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.

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

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!

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!

]]>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?]]>

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

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. ]]>

(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.]]>

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,...]]>