DIRECTORATE OF KINDERGARTEN AND PRIMARY DEVELOPMENT DIRECTORATE GENERAL OF PRIMARY AND SECONDARY EDUCATION MANAGEMENT MINISTRY OF NATIONAL EDUCATION
6th INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOL
IMSO 2009
ESSAY 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
Essay
IMSO2009
Yogyakarta, 8-14 November
General Remarks
1. A full mark for each problem is 3 points.
2. When the student gives final answer only, the mark is 1
point.
1. Bob bought a coat and a shirt. The normal prices of both items are the same,
but when Bob bought them, the shirt was discounted by 50% and the coat was
discounted by 25%. If he bought them for $130 what was the normal price of the
shirt?
Answer: $104
Let price = original price of one coat or one shirt.
50% x price + (100% - 25%) x price = 130.......................................... 1 point
50% x price + 75% x price = 130
125% price = 130
(5/4) price = 130...................
....................................................... 1 point
Price = (4/5) x 130
= 104
The original price of the shirt is $104............................................ 1 point
2. The area of a rhombus is 36 cm2. One of its diagonals has length twice of the
other diagonal. What is the length of the smaller diagonal of the rhombus?
Answer: 6 cm
... 1 point
Area of rhombus 36 cm2
4a2 = 36
... 1 point
2
a =9
a=3
The length of the smaller diagonal is 6 cm
Page 2 of 8
... 1 point
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
3. I plan to travel by car from City A to City B. I travel the first half of the distance
with the speed of 30 km/hour. In order to get my average traveling speed of 40
km/hour, what speed should I take for the second half of the distance?
Answer:
Marking Scheme:
If the distance = 120 km.
2 hours
T hour
A o------------------o----------------------o B
60 km
C
60 km
A half of the distance = 120/2 = 60 km.
Time needed from A to C = 60/30 = 2 hour. ………………………………………. 1point
In order get 40 km//hour from A to B:
120
40 = ------------ ………………………………………………….……….………. ½ point
2+T
3
1 = -----------2+T
2+T=3
T = 1. ……………………………………………………………….…….. 1 point
Speed at the next half of the distance is
60 / 1 = 60 km/hour. …………………………………………………….…... ½ point
Page 3 of 8
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
4. The mean value of the mathematics marks of the nine students is 70. When
Lambert’s mark is added, the mean value will be 69. When John’s mark is
added, the mean value will be 72. What will the mean value be, if Lambert’s and
John’s marks are both added?
Answer :
Scheme:
Finding Lambert’s mark (1 point)
Finding John’s mark (1 point)
Calculating the total mark sum of eleven students (½ point)
Calculating the mean (½ point)
Sum=70*9
Sum+L=10*69
Sum+J=10*72
2*Sum+L+J=10*141 or Sum+L+J=10*141-70*9=780
Thus, (Sum+L+J)/11=780/11
5. Andy bought three packages of goods, each worth $ 35, $ 30, and $ 40. The first
package contains 2 books, 1 pencil, and 1 eraser. The second package contains
1 book, 1 pencil, 2 erasers. The third package contains 3 books and 2 erasers.
Andy wants to buy the fourth package containing 2 books, 1 pencil, and 3
erasers. What is the price of the fourth package?
Answer :
Marking scheme
Formulating the system of equations (1 point)
Finding the prices of a book, a pencil, and an eraser (½ point for each)
Calculating the final answer (½ point)
Package
# Books
# Pencils
# Erasers
Price
I
2
1
1
$35
II
1
1
2
$30
III
3
0
2
$40
IV
2
1
3
?
Firstly, comparing (I) & (II): 1 Books - 1 Eraser = $5 or 2 Books = $10 + 2 Erasers
Secondly, combining (II) & (III) and comparing with (I): 2 Books + 3 Erasers = $35
Combining first & second results: 5 erasers = $25 so Eraser = $5.
The price of Package IV is $35 + 2 x price of eraser
Therefore, the price of package IV is $35 + 2 x $5 = $ 45.
