Final IMSO 2009 Short Answer
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
IMSO 2009
Short Answer
IMSO2009
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.
Answer:
Page 2 of 12
….
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
= 3 + 6 + 9 + 12 + … + 75
5. In the following
grid, the area of the shaded region is … unit square.
Answer: 21 square units
from a sheet of
6. Danny wants to create a set of cards of sizes
paper of size
. The number of cards that can be made by Danny
is at most … .
Answer: 23
Page 3 of 12
IMSO2009
IMSO 2009
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))=62
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 ... .
Answer: 36
P = 523
Q = 487
-------------- -36
Page 4 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
11. The natural numbers bigger than 1 are arranged in five columns as given by the
following figure. The number 2009 appears in the … column and the … row.
I
9
17
II
III
IV
V
2
3
4
5
8
7
6
10
11
12
16
15
14
13
Solution: 1st column, 502nd row
Observe that, the numbers appear on the first column are
occupied are the even rows, starting from 2. Since
column and
, with
, where the rows
then 2009 appears on the first
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 = 33.
Half of it is 4, which is also a square number, i.e., 4 = 22.
The next natural number which has the same properties is … .
The next natural number which has the same properties is … .
Answer: 288
Number
Propety 1
Property 2
Satisfy
(Even)
8
24
48
80
120
168
Page 5 of 12
8+1 = 9 = 3
2
8/2 = 4 = 2
2
1 and 2
24+1 = 25 = 5
2
24/2 = 12
1
48+1 = 49 = 7
2
48/2 = 24
1
80+1=81=9
2
80/2=40
1
120+1=121=11
2
120/6=60
1
168+1=169=13
2
168/2=84
1
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
224
288
224+1=225=15
2
288+1=289=17
2
224/2=112
1
288/2=144=12
2
1 and 2
The number = 288.
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 …:….
Answer: ¾ or 3 : 4
Wherever the position of P on CD, the ratio of
the areas of the triangle and the trapezoid is ¾
P
D
C
C
A
B
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: 7 boxes
Case
Box 1
Box 2
Box 3
…
Box (n-2)
Box (n-1)
Box n
Remaining
marbles
I
6
6
6
6
6
6
3
II
9
9
9
9
0
0
0
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.
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 ... .
4
1
X
3
2
Page 6 of 12
IMSO2009
IMSO 2009
Short Answer
Yogyakarta, 8-14 November
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: 4,1,5,4,1,3,2,5,3,2. So, X=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
0
1
0
Draw-Score
0
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 several 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.
So, overall 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 ....
Page 7 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
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
9
3
6
20
0
8
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.
Answer: 21
From dot P, triangles PBR,PBC, PAT, PAD
From dot Q, triangles QBR, QBC, QAT, QAD
From dot R, triangles RCS,RCD,RCT, RST, RDT, RTA,RAB,RBT
Page 8 of 12
(4)
(4)
(8)
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
From dot S, triangles SDT,SDA,SCB
From dot T, triangles TDC,TAB
(3)
(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
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
Page 9 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
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 … cm2.
Answer:
IV
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
I
II
= 77 49
2
2
III
=14
Thus, the shaded area = 14 x 4= 56 cm2
24. Replace the asterisks with digits so that the multiplication below is correct:
The product is ….
Answer: 182720097
3 3 3 3 7
A B C 1
3 3 3 3 7
Page 10 of 12
2
* *
* * *
* * *
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 6 9 6
* * * E
* * *
* 2 0 0 9 7
2
1 3
* * *
* * *
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 6 9 6
3 3 4 8
* * F
* 2 0 0 9 7
2
1 3
1 6 6
1 8 2
6
3
6
7
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 9 6
3 4 8
8 5
2 0 0 9 7
IMSO2009
IMSO 2009
Short Answer
* * *
* * * *
* * * * *
* * * * 2
Yogyakarta, 8-14 November
* * D
* *
*
0 0 9 7
25. The areas of 3 sides of a block are 44 cm2, 33 cm2, and 48 cm2. The volume of
the block is … cm3.
Answer : 264 cm
3
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, respectively 44 and 33. However, these measures are not suitable for the wide side.
12?
4
44
3
33
11
Because 11 is the Greatest Common Divisor of 33 and 44, so we can modify the measures so that the
vertices are in accordance with the broad sides of the unknown.
48
4x2
44
3x2
11
2
33
11
The size of the vertices block row are 8 cm, 6 cm, and 2
Thus, the block volume is 8 x 6 x
11
3
= 264 cm .
2
Method 2
For example the size of the vertices block is x, y, and z as shown in the following figure.
Retrieved
(xy) x (xz) x (yz) = 48 x 44 x 33 = (4 x 12) x (4 x 11) x (3 x 11)
or
x 2 y 2 z 2 = (4 x 4 ) x (11 x 11) x (12 x 3) = (4 x 4 ) x (11 x 11) x (6 x 6)
Page 11 of 12
y
48
Can be written: xy = 48, xz = 44, and yz = 33.
33
x
44
z
IMSO 2009
Short Answer
Yogyakarta, 8-14 November
or
xyz = 4 x 11 x 6 = 264.
3
Thus, the block volume is 264 cm .
Page 12 of 12
IMSO2009
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
IMSO 2009
Short Answer
IMSO2009
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.
