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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Well, we all know that maths can sometimes be a little too difficult for us. So why not ask help from those who can help? I am 12 and since I am going for inter-school competiton, I think that you can help me in my work, as well as anyone who needs your help.

Ideas are funny little things, they won't work unless you do.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

This is a question from Raffles Instituition Primary Mathematics World Contest 2006 Round One.

Q10, The 100th term of the sequence 1/3, 1/2, 5/9, 7/12, 3/5, 11/18, ... is (A) 167/270

(B) 199/300

(C) 67/101

(D) 287/312

(E) None of the above

Ideas are funny little things, they won't work unless you do.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Q15, If 5 different lines drawn randomly on a plane have k points of intersection, the number of possible value of k is

(A) 6

(B) 9

(C) 12

(D) 15

(E) None of the above

Ideas are funny little things, they won't work unless you do.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Q10 is confusing because the fractions have been simplified.

Putting them back into how they were when the sequence was made makes them look like this:

1/3, 3/6, 5/9, 7/12, 9/15, 11/18. And hopefully you can see a clear sequence from that.

Q15 just requires a bit of lateral thinking. If you think about it, it's obvious that two lines can't cross each other more than once, so that puts a limit on the maximum number of intersections.

Try starting with 1 line and working your way up, seeing if you can find a pattern.

So, try both of the questions again with those hints, and then check your answers against

Feel free to come back if you're still stuck or if your answers don't agree with mine. Other members may want to double-check Q15, as I'm not completely certain on that.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

Q10: Great thinking, mathsy! I agree with your answer.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Q15 is especially hard

**X'(y-Xβ)=0**

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Thanks so much for helping me. I definitely understand better now!

Ideas are funny little things, they won't work unless you do.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

It looks like I have another problem here. It is taken from the Singapore Mathematical Olympiad 2002.

Q22, Find the sum of the first 100 No. in the following sequence.

1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,...

Ideas are funny little things, they won't work unless you do.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Q30, Four football teams A,B,C and D are in the same group. Each team plays 3 matches, one with each of the other 3 teams. The winner of each match scores 3 points; the loser scores 0 points; and if a match is a draw, each team scores 1 point. After all the matches, the results are as follows:

(1) The total score of 3 matches for the 4 teams are consecutive odd numbers.

(2) D has the highest total score.

(3) A has exactly 2 draws, one of which the match with C.

Find the total score for each team.

Ideas are funny little things, they won't work unless you do.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,421

Jenilia,

I shall try Q22 first.

The question is to find the total of the first 100 numbers of the series.

1 to 9 is 9 numbers. It can be seen that thereafter, 10, 11, 12 etc have been given as two separate digits.

10 to 99 would be 2 digits each, therefore, 180 numbers.

But we require only 99 more of these.

Hence, 10 to 54 is 45 numbers and the 5 in 55 is to be taken.

1 occurs 16 times, 2 occurs 16 times, 3 occurs 16 times, 4 occurs 16 times, 5 occurs 11 times, 6, 7, 8, and 9 occur 5 times each.

Therefore, the total would be

16(1+2+3+4) + 11(5)+5(6+7+8+9)

=16(10) + 55 + 5(30) = 160 + 55 + 150 = 365.

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,421

Q 30:-

Team D wins against A and B, draws with C and has a total of 7 points.

Team C wins against B, draws with D and A, and has a total of 5 points.

Team B wins against A, losses to both D and C, and has a total of 3 points.

Team A draws with C, losses to D and B, and has a total of 1 points.

Points Tally:-

Team Played Won Lost Drawn Points

D 3 2 0 1 7

C 3 1 0 2 5

B 3 1 2 0 3

A 3 0 2 1 1

Character is who you are when no one is looking.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Thank you sooo much ganesh!

Ideas are funny little things, they won't work unless you do.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Another problem:

Find the fraction in its simplest form of 1/5 + 1/50 + 1/500 + 1/5000...

I do know how to work this out if it is correct,

I can simplify it as 0.2+0.02+0.002...=0.2222...

I'm using Algebra,

Let x be 0.222...

10x=2.222

9x (10x-x)=2

x=2/9

Now, If I have a No. like 0.999999...,is there anyway to simplify it?

Ideas are funny little things, they won't work unless you do.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,421

Jenilia,

To find the sum of 1/5 + 1/50 + 1/500 + 1/5000...,

rewrite it as 1/5(1+1/10+1/100+1/1000....)

The series of numbers inside the bracket forms a Geometric Progression. The sum would be 1/(1-1/10)=1/(9/10)=10/9.

Therefore, the sum 1/5 + 1/50 + 1/500 + 1/5000... = 1/5(10/9)=2/9=0.2222......

This is the answer you got!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,421

All recurring decimals, that is decimal numbers where numbers are repeated without ending after the decimal, can be converted into fractions.

For example, 0.21212121..... can be coverted into a fraction this way.

Let x = 0.212121....

100x = 21.212121....

Finding the difference of the two,

99x=21, therefore, x=21/99 or 7/33.

When we try to convert 0.99999.. this way, this is what happens:-

Let x = 0.99999999....

10x = 9.99999999....

Finding the difference of the two,

9x=9 or x=1.

Yes, 0.999999......... cannot be expressed as a fraction and can only be written as 1!

Character is who you are when no one is looking.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

But then it ist accurate right?

Ideas are funny little things, they won't work unless you do.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,421

How else can 0.9999......recurring indefinitely be expressed as a fraction?

It may appear inaccurate, but think of it, you may never encounter the number 0.999999......... in any area of mathematics!

Character is who you are when no one is looking.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Yeah, as Mr Hollis put in his Caculus book, we are using sum of infinite series IMPLICITLY.

Numbers are a theory, "theory" means approximation

0.99... is accurate to leave perhaps only one particle uncollected from recursively cutting a cake into equally 10 pieces, then collecting 9 and leaving the other one to be next round cut-ee.

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

By the way, is SAT that HaaaaarD?

**X'(y-Xβ)=0**

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I'm 19 and studying a-level mathematics. You're 12 you say?

Aloha Nui means Goodbye.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

Yes

Ideas are funny little things, they won't work unless you do.

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