2026 GDLSS Emath P2 MS

Marker’s Report Question Comments 1(a) 4 6 34 25 3 35 8 7 16 15 6 xy xz y z yz x x×= Do not accept negative index A number of students were able unable to simplify correctly or did not simplify completely. 1(b) 113 63 33 43 21 64 2 24 mn m nn m n mn −− −    =     = = Do not accept negative index A number of students did not apply the negative index rule correctly which only applies to “m” while some did not simplify completely e.g. leaving answer in negative index. 1(c) 3716 25 2 32 3 7 50 39 3 57 13 19 17,19 x x x x x −<≤ < −≤ <≤ <≤ = There were students who did not solve the inequalities but obtained the answers which resulted in loss of mark due to omission of essential working. 1(d) ( )( ) ( ) ( )( ) ( )( ) ( )( ) 2 14 5 2 20 5 14 5 2 54 5 14 5 2 4 54 14 5 2 8 54 67 54 x xx x x xx x xx xx xx xx x xx − +−− − −= −−+ − −− += −+ −−−= −+ −= −+ Most students who factorised the denominator were able to do so correctly. Some were unable to find the common denominator. 2(a) 225 8 3 25 22.51.5 11 4 3.5 5.5 3 71 1.5   = =   + + +   T xx x x Generally well done with a few who made error in the calculation. 2(b) The total cost of Kai’s snacks for store and online purchases respectively. Poor use of language as students followed the use of store

Question Comments pricing and online pricing. There were students who mentioned both Kai and Lina. 2(c) 3 7 22.50 5.50 3 21 7 x x x += + = = Well done. 2(d) 17.55 100% 78%22.50×= Percentage discount = 100% 78% 22%−= Generally well done. 2(e) : 3:4KL = : 2:7 4:14LA = = : 3:14KA = Well done. 3a 224 ( 7) 8.06unitsAB →= +− = Well done. 3b 7 4 ABm =− 7 , sub (4, 2),4 72 (4) 4 5 7 54 y xc c c yx = −+ − −= − + = = −+ Students were able to obtain the gradient correctly but there is no need to find the coordinates of B. There were students who substituted the vector into the equation instead which shows poor understanding of vectors. 3c 74 2 4 72 71 42 xy yx yx += = −+ = −+ Since both lines have the same gradient 74− and have different y- intercepts, they are parallel. Therefore line p does not intersect the line AB. A number of students were unable to recognise the need to express the equation in the form of y=mx+c to find gradient. A number of students did not mention about different y- intercept. 3d 448 27 9OB OA AB  →  →  →  =+= + =   −−−   Coordinates of (8, 9)B − A number of students were unable to find 2 possible coordinates. Some tried to find A D1

Question Comments ABCD1 11 1 14 4 10 87 1 (10, 1) OD OC CD D  →  →  → −   = += + =    −−    − ABD2C 22 2 14 4 18 8 7 15 (18, 15) OD OC CD D  →  →  →    = += + =    −−−    − ACBD3 23 3 8 14 6 9 81 4 62 213 ( 2, 3) BC BO OC OD OA AD D  →  →  →  →  →  → −   =+= + =    −    −−  = += + =   −−−   −− equation of lines or length of line segments which were often unsuccessful. Students should be more careful with the presentation of working as they tend to combine coordinates and vectors. 4(a) -5.5 Well done. 4(b) Many students were unable to recognise that the curves disjoint. 4(ci) Draw line Line cuts through the curve Some students made error in the manipulation of the equation and were unable to draw the straight line correctly. D2 A D3

Question Comments 4(cii) 0.65( 0.1)−± or 0.70( 0.1)± Some students were lucky in obtaining the solutions from their graphs. 4 (ciii) 2 2 32 13 2 322 1 92 022 9 40 9, 4 xx x x x xx AB − −=−− −+ = − += = −= A number of students tried to substitute the solutions from cii which they need to know that the solutions from the graphs are just estimates. 5 a 256 216 40 CAB∠ = °− ° = ° sin sin40 450 330 61.226 ABC ABC ∠° = ∠= ° 180 144(int ) 36 61.226 36 25.226 x y ∠= − ∠ = ° ∠ = °− ° = ° bearing 360 25.226 334.8 = °− ° = ° Most students were able to find angle ABC. A number of students rounded their answers to 3sf instead of 1dp. 5b sin 40 450 289.254 289m (3sf) CT CT °= = = More students tried to use area of triangle method to find the shortest distance, which is less efficient and there is tendency to make mistakes in calculations. 5c tan34 330 222.587 223m (3sf) DC DC °= = = Students who were unable to obtain the answer mainly made mistake in identifying the angle of depression. North

