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Warning: There are more than one versions of the test.
Instructions
Test period 90 minutes for Part I and Part II as a whole. We recommend that you use atthe most 45 minutes to work with Part I. You must not use the calculator
before you have handed in Part I. Use a separate sheet for part I. Tools Formula sheet and ruler.
Part I For most items a single answer is not enough. It is also expected• that you write down what you do• that you explain/motivate your reasoning• that you draw any necessary illustrations.
After every item is given the maximum mark your solution can receive.(2/3) means that the item can give 2 g-points (Pass level) and 3 vg-points(Pass with distinction level).
Items marked with (Problems 1f, 4, 5, 7, 15 and 16) give you a possibilityto show MVG-quality (Pass with special distinction quality) This means that
you use generalised methods, models and reasoning, that you analyse your
results and account for a clear line of thought in a correct mathematical
language.
Try all of the problems. It can be relatively easy, even towards the end of
the test, to receive some points for partial solutions. A positive evaluationcan be given even for unfinished solutions.
Mark limitsMaximum score: TOTAL 63 out of which 34 vg-points, and 7 MVG pointsG: 21 pointsVG: 42 points/ at least 11 VG points:MVG: 46 points/ at least 23 VG points; MVG-quality work ¤
1 2 3a 3b 3c 3d 3e 3f 4a 4b 4c 4d 4e 4f 4g 4g¤ 5a 5b 5c 5c¤ 6a 6b 6c 6d 7a 7b 7b¤
In each case, show how you arrived at your answer by clearly indicating all of
the necessary steps, formula substitutions, diagrams, graphs, charts, etc.
Part I
You are not allowed to use calculator in this part. Write your answer to the first part on a separate paper. Only after the submission of your solutions to the partone you may use your calculator. You may start working with part II beforesubmission of your solutions to part I.
In the multi-choice problems below just circle the correct answer (s) in the test paper andwrite the answer in the space provided as “Answer: _____________”
1. A member of the solution set of 32 <≤− x is (1/0)
a. 5−
b.
4−
c. 1 d. 3
Answer: Alternative: _____________
Suggested solutions:
32 <≤− x means x is larger than or equal to 2− but less than 3 . Therefore only 1 is a
member of − .32 <≤ x
Answer: Alternative (c)
2. Which statement is true when 01 <<−
< x
3
x <
x ? (1/0)
a. 32 x< x x b. x x << 23
c. x x << 2 x
d. 3 x x < 2
Answer: Alternative: _____________
Why? Explain in sufficient detail. (0/2)Suggested solution:
23
3
2
001
1001 x x x
x x
x x <<<⇔
⎪
⎩
⎪⎨⎧
<<<−
<<⇔<<−
1001 2 <<⇔<<− x x . This is due to the fact that the square of any real number is positive!On the other hand if the magnitude of the number is smaller than one, the square of thenumber will be a number smaller than one.On the other hand, if multiply both sides of an equality with a negative number, only the
direction of the inequality reverses: x x x x x x x >>⇔⋅>⋅>⇔<<⇔<<− 322 0101001
13. In a survey of 200 seniors at a high school it was found that 75 students play soccer but
not softball, 45 students play softball but not soccer, and 28 students do not participate ineither sport. Of the seniors surveyed, what percent play both soccer and softball? (1/2)
Suggested solutions:We may draw a Venn graph usingCircle A : 200 seniorsCircle B : Students playing soccerCircle C : Students playing softball
From the Venn diagram illustrated we maydeduce that:The number of students who play both soccerand football is:
( ) 52148200284575200 =−=++−=C B
N I
%2626.020052 ===
tot
C B
N
N I
Answer: 26% of students play both soccer and football.
At the aspect assessment of your work with problems 9-11 the teacher will consider
• what mathematical knowledge you have demonstrated and how well you have carriedthrough the task
• how well you have explained your work and motivated your conclusions
• how well you have written your solutions.
This means that you use generalised methods, models and reasoning , that you analyse your
results and account for a clear line of thought in a correct mathematical language .
14. Choose any arbitrary three-digit-number. Multiply the number by 7. Multiply the result by11 and then by 13. What is the result? Repeat the procedure for any other arbitrary three-digit number. What is the result? Is there any pattern? Why? Generalize and Explain.
(2/2/¤)
Suggested solutions:
Answer: I get a six-digit-number which is my number repeated twice. Ex:12312313117123 =×××
36436413117364 =×××
57957913117579 =×××
99999913117999 =×××
Reason:Assume the number I choose is . Due to the fact that ,
when I multiply by
abc 110013117 =××
abc 110013117 =×× , I get abcabcabc +×=× 10001001 . My
three-digit number multiplied by 1000 looks like and when I add to
it my number: . This is true for any 3-digit number.
15. Quadratic small tables, that has only three places for three people at each side, is arrangedas a long dining table, as illustrated below:
¤ How many persons, at the most, may be placed at along dining table consisted of five small squaretable?
¤ How many small square tables are needed in order to be able to place 60 persons, at a long dining tablearranged by according to the rules above?
¤ Find a formula that describes the relationship between the number of persons, p , that will be
placed at a dining table arranged as above of n
number of small tables.
¤ Describe your formula above in words. (2/4/¤)
Suggested solutions:
¤ 36 persons, at the most, may be placedat a long dining table consisted of fivesmall square table. This is due to the factthat 9 people will be able to take a seatat each of the two corner tables, theremaining 3 tables have 6 place each:
366392 =×+×=P
¤ 9 small tables are needed to place 60 people around it.This is due to the fact that as mentioned above 9 people have placearound each of the two corner tables. The middle tables have each a
maximum of 6 places available. Therefore 7 middle tables are neededto place 42 people at them.¤ Using the logic above, we may develop a formula for P people and n ,
number of small tables:( ) ( )1666126182692 +=+=−+=−×+×= nnnnP
16. Arrange the following numbers in order of magnitude without the aid of a calculator:
, , , . (2/2/¤)16525 5564 11064 66243
Suggested solutions:
In order to compare numbers in exponent form either we should find amaximum common exponent, MCE, to all of them and then compare thebases of the numbers, or find a maximum common base and compare theexponents of the numbers with each other. In this case we may choose amaximum common exponent. In the first glance we may assume that theMCE of the numbers 55, 66, 110, and 165 is 11, which is fine, but we mayrewrite the numbers and try to get the smallest possible integer base foreach number using the prime factors of the exponents, and lowest base,as follows:
( ) 115321652165 5525 ×××== using and2525= 1153165 ××=
( ) 11532115655 2264 ××××== using and6264 = 11555 ×=
( ) 1153211532211526110 42264 ×××××××××=== using ,6264 = 422 = and 1152110 ××=