Matematika | Középiskola » Matematika angol nyelven középszintű írásbeli érettségi vizsga megoldással, 2012

ÉRETTSÉGI VIZSGA 2012. május 8 Név: . osztály: MATEMATIKA ANGOL NYELVEN KÖZÉPSZINTŰ ÍRÁSBELI VIZSGA 2012. május 8 8:00 I. Időtartam: 45 perc Pótlapok száma Tisztázati Piszkozati NEMZETI ERŐFORRÁS MINISZTÉRIUM Matematika angol nyelven középszint írásbeli vizsga 1111 I. összetevő Matematika angol nyelven középszint Név: . osztály: Instructions to candidates 1. The time allowed for this examination paper is 45 minutes When that time is over, you will have to stop working. 2. You may solve the problems in any order 3. In solving the problems, you are allowed to use a calculator that cannot store and display verbal information. You are also allowed to use any book of four-digit data tables The use of any other electronic devices, or printed or written material is forbidden. 4. Write the final answers in the appropriate frames You are only required to write down details of the solutions if you are instructed by the problem to do so. 5. Write

in pen The examiner is instructed not to mark anything in pencil, other than diagrams. Diagrams are allowed to be drawn in pencil If you cancel any solution or part of a solution by crossing it over, it will not be assessed. 6. Only one solution to each problem will be assessed In the case of more than one attempt to solve a problem, indicate clearly which attempt you wish to be marked. 7. Do not write anything in the grey rectangles írásbeli vizsga, I. összetevő 1111 2/8 2012. május 8 Matematika angol nyelven középszint 1. Név: . osztály: 1 on the set of real numbers not x −3 1 equal to 3. For what real number x is the value of function f equal to ? 20 The function f is defined by the formula f ( x) = x= 2. 2 points The vectors a and b are the two vectors drawn from an acute-angled vertex of a rhombus to the two adjacent vertices. Express in terms of a and b the vector of the diagonal drawn from the same vertex. The vector in question is 2 points 3. Find

the real number x for which the following equation is true: 2−x = 8 x= írásbeli vizsga, I. összetevő 1111 2 points 3/8 2012. május 8 Matematika angol nyelven középszint 4. Név: . osztály: Which of the graphs below represents the function g: R R , g ( x ) = 2 x + 1 ? What is the zero of the function g? y y 1 y 1 1 1 x 1 A x 1 B C The letter marking the graph of g: 2 points The zero of the function is at 1 point 5. In how many different ways is it possible to select exactly four out of six books recommended for reading? The number of possible selections: 6. 2 points Given the information below, represent the sets A and B in a diagram, and list the elements of the set A ∩ B . A ∪ B = { x; y; z; u; v; w }, A B={ z; u }, B A={ v; w }. 1 point A∩ B ={ írásbeli vizsga, I. összetevő 1111 } 4/8 1 point 2012. május 8 x Matematika angol nyelven középszint 7. Név: . osztály: The percentage increase of the value of an

investment paper is 10% every year. If its present value is 50 000 forints, how much will it be worth in two years? Explain your answer. 2 points The value of the paper will be 1 point 8. N=437y51 denotes a six-digit number in decimal notation. Given that N is divisible by three, find the possible values of the digit y. The possible values of the digit y: 2 points írásbeli vizsga, I. összetevő 1111 5/8 2012. május 8 Matematika angol nyelven középszint 9. Név: . osztály: At which point of the domain does the function f: R R, f ( x) = −( x − 6) 2 + 3 have its maximum, and what is the maximum value? The maximum occurs at 1 point Maximum value: 1 point 10. There are five passengers travelling in a railway compartment One passenger knows three of the others, there are three who know two fellow passengers each, and there is one who knows only one other passenger. (Acquaintances are mutual) Draw a possible graph representing the acquaintances in this group of

