All About Math Olympiad Training & Questions

[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.

[Maths Hub]

pinky88:

Hi,


I need help for the following questions:

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.

2. There are 2 possible cases.
Case 1: 3 wigs, 5 sunglasses
10C3 X 12C5 = 120 X 792 = 95040
Case 2: 4 wigs, 4 sunglasses
10C4 X 12C4 = 210 X 495 = 103950

Total: 95040 + 103950 = 198990

Cheers.

[pinky88]

Many thx, Maths Hub

[Maths Hub]

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

[BeContented]

Maths Hub:

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

Thank you.....been waiting until neck long long liao.
DD will be delighted

[TanJ]

Maths Hub:

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

Does gettng a gold qualify for invitation round? How will my kid know?
But then.... second round heavyweights.... chances as good as nil....