2025 GDLSS Emath P2 MS

Marking Scheme 4 Exp/5NA Prelim P2 2025 Question Markers’ comments 1(a)(i) 13 610 404 = 13 600 000 (3 sf) = 1.36 × 107 Generally well done. Some either omitted the standard form or leaving answers to 3 significant figures. 1(a)(ii) Percentage increase = %16 526 312 6 305 7446 305 744 100×− = 162% (3 sf) Generally well done. Some took the values from the incorrect years and a few applied the percentage formula wrongly. 1(b) Price after 15% discount = 85 $175100× = $148.75 Members discount = 12 $148.75100× = $17.85 Total discount = $175 − $148.75 + $17.85 =$44.10 Generally well done. Some students found the members discount but did not answer the question. They went to calculate how much was paid in the end. 1(c)(i) 52.08 cm2 : 13.02 km2 = 1 cm2 : 0.25 km2 221 cm : 0.25 km 1 cm : 0.5 km 1: 50000 = = Some students went to convert km2 to cm2 wrongly, thereby affecting the whole question. A few students left n to be 0.5. 1(c)(ii) 1 cm : 0.5 km 65 cm : 0.5 65 km 32.5 km×= Generally well done. 1(c)(iii) 2225 13.02k0 m 3. 510 2 5km×= No. of trees = 2 2 3.255km 5m = 22 2 3.255 1000 m 5m × = 651000 Generally well done. Some students converted the units wrongly as per part (c)(i). 2(a) ( ) ( )4 3 6 7 23 4 1 26 723 4 18 5 3 5 18 23 xx xx xx x x − −=− − − −=−+ −= + = + = Generally well done.

Question Markers’ comments 2(b) 93 5 95 3 42 2 2 xx xx x x x − >− − >− + > > < Generally well done. 2(c)(i) ( ) ( ) 2 2 2 2 2 2 23 2 3 2 9 29 18 nRm t Rn mt Rn tm nm t R mnt R = = = = = Many students did not simplify the answer at the end. 2(c)(ii) ( ) ( ) 2 2 2 2 18 118 5 2 2 45/5.6258 mnt R t t = −= = Well done. 2(d) ( ) ( ) ( )( ) 2 2 2 2 22 2 2 2 13 15 5 23 1 13 15 5 62 115 3 5 2 13 16 5 2 3 16 5 2 17 7 0 ( 17) ( 17) 4(2)(7) 2(2) 8.06608 or 0.433915 8.07 or 0.43 x xx xx x xx xx x xx x xx xx xx x x xx x xx xx −=−− −− − =−− −−+ =− −+ − −+ =−+− − −+= − + − − += −− ± − −= = = = = Well done. A reminder to give the formula with values as working.

Question Markers’ comments 3(a) Let the height of the smaller cone that was removed be . 10 18 26 26 10 180 16 180 11.25cm (shown) x x x xx x x =+ = + = = Generally well done. 3(b) 22 3 2 Volume of frustum 11= (26) (18 11.25) (10) (11.25)33 19528.13993 Volume of hemisphere 2= (26)3 36811.08832 Volume of the empty space = (26) (18 26) 19528.13993 36811.08832 93443.53189 19528.13993 36811.0 ππ π π +− = = +− − =−− 3 8832 37100 cm= Generally well done. Many students assumed the height of cylinder is twice of 18cm. 3(c) 2 2 22 22 2 CSA frustum (26)( 29.25 26 ) (10)( 11.25 10 ) 2723.7444 TSA of toy (10) 2723.7444 2 (26) 7290 cm ππ ππ +− + = = ++ = Generally well done. Some students missed out the area of the small circle. 4(a) 40p = Well done. 4(b) Refer to last page. Draw graph. 4(c) Draw 10y = , 2.4x > Well done. Some got the inequality sign wrong. 4(d)(i) Draw straight line 10 6yx= − x 4− 0 4 y 46− 6− 34 Reminder to draw the full line as required

Question Markers’ comments 4(d)(ii) 3 3 3 3 3 4 10 62 3 4 10 6 02 7 202 14 4 0 x xx x xx x x xx + −= − + −− += − += − += As it is a show question, some students started from what is needed to be shown to get the line and the curve. This is not answering the question. 4(d)(iii) ( )3.9, 0.25, 3.6 0.1xxx= −= = ± Well done. 5(a)(i) 180 64 64 (base of isos. triangle) 52 FOG∠ = °− °− ° ∠ = ° Generally well done. 5(a)(ii) 52 ( at centre = 2 at circumference)2 26 90 ( in a semicircle) 180 90 26 25 ( s in opposite segments) =39 180 90 39 (adj s on st. line) =51 FEG FGE EGI IGH ∠ °= ∠∠ = ° ∠ = °∠ ∠ = °− °− °− °∠ ° ∠ = ° −° −° ∠ ° Poorly attempted by many students who incorrectly assumed EI = OI. Other incorrect responses assumed EH and FH were tangents to the circle. A significant number of students also wrongly used angles in the same segment to conclude that angle IOG is 25°