five people. A possible graph of acquaintances: 3 points írásbeli vizsga, I. összetevő 1111 6/8 2012. május 8 Matematika angol nyelven középszint Név: . osztály: 11. Determine the coordinates of the centre of the circle of equation x 2 + y 2 − 4 x + 2 y = 0 What is the radius of the circle? Explain your answer. 2 points Centre: 1 point Radius: 1 point 12. Decide about each of the statements below whether it is true or false A: Out of two real numbers, the one with a larger square is larger. B: If a number is divisible by both 5 and 15, then it is also divisible by their product. C: The cosine of the smaller angle out of two different acute angles is larger. írásbeli vizsga, I. összetevő 1111 A: 1 point B: 1 point C: 1 point 7/8 2012. május 8 Matematika angol nyelven középszint Part I Név: . osztály: maximum score Question 1 2 Question 2 2 Question 3 2 Question 4 3 Question 5 2 Question 6 2 Question 7 3 Question 8 2 Question 9 2

Question 10 3 Question 11 4 Question 12 3 TOTAL 30 date points awarded examiner elért pontszám egész programba beírt számra kerekítve / egész pontszám / score rounded to integer score entered in program integer I. rész / Part I javító tanár / examiner jegyző / registrar dátum / date dátum / date Megjegyzések: 1. Ha a vizsgázó a II írásbeli összetevő megoldását elkezdte, akkor ez a táblázat és az aláírási rész maradjon üresen! 2. Ha a vizsga az I összetevő teljesítése közben megszakad, illetve nem folytatódik a II. összetevővel, akkor ez a táblázat és az aláírási rész kitöltendő! Remarks. 1. If the candidate has started working on Part II of the written examination, then this table and the signature section remain blank. 2. Fill out the table and signature section if the examination is interrupted during Part I or it does not continue with Part II.

írásbeli vizsga, I. összetevő 1111 8/8 2012. május 8 ÉRETTSÉGI VIZSGA 2012. május 8 Név: . osztály: MATEMATIKA ANGOL NYELVEN KÖZÉPSZINTŰ ÍRÁSBELI VIZSGA 2012. május 8 8:00 II. Időtartam: 135 perc Pótlapok száma Tisztázati Piszkozati NEMZETI ERŐFORRÁS MINISZTÉRIUM Matematika angol nyelven középszint írásbeli vizsga 1111 II. összetevő Matematika angol nyelven középszint írásbeli vizsga, II. összetevő 1111 Név: . osztály: 2 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: Instructions to candidates 1. The time allowed for this examination paper is 135 minutes When that time is over, you will have to stop working. 2. You may solve the problems in any order 3. In part B, you are only required to solve two out of the three problems When you have finished the examination paper, write in the square below the number of the problem NOT selected. If it is not clear for the examiner which problem you do

not want to be assessed, then problem 18 will not be assessed. 4. In solving the problems, you are allowed to use a calculator that cannot store and display verbal information. You are also allowed to use any book of four-digit data tables The use of any other electronic devices, or printed or written material is forbidden. 5. Always write down the reasoning used in obtaining the answers, since a large part of the attainable points will be awarded for that. 6. Make sure that the calculations of intermediate results are also possible to follow 7. In solving the problems, theorems studied and given a name in class (eg the Pythagorean theorem or the altitude theorem) do not need to be stated precisely. It is enough to refer to them by the name, but their applicability needs to be briefly explained. 8. Always state the final result (the answer to the question of the problem) in words, too 9. Write in pen The examiner is instructed not to mark anything in pencil, other than diagrams.

Diagrams are allowed to be drawn in pencil If you cancel any solution or part of a solution by crossing it over, it will not be assessed. 10. Only one solution to each problem will be assessed In the case of more than one attempt to solve a problem, indicate clearly which attempt you wish to be marked. 11. Do not write anything in the grey rectangles írásbeli vizsga, II. összetevő 1111 3 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: A 13. The tenth term of an arithmetic progression is 10 and its common difference is 4 a) Paul states that the tenth term of the sequence in binary notation is 1011. Is he right? Prove or disprove his statement. b) What is the first term of the sequence? c) Find the smallest term in the sequence that is a three-digit number. What is the index of this term? d) Consider the set of two-digit positive terms of the sequence. elements does the set have? írásbeli vizsga, II. összetevő 1111 4 / 16 a) 3

