All About Math Olympiad Training & Questions
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Atan:
The boy should have gotten 3 certificates :
I know of a boy who took AMC8 last yr in P3 and got Silver or 2nd Place. How is this relevant to the above 'Distinctions'?
AMC 8 Distinguished Honor Roll Certificate
AMC 8 Achievement Roll Certificate
AMC 8 Certificate - 2nd Place -
dagong99:
Partial AMC 8 Results is out.
The boy should have gotten 3 certificates :Atan:
I know of a boy who took AMC8 last yr in P3 and got Silver or 2nd Place. How is this relevant to the above 'Distinctions'?
AMC 8 Distinguished Honor Roll Certificate
AMC 8 Achievement Roll Certificate
AMC 8 Certificate - 2nd Place
This year's worldwide statistics :
Cheng Puhua got perfect score !
Top 1% Score - Distinguished Honor Roll : Score => 22
Top 5% Score - Honor Roll : Score => 18
Top 25% Score : Score => 13
Top 50% Score : Score => 10
Top 75% Score : Score => 8
High Score = 25
Perfect Scores = 211 students
Average Score = 10.65
Standard Deviation = 4.35
Total Students 148,557
Total Schools = 2,359 -
dagong99:
Partial AMC 8 Results is out.
The boy should have gotten 3 certificates :dagong99:
[quote=\"Atan\"]
I know of a boy who took AMC8 last yr in P3 and got Silver or 2nd Place. How is this relevant to the above 'Distinctions'?
AMC 8 Distinguished Honor Roll Certificate
AMC 8 Achievement Roll Certificate
AMC 8 Certificate - 2nd Place
This year's worldwide statistics :
Cheng Puhua got perfect score !
Top 1% Score - Distinguished Honor Roll : Score => 22
Top 5% Score - Honor Roll : Score => 18
Top 25% Score : Score => 13
Top 50% Score : Score => 10
Top 75% Score : Score => 8
High Score = 25
Perfect Scores = 211 students
Average Score = 10.65
Standard Deviation = 4.35
Total Students 148,557
Total Schools = 2,359[/quote]Where can I see the results? -
dyh:
[/quote]http://amc-reg.maa.org/Reports/GeneralReports.aspx
Where can I see the results?dagong99:
[quote=\"dagong99\"]
Partial AMC 8 Results is out.
This year's worldwide statistics :
Cheng Puhua got perfect score !
Top 1% Score - Distinguished Honor Roll : Score => 22
Top 5% Score - Honor Roll : Score => 18
Top 25% Score : Score => 13
Top 50% Score : Score => 10
Top 75% Score : Score => 8
High Score = 25
Perfect Scores = 211 students
Average Score = 10.65
Standard Deviation = 4.35
Total Students 148,557
Total Schools = 2,359 -
dagong99:
Thanks.
http://amc-reg.maa.org/Reports/GeneralReports.aspxdyh:
Where can I see the results? -
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A -
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1.
Thus total area will be (sqrt3/4X4+3)A=(3+ sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS! -
Absolutely Bo Chap:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
given answers are correct.
For Q1, you need to move the areas around to form a cross with 5 squares. Each square is 1/5 of the area of the original square = 20 cm2
For Q2 and Q3, not necessary to use algebra. Concepts of speed ratio/distance ratio can be applied to solve these.
For Q4, secondary school students should be able to solve it using trigo and pythagoras theorem.
cheers. -
kiasuMathsOlympiadtrainer:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
Answer is 20 square cm. Using cut and paste, we will form a cross with 5 squares and each square will be 1/5. My apologies for missing out this qn earlier.
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
The part in blue below is slightly wrong previously:
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1=1 : 4/sqrt 3 = A : 4/sqrt3 A
Thus total area will be (4/sqrt3X4+3)A=(3+ 16/sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS! -
kiasuMathsOlympiadtrainer:
nothing wrong with Q4. Radius of circle is sqrt(7/3).kiasuMathsOlympiadtrainer:
Keban Baru CC is organising Maths Olympiad Training for upper primary / lower secondary at $50 for 2 x half-day sessions next week, just after Christmas. Some problems with numerical answers were given in the posters in the CC.