Page 4 of 8
IMSO2009
IMSO 2009
Essay
Yogyakarta, 8-14 November
6. The following figure shows a regular hexagon. On each side of the hexagon,
there is an isosceles right triangle. One side forming a right angle is a side of the
hexagon. Determine the angle x.
x
Answer:
Internal angle of hexagonal is 1200
... 1 point
0
Internal angle of right triangle are 90 dan 45
0
0
0
0
Determining of angle x is 360-120 -90 - 45 =105
... 1 point
0
... 1 point
7. Bob and Ivan have tasks of mowing (cutting) grass in their yard. One day they
do their work at the same time. After that, Bob works every 8 days, while Ivan
every 6 days. Each time Bob works, he is paid $15 while Ivan gets $17.5. If they
work again on the same day, how much money will each of them earn at the end
of the day?
Answer:
Day
Bob
Ivan
0
v
v
1
2
3
4
5
6
7
8
v
9
v
10
11
12
v
13
14
15
16
v
17
18
19
20
21
v
th
They meet again at 24 day. At this time: ………………………………………1 point
Bob’s work 4 times ………………………………………………………… ½ point
Ivan’s work 5 times ………………………………………………………… ½ point
Hence
Bob’s earn = 4 x $15 = $ 60 ………………………………………………… ½ point
Ivan’s earn = 5 x $17.5 = $ 87.5 ……………………………………………..½ point
Page 5 of 8
22
23
24
v
v
IMSO2009
IMSO 2009
Essay
Yogyakarta, 8-14 November
8. Mum’s kitchen scale is set incorrectly, but otherwise it works fine. When she
weighs a bag of sugar, it shows 1.5 kg. When she weighs a bag of flour, it
shows 1.3 kg. However, when she weighs both items together, it shows 2.5 kg.
If she weighs a piece of butter of weight 0.3 kg, what number does the scale
show?
Answer:
Solution: 0.6 kg
kg is the real weight of the bag of sugar. In a similar way,
Observe that
kg is the real weight of the bag of flour. So the scale shifts 0.3 kg to the
kg for the butter.
right. Hence, the scale shows
Scheme:
Finding the real weight of sugar or flour ………………..1 point
Finding the shift of the scale ……………………………..1 point
Finding the weight of the butter on the scale …………..1 point
9. Alan, Billy, Candy, and David are queuing (lining up) in alphabetical order. Alan
is in the 7th position from the front while David is in the 9th position from the
back. The number of persons between Alan and Billy is the same as those
between Candy and David. In total, there are 48 persons in the queue, and six
of them are between Billy and Candy. How many persons are there between
Alan and Candy?
Answer:
6 people
A
B
6 people
C
D
8 people
From A to D, there are all together 48-14=34 persons. ……………………………… 1 point
Since the number of persons between A and B is the same as those between C and D, the number
of persons between A and B is:
34 - 6
- 2 =12 ……………………………………... 1 point
2
The number of persons between A and C is 12+7=19 ………………………………… 1 point
10. A rectangle has two axes of symmetry, the vertical axis and the horizontal axis.
If the rectangle is folded with respect to the vertical axis, we obtain a rectangle
Page 6 of 8
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
with perimeter 40 cm. If the rectangle is folded with respect to the horizontal
axis, we obtain a rectangle with perimeter 50 cm. What is the original perimeter
of the rectangle?
Answer:
2a + 4b = 40
4a + 2b = 50
... 1 point
Then,
2a + 4b = 40
2a + b = 25
3b = 15
b =5
……1 point
... ½ point
a = 10
The original perimeter is 4x10 + 4x5 = 60 cm
... ½ point
11. Brad and Jake traveled 9 km from A to B. First Jake used his bicycle at 8
km/hour and left the bicycle at a certain place. Then he walked 5 km/hour to
reach B. On the other hand, Brad walked at 4 km/hour, then took Jake's bicycle,
and rode at 10 km/hour to reach B. If they started to travel and arrived at B at
the same time, how many minutes did Jake leave his bike before it was used by
Brad?
