Mathematics > MARK SCHEMES > Level 2 Certificate in Further Mathematics Practice Paper Set 2. Paper 1 8360/1. (All)
Level 2 Certificate in Further Mathematics Practice Paper Set 2. Paper 1 8360/1.
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Glossary for Mark Schemes These examinations are marked in such a way as to award positive achievement wherever possible. Thus, for these papers, marks are awarded under various categories. M Metho... d marks are awarded for a correct method which could lead to a correct answer. A Accuracy marks are awarded when following on from a correct method. It is not necessary to always see the method. This can be implied. B Marks awarded independent of method. M Dep A method mark dependent on a previous method mark being awarded. BDep A mark that can only be awarded if a previous independent mark has been awarded. ft Follow through marks. Marks awarded following a mistake in an earlier step. SC Special case. Marks awarded within the scheme for a common misinterpretation which has some mathematical worth. oe Or equivalent. Accept answers that are equivalent. eg, accept 0.5 as well as 12Practice Paper - Paper 1 - Set 2 - 8360/1 - AQA Level 2 Certificate in Further Mathematics 4 Paper 1 - Non-Calculator Q Answer Mark Comments 1 6x2 5 B2 B1 For one term correct 2(a) 7n 3 B2 B1 For 7n ? 2(b) Their 7n 3 150 M1 oe Allow 7n 3 = 150 21 A1 Alt 2(b) Substitutes a value of n 20 M1 Works out 15 or more terms 21 A1 3 5x + k = 3 M1 3x k = 1 M1 8x = 4 M1 (x = ) 0.5 A1 oe 4 3a + 8 = 2 or –3 + 4 = b M1 Sight of b = 1 or a = 2 earns this mark 3a + 8 = 2 and –3 + 4 = b M1 a = –2 and b = 1 A1 5(a) 12 4 or 13 M1 oe 5(b) y7 = 12 4 M1 oe (y =) 2 13 A1 oe 6 20x2 + 15xy 8xy 6y2 M1 oe Must have 4 terms with at least 3 correct 20x2 + 15xy 8xy 6y2 A1 20x2 + 7xy 6y2 A1ft ft From M1 A0AQA Level 2 Certificate in Further Mathematics - 8360/1- Set 2 - Paper 1 - Practice Paper 5 Q Answer Mark Comments 7(a) 3(42) + 4 (3)2 (3) M1 46 A1 7(b) 3x2 + x – 52 – 5 (= 0) M1 3x2 + x – 30 (= 0) M1 (3x ± a)(x ± b) (= 0) M1 ab = 30 or a + b = 1 3 10 and 3 A1 oe 8(a) Gradient = –2 M1 y + 6 = – 2(x – 1) M1 y = – 2x – 4 A1 Alt 8(a) Gradient = –2 M1 6 = 2(1) + c or c = 4 M1 y = 2x 4 A1 8(b) Gradient = 12 M1 ft Their gradient in (a) y – 4 = 12 (x + 5) M1 Substituting y = 0 M1 Dep Dep on attempt at straight line equation x = –13 or (–13, 0) A1 Alt 8(b) Gradient = 12 M1 ft Their gradient in (a) 4 = 12 (5) + c or c = 6 12 M1 y = 12 x + 6 12 and substituting y = 0 M1 Dep Dep on attempt at straight line equation x = 13 or (13, 0) A1 [Show More]
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