Question Comments 6(a) Tina: 800x Bella: 8001x− or 800x +62 Most students are able to do (ai), but some wrote new speed as x+1 instead of x–1 for (aii). 6(b) 2 2 800 800 621 62 62 800 0 31 31 400 0 xx xx xx −=− −−= −−= Some students are not able to construct equation. Wrote 800 𝑥𝑥 − 800 𝑥𝑥−1 = 62 instead. 6(c) ( ) ( ) ( )( ) ( ) 2 2 31 31 400 0 31 31 4 31 400 2 31 3.12673 4.12673 3.127 4.127 xx x x or x or −−= −− ± − − −= =− =− Answer must be given to 3 d.p. Most students are able to apply quadratic formula, but did not leave answer in 3d.p. 6(d) Total time for Tina 800 193.89 sec4.126= = Total time for Bella 400 400 207.95 sec4.126 1 5= += − Bella will not be able to catch up with Tina. Some students read “increases her speed to 5 m/s” as “increase her speed by 5m/s”, hence wrong total time for Bella. 7a It is NOT conclusive that the coin is biased because the number of experiments conducted is too small. Most students are not able to state that no. of expt is too small. 7b(i) 2 23 2 34 3 +× =× Some students are able to do this question. 7b(ii) 2563 12 12 25 96 ×× = Few students are able to do this question. Some did not x3. Some add instead of multiply. Some student did “without replacement” 7c(i) 30.222 ≈ 30.2 Most are able to do this question.

Question Comments 7c(ii) 9.9138 ≈ 9.91 Most are able to do this question. 7c(iii) The mean will increase because the new plants are taller than the existing 90 plants. The standard deviation will increase because the difference between the heights of the new plants and the mean height is more than the current standard deviation. Most are not able to explain, only state both increase. 8a Triangle BAT All students who answer this question got this correct. 8b (vert. opp s) ( s in same seg) ( s in same seg) CED BEA DCE ABE EDC EAB ∠= ∠ ∠ ∠= ∠∠ ∠= ∠∠ By AA property, and are similar.CED BEA∴ ∆∆ Most are able to do this, but got deduction of mark because of reasoning and sequencing of letters. 8c Angle OBT = rad2π (tan ⊥rad) Angle OTB = 0.4 rad (tan fr ext. pt) tan 0.4 12 5.0735 5.07(3sf) OB OB = = = Some students did not attempt this question. 8d 0.8 (tan fr ext pt) 2.3415 radBOA π∠= − = Arc 5.0735( 0.8) 11.8801AB π= −= 1 2sin0.4 12 9.3460 AB AB = = Perimeter =11.8801 9.3460 21.226 21.2 cm+== Some students did not attempt this question. 9a Height of cylindrical tube = height of 1 shuttlecock + 7 x base =7.5 7 1.5+× =18cm [shown] Most students are able to show how they got 18 cm. 9b Height of training shuttlecock Most students are able to apply ratio of length. O 0.4 rad 12 cm

Question Comments 7.5 43 10 cm = × = 9c Cost (Standard shuttlecock) 10.5 $4.80100 $0.504 = × = Cost (Training shuttlecock) 3 3 $0.504 43 $1.1946 $1.19 (shown) = × = = Most students are not able to see the relation between ratio of volume vs ratio of mass. Did not apply 3 3 4 3 . 9d Standard Training Shuttlecocks $0.504 8 $4.032 × = $1.19 8 $9.52 × = or $1.1946 8 $9.5568 × = Cardboard Total surface area 22 (3)(18) 2 (3) 395.84 ππ= + = Cost 395.84 $1.1875 = = ÷ 100×$0.30 Cost 22 22 1.19 1.19464 or 433 $2.1155 or $2.1237 = ×× = Or 22 (4)(24) 2 (4