GESS 2025 EM P2 MS

GESS 4E5N EM P2 PRELIM 2025 SZK Solutions 1(a) 2 1 3 1 35 2 1 3 1 35 xx xx +−− − +−− − 5(2 1) 3(3 1)xx+  − 10 5 9 3xx+  − 8x− M1 A1 1(b) 2 7 1 3 (3 1) x xx−+−− = 2 7 3 1 (3 1) x xx +−− = 22 (3 1) 7 (3 1) (3 1) xx xx − +−− = 2 (3 1) 7 (3 1) xx x −+ − = 2 2 37 (3 1) xx x −+ − Alt method 2 7 1 3 (3 1) x xx−+−− = 2 7 1 3 (1 3 ) x xx−+−− = 22 7 (1 3 ) (1 3 ) (1 3 ) xx xx −−−− = 2 2 73 (3 1) xx x −+ − = 2 2 37 (3 1) xx x −+ − M1 M1 A1 M1 M1 A1

3 GESS 4E5N EM P2 PRELIM 2025 SZK 1(c) 5 49 4 (1) 4 5 10 (2) From (1) : 5 4 49 (3) (3) 4 : 5 4 49 (4) (2) 5 : 20 25 50 (5) (4) (5) : 16 ( 25 ) 196 ( 50) 41 246 6 Subt 6 into (1) 5 49 4(6) 5 25 5 xy xy xy xy xy y y y y x x x = − −−−−−−−−− − =− −−−−−−−− + = −−−−−−  + = −−−−−−−−  − =− −−−−−− − − − = − − = = = =− = = Ans x = 5, y = 6 M1 A1 A1 1(d) 22 15 12 33 12 15 2 12 3 15 8 10 64 64 64 = 16 = hf fh f h f h −     =         M1 A1 2(a) 164 144 50 128 90 40   B1 2(b) 30 25 20    B1 2(c) AP = 30164 144 50 25128 90 40 20     = 9520 6890   ( ) ( ) 21 0 1   8 9511 6 90   =AP = (16 410) M1 A1

4 GESS 4E5N EM P2 PRELIM 2025 SZK The element 16410 represents the total amount collected from the sales of tickets for Saturday and Sunday for the 1st weekend. B1 2(d)(i) 1.25 0 0 0 1.5 0 0 0 0.8    B1 2(d)(ii) Each element in matrix FP represents the total amount collected from the sales of tickets for Saturday and Sunday respectively for the 2nd weekend. B1 3(a) By sine rule, 9 sin 85 sin 30 BD = 17.9315 (6s.f)BD= BD = 17.9 m (3 s.f) M1 A1 3(b) 2 2 26 17.9315 14.5cos 2(6)(17.9315)BDA +−= = 46.8039º (4 d.p) Bearing = 180º + 46.8039º = 226.804º = 226.8º M1 M1 A1 3(c) Note : Greatest angle of elevation/depression occurs at shortest distance CX. In triangle DCX, sin 65 9 9sin 65 CX CX = = Let  be the greatest angle of elevation 12tan 9sin 65 =  55.7948 (4 d.p)= 55.8 (1 d.p)= Students may use area of triangle to find CX. M1 M1 A1 4(a) Vol. of container = 22 1(1.5) (10) (1.5) (2)2 + = 75.3982 cm3 (4 s.f) = 75.4 cm3 (3 s.f) M1, M1 for each working A1

5 GESS 4E5N EM P2 PRELIM 2025 SZK 4(b) Vol. of water = 75 75.3982100 =56.5487 (4 s.f) Height of water = 2 56.5487 (1.5) = 8.00 cm d = 10 + 2 – 8.0000 = 4.00 (3 s.f) M1 M1 A1 4(c) Let the height of pyramid be h cm. 1 4 4 56.54873 10.6029 (6 s.f) 10.6 cm (3 s.f) h h h    = = = M1 A1 4(d) Time = 3 21 3 x  −  = 19 27 x minutes Accept 1 27 x minutes (as students may see it in another perspective) B1 5(a) P = – 0.05 B1 5(b) Refer to graph paper 5(c) The graph 2 2 27 xy x= + − cuts the x-axis at 2 points. Hence, 2 2 207 x x+ − = has 2 solutions B1 5(d) Accurate tangent line drawn Gradient = 1.8 1.8 = 1 (accepts 1.02 0.05 ) M1 A1 5(e) 32 2 2 2 7 28 14 0 147 28 0 2407 2 227 x x x xx x x x x x xx + − + = + − + = + − + = + − =− + Line y = 2 – x drawn From graph, x = 2.35 0.05 M1 M1 A1

