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Matematika angol nyelven középszint 0511 ÉRETTSÉGI VIZSGA 005. május 10. MATEMATIKA ANGOL NYELVEN MATHEMATICS KÖZÉPSZINTŰ ÉRETTSÉGI VIZSGA STANDARD LEVEL FINAL EXAMINATION Az írásbeli vizsga időtartama:

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Matematika angol nyelven középszint 0511 ÉRETTSÉGI VIZSGA 005. május 10. MATEMATIKA ANGOL NYELVEN MATHEMATICS KÖZÉPSZINTŰ ÉRETTSÉGI VIZSGA STANDARD LEVEL FINAL EXAMINATION Az írásbeli vizsga időtartama: 180 perc Time allowed for the examination: 180 minutes JAVÍTÁSI-ÉRTÉKELÉSI ÚTMUTATÓ MARKSCHEME OKTATÁSI MINISZTÉRIUM MINISTRY OF EDUCATION Instructions to examiners Formal requirements: Mark the paper in ink, different in colour from the one used by the candidate. Indicate the errors, incomplete solutions, etc. in the conventional way. The first one of the rectangles under each problem shows the maximum attainable score on that problem. The points given by the examiner are to be entered in the rectangle next to that. If the solution is perfect, it is enough to enter the maximum scores in the appropriate rectangles. If the solution is incomplete or incorrect, please indicate the individual subtotals on the paper, too. Assessment of content: The markscheme contains more than one solution for some of the problems. If the solution by the candidate is different, allocate the points by identifying the parts of the solution equivalent to those of the one given in the markscheme. The subtotals in the markscheme can be further divided, but the scores awarded should always be whole numbers. If it is clear that the reasoning and the final answer are both correct, you may award the maximum score even if the solution is less detailed than the one in the markscheme. If there is a calculation error or inaccuracy in the solution, only take off the points for that part where the error is made. If the reasoning remains correct and the error is carried forward, the points for the rest of the solution should be awarded. In the case of a principal error, no points should be awarded at all for that section of the solution, not even for steps that are formally correct. (These logical sections of the solutions are separated by double lines in the markscheme.) However, if the wrong information obtained owing to the principal error is carried forward to the next section or in the next part of the problem and used correctly, the maximum score is due for the next part. Where the markscheme shows a unit in brackets, the solution should be considered complete without that unit as well. If there are more than one different approaches to a problem, assess only one of them (the one that is worth the largest number of points). Do not give extra points (i.e. more than the score due for the problem or part of problem). Do not take off points for steps or calculations that contain errors but are not actually used by the candidate in the solution of the problem. Assess only two out of the three problems in part B of Paper II. The candidate was requested to indicate in the appropriate square the number of the problem not to be assessed and counted in their total score. Should there be a solution to that problem, it does not need to be marked. However, if it is still not clear which problem the candidate does not want to be assessed, assume automatically that it is the last one in the question paper, and do not assess that problem. írásbeli vizsga 0511 / május 10. 1. F ; 1. I. should be awarded if only one coordinate is correct.. B.. [; 6] Or: y 6. points should be subtracted if the left or right end of the interval is wrong or if the interval is open or partly open. points 4. A: false. B: true. C: false. points 5. ( + ) + ( y 5) = 16 x. Or: x + y + 6x 10y + 18 = Any form of the answer is or 14% or acceptable. írásbeli vizsga 0511 / május 10. 7. tan 18.5 =. x The other leg is x (cm). points is given for indicating the data in the diagram. It is also correct without rounding a 5 =. 9. The number of edges is 4. should be awarded if there is only a correct sketch. 10. If the graph is good but it is not restricted to the given interval, should be awarded. írásbeli vizsga / május 10. 11. a) b) = ! = 10. The should also be awarded if the binomial coefficient is not evaluated. The should also be awarded if the factorial is not evaluated. 1. 4r π V =. 4 1 π V =. V 90.8 (cm ). There is 9. litres of air in the ball. The is for the conversion. points írásbeli vizsga / május 10. 1. ( cos x) II/A cos x + 4cos x = 1. Rearranged: 4cos x + 4cos x = 0. The roots of this equation are 1 cos x = or cos x =. 1 π If cos x =, then x 1 = + kπ, 5π or x = + kπ, points where k Z. If cos x =, then there is no solution, since cos 1 for all x. Since the transformations have been equivalent, both sets of roots are solutions of the original equation. 