Answer:
Suppose Jake was cycling for x km, then walked for (9-x) km
Jake was cycling at 8 km/h and walking with speed of 5 km/h, so the time needed by Jake to get the
town B is
x 9 x
8
5
x 9 x
4
10
hours. ………………………………………. (½ point)
Brad walked at 4 km/h and cycling at 10 km/h, so the time needed Brad to get to the town B is
hours …………………………………………………... (½ point)
Because of Brad and Jake start traveling and arrive at B at the same time, then it must satisfies
x 9 x x 9 x
8
5 =4
10 or x = 4…………………………………….. (½ point)
This means that Jake has cycled for 4 km before leaving his bike and then walk for 5 km to reach B.
Because he was cycling with a speed of 8 km/hour, so it take 4/8 = 1/2 hours. (½ point)
Because Brad walk at 4 km/hour, so he walked on foot for 4/4 = 1 hour before he was cycling to
reach B …………………………………………………………… (½ point)
Page 7 of 8
So, Jake has left his bike for 60-30 = 30 minutes before it was used by Brad (½ point)
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
12. In the figure, BC = 25 cm, BE = 8 cm, and AD = 4 cm. What is the area of the
triangle CDF?
A
F
D
B
C
E
Answer:
Area ADC = ½ x AD X BC = ½ x 4 x 25 = 50 cm
Area ADF = ½ x AD x BE = ½ x 4 x 8 = 16 cm
Area DFC = area ADC – area ADF = 34 cm
2
2
2
... 1 point
... 1 point
... 1 point
13. Tom has a contract to dig out some foundations and it must be done in 30 days.
His own machine, which he wishes to use as much as possible, would take 50
days to do all the work. He can hire a bigger machine which would finish the job
in 21 days. There is only enough room for one machine at a time. What is the
least number of days for which he must hire the bigger machine?
Answer:
Performance of small machine is 1/50
Performance of big machine is 1/21
……………………………..…… 1 point
big machine
rest of work
14 days
1 – 14/21 = 1/3
15 days
1 - 15/21 = 2/7
small machine
needed
1/3 . 50 = 16,66
2/7. 50 = 14,3 ………… 1 point
total day needed
14+16,66 = 30,66 days 15+14,3 = 29,3 days
so Tom should hire a big machine for 15 days. ………………..………… 1 point
If the answer is at least 14.5 and less than 15 (½ point)
Page 8 of 8
IMSO 2009
ESSAY 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
Essay
IMSO2009
Yogyakarta, 8-14 November
General Remarks
1. A full mark for each problem is 3 points.
2. When the student gives final answer only, the mark is 1
point.
1. Bob bought a coat and a shirt. The normal prices of both items are the same,
but when Bob bought them, the shirt was discounted by 50% and the coat was
discounted by 25%. If he bought them for $130 what was the normal price of the
shirt?
Answer: $104
Let price = original price of one coat or one shirt.
50% x price + (100% - 25%) x price = 130.......................................... 1 point
50% x price + 75% x price = 130
125% price = 130
(5/4) price = 130...................
....................................................... 1 point
Price = (4/5) x 130
= 104
The original price of the shirt is $104............................................ 1 point
2. The area of a rhombus is 36 cm2. One of its diagonals has length twice of the
other diagonal. What is the length of the smaller diagonal of the rhombus?
Answer: 6 cm
... 1 point
Area of rhombus 36 cm2
4a2 = 36
... 1 point
2
a =9
a=3
The length of the smaller diagonal is 6 cm
Page 2 of 8
... 1 point
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
3. I plan to travel by car from City A to City B. I travel the first half of the distance
with the speed of 30 km/hour. In order to get my average traveling speed of 40
km/hour, what speed should I take for the second half of the distance?
Answer:
Marking Scheme:
If the distance = 120 km.
2 hours
T hour
A o------------------o----------------------o B
60 km
C
60 km
A half of the distance = 120/2 = 60 km.
Time needed from A to C = 60/30 = 2 hour. ………………………………………. 1point
In order get 40 km//hour from A to B:
120
40 = ------------ ………………………………………………….……….………. ½ point
2+T
3
1 = -----------2+T
2+T=3
T = 1. ……………………………………………………………….…….. 1 point
Speed at the next half of the distance is
60 / 1 = 60 km/hour. …………………………………………………….…... ½ point
Page 3 of 8
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
4. The mean value of the mathematics marks of the nine students is 70. When
Lambert’s mark is added, the mean value will be 69. When John’s mark is
added, the mean value will be 72. What will the mean value be, if Lambert’s and
John’s marks are both added?
