MATHEMATICS Grade 10

AEO “Nazarbayev Intellectual Schools” Cambridge International Examinations

Paper 2 May 2017

Maximum Mark: 80

Marks awarded · The number of marks awarded for each part of the question should be recorded in the ‘For Examiner’s Use’ column at the right side of the page using the annotations indicated in the mark scheme e. g. M1 A1 · Half marks cannot be awarded. · If a question instructs the candidates to use a particular method then that method must be used. · In other questions any valid alternative method is acceptable, and candidates should be awarded equivalent marks for reaching a comparable stage in their solution. · Particular care should be taken when marking questions where the working leads to a given solution – the candidate must provide a full justification of the result. · If a question requires an exact solution then the candidate must use exact values throughout their working. Annotations and abbreviations M Marks are awarded for using a correct method and are not lost for purely numerical errors. A Marks are awarded for an accurate answer and depend on the preceding M marks. Therefore M0 A1 cannot be awarded. B Marks are independent of M marks and are awarded for a correct final answer or correct intermediate stage. DM or DB (or dep) is used to indicate that a particular M or B mark is dependent on earlier M or B mark in the mark scheme. Where follow through (ft) is indicated in the mark scheme, marks can be awarded where the candidate’s work follows correctly from a previous answer, whether or not it was correct.

Question	Answer	Marks	Additional Guidance
1(a)		B1 [1]	Accept any unambiguous indication of the correct answer
1(b)		B1 [1]	Accept any unambiguous indication of the correct answer
Total: 2 marks
2(a)	0. 56141579…	B1 [1]	Allow answer rounding to 0. 5614158
2(b)	57857. 476…	B1 [1]	Allow answer rounding to 57857. 48
Total: 2 marks
		B2 [2]	Accept for 2 marks Award B1 either for a parallelogram (not a rectangle) with height 3cm or for D being marked such that DM is perpendicular to AB
Total: 2 marks

Question	Answer	Marks	Additional Guidance
4(a)	or equivalent	M1	If correct formula is stated, allow one error when substituting a or r or n Accept 20 + 18 + 16. 2 + 14. 58 + 13. 1(22) + 11. 8(098)
	93. 7(118)	A1 [2]
4(b)		M1
		A1 [2]
Total: 4 marks
5(a)	(Σ x =) 30. 54 or 5. 09	B1
	Attempt to find 5. 65² + … + 5. 34² or 156(. 4614)	M1	Attempt to square and add
	or 0. 168 (8…)	M1 dep	Condone missing brackets. Dependent on previous M mark
	An answer rounding to 0. 411 (m)	A1 [4]	Accept 0. 41 (m) A correct answer with no working scores B2
	Alternative method (Σ x =) 30. 54 or 5. 09	B1
	(5. 65 –their 5. 09)²and (4. 83 – their5. 09)²and …	M1	Attempt to subtract mean from each value and to square
	or0. 168(8…)	M1 dep	Sum their squared values and divide by 6 Dependent on previous M mark
	An answer rounding to 0. 411 (m)	A1 [4]	Accept 0. 41 (m) A correct answer with no working scores B2
5(b)	Ticks Aisha and gives a correct explanation, e. g. · Aisha’s standard deviation is smaller · 0. 34 is less than 0. 41(1)	B1ft [1]	Follow through from their answer to (a) provided it is a positive number Accept equivalent responses
Total: 5 marks

Question	Answer	Marks	Additional Guidance
	or or equivalent or or or equivalent	M1	Attempt subtraction of vectors either way round Accept differences in coordinates for A and B seen or implied on diagram
	or equivalent or or equivalent	M1 dep	Dependent on previous M mark Correct method to find the position vector of C Accept difference in coordinates on diagram applied from B to C.
	(‒ 6, ‒ 3)	A1 [3]
Total: 3 marks
	π × 15²or706. 858… or707or706. 5	M1	All method marks could be earned by a single line of working
	(Area of sector =) π × 15²or 176. 7(…) or 177 or 176. 6(…)	M1
	(Area of triangle =) or112. 5	M1
	Answer between 64. 1 (cm²) and 64. 2 (cm²) inclusive	A1 [4]
Total: 4 marks

Question	Answer	Marks	Additional Guidance
8(a)	17 – (4)(5) < 0 so the function is not defined	B1 [1]	Or equivalent, e. g. and cannot square root a negative value
8(b)	or seen	M1
	≤ x ≤ aor ‒ b ≤ x ≤ or equivalent where a and b are positive numbers	M1	Accept use of < Accept region defined by two separate inequalities, e. g. x ≥ , x ≤ a
	or equivalent	A1 [3]	Now not allowing < signs
Total: 4 marks
9(a)	(3, 0)	B1
		B1 [2]	Accept
9(b)	x-coordinate of S= 8	B1ft	Follow through as their 3 + their 5. Check diagram for evidence of this mark
		M1	Attempt gradient of radius as (difference in y) / (difference in x). Can be implied as
		A1	Can be implied by an equation of the form or 4x + 3y + k = 0 or equivalent
	An equation through theirS with any non-zero, finite gradient, e. g. · y – 0 = m(x – their 8) · y = mx + c for any m with substitution of y = 0 and x = their 8 with an attempt to find c	M1
	4x + 3y – 32 = 0	A1 [5]	Allow 4kx + 3ky – 32k = 0 for any integer k
Total: 7 marks

