IMSO 2009 IMSO2009 Short Answer Yogyakarta, 8-14 November

  

6th INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY

SCHOOL

  

IMSO 2009

SHORT ANSWER PROBLEMS

Yogyakarta, 8 – 14 November 2009

  

DIRECTORATE OF KINDERGARTEN AND PRIMARY DEVELOPMENT

DIRECTORATE GENERAL OF PRIMARY AND SECONDARY EDUCATION MANAGEMENT

MINISTRY OF NATIONAL EDUCATION

  IMSO20

  IMSO 2009

  09 Short Answer Yogyakarta, 8-14 November

  

1. A big square is formed from twenty five small squares. Some of the small

squares are black. To make the big square symmetric about both diagonals, at least … additional small squares are needed to be colored black.

  Answer: 4 2. If , then =….

  Answer: 4

  

3. In a training program, an athlete must eat 154 eggs, during a period of time from

November 8th till November 14th. Every day in this period he must eat 6 more eggs than the previous day. The number of eggs he eats on November 13th is …

  Solution: 34 eggs Let a be the number of eggs eaten on November 8th. Then 154=7a + (1+2+3+4+5+6)6=7a+126.

  So a=4. So the number of eggs eaten on November 13th is 4+5x6=34.

  4.

  ….

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  09 Short Answer Yogyakarta, 8-14 November

  Answer:

  = 3 + 6 + 9 + 12 + … + 75 5. In the following grid, the area of the shaded region is … unit square.

  Answer: 21 square units

  

6. Danny wants to create a set of cards of sizes from a sheet of

paper of size . The number of cards that can be made by Danny is at most … .

  Answer

  : 23

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  09 Short Answer Yogyakarta, 8-14 November

  

7. Mrs. Anna has 4 children: Alex, Brad, Christine, and Dennis. Alex is not the

youngest, but he is younger than Dennis. If Brad’s age is the same as the mean of the ages of Alex and Dennis, then the oldest one is … .

  Solution: Dennis

Since Brad’s age is the same as the mean of ages of Alex and Dennis, then Brad is

between Dennis and Alex. Alex is not the youngest, so Christine is. Thus the oldest is

Dennis.

  

8. In a math test, a correct answer will be marked 5 points and a wrong answers

points. Tom answered all of the 35 questions and got a total score of 140.

  The number of questions Tom answered correctly is … .

  Answer: 30 # correct answer:

  25

  26

  27

  28

  29

  30

  31

  32 Total score: 105 112 119 126 133 140 147 154

  

9. The following shape is made from horizontal and vertical lines. The lengths of

some of the lines are given. The perimeter of the shape is … unit.

  Answer : The perimeter is 2x((6+12)+(5+8))=6

  2

  

10. Use numbers 2, 3, 4, 5, 7, and 8 exactly once to form two three-digit numbers P

and Q. If is a positive number; the smallest possible value of is ... .

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  09 Short Answer Yogyakarta, 8-14 November

  Answer: 36

  

P = 523

Q = 487

• -------------- --

  

36

  

11. The natural numbers bigger than 1 are arranged in fie columns as giien by

the following fgure. The number 2009 appears in the … column and the … row.

  I II

  III

  IV V

  2

  

3

  4

  5

  9

  8

  

7

  6

  10

  

11

  12

  13

  17

  16

  

15

  14 Solution: 1st column, 502nd row

  Observe that, the numbers appear on the first column are , with , where the rows occupied are the even rows, starting from 2. Since then 2009 appears on the first column and nd row.

  12.The integer 8 has two properties:  If the number 1 is added, we get the number 9, which is a square number, i.e., 9 = 33.

   Half of it is 4, which is also a square number, i.e., 4 = 22. The next natural number which has the same properties is … .

  The next natural number which has the same properties is … . Answer: 288

  A B C C D P

  marbles

  9 From case I, there should be 15 marbles that must be distributed into a multiple of 3 number of boxes to get the case II. Because 15 is a multiple of 3, then the only possible number of boxes that each has 9 marbles is 5 (and the other 2 boxes are empty). So the number of boxes are 5+2 = 7.