Answer :
Scheme:
Finding Lambert’s mark (1 point)
Finding John’s mark (1 point)
Calculating the total mark sum of eleven students (½ point)
Calculating the mean (½ point)
Sum=70*9
Sum+L=10*69
Sum+J=10*72
2*Sum+L+J=10*141 or Sum+L+J=10*141-70*9=780
Thus, (Sum+L+J)/11=780/11
5. Andy bought three packages of goods, each worth $ 35, $ 30, and $ 40. The first
package contains 2 books, 1 pencil, and 1 eraser. The second package contains
1 book, 1 pencil, 2 erasers. The third package contains 3 books and 2 erasers.
Andy wants to buy the fourth package containing 2 books, 1 pencil, and 3
erasers. What is the price of the fourth package?
Answer :
Marking scheme
Formulating the system of equations (1 point)
Finding the prices of a book, a pencil, and an eraser (½ point for each)
Calculating the final answer (½ point)
Package
# Books
# Pencils
# Erasers
Price
I
2
1
1
$35
II
1
1
2
$30
III
3
0
2
$40
IV
2
1
3
?
Firstly, comparing (I) & (II): 1 Books - 1 Eraser = $5 or 2 Books = $10 + 2 Erasers
Secondly, combining (II) & (III) and comparing with (I): 2 Books + 3 Erasers = $35
Combining first & second results: 5 erasers = $25 so Eraser = $5.
The price of Package IV is $35 + 2 x price of eraser
Therefore, the price of package IV is $35 + 2 x $5 = $ 45.
Page 4 of 8
IMSO2009
IMSO 2009
Essay
Yogyakarta, 8-14 November
6. The following figure shows a regular hexagon. On each side of the hexagon,
there is an isosceles right triangle. One side forming a right angle is a side of the
hexagon. Determine the angle x.
x
Answer:
Internal angle of hexagonal is 1200
... 1 point
0
Internal angle of right triangle are 90 dan 45
0
0
0
0
Determining of angle x is 360-120 -90 - 45 =105
... 1 point
0
... 1 point
7. Bob and Ivan have tasks of mowing (cutting) grass in their yard. One day they
do their work at the same time. After that, Bob works every 8 days, while Ivan
every 6 days. Each time Bob works, he is paid $15 while Ivan gets $17.5. If they
work again on the same day, how much money will each of them earn at the end
of the day?
Answer:
Day
Bob
Ivan
0
v
v
1
2
3
4
5
6
7
8
v
9
v
10
11
12
v
13
14
15
16
v
17
18
19
20
21
v
th
They meet again at 24 day. At this time: ………………………………………1 point
Bob’s work 4 times ………………………………………………………… ½ point
Ivan’s work 5 times ………………………………………………………… ½ point
Hence
Bob’s earn = 4 x $15 = $ 60 ………………………………………………… ½ point
Ivan’s earn = 5 x $17.5 = $ 87.5 ……………………………………………..½ point
Page 5 of 8
22
23
24
v
v
IMSO2009
IMSO 2009
Essay
Yogyakarta, 8-14 November
8. Mum’s kitchen scale is set incorrectly, but otherwise it works fine. When she
weighs a bag of sugar, it shows 1.5 kg. When she weighs a bag of flour, it
shows 1.3 kg. However, when she weighs both items together, it shows 2.5 kg.
If she weighs a piece of butter of weight 0.3 kg, what number does the scale
show?
Answer:
Solution: 0.6 kg
kg is the real weight of the bag of sugar. In a similar way,
Observe that
kg is the real weight of the bag of flour. So the scale shifts 0.3 kg to the
kg for the butter.
right. Hence, the scale shows
Scheme:
Finding the real weight of sugar or flour ………………..1 point
Finding the shift of the scale ……………………………..1 point
Finding the weight of the butter on the scale …………..1 point
9. Alan, Billy, Candy, and David are queuing (lining up) in alphabetical order. Alan
is in the 7th position from the front while David is in the 9th position from the
back. The number of persons between Alan and Billy is the same as those
between Candy and David. In total, there are 48 persons in the queue, and six
of them are between Billy and Candy. How many persons are there between
Alan and Candy?
