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If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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21 Mar 2012, 09:04

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If \(t = \frac{1}{(2^9*5^3)}\) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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15 Apr 2012, 19:43

1

This post was BOOKMARKED

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Answer: B.

Can someone please explain how is multiplying 5^6 to the denominator (2^9 * 5^3) get 10^9?

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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15 Apr 2012, 19:50

4

This post received KUDOS

catty2004 wrote:

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Answer: B.

Can someone please explain how is multiplying 5^6 to the denominator (2^9 * 5^3) get 10^9?

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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15 Apr 2012, 21:00

1

This post was BOOKMARKED

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Answer: B.

1st method is really awasome to follow...thanks Bunuel
_________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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20 Oct 2012, 02:58

With a question like this always try to convert the numbers to 10^(x) times something so that you can see the shift of the decimal point. As stated above, the answer is B.
_________________

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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07 Jan 2013, 21:58

Hi, I still dont understand why we have to multiply by 5^6/5^6 i understand that this equals one but what is the general rule for this? how did you know to pick 5^6?

Thanks

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Hi, I still dont understand why we have to multiply by 5^6/5^6 i understand that this equals one but what is the general rule for this? how did you know to pick 5^6?

Thanks

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Answer: B.

Welcome to GMAT Club shahir16.

We want the denominator of the fraction to be written as some power of 10. We need that in order to transform the fraction into decimal easily.

Now, the denominator = 2^9 * 5^3, hence we need to multiply it by 5^6 to get 10^9.

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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08 Jan 2013, 06:49

Can you please show me step-by-step how to convert 2^9 * 5^3 to a power of 10? I am unclear on the concept of converting an expression with exponents to a power of 10. I appreciate your help

Answer: B.[/quote][/quote]

Welcome to GMAT Club shahir16.

We want the denominator of the fraction to be written as some power of 10. We need that in order to transform the fraction into decimal easily.

Now, the denominator = 2^9 * 5^3, hence we need to multiply it by 5^6 to get 10^9.

Can you please show me step-by-step how to convert 2^9 * 5^3 to a power of 10? I am unclear on the concept of converting an expression with exponents to a power of 10. I appreciate your help

Here it goes: \((2^9 * 5^3)*5^6=2^9 * (5^3*5^6)=2^9*5^9=(2*5)^9=10^9\).
_________________

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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11 Sep 2014, 11:35

1

This post received KUDOS

TomB wrote:

If t = 1/(2^9*5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

t = 1/(2^9*5^3)

2^9 = 8^3

8^3 *5^3 = 40^3

1/64000 = there will be at least 3 zeros

1/64=0.0something ,,, adding another 3 zeros then 4 zeros

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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11 Sep 2014, 18:37

Bunuel wrote:

TomB wrote:

If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?

A. Three B. Four C. Five D. Six E. Nine

bunnel , can you please explain this problem

Given: \(t=\frac{1}{2^9*5^3}\).

Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.

Answer: B.

Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).

Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).

Re: If t = 1/(2^9*5^3) is expressed as a terminating decimal, ho [#permalink]

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18 Sep 2014, 10:41

I used a slightly different method..

1/2^9*5^3= 1/10^3*2^6 ....then,1/64=0.01 (dont need to calculate further since we have a non zero digit)....Now, 0.01/10^3 =0.00001 (shifting 3 decimals aside) Hence,4 zeros..
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ITS NOT OVER , UNTIL I WIN ! I CAN, AND I WILL .PERIOD.