  9

  9

  9

  3 II

  6

  6

  6

  6

  6

  6

  I

  Answer: 7 boxes Case Box 1 Box 2 Box 3 … Box (n-2) Box (n-1) Box n Remaining

  IMSO 2009

  

14. I have some marbles and some empty boxes. If I try to put 9 marbles on each

box, then there will be 2 empty boxes. If I try to put 6 marbles in each box, then there will be 3 remaining marbles. I have … boxes.

  Answer: ¾ or 3 : 4 Whereier the position of P on CD, the ratio of the areas of the triangle and the trapezoid is ¾

  

13. ABCD is a trapezoid (trapezium) with AB parallel to CD. The ratio of AB : CD is

3 : 1. The point P is on CD. The ratio of the area of triangle APB to the area of trapezoid ABCD is …:….

  1 288 288+1=289=17 ² 288/2=144=12 ² 1 and 2 The number = 288.

  1 224 224+1=225=15 ² 224/2=112

  1 168 168+1=169=13 ² 168/2=84

  1 120 120+1=121=11 ² 120/6=60

  1 80 80+1=81=9 ² 80/2=40

  1 48 48+1 = 49 = 7 ² 48/2 = 24

  Propety 1 Property 2 Satisfy 8 8+1 = 9 = 3 ² 8/2 = 4 = 2 ² 1 and 2 24 24+1 = 25 = 5 ² 24/2 = 12

  Number (Even)

  09 Short Answer Yogyakarta, 8-14 November

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15.The ten numbers 1,1, 2, 2, 3, 3, 4, 4, 5, 5, are arranged in a row (see the figure),

so that each number, except the first and last, is the sum or the difference of its two adjacent neighbors. The value of X is ... .

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  09 Short Answer Yogyakarta, 8-14 November

  4

1 X

  3

  2 The ten numbers 1,1,2,2,3,3,4,4,5,5, are arranged in a row (see figure), so that each number, except the first and last, is the sum or the difference of its two adjacent neighbors. The value of X is ....

  Answer: 4,1,5,4,1, ,2,5,3,2. So, X=3.

  3

  

16.In a football competition, if a team wins it will get 3 points. If it draws it will get 1

point, and if it loses it will get 0 points. After playing 20 times, Team B gets the total score of 53. Team B loses at least … times.

  Answer:1

  # Wins

  20

  19

  18

  17

  16

  15

  14

  13

  12

  11

  10

  9

  8

  7

  6 Win-Score

  60

  57

  54

  51

  48

  45

  42

  39

  36

  33

  30

  27

  24

  21

  18 # Loses

1 Draw-Score

  2

  4

  5

  6

  7

  8 # Scores

  60

  53

  52

  50

  48

  

17.Among 8 points located in a plane, five of them lie on one line. Any three points

are selected from those 8 points as corner points of a triangle. There are at most … triangles that can be formed.

  Answer: 46 There are seieral possibilities triangle formed.

   Triangle formed by points outside the line. In this case there is exactly one triangle.

   Triangle formed by a point outside the line and two points on the line. There

are 10 pairs of points from 5 points located on the line. Since there are 3

points out of line, then the number of triangles that might be formed is 3 x 10 = 30 triangles. Triangle formed by a point on the line and two points outside the line. There  are 3 pairs of points from beyond the 3 point line. Since there are 5 points on the line, there is a 5 x 3 = 15 triangle that may be formed.

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  IMSO 2009

  09 Short Answer Yogyakarta, 8-14 November So, oierall there is (1 + 30 + 15 = 46) triangles that may be formed.

  

18. Fill in all the numbers 0,1,2,3,4,5,6,7,8,9 on the ten squares below, so that the

sum of numbers located on each arrowed line is 20. Two numbers are already filled in. The number on the square with a question mark ("?") is ....

  Answer: 7 2.? + 2.8 + 2 +5 +6 +9 +3 +1 +4 =60 2.? + 46 = 60 ? = 7 One of the suitable arrangements is given in the following figure:

  7

  2

  5

  

6

  20

  8

  9

  3

  20

  1

  4

  20

19. Nine dots are arranged as shown in the figure below, where ABCD is a square.

  AT=TD, DS=SC, CR=BR, and AP=PQ=QB. A triangle can be constructed by lining from dot to dot. At most, there are … different right triangles that one can construct so that at least one of the dots P,Q,R,S, and T is its vertex.

