I think that's the best explanation that there is. You only have to work on the division to four - at most - decimal places to realize it's going to be a repeating sequence. The GMAT would not test you on a number that doesn't have this sequence, since it would take forever to do long division by hand out to the 25th decimal place.

As far as a general technique, look for patterns and shortcuts. Remember that there is almost always an easier way than doing the entire calculation.

i dunno you can try this...but i could be wrong...

to find a specific number in a list of patterned numbers:
1) count the number of items in the pattern = the 2 items in the pattern are "5" and "4"

2) divide the number of items into the place number you are looking for AND NOTE THE REMAINDER = your place number here is 25

25/2 = 12 with remainder of 1

3) count the remainder from the left
6/11 = 0.54545454
- since you only count 1 spot, you land on the number 5

Here's another example of how you can apply this again...
39713971.....find the 87th number on this list

1) how many items do you see? 4 (3971)
2) what is your place number = 87
3) divide and note remainder 87/4 = 21 with remainder of 3
4) count 3 places from the left = 7

to illustrate further - try looking for the 5th number on this list - its obvious that in 39713971, the fifth number is 3. but even if you apply the same method above, you'll still land on 3.

thanks guys - i agree identifying the pattern here is key. But typically for me it is the simple things i have problems with and not the more complicated one.

Such as what is the best process to divide 6 by 11?

i can 60 by 55 which 5 then 5 remainder and try to play with that, but this takes forever and not always accurate, dividing samller numbers by larger ones always gives me problems..... any techniques are welcome...

Hey Has..

I think we are over complicating this...You have already done half of job..

Has wrote:Hello folks,

OG PS section question 219 has a really poor explanation to the answer - can anyone do a better job?
6/11 = 0.54545454…

I would do only upto 6/11 = 0.5454....since its gonna be repetitive...

No only question is to find 25th digit...so here pattern is 2 digits are repetiting...so 25/2 gives 1 remainder...that means it is first digit of repetitive cycle...which is 5..

I hope I am clear and correct
Question
What is the 25th digit to the right of the decimal point in the form of 6/11 (6 divided by 11)?

Answer is 5

Explanation 6/11 = 0.54545454… every odd numbered digit is 5 therefore 25th digit is 5.


Can someone please do a better job of explaining this to me, also what technique can I apply generally to work out such questions, in under 2 mins…..


Many Thanks
H

cool - i'm sticking to my method, 60 over 11, 5 times (i decimal diff) thus 0.5 reminder 5 (thus 0.05) is 50 over 11 which is 4 times (44) (the 4 is 2 decimal diff thus 0.04 total now 0.54) and remainder 6 which is actuallly 0.006 which again becomes 60 but with 3 decimal difference and the cycle continues.

i will use this pricess until i identify the pattern thus answer. Of course, if anyone has a quiclker method, then, by all mean let me knwo.

Thanks ALL

an easy way to find 6/11 in decimal terms

expand it by 9

(6*9)/(11*9) = 54/99

note that 54/99 is the fractional representation of circulating (recurring) decimal 0,545454...

Thanks

Has wrote:Hello folks,

OG PS section question 219 has a really poor explanation to the answer - can anyone do a better job?

Question
What is the 25th digit to the right of the decimal point in the form of 6/11 (6 divided by 11)?

Answer is 5

Explanation 6/11 = 0.54545454� every odd numbered digit is 5 therefore 25th digit is 5.


Can someone please do a better job of explaining this to me, also what technique can I apply generally to work out such questions, in under 2 mins�..


Many Thanks
H

Solution:

To solve this question, we first have to use some long division. This long division allows us to get 6/11 in decimal form, which is 0.545454... where "54" is repeating.

We can see that the 1st, 3rd, 5th (and so on) digit to the right of the decimal point is a 5 and that the 2nd, 4th, 6th (and so on) digit to the right of the decimal point is a 4. In other words, each odd-positioned digit is a 5, and each even-positioned digit is a 4.

Because we are being asked about the 25th digit to the right of the decimal point and we see that 25 is odd, we know that the 25th digit is a 5.

The Answer is C