Question		Answer	Marks	Additional Guidance
		1 + 2(7 – 1)/3 or 4 + 2(16 – 4)/3	M1	Correct method to find either the x or y coordinate of S. Allow equivalent methods using vectors.
		S is (5, 12)	A1
		(4 – their 5)² + (3 – their 12)²or 82	M1
		or 9. 06	A1 [4]
Total: 4 marks
	y + 2 = (y – 4)² with an attempt to expand brackets on right hand side		M1	Attempt to square (y – 4) should involve at least 2 of y², –8y, 16
	Obtain or equivalent 3 term quadratic equation		A1
	(y – 2)(y – 7) or or equivalent		M1 dep	Dependent on previous M mark. Attempt to solve a three term quadratic equation by either factorising (at least 2 correct terms in the quadratic should be obtained when their factors are multiplied) or by using the quadratic formula (allow + instead of ± and condone missing bracket around –9 under the square root)
	7 only		A1 [4]	If 7 is the answer but there is no working, award B1 as a special case
Total: 4 marks

Question	Answer	Marks	Additional Guidance
12(a)	0. 5(2x + 1 + x + 5)(2x + 6) or equivalent	B1	Accept equivalent correct expressions using the area of a triangle and rectangle
	Attempt to expand brackets to obtain a quadratic expression	M1	Expanded expression does not need to be simplified Mark can be awarded for expanding two brackets if their expression would give two correct terms when simplified to a three term quadratic
	Correct working leading to	A1 [3]	Award 0 marks if working is not seen
12(b)	State 2x + 6 < 32 or x< 13	B1
	3x² + 15x – 252 (> 0) or equivalent	M1
	x> 7	A1	allow statement in words Condone use of ≥
	7 < x< 13 or equivalent	A1 [4]
Total: 7 marks

Question	Answer	Marks	Additional Guidance
13(a)	or or	M1
		A1 [2]
13(b)	6! or 720	B1
	Sight of 4! or 24	M1
	or48	A1
	or equivalent	A1 [4]	Accept equivalent methods
	Alternative method and or and or and	B2	Can be implied by subsequent working Award B1 for sight of or or equivalent.
	× or × or 2 × ×	M1
	or equivalent	A1 [4]	An answer of implies B2
Total: 6 marks

Question	Answer	Marks	Additional Guidance
14(a)	State or imply	B1
	Correct use of either the addition law or subtraction law for logarithms	M1	e. g. or or
		A1 [3]
14(b)		M1	Allow = instead of >
		M1	For correct removal of logs in an inequality of the form log₈(…) > a Allow = instead of >
	t> 16	A1 [3]	Accept alternative methods
Total: 6 marks

Question	Answer	Marks	Additional Guidance
15(a)	or 9i + 2jor equivalent and or 8i – 5jor equivalent	B2	Or B1 for or or or or equivalent
	(their 9)(their 8) + (their 2)(their ‒ 5)	M1	Calculate scalar product of their attempts at the two vectors
	Confirm 62	A1 [4]	Answer given. Must see correct vectors and evidence of scalar product, e. g.
15(b)	or	M1	Attempt magnitude of at least one vector
	Obtain and	A1ft	Follow through on their and
	62 = (their )(their )cosθ	M1	Or equivalent
	An answer that rounds to 44. 5°found using the scalar product	A1 [4]	Condone answer given in radians (0. 777 radians). If the cosine rule is used, the first M1 A1 are the only marks that may be awarded
Total: 8 marks

Question	Answer	Marks	Additional Guidance
16(a)	common difference = 0. 15	B1	Seen or implied
	Attempt expression of form 0. 6 + (30 – 1)d	M1
	4. 95 (kg)	A1 [3]
16(b)	or	M1	Attempt to find total mass of fertiliser
	83. 25	A1ft	Seen or implied by later work. Follow through their 4. 95 or theird
	11 sacks needed	A1ft	Follow through by rounding their 83. 25 ÷ 8 up to the nearest whole number
	19800 (Tenge)	A1ft [4]
Total: 7 marks

Question	Answer	Marks	Additional Guidance
	AngleADB = 4°	B1
	or	M1	their 4 must be an attempt at angle ADB. Check for evidence on diagram.
	or 99. 7… or or 149(. 02…)	M1
	or	M1	Correct trigonometry statement involving CD in either triangle BCD or triangle ACD
	21 (m)	A1 [5]	An answer of 20. 7(4…) implies 4 marks.
	Alternative method	B1
		B1	Not tan8 =
	Attempt to eliminate BC, e. g.	M1	tan8 and tan12 may be correct numerical values
	Correct linear equation involving CD, e. g.	A1
	21 (m)	A1 [5]	An answer of 20. 7(4…) implies 4 marks.
Total: 5 marks

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