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)