I cannot see how any of these problems can be solved using upper primary / lower secondary mathematics, especially the last problem. And the posters/banner claimed that only simple algebra is needed.
I had tried giving these problems to some JC students. They can solve Problems 1 to 3, but the methods are quite complicated and involve messy algebra that are beyond any primary school kids. And none of them can solve Problem 4, though it is also true that none of them were Maths Olympiad students.
Can anyone help?
1. Let ABCD be a square with area 100 cm square. Let points P, Q, R and S be the midpoints of sides CD, DA, AB and BC respectively. Determine the area of the square formed by lines AP, BQ, CR and DS.
Ans: \t20 cm square
Answer is 20 square cm. Using cut and paste, we will form a cross with 5 squares and each square will be 1/5. My apologies for missing out this qn earlier.
2. Road Runner is walking over a railway-bridge. When he is 10 m away from the middle of the railway-bridge, Wile E. Coyote approaches him from behind on a rocket. The rocket travels at a speed of 100 km/h. The distance of the rocket from the railway-bridge is the same as the length of the railway-bridge. If the Road Runner runs towards the rocket to reach the beginning of the railway-bridge, he will miss the rocket by 4 m. If Road Runner runs in the other direction, the rocket will reach him 8 m before the end of the bridge. What is the length of the railway-bridge?
Ans: \t44 m
Speed of rocket = 100km/h
Speed of Road runner = v km/h
Let the length of half the bridge be L km.
Scenario 1: Run towards each other
we form a equation using Distance moved by rocket:
100 X (L-10)/v = 2L-4
Scenario 2: Rocket chase after road runner
100 X (L+10-8)/v = 4L-8
Take scenario 2 divide by scenario 1 equation (simulataneous eqn):
(L+2)/(L-10)=2
L+2=2L-20
L=22
2L=44m
3. Two boats are docked at the different sides of a river, which we labelled as side A and side B. They start to travel to the opposite sides at the same time. When they reach the opposite sides, they immediately reverse and travel back to their original positions. Both boats travel at constant but different speeds. The first time they meet, they are at 21 m away from side A. When they meet on their way back, they meet 28 m away from side B. What is the width of the river?
Ans: \t35 m
Let the distance of the river be 21+X m, with first boat travelling 21 m and second boat travel X m. When they meet the second time, a total of 3 times the length is travelled. You need to plot out the path way of each boat to be able to see that altogether 3 times the length of the river is covered.
Looking at the second time they meet, first boat will travelled 63 m, while second boat has travelled 3X m. When they meet, first boat has covered 1 length of the river and 28 m.
1 Length of river + 28 m = 63 m
1 Length of river = 35 m
4. Let the area of a 1 cm x 1 cm x 1 cm triangle be A. Find the area of a 2 cm x 2 cm x 2 cm x 1 cm x 1 cm x 1 cm hexagon that is inscribed in a circle in terms of A.
(This was the problem on the banner, but the banner was taken down. Is it because the problem was too difficult?)
Ans: \t13 A
The answer is wrong as well. If you try to draw out the hexagon, it will comprise of 3 1X1X1 triangles and 1 2X2 square.
Area of 2X2 squares= 4 X Area of 1X1 square
The part in blue below is slightly wrong previously:
Using pythagoras theorem, we can see that the ratio of a 1X1X1 triangle to 1X1 square will be sqrt3/4 : 1=1 : 4/sqrt 3 = A : 4/sqrt3 A
Thus total area will be (4/sqrt3X4+3)A=(3+ 16/sqrt3)A
I think they realised the mistakes that's why they take that down. Nowadays there are alot of \"Maths Olympiad Trainer\" or \"best tuition teacher\" (as they claim to be) around. Just be careful of what you sign up for.
Have a MERRY XMAS!
Area of triangle of 1 cm side = sqrt(3)/4 cm2 = A
Area of hexagon = 13 * sqrt(3)/4 cm2 = 13 A
cheers.
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