CHIJ SJC 4048 01 AS

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 CHIJ St Joseph’s Convent Solutions for 4 EXP Elementary Math Preliminary Exam Paper 1 2021 Qn Marking Point 1 6− 2 3 4 12 24 2 3 16 15 3 125 3 xx x x and x x x  −  − −     3, 4, 5 3(i) (2 5)( 2)xx+− 3(ii) [2(2 3) 5][(2 3) 2] (4 1)(2 5) yy yy − + − − = − − 4(a) 332 3 5 4(b) 2 3 2 a b c = = = 4(c) 36 5(i) (2a + 2b)2 = [2(a + b)]2 = 4(a + b)2 = 4(a2 + b2 + 2ab) = 4[(a – b)2 + 2ab + 2ab] = 4[(a – b)2 + 4ab] = 4(196) = 784 5(ii) (2a)2 – (2b)2 = 4a2 – 4b2 = 4(a2 – b2) = 4(a + b)(a – b) Since a > 0 and b > 0, a + b > 0. From (i), (a + b)2 = 196  a + b = 196 = 14  (2a)2 – (2b)2 = 4(a + b)(a – b) = 4(14)(–6) = –336

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point 6 Original area of triangle = ( )( )ACAB2 1 New area of triangle = ( ) )100+1(902 1 ACxAB.  ( ) ( )( ) ( )( ) ( )( )ACAB ACABACABx. 2 1 2 1 - )100+1( 902 1 = 100 125 ( )( ) ( )( )ACAB x.ACAB 2 1 ]1 - )100+1(90[2 1 = 4 5 0.9 + 1000 9x − 1 = 4 5 1000 9x = 4 5 + 0.1 1000 9x = 20 27  x = 9 1000×20 27 = 150 7(i) 14 7(ii) 35 7 12 25 12 37 t t t =− =− = 8(a)(i) 1426 115 15 $186 8(a)(ii) 1426 100115  or 186 10015  $1240 8(b) 1240 1007.50 100%1240 18.75% −  = 9(a)(i) 1 9(a)(ii) 2

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point 9(b) 2 2 2 54 9 1 ()9 1 (54)9 6 k r kF r kF r F F = = = = = 10(a) 'AB 10(b) (i)  (ii)  (iii)  11(a) 168.3 cm 11(b) The key here is that the response should refer to the “change” of the gradient of the graph for female. This can be done explicitly or implicitly. Explicitly mentioning about the steepness of the curve of the graph: • It does no longer go straight up, it straightens out. • The curve levels off. • It is more flat after 12. • The line of the girls starts to even out and the boys line just gets bigger. • It straightens out and the boys graph keeps rising. • You can see the gradient is less. • The rate of change of the graph decreases from 12 years on. • [The student computed the angles of the curve with respect to the x-axis before and after 12 years.] Implicit comparison using the actual amount of growth before 12 years and after 12 years of age: • From 10 to 12 the growth is about 15 cm, but from 12 to 20 the growth is only about 17 cm. • The average growth rate from 10 to 12 is about 7.5 cm per year, but about 2 cm per year from 12 to 20 years. 12(a) 21 (4) (2 0.125)2  = 2cm2 12(b) tan 0.25 4 1.0213 CT CT =  1Area of triangle 1.0213 42OCT =   Area of region BCT = 0.0426 cm2

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point 13(a) Acceleration when t = 6 s = 40 20 = 2 ms-2 13(b) ( )( ) ( )1 40 20 40 20 9202 k+ − = k-20 = 13 k = 33 13(c) 14(a) 14(b) Time (t seconds) Distance (metres) 400 920 20 k

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point 15(a)(i) 15(a)(ii) x = -0.5 15(b) 16 -4 3 12 (-0.5, 12.25)

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point 17(a) Amount accumulated at the end of 2 years = 5000 2 100 41      + + 5000      + 100 41 = $10 608 Interest earned at the end of 2 years = $(10 608 − 10 000) = $608 17(b) Amount accumulated at the end of 10 years = 5000 10 100 41      + + 5000 9 100 41      + + ... + 5000      + 100 41 = 5000(1.04 + 1.042 + … + 1.0410) = 5000(1.04)(1 + 1.04 + … + 1.049) = 5200       − − 04.11 04.11 10 = $62 431.76 (correct to 2 d.p)

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point Total interest = $62 431.76 − 10($5000) = $12 431.76 = $12 432 (correct to the nearest dollar) 18(a) BCD = 120 (Angles in opposite segments) 18(b) CDB = 30 (base angle of isosceles triangle) EDB = 74 (angle sum of triangle) CDE = 104 18(c) ADC= 90 (right angle in semicircle) ADE= 14 19(a) Gradient of BC = 53 62 − − = 1 2 19(b)(i) Form a rectangle ADBECF Area of ABC = Area of ADBECF − Area of ADB − Area of BEC − Area of AFC 10 = (8 − 2)(5 − k) − ( )( )k -32 -82 1 − ( )( )3-52-62 1 − ( )( )k-56 -82 1 or 10 = 6(5 − k) − 3(3 − k) − 4 − (5 − k) 19(b)(ii) 6(5 − k) − 3(3 − k) − 4 − (5 − k) = 10 30 − 6k − 9 + 3k − 4 − 5 + k = 10 12 − 2k = 10 2k = 2 k = 1 20(a)(i) 27 20(a)(ii) 11.1 20(b) Class Integrity performed better as its mean mark is higher than that of class Respect. There is a narrower spread in the marks in Class Integrity as its standard deviation is lower than that of Class Respect. (or The marks in Class Integrity are more consistent as the standard deviation of the marks in Class Integrity is smaller than that of Class Respect.)

Updated as at 11 Aug 2021 4048 4EXP/5NA EM Prelim 2021 Qn Marking Point