All About Math Olympiad Training & Questions

[dagong99]

clover18:

Hi, can someone advise if competitions like Gauss contest and AMC carry much weight in DSA applications?

Besides Maths Hub, do you know any other centre holding AMC ? Maths Hub used to hold it in Bt Timah Branch too but this year, they don't.

[clover18]

dagong99:

clover18:

Hi, can someone advise if competitions like Gauss contest and AMC carry much weight in DSA applications?

Besides Maths Hub, do you know any other centre holding AMC ? Maths Hub used to hold it in Bt Timah Branch too but this year, they don't.

I had the impression Mathshub hold it at all their branches. Perhaps you can also check Maths Oasis?

[MathsOlympiadtrainer]

empressplace:

I just chanced upon this thread and post. Sorry for my ignorance but could someone please explain to me why there is only one combination for 4 in one box? There are 4 boxes. So shouldn't there be 4 ways to put the 4 marbles in one box?

quekcc:

[quote=\"trileets\"]May I know how this qn is done?

Find the number of ways to put 4 different coloured marbles into 4 identical empty boxes.

Ans: 15 (my ans of 24 is wrong)



TIA

Your answer 24 is the number of ways to arrange 4 different coloured marbles in order. But the question is looking for number of ways to group the 4 different coloured marbles.

There are 5 possible ways of grouping.

1) all in 1 box - 1 possible combination only
2) 3 in 1 box and 1 in another box - 4 possible combinations
3) 2 in 1 box and 2 in another box - 6 possible combinations
4) 2 in 1 box, 1 in 2nd box and 1 in another box - 3 possible combinations
5) 1 in each box - 1 possible combination only

So total = 1 + 4 + 6 + 3 + 1 = 15

[/quote]the boxes are identical so the way to put 4 different marbles in 4 identical is only 1.

[dagong99]

clover18:

dagong99:

[quote=\"clover18\"]Hi, can someone advise if competitions like Gauss contest and AMC carry much weight in DSA applications?

Besides Maths Hub, do you know any other centre holding AMC ? Maths Hub used to hold it in Bt Timah Branch too but this year, they don't.

I had the impression Mathshub hold it at all their branches. Perhaps you can also check Maths Oasis?[/quote]I went to Maths Hub last week, they mentioned only Bt Batok branch this year. As for Maths Oasis, they dun hold AMC, they hold the other AMC, i.e. AMC 8, 10 & 12.

[empressplace]

MathsOlympiadtrainer:

the boxes are identical so the way to put 4 different marbles in 4 identical is only 1.

Thanks! Finally got it. I missed the key words. 4 different colours.

Is this not permutations and combinations taught in A levels in the past? What would be the workings if done in the permutations and combinations' way for exams?

[kiasiparent]

empressplace:

MathsOlympiadtrainer:

the boxes are identical so the way to put 4 different marbles in 4 identical is only 1.

Thanks! Finally got it. I missed the key words. 4 different colours.

Is this not permutations and combinations taught in A levels in the past? What would be the workings if done in the permutations and combinations' way for exams?

4C4 = 1
4C3 x 1C1 = 4 x 1 = 4 (because only 1 marble remains after you pick 3 marbles out)
4C2 x 2C2/2 = 6/2 = 3
4C2 = 6
1x1x1x1= 1

Total = 15

[kiasiparent]

quekcc:

trileets:

May I know how this qn is done?


Find the number of ways to put 4 different coloured marbles into 4 identical empty boxes.

Ans: 15 (my ans of 24 is wrong)



TIA

Your answer 24 is the number of ways to arrange 4 different coloured marbles in order. But the question is looking for number of ways to group the 4 different coloured marbles.

There are 5 possible ways of grouping.

1) all in 1 box - 1 possible combination only
2) 3 in 1 box and 1 in another box - 4 possible combinations
3) 2 in 1 box and 2 in another box - 6 possible combinations
4) 2 in 1 box, 1 in 2nd box and 1 in another box - 3 possible combinations
5) 1 in each box - 1 possible combination only

So total = 1 + 4 + 6 + 3 + 1 = 15

3) 2 in 1 box and 2 in another box - 3 possible combinations
4) 2 in 1 box, 1 in 2nd box and 1 in another box - 6 possible combinations

[kiasiparent]

Those interested in Maths Olympiad classes, its best to go Maths Oasis as they have good trainers and small class. Another alternative would be learning interactive. (Take note that I am not related/affiliated to them in anyway)


However, if the class size is too big (>10 students), then it might not be worth it as the teacher is unable to give enough 1-1 guidance. You might want to consider 1-1 home enrichment too.

[pinky88]

Hi,


I need help for the following questions:

1.\tIn a competition consisting of 30 problems. Lydia was given 12 points for each correct solution & 7 points were subtracted from her score for each incorrect solution. Problems not attempted contributed 0 points. How many correct solutions did Lydia have if her score was 209? (Hint : 19 X 11 = 209).

2.\tDiana is to go for a business trip. She needs a collection of wigs and sunglasses, 8 items altogether. At least 3 of these have to be wigs and at least 4 sunglasses. She has 10 different wigs and 12 different sunglasses to choose from. In how many ways can she organize the collection of wigs and sunglasses she needs for the trip.

TIA.

[Maths Hub]

pinky88:

Hi,


I need help for the following questions:

1.\tIn a competition consisting of 30 problems. Lydia was given 12 points for each correct solution & 7 points were subtracted from her score for each incorrect solution. Problems not attempted contributed 0 points. How many correct solutions did Lydia have if her score was 209? (Hint : 19 X 11 = 209).

TIA.

1. Assume all are correct.
Total Score will be 30 X 12 = 360
For each wrong question, a total of 12 + 7 = 19 marks is deducted from the total marks.
For each unanswered question, 12 marks are deducted from the total marks.
Now, there is a difference of 360 - 209 = 141 from the total score.
We try to find 141 as a sum of multiples of 19 and 12.
141 = 1 X 19 + 122 (not a multiple of 12)
= 2 X 19 + 103 (not a multiple of 12)
= 3 X 19 + 84 (a multiple of 12)
= 4 X 19 + 65 (not a multiple of 12)
= 5 X 19 + 46 (not a multiple of 12)
= 6 X 19 + 27 (not a multiple of 12)
= 7 X 19 + 8 (not a multiple of 12)

The only case is 3 X 19 + 7 X 12. So there must be 3 wrong, 7 unanswered and thus 30 - 3 - 7 = 20 correct.

Hope that helps.