points b) 2 points c) 4 points d) 3 points T.: 12 points How many 2012. május 8 Matematika angol nyelven középszint írásbeli vizsga, II. összetevő 1111 Név: . osztály: 5 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: 14. The population of Ailington is 12 320 According to statistics issued by Ailington Hospital, 1978 of the inhabitants were hospitalised during the previous year. a) What is the probability that an inhabitant of Ailington selected at random was treated in the hospital of the town in the previous year? Round the probability to two decimal places. Among those treated in hospital in that year, there were 138 people under the age of 18 years, 633 people between 18 and 60 years, and the rest of them were older. 24% of the population of the town are over 60 years, 18% are under 18. (In calculations, you can assume that there was no significant change in the given data during the course of the year.) b) Draw a

pie chart representing the age distribution of the people treated in hospital. Write down all calculations used in making the pie chart. c) How much will the probability calculated in a) increase or decrease if the random person is selected from the inhabitants over the age of 60 years? írásbeli vizsga, II. összetevő 1111 6 / 16 a) 3 points b) 5 points c) 4 points T.: 12 points 2012. május 8 Matematika angol nyelven középszint írásbeli vizsga, II. összetevő 1111 Név: . osztály: 7 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: 15. The diagram below was created by surveyors It represents points in a horizontal plane Point Q is separated from the other points by a river. The surveyor working in point A was 720 metres away from point P, and he saw the points P and Q in a straight line. He measured the angle PAB to be 53° The surveyor standing at point B, 620 metres away from A, measured the angle ABQ to be 108°. Use

the given information to calculate the distances BP, PQ and BQ. Round your answers to the nearest metre. Q P A T.: B írásbeli vizsga, II. összetevő 1111 8 / 16 12 points 2012. május 8 Matematika angol nyelven középszint írásbeli vizsga, II. összetevő 1111 Név: . osztály: 9 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: B You are required to solve any two out of the problems 16 to 18. Write the number of the problem NOT selected in the blank square on page 3. 16. Teams A and B are the national teams of chess players of two countries. They are training together for the world tournament. During the first week of the training, each national team play a round robin, that is, every player plays a game against every other player in the same team. Team A consists of 7 players, team B plays 55 games altogether. a) How many games are played in team A, and how many members does team B have? During the second week, each of 6

selected members of team A opposes each of 8 members of team B once. b) How many games are played during the second week altogether? At the end of the training, four identical gifts are given to players selected at random from all participants of the training. Each player may only get at most one gift c) What is the probability that one gift is received by a member of team A and three are received by members of team B? írásbeli vizsga, II. összetevő 1111 10 / 16 a) 7 points b) 3 points c) 7 points T.: 17 points 2012. május 8 Matematika angol nyelven középszint Név: . osztály: írásbeli vizsga, II. összetevő 1111 11 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: You are required to solve any two out of the problems 16 to 18. Write the number of the problem NOT selected in the blank square on page 3. 17. a) Solve the following equation on the set of real numbers. lg(2 x − 1) + lg(2 x − 3) = lg 8 b) An angle x

of a triangle satisfies the equation 4 cos2 x − 8 cos x − 5 = 0 . Find the angle x. c) Solve the following equation on the set of real numbers. 4y − 5 = 8 y d) Seven different real numbers are given. One of them is a solution of the equation in question c). The numbers are listed in some order How many possible orders of the given numbers are there in which the number from c) stands in the middle position? írásbeli vizsga, II. összetevő 1111 12 / 16 a) 6 points b) 4 points c) 4 points d) 3 points T.: 17 points 2012. május 8 Matematika angol nyelven középszint Név: . osztály: írásbeli vizsga, II. összetevő 1111 13 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: You are required to solve any two out of the problems 16 to 18. Write the number of the problem NOT selected in the blank square on page 3. 18. The middle part of a water tank is a right circular cylinder of inner diameter 6 m and height 8 m. The

