write 3 5x in index form. Simplify coefficient in the numerator and simplifying power of x in the numerator.

Question Answer Marks Guidance 13a Let n be the number of sides of regular polygon P. 2n be the number of sides of regular polygon Q. Interior angle of polygon P = ( )180 2 n n − Interior angle of polygon Q = ( )180 2 2 2 n n − ( ) ( ) 180 2 7 180 2 2 8 2 n n n n − =− ( ) ( )180 2 180 2 2 7 28 nn nn  −  −= ( ) ( ) 180 2 27 180 2 2 8 n n nn − = − ( ) ( ) 2 27 1 2 2 8 n n − = − ( ) ( ) 2 27 1 2 1 8 n n − = − ( ) ( ) 2 7 18 n n − =− ( ) ( )7 1 8 2nn− = − 7 7 8 16nn− = − 9n= Polygon P is 9-sided. M1 M1 A1 Able to apply ratio equivalence by forming an equation. Simplifying the equation to algebraic fractions. Guess and check is not accepted. 13b Interior angle of polygon Q = ( )180 18 2 18 − = 160° Exterior angle of polygon Q = 180°−160° = 20° M1 A1 Or Exterior angle of polygon Q = 360 29

Question Answer Marks Guidance 14 Volume of cone B = ( ) ( ) 21 1.5 63 rh = ( ) ( ) 221 1.5 63 rh  = 21 27 32 rh  = 274 2 = 54 cm3 M1 A1 Question Answer Marks Guidance 15 2V kr= where k – constant When r2 = 1.2r 2 22V kr= = ( ) 2 1.2kr = 21.44kr Percentage increase in volume = 21 1 100%VV V −  = 22 2 1.44 100%kr kr kr −  = 44% M1 A1 For ( ) 2 1.2kr No A1 if student shows this working: 22 2 1.44 100 44%kr kr kr − = Finding the answer by substituting values is not accepted

Question Answer Marks Guidance 16a  B1 B1 B1 For elements in set A ∩ B′ = {14, 15, 16}. For elements in set (A U B )′ = {18, 20, 21, 22}. For elements in set (A ∩ B ) = {13, 17} and A′ ∩ B = {19, 23}. 16b A′ ∩ B′ = {18, 20, 21, 22} √A1 With curly brackets. 16c (A ∩ B′ ) U (A′ ∩ B ) = {14, 15, 16, 19, 23} = 5 √A1 A B 18 14 13 19 20 15 17 23 21 16 22

Question Answer Marks Guidance 17a Height P to QR = 22 4k − cm 2 16sin kPRQ k −= M1 A1 Must show QR = 22 4k − (need not state Pythagoras Theorem) 17b Let h cm be the height of the triangles. Area of PRQ = 1 2 kh cm2 Area of QSP = ( )1 0.752 kh 30 = 10.75 2 kh 10.75 2 kh = 30 1 2 kh = 30 0.75 Area of PRQ = 1 2 kh = 40 cm2 M1 A1

Question Answer Marks Guidance 18a 2 10 15xx−+ = 22 2 10 1010 15 22xx    − + − +       = 22 10 5 25 15xx− + − + = ( ) 2 5 10x−− M1 A1 18b ( ) 2 5 10 0x− − = ( ) 2 5 10x−= 5 10x− = x = 1.8837722 or 8.16227766 = 1.84 or 8.16 M1 A1 A1 for both correct answers. 18c B1 B1 B1 Correct shape and symmetrical. Correct intercepts indicated. (must be intercept values, not coordinates) Correct turning point indicated. (can be coordinates or indicate the axis values) y x 1.84 8.16 15

Question Answer Marks Guidance 19a P = 25423 52 5 1 5 100 2 2 50x       = 200 289 110 10 x    + B1 B1 For first two elements. For the third elements in terms of x. 19b Amount of money Patrick, Quincy and Rachel save, respectively. B1 19c 110 10 200x+= 9x= B1 19d Total amount that Patrick, Quincy and Rachel save in total = ( ) 200 1 1 1 289 200    = ( )689 Or Total amount that Patrick, Quincy and Rachel save in total = ( ) 2 57 11 12 10 10 50       = ( )689 M1 A1