6 GESS 4E5N EM P2 PRELIM 2025 SZK 6(a)(i) 90 (tan perpendicular rad) 90 (tan perpendicular rad) is a common line. (radii of the same circle) OBI OCI OI OB OC = = = By RHS, triangle OBI is congruent to triangle OCI. M1 A1 6(a)(ii)(a) 180 (62 70) ( in opposite segment) 48 DCB= − + = B1 6(a)(ii)(b) 62 ( in the same segment) (radii of the same circle) 76 (base of isos triangle) 76 62 14 GCE GAB OC OE OCE OEC GOC = = = = =  =− = Accepts other methods ( eg. Find angle OBC first) M1 A1 6(a)(ii)(c) 90 (tan perpendicular rad) 90 14 104 OCH BCH = =+ = B1 (b) Major arclength = (2 2.18) 8 − = 16 17.44 − 2 2 2 8 (8 3.6) 2(8)(8 3.6)cos 2.18AD = + + − + AD = 17.4575 (6s.f) Perimeter = 16 17.44 − +3.6 + 17.4575 = 53.9 cm (3 s.f) M1 M1 M1 A1 (c) area of sector ABCO area of triangle AOD = 21 (8) (2 2.18)2 1 (8) (8 3.6)sin 2.182   −   + = 3.45052 (5 d.p) = 3.45 (2 d.p) M1 for denominator M1 for numberator A1 7(a) Mean = 22(8) 26(10) 30(21) 34(7) 38(4) 50 + + + + =29.12 minutes 2 2 2 2 2 28(22) 10(26) 21(30) 7(34) 4(38) 50 fx N + + + += = 868 Standard deviation = 2868 29.12− = 4.474997 M1 M1 A1 7(b) The mean waiting time for hospital A and hospital are the same (29.12 min). Hence, the average waiting time is generally the same for both hospital. B1

7 GESS 4E5N EM P2 PRELIM 2025 SZK However, the standard deviation for waiting times in hospital B (3.2 min) is lower than the waiting times in hospital A (4.47 min). Hence, the waiting times in hospital B are more consistent. B1 7(c) P(required) = 8 11 11 8 50 49 50 49 +  = 88 1225 Accepts other relevant method M1 A1 8(a)(i) ab−+ B1 8(a)(ii) CD CO OD=+ = 21( ) ( )32 ab−+ = 21 32ab−+ 1 2 1 2 3 2 11 34 CE a b ab = − + =− + M1 A1 8(b)(i) OF OA AF=+ ()a m a b= + − + (1 )m a mb= − + M1 A1 8(b)(ii) 3 5OE OF= Since OE is a scalar multiple of OF and there is a common point O, O, E and F lie on a straight line. M1 A1 8(c)(i) 1 1area of triangle AEC 2 1area of triangle OEC 22 h h  =  1 2= B1 8(c)(ii) area of triangle OCD area of triangle OCD area of triangle OAD area of triangle OAD area of triangle OAD area of triangle OAB= = 21 32 = 1 3 Accepts other methods B1 9(a) Time taken by Ali = 42 x h Time taken by Robert = 42 2x− h

8 GESS 4E5N EM P2 PRELIM 2025 SZK 42 42 1 22xx −=− 42 42( 2) 1 ( 2) 2 xx xx −− =− 84 1 ( 2) 2xx =− 2 ( 2) 168 2 168 0 xx xx −= − − = M1 ( 42 x and 42 2x− seen) M1 M1 M1 (able to reduce) 9(b) 2 2 168 0xx− − = ( 12)( 14) 0xx+ − = 12 14x or x=− = Accept by Formula method M1 A1 9(c) Since x represents speed, x cannot be negative. B1 9(d) Time taken = 42 14 2− = 3.5 h B1 10(a) Total time taken = 13 h 50 min + 2 h = 15h 50 min Time = 8.05 am B1 10(b) Total per day = £25 + £10 + £15 = £50 Total for 6 days = 6 50 = £300 Assuming that he needs to save 10% of the total amount of $700 Total planned spending = 300(1.65) + 10 700100 = $565 Sufficient Alternative Solutions : Assuming that he needs to save 10% of the total amount planned for 6 days : Total planned spending = 300(1.65) + 10 (300 1.65)100 = $544.50 M1 M1 A1 M1*

9 GESS 4E5N EM P2 PRELIM 2025 SZK 10(c) Budget = 15 4 =£60 Option A Full price = £60 Discount = 5 90100 = £4.50 Final cost = 60 - 4.50 = £55.50 Exceed budget Option B 8 attractions 12.50 =£100 Exceed budget Option C 2 d