lower part of the tank is a hemisphere, and the upper part is a right circular cone. The height of the cone is 3 m The water tank is standing in a vertical position, a longitudinal section through its axis of rotation is shown below. a) The entire interior surface is to be repaired. Calculate the area in square metres that needs to be coated with waterproof material. b) The tank is filled with water up to 85% of its total height. Find the volume of the water in cubic metres. Ignore the thickness of the waterproof coating Round your answers to the nearest whole number. írásbeli vizsga, II. összetevő 1111 14 / 16 a) 6 points b) 11 points T.: 17 points 2012. május 8 Matematika angol nyelven középszint Név: . osztály: írásbeli vizsga, II. összetevő 1111 15 / 16 2012. május 8 Matematika angol nyelven középszint Név: . osztály: number of problem maximum score 13. 12 14. 12 15. 12 Part II.A points awarded total 17 Part II.B 17 ←

problem not selected TOTAL 70 maximum score Part I 30 Part II 70 Total score on written examination 100 date points awarded examiner elért pontszám egész számra kerekítve / score rounded to integer programba beírt egész pontszám / integer score entered in program I. rész / Part I II. rész / Part II javító tanár / examiner jegyző / registrar dátum / date dátum / date írásbeli vizsga, II. összetevő 1111 16 / 16 2012. május 8 ÉRETTSÉGI VIZSGA 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 1111 MATEMATIKA ANGOL NYELVEN KÖZÉPSZINTŰ ÍRÁSBELI ÉRETTSÉGI VIZSGA JAVÍTÁSI-ÉRTÉKELÉSI ÚTMUTATÓ NEMZETI ERŐFORRÁS MINISZTÉRIUM Matematika angol nyelven középszint Javítási-értékelési útmutató Instructions to examiners Formal requirements: 1. 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 2. 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. 3. If the solution is perfect, it is enough to enter the maximum scores in the appropriate rectangles. 4. If the solution is incomplete or incorrect, please indicate the individual partial scores in the body of the paper, too. 5. Do not assess anything, except diagrams, that is written in pencil Assessment of content: 1. The markscheme may contain 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 solution(s) given in the markscheme. 2. The subtotals in the markscheme can be further divided, but the scores awarded should always be whole numbers. 3. 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. 4. If there is a calculation error or inaccuracy in the solution, only take off the points for that part where the error occurs. If the reasoning remains correct and the error is carried forward, the points for the rest of the solution should be awarded. 5. 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 is used correctly, the maximum score is due for the next part, provided that the error has not changed the nature of the task to be completed. 6. Where the markscheme shows a unit or a remark in brackets, the

solution is considered complete without that unit or remark as well. 7. If there are more than one different approaches to a problem, assess only the one indicated by the candidate. 8. Do not give extra points (ie more than the score due for the problem or part of problem). 9. 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. 10. 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 1111 2 / 13 2012. május 8 Matematika angol nyelven középszint

Javítási-értékelési útmutató I. 1. x − 3 = 20 1 point Total: Award full mark for a 1 point correct answer without explanation. 2 points Total: Award 1 point if the answer 2 points does not make it clear that a and b are vectors. 2 points Total: 2 points 2 points The letter marking the graph of function g: B. The zero of the function: ( x =) − 1. Total: 2 points 1 point 3 points x = 23 2. a+b 3. x = −3 4. 5. ⎛ 6⎞ There are 15 possibilities. Total: 2 points Accept ⎜⎜ ⎟⎟ as well. ⎝ 4⎠ 2 points 6. Correct diagram. A z u B x y 1 point v w A ∩ B = {x; y} Total: 1 point 2 points Total: These two points are due if 50 000 ⋅1.12 is stated without 1 point referring to a formula. Award 1 mark if the value in one year is calculated 1 point correctly but the final result is wrong. 3 points 7. t 2 = t0 ⋅ q 2 1 point t2 = 50 000 ⋅1.12 The value of the investment paper is 60 500 forints. írásbeli vizsga 1111 3 / 13 2012. május 8

