TKGS 2024 S4 Math Prelim P2_Solutions

2024 TKGS PRELIM MATH P2 MARK SCHEME Qn Solution Content/Success Criteria 1 (a) 3 2 1 (2 1)73 x x− =− 3(3 2 ) 7(2 1)xx− = − 9 6 14 7xx− = − 20 16x= 40.8 / 5x = I can multiply LCM to remove the denominators on both sides of a linear equation involving fractions. I can solve a linear equation. Content L Complexity L Context L Response Strategy Simple Assessment Objective AO1 (b) (i) 22 2 25 9 15 10 6 9 ab a ab b b − − + − 22(5 ) (3 ) 5 (3 2 ) 3 (2 3) ab a b b b −= − + − (5 3 )(5 3 ) 5 (3 2 ) 3 (3 2 ) a b a b a b b b −+= − − − (5 3 )(5 3 ) (5 3 )(3 2 ) a b a b a b b −+= −− 53 32 ab b += − I can factorise algebraic expressions by identity and grouping. Content L Complexity L Context L Response Strategy Simple Assessment Objective AO1 (ii) 3 2 3 4 12 cc ab ab 2 3 3 12 4 c a b ab c= 2 3 36 4 a bc abc= 2 9a c= I can take reciprocal to convert division to multiplication. I can apply laws of indices to simplify algebraic expressions. Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1

4052/S4Prelim/P2/2024 Qn Solution Content/Success Criteria (c) (i) 2 790 2xx− + − = 2 790 2xx− + = 22 9 7 9 02 2 2x   − + − − =       2 9 67 024x − − = 2 9 67 24x −= 67 9 42x =+ 8.59268 or 0.407324x= 8.59 or 0.41x= I can complete the square. I can solve equation using complete the square method Content L Complexity L Context L Response Strategy Routine Assessment Objective AO1 (ii) 22 7 679 ( 4.5) 16.75 /24x x x− + − = − − + Since the maximum value of 2 79 2y x x= − + − is 16.75/ 67 4 < 18. There is no solution when 18.y = I know that 2 79 2y x x= − + − has max. value. I can find max. value with complete square form of quadratic equation. Content L Complexity M Context M Response Strategy Routine Assessment Objective AO3 OR 2 2 79 18 2 4390 2 xx xx − + − = − + − = Discriminant 2 43(9) 4( 1) 2 = − − −  5=− Since discriminant < 0, 2 79 18 2xx− + − = has no solution.

Qn Solution Content/Success Criteria 2 (a) P(man who passed test) 1 3 3 7 3 21 ==  the number of men who passed the test 3= Let the number of women who failed the test be x P(both women failed the test) 11 21 20 10 xx −=  = 2 420 10xx−= 2 42 0xx− − = ( 7)( 6) 0xx− + = 7 or 6 (rejected)x=− Women who passed the test 21 9 7 3= − − − 2= Passed Failed Men 3 9 Women 2 7 I can find the probability of a single event. I can use multiplication of probabilities for simple combined events. Content M Complexity M Context M Response Strategy Unfamiliar Assessment Objecti