38 Area of large square = 72 = 49 B1 2 B1 for area of large square M 3 To find the length (z) of the side of the M1 M1 for use of theorem to help find the side Pythagoras’ small square: length of inner square z2 = z2 = z2 = 5 A1 A1 cao So the area of the small square is 25. A1 A1 cao Shaded area is equal to A1 A1 for area of shaded part area large square area small square. C2 C1 for clear explanation – 49 25 = 24 C1 for complete explanation with correct mathematical – So less than half is shaded, as notation throughout required. 7 39 Set up two simultaneous P1 2 P1 for clear initial explanation M equations, using the information given. 5 is the first term; from the rule for the sequence, the next term is: a b, 5 × which equals 23 – B1 B1 for first equation b Hence 5a = 23 (1) B1 – …… Doing the same to the next term M1 B1 for second equation gives: b 23a = (2) – 113 ……. M1 for clear method of elimination A1 Subtract (1) from (2) to give: 18a = 90 M1 A1 cao a = = 5 Substitute in (1): A1 M1 for substitution b 25 = 23 – A1 cao b = 2 C1 a b So = 5, = 2 8 C1 for next two terms correctly found Hence, the next 2 terms are 563 and 2813.
38 Area of large square = 72 = 49 To find the length (z) of the side of the small square: z2 = z2 = z2 = 5 So the area of the small square is 25. Shaded area is equal to area large square – area small square. 49 – 25 = 24 So less than half is shaded, as required. B1 M1 A1 A1 A1 C2 2 3 B1 for area of large square M1 for use of Pythagoras’ theorem to help find the side length of inner square A1 cao A1 cao A1 for area of shaded part C1 for clear explanation C1 for complete explanation with correct mathematical notation throughout M 7 39 Set up two simultaneous equations, using the information given. 5 is the first term; from the rule for the sequence, the next term is: 5 × a – b, which equals 23 Hence 5a – b = 23 …… (1) Doing the same to the next term gives: 23a – b = 113 ……. (2) Subtract (1) from (2) to give: 18a = 90 a = = 5 Substitute in (1): 25 – b = 23 b = 2 So a = 5, b = 2 Hence, the next 2 terms are 563 and 2813. P1 B1 B1 M1 A1 M1 A1 C1 2 P1 for clear initial explanation B1 for first equation B1 for second equation M1 for clear method of elimination A1 cao M1 for substitution A1 cao C1 for next two terms correctly found M 8
Higher_Skills_Algebra_MarkScheme_Edexcel.docx