Matematika angol nyelven középszint Javítási-értékelési útmutató 8. Total: 1 point for one or two correct values. No point 2 points if any incorrect value is stated. 2 points Total: 1 point 1 point 2 points The possible values of y are 1; 4; 7. 9. The maximum occurs at 6. The maximum value is 3. 10. The diagram contains exactly one point of degree three, exactly three points of degree two, exactly one point of degree one. 1 point 1 point 1 point All three points are due for a correct diagram. Total: 3 points Total: These 2 points are also due if the formulae given 2 points in the data tables book are used correctly. 1 point 1 point 4 points Total: 1 point 1 point 1 point 3 points 11. (x − 2 )2 + ( y + 1)2 = 5 The centre is point O(2; –1), the radius is 5 . 12. A: false. B: false. C: true. írásbeli vizsga 1111 4 / 13 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató II. A 13. a) 10112=11, Paul’s

statement is false. Total: 2 points 1 point 3 points Total: 1 point 1 point 2 points 13. b) 10 = a1 + 36 a1 = −26 13. c) Solution 1 − 26 + (n − 1) ⋅ 4 ≥ 100 2 points n ≥ 32.5 , therefore the index of the term is 33 The term in question is a33 = 102 . Total: Award 1 point if a stricter relation is used. 1 point 1 point 4 points 13. c) Solution 2 The sequence consists of numbers leaving a remainder of 2 when divided by 4. The smallest such number of three digits is 102. 10 + k .4 = 102 ; k = 23 Thus the index of the term in question is 10 + 23 = 33. Total: 1 point 1 point 1 point 1 point 4 points 13. d) The first two-digit positive term is a10 = 10 , 2 points and the last one is a32 = 98 , so the set has 22+1=23 elements. Total: írásbeli vizsga 1111 5 / 13 1 point 3 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 14. a) p= p= This point is also due if the 1 point idea is only reflected by the

solution. k ⎛ number of favourable cases ⎞ ⎜= ⎟ n⎝ number of all cases ⎠ 1978 ≈ 12320 1 point ≈ 0.16 Total: 1 point ≈16,06% 3 points 14. b) between 18 and 60 under 18 60 év feletti over 60 The number of those over 60 treated in hospital is 1978 − 138 − 633 = 1207 . The 138 people under 18 correspond to a central 138 angle of ⋅ 360° ≈ 25o . 1978 The 633 between 18 and 60 correspond to a central 633 ⎞ angle of ⎛⎜ ⋅ 360° ≈ ⎟115° . ⎝ 1978 1 point 1 point 1 point ⎠ The 1207 over 60 correspond to a central angle of ⎛ 1207 ⎞ ⋅ 360° ≈ ⎟ 220° . ⎜ 1978 ⎝ ⎠ 1 point If the correct method of calculating the central angle is not shown at all, award only 1 point even for numerically correct answers. If one detailed calculation is shown only but all three results are correct, award 2 points. Correct pie chart (with correct angles by eye, sectors 1 point labelled). Total: 5 points 14. c) 12320⋅ 0.24 = = 2956.8(≈

2957) of those living in Ailington are over 60. The number of those over 60 treated in hospital is 1207 (≈ 0.41) 2957 The probability increased by 0.41 − 016 = 025 Total: 1207, so the probability in question is írásbeli vizsga 1111 6 / 13 1 point 1 point Accept 2956, too. 1 point 1 point 4 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 15. Applying the cosine rule to triangle ABP: BP2 = 6202 + 7202 − 2 ⋅ 620 ⋅ 720 ⋅ cos53° , BP ≈ 605 Angle AQB is 19º. Applying the sine rule to triangle ABQ (twice): 620 AQ = , sin 19° sin 108° AQ ≈ 1811 PQ ≈ 1811 − 720 = 1091 620 BQ = , sin 19° sin 53° BQ ≈ 1521 The distances, rounded to the nearest metre, are PQ = 1091 m, BQ = 1521 m and BP = 605 m. This point is also due if 1 point the idea is only reflected by the solution. 1 point 2 points * 1 point This point is also due if 1 point the idea is only reflected by the solution. 1 point 1 point * 1 point * 1