  IMSO20

  IMSO 2009

  09 Short Answer Yogyakarta, 8-14 November Answer: 21 From dot P, triangles PBR,PBC, PAT, PAD (4) From dot Q, triangles QBR, QBC, QAT, QAD (4) From dot R, triangles RCS,RCD,RCT, RST, RDT, RTA,RAB,RBT (8) From dot S, triangles SDT,SDA,SCB (3) From dot T, triangles TDC,TAB (2) Total:

  21

20.ABCDE is a five-digit positive number. ABCDE1 is three times 1ABCDE.

  ABCDE is … .

  Answer: 42857 Let x = ABCDE 3(100000 + x) = 10x+1 10x+1 = 3(100000 + x) 10x+1 = 300000 + 3x 10x = 299999 + 3x 7x = 299999 x= 299999/7 = 42857

21. In the diagram below, BC=5, DE=1 and DC=20, where D lies on AC and E lies on AB. Both ED and BC are perpendicular to AC. The length of AD is … .

  (Note: the figure is not in proportional scale) Answer : 5

  

22. The number N consists of three different digits and is greater than 200. The

digits are greater than 1. For any two digits, one digit is a multiple of the other or the difference is 3. For example, 258 is one of such numbers. There are at most … possible N.

  Answer: 18 numbers

  IMSO 2009

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  09 Short Answer Yogyakarta, 8-14 November

  Possibilities: 248;284;428;482;824;842  6 numbers 258;285;528;582;825;852  6 numbers 369;396; 639;693;936;963  6 numbers

• 18 numbers

  

23.In the figure, two half-circles are inscribed in a square. These two half-circles

intersect at the center of the square. If the side of the square has length 14 cm, then the area of the shaded region is … cm

  2 .

  Answer: 56 OR ½(2π-1)49 OR 55.86 See the picture on the side. The shaded area consists of four parts which are congruent. The area I = area of a quarter circle - area of triangle =

  2

  49

  2 77  =14 Thus, the shaded area = 14 x 4= 56 cm

  2

  I II

  III

  IV

24. Replace the asterisks with digits so that the multiplication below is correct: The product is ….

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  09 Short Answer Yogyakarta, 8-14 November

  Answer: 182720097

   3 3 3 3 7 3 3 3 3 7 3 3 3 3 7 3 3 3 3 A B 8 1 A B 8 1 A B 8 1

  7 3 3 3 3 7 3 3 3 3 7 3 3 3 3 7 A B 2 6 6 6 9 6 2 6 6 6 9 6 2 6 6 6 9 6 C 1

• E 1 3 3 3 4 8 1 3 3 3 4 8 3 3 3 3
• F 1 6 6 6 8 5

  7

• 2 0 0 9 7 * * * * 2 0 0 9 7 1 8 2 7 2 0 0 9 7
• D * * * * * *
• 2 0 0 9 7

  2

  2

  2

25.The areas of 3 sides of a block are 44 cm , 33 cm , and 48 cm . The volume of

  3 the block is … cm .

3 Answer : 264 cm

  Method 1 If the size of the edges block row is 4 cm, 3 cm and 11 cm, the area of the front and side would be appropriate, ie, respectiiely 44 and 33. Howeier, these measures are not suitable for the wide side.

  12?

  3

  33

  4

  1

  44

1 Because 11 is the Greatest Common Diiisor of 33 and 44, so we can modify the

  measures so that the iertices are in accordance with the broad sides of the unknown.

  3x2

  48 4x2

  33

  11

  44

  2

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  IMSO 2009

  09 Short Answer Yogyakarta, 8-14 November

  11

  2 The size of the iertices block row are 8 cm, 6 cm, and

  11

  3

8 x 6 x = 264 .

  Thus, the block iolume is cm

2 Method 2

  For example the size of the iertices block is x, y, and z as shown in the following fgure.

  y

  48 xy = 48, xz = 44, yz = 33.

  Can be written: and

  x

  33 Retrieied z

  44 (xy) x (xz) x (yz) = 48 x 44 x 33 = (4 x 12) x (4 x 11) x (3 x 11)

  or _{2 2 2}

  x y z = (4 x 4 ) x (11 x 11) x (12 x 3) = (4 x 4 ) x (11 x 11) x (6 x 6)

  or xyz = 4 x 11 x 6 = 264.