Answer:
6 people
A
B
6 people
C
D
8 people
From A to D, there are all together 48-14=34 persons. ……………………………… 1 point
Since the number of persons between A and B is the same as those between C and D, the number
of persons between A and B is:
34 - 6
- 2 =12 ……………………………………... 1 point
2
The number of persons between A and C is 12+7=19 ………………………………… 1 point
10. A rectangle has two axes of symmetry, the vertical axis and the horizontal axis.
If the rectangle is folded with respect to the vertical axis, we obtain a rectangle
Page 6 of 8
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
with perimeter 40 cm. If the rectangle is folded with respect to the horizontal
axis, we obtain a rectangle with perimeter 50 cm. What is the original perimeter
of the rectangle?
Answer:
2a + 4b = 40
4a + 2b = 50
... 1 point
Then,
2a + 4b = 40
2a + b = 25
3b = 15
b =5
……1 point
... ½ point
a = 10
The original perimeter is 4x10 + 4x5 = 60 cm
... ½ point
11. Brad and Jake traveled 9 km from A to B. First Jake used his bicycle at 8
km/hour and left the bicycle at a certain place. Then he walked 5 km/hour to
reach B. On the other hand, Brad walked at 4 km/hour, then took Jake's bicycle,
and rode at 10 km/hour to reach B. If they started to travel and arrived at B at
the same time, how many minutes did Jake leave his bike before it was used by
Brad?
Answer:
Suppose Jake was cycling for x km, then walked for (9-x) km
Jake was cycling at 8 km/h and walking with speed of 5 km/h, so the time needed by Jake to get the
town B is
x 9 x
8
5
x 9 x
4
10
hours. ………………………………………. (½ point)
Brad walked at 4 km/h and cycling at 10 km/h, so the time needed Brad to get to the town B is
hours …………………………………………………... (½ point)
Because of Brad and Jake start traveling and arrive at B at the same time, then it must satisfies
x 9 x x 9 x
8
5 =4
10 or x = 4…………………………………….. (½ point)
This means that Jake has cycled for 4 km before leaving his bike and then walk for 5 km to reach B.
Because he was cycling with a speed of 8 km/hour, so it take 4/8 = 1/2 hours. (½ point)
Because Brad walk at 4 km/hour, so he walked on foot for 4/4 = 1 hour before he was cycling to
reach B …………………………………………………………… (½ point)
Page 7 of 8
So, Jake has left his bike for 60-30 = 30 minutes before it was used by Brad (½ point)
IMSO 2009
Essay
IMSO2009
Yogyakarta, 8-14 November
12. In the figure, BC = 25 cm, BE = 8 cm, and AD = 4 cm. What is the area of the
triangle CDF?
A
F
D
B
C
E
Answer:
Area ADC = ½ x AD X BC = ½ x 4 x 25 = 50 cm
Area ADF = ½ x AD x BE = ½ x 4 x 8 = 16 cm
Area DFC = area ADC – area ADF = 34 cm
2
2
2
... 1 point
... 1 point
... 1 point
13. Tom has a contract to dig out some foundations and it must be done in 30 days.
His own machine, which he wishes to use as much as possible, would take 50
days to do all the work. He can hire a bigger machine which would finish the job
in 21 days. There is only enough room for one machine at a time. What is the
least number of days for which he must hire the bigger machine?
Answer:
Performance of small machine is 1/50
Performance of big machine is 1/21
……………………………..…… 1 point
big machine
rest of work
14 days
1 – 14/21 = 1/3
15 days
1 - 15/21 = 2/7
small machine
needed
1/3 . 50 = 16,66
2/7. 50 = 14,3 ………… 1 point
total day needed
14+16,66 = 30,66 days 15+14,3 = 29,3 days
so Tom should hire a big machine for 15 days. ………………..………… 1 point
If the answer is at least 14.5 and less than 15 (½ point)
Page 8 of 8