point 1 point * This point is awarded for 1 point * the correct unit (m) in the answer. Total: 12 points The points marked with * are also due if the results differ by at most 3m from those in the markscheme, provided that there is evidence for rounding correctly during the calculations. írásbeli vizsga 1111 7 / 13 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató II. B 16. a) (In team A, each of the 7 players plays a game against each of the other 6. That means every game is counted twice.) 7⋅6 There were = 21 games in team A. 2 (Team B has n members,) n ⋅ (n − 1) the number of games played is = 55 . 2 This leads to the equation n2 − n − 110 = 0 . The positive root is 11 (the roots are –10 and 11). Team B has 11 members. Total: 1 point 2 points 1 point 2 points 1 point 7 points 16. b) Each of the 6 players from team A plays 8 games. That makes a total of 6·8 = 48 games played during the second week. Total: 1 point 2

points 3 points 16. c) (The classical model of probability can be applied.) number of favourable cases p= number of all cases ⎛18 ⎞ This point is also due if 1 point the idea is only reflected by the solution. The winners may be selected in ⎜⎜ ⎟⎟ ways. 4 1 point There are 7 ways to choose 1 out of the 7 members of team A, 1 point ⎝ ⎠ 11 and ⎛⎜⎜ ⎞⎟⎟ ways to choose 3 out of the 11 members of ⎝3⎠ team B. (The two selections are independent of each other.) ⎛11⎞ The number of favourable cases is 7 ⋅ ⎜⎜ ⎟⎟ . ⎝3⎠ The probability in question is ⎛11⎞ 7 ⋅ ⎜⎜ ⎟⎟ 3 p= ⎝ ⎠ ⎛18 ⎞ ⎜⎜ ⎟⎟ ⎝4⎠ ⎛ 7 ⋅165 ⎞ ⎜= ⎟≈ ⎝ 3060 ⎠ 1 point ≈ 0.377 ≈ 38% Total: írásbeli vizsga 1111 These points are also due 1 point if the candidate only expresses the number of favourable cases correctly. 1 point 8 / 13 Award the 1 point for the 1 point correct probability expressed in any form. 7 points 2012.

május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 17. a) This point is also due if the extraneous root is 1 point rejected at the end by substitution. 2 x − 1 > 0 and 2 x − 3 > 0 , therefore x > 1.5 By the identities of logarithms: lg(2 x − 1)(2 x − 3) = lg 8 (The logarithm function is one-to-one,) therefore (2 x − 1)(2 x − 3) = 8 , that is 4 x 2 − 8x − 5 = 0 . The roots are 5 1 x1 = and x 2 = − . 2 2 5 Only x1 = is in the domain, and that is really a 2 solution. Total: 1 point 1 point 1 point 1 point 1 point 6 points 17. b) By solving the equation for cos x, the roots of the quadratic equation in a) are obtained. 5 1 ( (cos x )1 = and (cos x )2 = − ) 2 2 5 cos x = is not possible. 2 1 The only angle x obtained from cos x = − that may 2 2 π be an angle of a triangle is x = 120o = , and that 3 is really a solution. Total: 2 points 1 point Award the point for either correct value of the 1 point angle. Award no

point if any other angle is listed. 4 points 17. c) Solution 1 With the new variable y = z introduced, only solutions 0 ≤ z are sought. The only non-negative root of the quadratic equation 5 4 z 2 − 8z − 5 = 0 is z = . 2 It follows that the solution of the original equation is 25 y= , and that is really a solution. 4 Total: írásbeli vizsga 1111 9 / 13 1 point 1 point 1 point 1 point 4 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 17. c) Solution 2 Squaring both sides: 16 y 2 − 40 y + 25 = 64 y The roots of the quadratic equation 25 1 16 y 2 − 104 y + 25 = 0 are y1 = , y2 = . 4 4 Checking by substitution, or considering the range of each side to show that only the first root satisfies the equation. Total: 1 point 2 points 1 point 4 pont 17. d) If the middle number is fixed, the other six numbers can be ordered in 6! ways, thus there are 720 possible orders of the seven numbers. Total: írásbeli vizsga 1111