Question Markers’ comments 5(b) 180 64 116 180 116 64 3.2 sin 51 sin 64 2.76689 25 2 50 Length of major arc 360 50= 2 (3.5)360 18.93682 Perimeter = 18.93682 2.76689 3.2 24.9037 24.9cm EIG GIH IH IH IOG GFEI π ∠ = °− °= ° ∠ = °− °= ° =°° = ∠ = °× = ° °− ° × = ++ = = Alternative method/solution: 2 22 3.5 3.5 2(3.5)(3.5) 2.958327 2.958327 sin 51 sin 65 2.5367 Perimeter = 18.93682 2.5367 3.2 24.6735 24.7cm IG IG IH IH =+− = = = ++ = = Many students were unsure what constitutes the perimeter of the shaded shape and included the perimeter of the circle as the final calculation. Students who gave incorrect answers did not realise that they needed the major arc GI. Other were unable to obtain the accurate answer as the angles found in the earlier part is incorrect. Many students also incorrectly assumed IH = GH. 6(a)(i) ( ) 73 2 53 7 32 53 3 2 7 ------(1) 3 5 ------(2) Fr (2), 5 3 ------(3) Subst (3) into (1): 32 7 53 3 2 35 21 21 2 35 3 19 38 2 Subst 2 i a ba ab ab ba ab ab bb bb bb b b b −  = +  −   −  =  −+   −= += − = −− − = −− −= −− + = −− =− =− =− ( ) nto (3): 53 2 =1 a= −− − Generally well done and most students are able to write the eqns 1 and 2. However some made arithmetic mistakes and were unable to obtain the accurate answers for a and b. 6(a)(ii) ( ) 22 2 75 74 8.60 units PQ  → = +− = = Generally well done.

Question Markers’ comments 6(b)(i) ( ) ( ) 2254 12 90 9.49 units AB= −− +−− = = Generally well done. Students who made mistakes here were the ones who were not able to recall the correct formula. 6(b)(ii) (−2, 0) 6(b)(iii) ( ) ( ) ( ) 21 45 1 3 1 3 136 3 5 1 53 L ABmm y xc c c yx = −−= −− = = + = −+ = ∴= + Generally well done. 6(b)(iv) When x = 1, ( )1 153 12533 y= + = ≠ The point does not lie on line L. Most students were able to do this part. However some students substituted the x value as 23 instead of 1. Some wrote mathematically incorrect statements ( )21 1533= + . 7(a)(i) No. The upper quartile for Model B is 36 hours, which is lower than that of Model A of 39 hours. Generally well done. Students who did not obtain the marks generally did not state the correct Q3 values or made reference to upper quartile. Other incorrectly stated that the top 25% of A is 6 hrs. 7(a)(ii) I will recommend Model B. The median lifetime of the battery in Model B is 32 hours, which is higher than that for Model A of 28 hours. The IQR of the battery life of Model B is 36 – 25 = 11 hours compared to that of Model A of 39 – 22 = 17 hours, which is lower, hence the battery life of Model B is more consistent. This part was poorly done as many students did not state the values of median and IQR. 7(b)(i) 1 40 Generally well done. 7(b)(ii) Outcomes: 8, 16, 24, 32, 40 54 1 40 39 78×= Generally well done. Students who did not obtain the marks assumed incorrectly that there is replacement.

Question Markers’ comments 7(b)(iii) 9 8 31 279 340 39 38 2470× × ×= Poorly attempted. Many students did not realise there are 3 cases or did not realise that there are 31 double digits instead of 30 (30 to 40 has 11 numbers instead of 10). 8(a) 20, 320pq= = Generally well done. 8(b) ( ) ( ) ( ) ( ) 2 42 41 8 24 4 888 16 8 xn n nn nn n =++ = ++ =++ = + ( )( ) 2 444 16 16 y nn nn = + = + Generally well done. 8(c) 16 8 104 16 96 6 Width = 4 6 = 24 n n n += = = × Generally well done. 8(d) ( ) ( )( ) 2 2 2 16 16 140 16 16 140 0 4 4 4 35 0 42 7 2 5 0 2.5 or 3.5 nn nn nn nn nn += +−= +− = + −= = =− n is not an integer and cannot be negative, hence Ming is correct that it is not possible for a rectangle in the sequence to have an area of 140cm2. Students who did not obtain the full marks did not mathematically show that there are no integer values which can give an area of 140. 9(a) Size of a football field = 66.7 10 18 60 24 365 × ××× [Accept 366 days as 2024 is a leap year] = 0