Answer:
Page 2 of 12
….
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
= 3 + 6 + 9 + 12 + … + 75
5. In the following
grid, the area of the shaded region is … unit square.
Answer: 21 square units
from a sheet of
6. Danny wants to create a set of cards of sizes
paper of size
. The number of cards that can be made by Danny
is at most … .
Answer: 23
Page 3 of 12
IMSO2009
IMSO 2009
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))=62
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 ... .
Answer: 36
P = 523
Q = 487
-------------- -36
Page 4 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
11. The natural numbers bigger than 1 are arranged in five columns as given by the
following figure. The number 2009 appears in the … column and the … row.
I
9
17
II
III
IV
V
2
3
4
5
8
7
6
10
11
12
16
15
14
13
Solution: 1st column, 502nd row
Observe that, the numbers appear on the first column are
occupied are the even rows, starting from 2. Since
column and
, with
, where the rows
then 2009 appears on the first
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 = 33.
Half of it is 4, which is also a square number, i.e., 4 = 22.
The next natural number which has the same properties is … .
The next natural number which has the same properties is … .
Answer: 288
Number
Propety 1
Property 2
Satisfy
(Even)
8
24
48
80
120
168
Page 5 of 12
8+1 = 9 = 3
2
8/2 = 4 = 2
2
1 and 2
24+1 = 25 = 5
2
24/2 = 12
1
48+1 = 49 = 7
2
48/2 = 24
1
80+1=81=9
2
80/2=40
1
120+1=121=11
2
120/6=60
1
168+1=169=13
2
168/2=84
1
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
224
288
224+1=225=15
2
288+1=289=17
2
224/2=112
1
288/2=144=12
2
1 and 2
The number = 288.
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 …:….
Answer: ¾ or 3 : 4
Wherever the position of P on CD, the ratio of
the areas of the triangle and the trapezoid is ¾
P
D
C
C
A
B
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: 7 boxes
Case
Box 1
Box 2
Box 3
…
Box (n-2)
Box (n-1)
Box n
Remaining
marbles
I
6
6
6
6
6
6
3
II
9
9
9
9
0
0
0
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.
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 ... .
4
1
X
3
2
Page 6 of 12
IMSO2009
IMSO 2009
Short Answer
Yogyakarta, 8-14 November
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: 4,1,5,4,1,3,2,5,3,2. So, X=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
0
1
0
Draw-Score
0
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 several 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.
So, overall 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 ....
Page 7 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
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
9
3
6
20
0
8
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.
Answer: 21
From dot P, triangles PBR,PBC, PAT, PAD
From dot Q, triangles QBR, QBC, QAT, QAD
From dot R, triangles RCS,RCD,RCT, RST, RDT, RTA,RAB,RBT
Page 8 of 12
(4)
(4)
(8)
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
From dot S, triangles SDT,SDA,SCB
From dot T, triangles TDC,TAB
(3)
(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
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
Page 9 of 12
IMSO 2009
Short Answer
IMSO2009
Yogyakarta, 8-14 November
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 … cm2.
Answer:
IV
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
I
II
= 77 49
2
2
III
=14
Thus, the shaded area = 14 x 4= 56 cm2
24. Replace the asterisks with digits so that the multiplication below is correct:
The product is ….
Answer: 182720097
3 3 3 3 7
A B C 1
3 3 3 3 7
Page 10 of 12
2
* *
* * *
* * *
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 6 9 6
* * * E
* * *
* 2 0 0 9 7
2
1 3
* * *
* * *
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 6 9 6
3 3 4 8
* * F
* 2 0 0 9 7
2
1 3
1 6 6
1 8 2
6
3
6
7
3 3 3 3 7
A B 8 1
3 3 3 3 7
6 6 9 6
3 4 8
8 5
2 0 0 9 7
IMSO2009
IMSO 2009
Short Answer
* * *
* * * *
* * * * *
* * * * 2
Yogyakarta, 8-14 November
* * D
* *
*
0 0 9 7
25. The areas of 3 sides of a block are 44 cm2, 33 cm2, and 48 cm2. The volume of
the block is … cm3.
Answer : 264 cm
3
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, respectively 44 and 33. However, these measures are not suitable for the wide side.
12?
4
44
3
33
11
Because 11 is the Greatest Common Divisor of 33 and 44, so we can modify the measures so that the
vertices are in accordance with the broad sides of the unknown.
48
4x2
44
3x2
11
2
33
11
The size of the vertices block row are 8 cm, 6 cm, and 2
Thus, the block volume is 8 x 6 x
11
3
= 264 cm .
2
Method 2
For example the size of the vertices block is x, y, and z as shown in the following figure.
Retrieved
(xy) x (xz) x (yz) = 48 x 44 x 33 = (4 x 12) x (4 x 11) x (3 x 11)
or
x 2 y 2 z 2 = (4 x 4 ) x (11 x 11) x (12 x 3) = (4 x 4 ) x (11 x 11) x (6 x 6)
Page 11 of 12
y
48
Can be written: xy = 48, xz = 44, and yz = 33.
33
x
44
z
IMSO 2009
Short Answer
Yogyakarta, 8-14 November
or
xyz = 4 x 11 x 6 = 264.
3
Thus, the block volume is 264 cm .
Page 12 of 12
IMSO2009