10 / 13 This point is also due if 1 point the idea is only reflected by the solution. 1 point 1 point 3 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 18. a) A E F 3m 3m 3m B 8m D G C 3m Understanding the problem. The surface of the lower part of the water tank (surface area of a hemisphere of radius r = 3 metres): 4r 2π A 1= = 2r 2π = 2 ⋅ 32 ⋅ π = 18π (≈ 56.5) 2 The surface of the middle part (the lateral surface area of a circular cylinder of radius r = 3 metres and height m = 8 metres): A 2 = 2rπ m = 2 ⋅ 3 ⋅ π ⋅ 8 = 48π (≈ 150.8) The surface of the upper part of the tank (the lateral surface area of a right circular cone of radius r = 3 metres and height m = 3 metres): The slant height is AB = a = 2r A 3 = raπ = 3 ⋅ 3 2 ⋅ π = 9 2 π (≈ 40 ) The total interior surface area is A = 18π + 48π + 9 2π = 66 + 9 2 π ≈ 247 .33 m2 Since the wording of the problem implies that the result

should be rounded upwards to make sure that there is enough material for coating the entire surface, the correct answer is 248 m2. Total: ( írásbeli vizsga 1111 1 point 1 point 1 point 1 point 1 point Award this point for the mathematical rounding to 247 m2 as well. ) 11 / 13 1 point 6 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató 18. b) A A Fig. 1 E F 0.9 m 3m 3m B 2.1 m r’ 0.9 m I E F H B 3m 8m A Fig. 2 G D C I 3m E The height of the water tank is (3 + 8 + 3 = ) 14 metres. 85% of the height is (14 ⋅ 0.85 = ) 119 metres, that is, the hemisphere and the cylinder are full, and the water level is 0.9 metres above the base of the cone The volume of the lower part of the water tank (volume of a hemisphere of radius r = 3 metres): V 1= 1 4r 3π ⋅ 2 3 ⎛ 2r 3π ⎞ ⎜⎜ = ⎟= 3 ⎟⎠ ⎝ 2 ⋅ 33 ⋅ π (= 18π ≈ 56.5) = 3 The volume of the middle part (volume of a circular cylinder of radius

r = 3 metres and height m = 8 metres): V 2= r 2π m = = 32 ⋅ π ⋅ 8 (= 72π ≈ 226.2) The remaining volume in the upper part is the volume of a truncated cone. The radius of the top circle of the truncated cone can be calculated by using the theorem about line segments cut out of parallel lines by the arms of an angle: (Figure 1) 3m F r’ H 0.9 m J 0.9 mB 1 point 1 point 1 point 1 point 1 pont 1 point 1 point * ⎛ IH ⎞ r 2.1 ⎛ AI ⎞ , =⎟ = ⎟ ⎜= ⎜ 3 ⎝ AF ⎠ ⎝ FB ⎠ 3 r = 2.1 π V 3= m(r 2 + r 2 + rr ) = 3 π = ⋅ 0.9 ⋅ (32 + 212 + 3 ⋅ 21) = (5913π ≈ 186) 3 The total volume of the water in the tank is A = 18π + 72π + 5.913π = 95913 π ≈ 301 m3 1 point * 1 point 1 point 1 point Total: írásbeli vizsga 1111 12 / 13 11 points 2012. május 8 Matematika angol nyelven középszint Javítási-értékelési útmutató Another way to calculate the results for the two points marked with *: The remaining volume in the upper part is

the volume of a truncated cone. The radius of the top 1 point * circle of the truncated cone can be calculated by observing that ∆AFB and ∆HJB both are isosceles right-angled triangles, (Figure 2) thus r = (FB − JB = 3 − 0.9 = ) 21 1 point * írásbeli vizsga 1111 13 / 13 2012. május 8