2026 GDLSS Emath P1 MS

Marking Scheme 4Exp/5NA Prelim P1 2026 Question Marker’s Report 1 ( ) ( )6 2 7 5 34 6 2 75 34 9 14 aa a aa a a − ++ −− = − −+ −− = − Well done. Common mistakes were not grouping the like terms accurately. 2 Well done. Errors were those who drew two lines (which is their working and not considered the final answer) 3(a) 855 24 198.75 $5625 +× = Well done. Errors were by those who keyed in values wrongly into the calculator. Working is correct but final answer is wrong. 3(b) Let $x be the cash price. 5625 100 25 5625 0.25 5625 1.25 4500 x x xx x x − ×= −= = = Alternative method: 5625 100 4500125 ×= Many made the error of stating that 75%  5625. 4 4 60 485f = ×= 91x= (accept 90.5) Poorly done. Many do not grasp the concept of the speed being at least x vs at most x. Common mistakes were 75 km/h. 5 ( ) 5423 9 18 6 823 31 5 18 3 12 18 4 6 xxx xx x x x +−−− − +− −= −= −= Common mistakes was simplifying this wrongly − 2𝑥𝑥−3 18 and not giving the final answer as simplest form. 6(a) ( ) 33 22 ab a b ab b a + = + Common mistake was confusing 𝑏𝑏2 + 𝑎𝑎2 to 𝑏𝑏2 − 𝑎𝑎2 x 0 -3 5

Question Marker’s Report 6(b) ( ) ( ) ( ) ( ) ( )( ) 29 623 33 2 2 3 3 32 32 3 32 cd ce de d cd e d e d c de d de cd d e −+− = −+ − = −− − = −− Generally ok. Common mistake was confusing the change of sign and having poor presentation. 7(a) 22 22 4 cm :1 km 1 cm : 0.25 km 1 cm : 0.5 km 1: 50000 Mistakes were made when student converted km to cm prematurely and had to deal with the many 0s. 7(b) 28.7 0.5 57.4 cm÷= Generally ok. 7(c) 20.52 4 0.13 km÷= Generally ok. 8(a) 10xa+= 4 22xa+= 3 12 4 a a = = 6 14 18 x y z = = = Generally ok. 8(b)(i) – 1, –8 Well done 8(b)(ii) 34 7 n− Generally ok. A few who used the general term formula and did not simplify their answer. 9 1(13 10)( ) 502 4.347 4.35 (3sf) h h += = = Generally ok. Mistakes were made by those who could not recall area of trapezium accurately. 10 1(7)(9)sin 252 52.528 (rej) or 127.471 ABC ABC ∠= ∠=  2 22 2 7 9 2(7)(9) cos127.471 206.655 14.375 14.4 (3sf) AC AC AC =+− = = =  Common mistake was not finding and using the obtuse angle as required by the question. 11a(i) 1.7p= , 2.8q= Generally ok 11a(ii) 2.4 1.7 0.7−= Generally ok 11b Since the median height is 1.9m, Candice is correct to say that half the trees are taller than 1.9m Common error was saying 50% was 1.9m but with no reference

Question Marker’s Report that 1.9m is the median. 12(a) ( ) ( )( ) 22 22 1443 35 3 21 rp qr −−−= = = −− −− − Some students did not show any working. Given it is a 2 mark question, they should. No marks were deducted. 12(b) ( ) 2 2 22 2 2 4 3 34 1 34 34 1 34 1 rp qr p pqr r r pq p pr pq pr pq −= − −= − += + += + += ± + Common error was confusing the signs, not leaving ± in their final answer. Remind students when balancing the equation to avoid dealing with so many negative signs. 13 (6 2) 1801 int angle of hexagon 6 120 −×= = ° 1ext angle of n-sided polygon 120 6 20 = ÷ = ° 360 20 18 n= = Common errors included finding only the exterior angle as the final answer, assuming total exterior angle is 180 instead of 360. 14 180 120 ( s in opp seg) 60 90 ( in semicircle) 180 120 (base s of isos )2 30 90 30 60 ADC BAD BCD BAC BCA CAD ACD ∠ = °− ° ∠ = ° ∠ =∠ = °∠ ∠= ∠ °− °= ∠∆ = ° ∠= ∠ = °− ° = ° Since 60CAD ACD ADC∠= ∠= ∠= ° , triangle DAC is an equilateral triangle. Explanation was generally ok. Common errors: - Incorrect reasons (remind students to memorise the given reasons) - Wrong reasons such as tangent from external point - No/insufficient reasoning given 15 3 2 24 36 yx xy −= = + ( )3 3 6 2 24 7 42 6 xx x x −−= = = Well done.

