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Re: repeating decimals [#permalink]
11 Feb 2009, 00:00

You can solve this pretty quickly using approximation.

A) 2/11 can be approx. to 2/10 = 1/5 = 0.2 B) 1/3 = 0.33 C) 41/99 can be approx. to 40/100 = 2/5 = 0.4 D) 2/3 = 0.66 E) 23/37... mmh, the weird one; if you want to double check you can proceed with the division (which yields 0.621, hence the longest sequence of different digits).

Re: repeating decimals [#permalink]
11 Feb 2009, 02:51

1

This post received KUDOS

Expert's post

There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.

1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.

2) cut numerators.

3) Can we compose 9 out of each denominators? Only for B and D. They are out.

4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.

If you've truly grasped this way, you can solve problems like that under 20 sec and without complex calculations.

Re: repeating decimals [#permalink]
11 Feb 2009, 07:10

walker wrote:

There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.

1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.

2) cut numerators.

3) Can we compose 9 out of each denominators? Only for B and D. They are out.

4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.

If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations.

Walker = Genius

+1

Thanks buddy for a great tip.

_________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: repeating decimals [#permalink]
11 Feb 2009, 07:41

2

This post received KUDOS

Expert's post

There is a prove. It is better to remember a formula when it is understandable

1/a=0.(b)=b*(10^{-n}+10^{-2n}+10^{-3n.....}+...)=b*\frac{10^{-n}}{1-10^{-n}}=b*\frac{1}{10^{n}-1}=\frac{b}{99...99_n} where n - the length of sequence (the number of digits of b)

Re: repeating decimals [#permalink]
11 Feb 2009, 09:01

walker, how can u b soooooo perfect in maths?Give us also sum tips. U r simply marvellous! needless 2 say +1 frm me

walker wrote:

There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.

1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.

2) cut numerators.

3) Can we compose 9 out of each denominators? Only for B and D. They are out.

4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.

If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations.

Re: repeating decimals [#permalink]
11 Feb 2009, 10:10

walker wrote:

There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.

1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.

2) cut numerators.

3) Can we compose 9 out of each denominators? Only for B and D. They are out.

4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.

If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations.

wow this is EXACTLY what I was looking for..gmatclub is blessed to have you math gurus here...thanks alot!

Re: repeating decimals [#permalink]
11 Feb 2009, 10:59

walker wrote:

There is a general rule: 0.(abcd)=abcd/9999. So, the more "9" we need to write out our fraction in such way, the longer sequence we have. (You can use search to find my prove of this rule). Moreover, if we have proper fraction x/y, x doesn't influence on length of a sequence.

1) First of all, let's take a fast look at our options. Are their proper fraction? Or do we need to change them? All fractions are proper fractions.

2) cut numerators.

3) Can we compose 9 out of each denominators? Only for B and D. They are out.

4) Can we compose 99 out of each denominators? Only for A and C. They are out. Only E remains. In other words, 1/37 (23/37) has sequence consisted of more than two digits.

If you've truly grasped this way, you can solve problems like than under 20 sec and without complex calculations.

Words are less to appreciate this explanation. wonderful GMAT Tip. +1 from me too.