56 a x y x B1 2 B1 for initial explanation M Take two numbers and where > y. B1 B1 cao B1 B1 cao First step: 5(x 2) = 5x 10 – – B1 B1 cao y) Second step: 2(5x 10 + – = 10x 20 + 2y – B1 B1 cao y Third step: 10x 20 + 2y + 9 – – M1 M1 for explanation of how this shows the final result y = 10x 11 + – y Fourth step: 10x 11 + + 11 = 10x + – y Hence where the first two numbers b might have been 7 and 3, the final A1 A1 for defining each digit outcome would be 70 + 3 = 73. x Let the single-digit number be and M1 M1 for expressing the first manipulation algebraically b. the two-digit number be 10a + b) First step: 10(10a + 9x = 100a + – A1 for showing it in terms of hundreds, tens and units A1 10b 9x – A1 for correct hundreds A1 x = 100a + 10b + 10x – A1 for correct tens A1 x = 100a + 10b 10x + A1 for correct unit explanation – A1 x) x B1 for clear explanation as to how the second = 100a + 10(b + B1 – manipulation gives the two-digit term a. The hundreds unit is x). The tens unit is (b – x, The units term is the same as the single digit we started with. 13 x) The split then becomes 10a + (b – Expand each square and add: M1 2 M1 for expanding brackets M 57 x. and A1 A1 for correct expansion of brackets n2 n2 n2 2n + 1 + + + 2n + 1 – b x Adding these two gives 10a + + – n B1 B1 for showing how the terms cancel = 3n2 + 2n 2n + 2 x – B1 = 3n2 + 2 as given. b, B1 for complete solution with no incorrect notation or which is 10a + the two-digit term. terminology 4
56 a b Take two numbers x and y where x > y. First step: 5(x – 2) = 5x – 10 Second step: 2(5x – 10 + y) = 10x – 20 + 2y Third step: 10x – 20 + 2y + 9 – y = 10x – 11 + y Fourth step: 10x – 11 + y + 11 = 10x + y Hence where the first two numbers might have been 7 and 3, the final outcome would be 70 + 3 = 73. Let the single-digit number be x and the two-digit number be 10a + b. First step: 10(10a + b) – 9x = 100a + 10b – 9x = 100a + 10b + x – 10x = 100a + 10b – 10x + x = 100a + 10(b – x) + x The hundreds unit is a. The tens unit is (b – x). The units term is x, the same as the single digit we started with. The split then becomes 10a + (b – x) and x. Adding these two gives 10a + b – x + x which is 10a + b, the two-digit term. B1 B1 B1 B1 B1 M1 A1 M1 A1 A1 A1 A1 B1 2 B1 for initial explanation B1 cao B1 cao B1 cao B1 cao M1 for explanation of how this shows the final result A1 for defining each digit M1 for expressing the first manipulation algebraically A1 for showing it in terms of hundreds, tens and units A1 for correct hundreds A1 for correct tens A1 for correct unit explanation B1 for clear explanation as to how the second manipulation gives the two-digit term M 13 57 Expand each square and add: n2 – 2n + 1 + n2 + n2 + 2n + 1 = 3n2 + 2n – 2n + 2 = 3n2 + 2 as given. M1 A1 B1 B1 2 M1 for expanding brackets A1 for correct expansion of brackets B1 for showing how the n terms cancel B1 for complete solution with no incorrect notation or terminology M 4
Higher_Skills_Algebra_MarkScheme_AQA.docx