Question Marker’s Report student ticket = $6 adult ticket = $12 16 4 158 532 4 282 2 10 5 xx xx x x ++−=++ −=+ = = Those who made errors were confused by the 5. 17(a) 43.4501 10 −× Many gave answer in 3sf. Question is in exact form. Leave it in exact form. 17(b)(i) 77 Generally ok. 17(b)(ii) ( ) ( ) ( )1999 1999 1999 1999 2000 77 10 2.4 10 77 2.4 10 74.6 10 7.46 10 × −× =−× = × = × Many errors. Students seem to lack understanding of standard form. Some took 7.7-2.4. 18(a) P = 2227× Some students did not leave the answer as a product of prime factors. 18(b) k = 10 Some students found k2 instead. 19a 3sin sin(180 ) 5θθ= °− = Quite a number of students were unable to apply the concept of sin sin(180 )θθ= °− . 19b 22534−= 4cos(180 ) 5θ°− =− Similar to 19a, students were unable to apply the concept of cos cos(180 )θθ=− °− . 20a ( )( ) 2 2 9 3 7 16 23 17 7 16 3 40 0 5 80 5 or 8 (rejected) pp pp pp pp ++ ++++= × +−= − += = =− Students were able to form the equation but there were some students who were unable to continue as they failed to recognise the quadratic equation and solve it. A few students did not show show how they solved the equation/ did not reject −8, resulting in loss of marks.

Question Marker’s Report 20b 1. Shereen received more comments than Joel as her mean of 16 comments per day is higher than Joel’s mean of 13. 2. The number of comments Shereen received each day is more consistent than that of Joel as her standard deviation of 6.12 is lower than Joel’s standard deviation of 8.35. Quite a number of students did not compare/ state the mean/ standard deviation as supporting reason for their conclusion. 21 ( )( ) ( ) 3 32 1 5 3 3 2 10 5 3 6 30 3 15 3 30 3 15 px p x px px p x px px p x px p x − +− = − − +− = − + −+ = − + −+ Most students were able to expand ( )( )21 5px+− correctly. A number of them continue to try to factorise. 22a 2 2 22 22 20 21 841 29 841 FA AB FB += + = = = Since 222FA AB FB+= , by converse of Pythagoras’ theorem, triangle FAB is a right-angled triangle Many students did not apply the converse of Pythagoras’ theorem correctly, some missing out on substitution and others applied Pythagoras’ theorem instead. There were students who had poor presentation such as 22 220 21 29+= which was not accepted as they should show that this is equal. 22b 2 5 18 cm BR BC= = 2 22 21 18 765 27.658 AR = + = = 20tan 765 35.870 35.9 FRA FRA ∠= ∠= = ° Many students were unable to find BR correctly – had misconception BR is 3 units. Some students were unable to identify the right-angled triangles correctly.

Question Marker’s Report 23 ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) 22 32 2 2 2 2 22 5 10 22 22 52 2 32 52 23 2 52 32 5 xx xx x xx x xx xx xx xx xx x x +− − +− ++    = − −+= − −+= − − += − Most students tried to expand the numerator but did not factorise the numerator and denominator. A number of students factorise numerator wrongly, missing out the negative sign 23 44xx−++ should be factorised into either ( )( )23 2xx−− + or ( )( )2 32xx−+ . 24 2 2 2 5 11 0 5 25 1124 5 69 24 5 69 24 6.65 1.65 xx x x x x or −−=  −= +  −= −= ± = − Some students were able to complete the square but unable to continue to solve the equation, while there were others who used the quadratic formula to solve which is not accepted. 25 4.5sin 7 40.005 2 80.010 θ θ θ = = = 1Area of (7)(7)sin80.0102 24.128 OAB∆= = 280.010Area of minor segment ( 7) 24.128360 34.2129 24.128 10.084 π= ×− = − = 2 2 Shaded Area (7) 2 10.084 133.769 134cm (3sf) π= −× = = A number of students were unable to add in the radii to the diagram and only tried to find area of circle or left the question blank. Some students tried to find the unshaded segment and only subtract one of it. A B 7 9 O

Question Marker’s Report 26 ( ) 10 66 10 10 10 6.11 10 6.11 10 570000 1 100 5711 100 554 5711 100 554 571 1 100554 0.984 r r r r ×= ×− +  += += = −× =