1 The for checking is also due if periods are not indicated but the two roots obtained are substituted into the equation. 14. a) a = 17 and a = 1. d = 4. The is due for the common difference. a 1 = 1. a 150 = 609. The value of a 150 is also accepted if it only appears in the summation formula S 150 = 150. S 150 = points b) The rule for divisibility by three can be applied. The digits of 5 86 add up to 4, thus it is divisible by three. The sum remains the same for any order, so the statement is true. points The are also due if the divisibility rule is not stated, only applied. írásbeli vizsga / május 10. c) The rule for divisibility by four can be applied. The is also due if the rule is not stated but there is evidence of its correct application. In this case, the condition is met if the last two digits are 8; ; 6; 5; 56; 68. If there are only four or five out of the six endings listed, award instead of. If there are fewer than that, award 0 points. Thus the digit in the tenths place may be,, 5 or 6. This point is only due if all solutions are listed. 4 points If none of the six endings are listed but the rule of divisibility is applied correctly and the answer is also correct, award 4 points. Award 4 points as well if the rule of divisibility is not stated but the endings are listed correctly and the answer is correct, too. 15. a) The arithmetic mean is = 15 = 61. Mode: 100. Median: points b) Grade excellent good satisfactory pass fail Number of students c) Excellent: 19. Good: 4. Satisfactory: 48. Fail: 96. The calculation of the central angles does not need to be shown but the angles have to be stated. írásbeli vizsga / május 10. exc e lle nt good fail pass points 5 points Award only if it is not clear from the pie chart which grade belongs to which sector. The diagram is only acceptable if the marked boundaries of sectors lie between the appropriate ten-degree divisions. írásbeli vizsga / május 10. II/B Out of problems 16 to 18, do not assess the one indicated by the candidate. 16. a) a = r. From the Pythagorean theorem: = r + ( 5 ) 4 = r + ( 5 ) * a. * r. * r = 5 cm. * a = 10 cm. * The longitudinal section containing the axis is an equilateral triangle. A = r π + rπa. A = 5π + 50π. A = 75π. Or A 5.6 cm. 9 points * Award the appropriate points as well if these results only appear in the answers to parts b) or c). b) r π m V =. 5π 5 V =. V 6.7 cm. c) Solution 1. The radius of the sector is a. The length of the arc is aπ. α aπ =. 60 aπ The central angle in question is α = 180. The point is also due for correct calculation with an approximate value. 6 points írásbeli vizsga / május 10. Solution. The radius of the sector is a. The length of the arc is aπ. The perimeter of the whole circle is aπ. The arc length is one half of it, i.e. it is a semicircle. Thus α = points 17. a) Let x denote the price of the magazine. This is also due if the unknown is not defined but its meaning is made clear by the verbal answer. Anna had 0.88x forints. 4 Zsuzsi had x forints. 5 The equation: x + x x = x = x = 94 and 4 x = The magazine cost 1050 forints. Anna originally had 94 forints and Zsuzsi had 840 forints. Checking: 10 points b) Solution 1. Anna receives a share of a forints, and Zsuzsi gets 714 a forints out of the money remaining. 94 a a = or = a a Hence a = 74; 714 a = 40. Thus Anna will have 74 forints and Zsuzsi will have 40 forints left after buying the magazine. Checking: 7 points Award 4 points altogether for setting up the equation. This is also due if the unknown is not defined but its meaning is made clear by the verbal answer. Either equation is acceptable. írásbeli vizsga / május 10. Solution. The two of them had 1764 forints. 94 Anna receives of the money remaining, that is 714 = 1764 = 74 forints, and 840 Zsuzsi gets of it, that is 714 = 1764 = 40 forints. 7 points 18. a) Solution 1. The number of differences that at least one of them noticed is = 19. Neither of them noticed 19 = 4 differences. Solution. The number of differences found can also be expressed without a set diagram: Thus the number of differences that at least one of them noticed is 19. Neither of them noticed 19 = 4 differences. b) 4 points 4 points If only one or two of the three numbers in the set diagram are correct, award only. Do not give partial credit here. 7 points c) There is a difference that Enikő did not find. OR: Enikő did not find every difference. OR: Enikő did not find all the differences. 7 points One point for each correct number in the diagram. Do not give partial credit here. írásbeli vizsga / május 10. d) The number of favourable cases is 14. The number of all cases is. 14 The probability in question is or 0.61 or 61%. 4 points These points are also due if the diagram in part b) is filled out with errors but those values are carried forward consistently here. The result is acceptable in any form, including values rounded correctly. írásbeli vizsga / május 10.

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