TKGS 2024 S4 Math Prelim P1 Solution

2024 TKGS PRELIM MATH P1 SOLUTION Qn Solution Content/Success Criteria 1 (a) 8700 I can perform calculations with a calculator. I can round off values to the nearest hundred. (b) 1050 Content L Complexity L Context L Response Strategy L Assessment Objective AO1 Qn Solution Content/Success Criteria 2 (a) 5 2(6 1)x−− I can expand and simplify algebraic expressions. I can apply simplify expression using laws of indices. 5 12 2x= − + 7 12 x=− Content L Complexity L Context L (b) 11 96 33 69 27 27 ba ab −     =       Response Strategy L 2 33 a b= Assessment Objective AO1

Tanjong Katong Girls’ School [Turn over 4052/S4Prelim/01/2024 Qn Solution Content/Success Criteria 3 Let amount of savings Albert and Chris have initially be 5x and 3x respectively. 5 30 2 3 30 1 5 30 2(3 30) 5 30 6 60 30 3 90 x x xx xx x x − =− − = − − = − = = The amount of savings Chris has at the start = $90 Content L Complexity L Context L Response Strategy M Assessment Objective AO2 Qn Solution Content/Success Criteria 4 (a) 321400 2 5 7=   I can express a number in its prime factors. (b) 321400 2 5 7=   22 3 7 84q=   = 2HCF 2 7= 84q= I can use the prime factors from HCF to find the original number. Content L Complexity H Context L Response Strategy M Assessment Objective AO1/ AO2

Qn Solution Content/Success Criteria 5 3333 1 1 22222a− = + + + 1 1 322a− = 1522 a− = Comparing indices, 15 a−= 4a =− I can solve equation using laws of indices. m n m na a a += 1n na a − = Content L Complexity H Context L Response Strategy H Assessment Objective AO2 OR 3333 1 1 22222a− = + + + 3 1 1 2 (1 1 1 1)2a− = + + + 32 1 5 1 ( 1) 5 1 2 (2 )2 1 22 22 15 4 a a a a a − − −− = = = −+= =− Qn Solution Content/Success Criteria 6 (a) 2 22 2(5 3 ) 2(25 30 9 ) mn m mn n + = + + I can use identity 2 2 2( ) 2a b a ab b+ = + + to expand algebraic expressions. 2250 60 18m mn n= + + (b) 23 2 2 3 2 24( ) 21 24 21 3 (8 7 ) mn mn m n mn mn m n − =− =− I can factorise algebraic expressions by taking out common factor. Content L Complexity M Context L Response Strategy M Assessment Objective AO1

Tanjong Katong Girls’ School [Turn over 4052/S4Prelim/01/2024 Qn Solution Content/Success Criteria 7 1 (4.5)(7.1)sin 12.62 15.975sin 12.6 12.6sin 15.975 56sin 71 XYZ XYZ XYZ XYZ = = = = I can use 1 sin2 ab C for area of triangle to find sine of acute and obtuse angles in radians. 11 12.6 12.6sin or sin15.975 15.975 0.909 radian 2.23 radian XYZ XYZ XYZ XYZ −−    =  = −        =  = Content L Complexity M Context L Response Strategy L Assessment Objective AO2 7.1 4.5 4.5 7.1

Qn Solution Content/Success Criteria 8 3 kP Q= , where k is a non-zero constant Q is reduced by 20%, substitute original Q with 0.8Q . new 3 3 (0.8 ) 0.512 kP Q k Q = = I can use inverse proportion to find percentage change. Content L Complexity M Context L Response Strategy L 1New percentage of 100%0.512 5195 %16 P= = Percentage change in 595 %16P= or 95.3125% Assessment Objective AO2 OR 80% newQ of Q= 4 5 Q= 3 3 new 4 5P Q PQ = new 125 64PP = Percentage change in 125 64 100% PP P P − = 61 100%64= 595 %16= or 95.3125%

Tanjong Katong Girls’ School [Turn over 4052/S4Prelim/01/2024 Qn Solution Content/Success Criteria 9 (a) (b) 21 ( 6)5 1 55 181 13 or 7 or 7.5424 24 p +− = − =− − − 2 2 5 ( 5) qrp q p q q r += − − = + I can evaluate algebraic expressions by substitution. I can change the subject of the formula. 2 2 2 5 5 ( 1) 5 pq p q r pq q r p q p r p − = + − = + − = + Content L Complexity L Context L 2 5 1 rpq p += − Response Strategy L Assessment Objective AO1

Qn Solution Content/Success Criteria 10 (a) sin 30 32 32sin 30 BD DB = = I can use TOA CAH SOH to find unknown sides in right-angled triangle. 16 cm= Content L Complexity L Context L Response Strategy L Assessment Objective AO1 (b) 2 2 2 2 12 16 400 AD BD+ = + = 22 20 400 AB = = I can use converse of Pythagoras Theorem to prove right angles. Since 2 2 2AD BD AB+= , angle 90ADB= by the converse of Pythagoras’ Theorem. Since angle 90ADB= , by the property right angle in semicircle, it is possible to draw diameter AB such that point D lies on the circumference of the circle. I can use right-angle in semicircle to determine that the points in triangle lie on the circumference of circle. Content H Complexity H Context L Response Strategy H Assessment Objective AO3 Essential step

Tanjong Katong Girls’ School [Turn over 4052/S4Prelim/01/2024 Qn Solution Content/Success Criteria 11 2 2 38 (1 2 ) 2 1 38 (1 2 ) 1 2 hh hh +−− =− −− Alternatively, 2 2 38 (1 2 ) 2 1 38 (2 1) 2 1 hh hh +−− =+ −− I can add two algebraic fractions with quadratic denominators. 2 2 2 3 8(1 2 ) (1 2 ) 3 8 16 (1 2 ) 5 16 (1 2 ) h h h h h h −−= − −+= − −+= − 2 2 2 3 8(2 1) (1 2 ) 3 16 8 (1 2 ) 16 5 (1 2 ) h h h h h h +−= − +−= − −= − Content L Complexity M Context L Response Strategy M Assessment Objective AO1 Qn Solution Content/Success Criteria 12 (a) 25k− I can form and solve linear inequalities in one variable. (b) 2 5 2 3 16 4 8 16 4 24 6 kk k k k − + −  −   Content L Complexity L Context L Response Strategy L Smallest possible value of the larger odd number 2(7) 3=− 11= Assessment Objective AO1

Qn Solution Content/Success Criteria 13 (radius of circle)OA OB OD== Angle BAO (base of isos )xs=    Angle AOB 180 2 ( sum of isosceles )x= −    Angle DAO (base of isos )ys=    Angle AOD 180 2 ( sum of isosceles )y= −    I can use angle properties of circles to find an unknown angle. Angle BCD  1 (180 2 ) (180 2 )2 ( at centre = 2 at circumference) xy= −  + −   1 (360 2 2 )2 xy= − −  Content L Complexity L Context L Response Strategy M 180 xy= − −  Assessment Objective AO2 OR Angle BAO (base of isos )xs=    Angle DAO (base of isos )ys=    Angle BCD 180 ( in opp. segment)x y s= − −   Qn Solution Content/Success Criteria 14 (a) (3 1)( 5)xx+− I can factorise quadratic expression in the form 2ax bx c++ . (b)    2 2 2 3( 1) 14 19 3( 1) 14 14 5 3( 1) 14( 1) 5 By observation with algebraic expression in part (a), it is observed that 1 . Using answer from part (a), 3( 1) 1 ( 1) 5 (3 4)( 4) yy yy yy yx yy yy + − − = + − − − = + − + − += + + + − = + − Content L Complexity M Context L Response Strategy M Assessment Objective AO1/ AO2

Tanjong Katong Girls’ School [Turn over 4052/S4Prelim/01/2024 Qn Solution Content/Success Criteria 15 Substitute 2x= and 3y= into equation of the curve 23 (2) (2) 2 3 4 2 2 ab ab = + + = + + 1 4 2ab=+ ------------------------(1) I can apply the concept of substituting coordinates of points to form and solve linear Substitute 1x=− and 3y=− into equation of the curve 23 ( 1) ( 1) 2 32 ab ab − = − + − + − = − + 5 ab− = − -------------------------(2) equations in two varia