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If there are fewer than 8 zeroes between the decimal point and the
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15 Jun 2016, 01:34

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Question Stats:

54% (02:06) correct 46% (02:06) wrong based on 1966 sessions

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If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

If there are fewer than 8 zeroes between the decimal point and the
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15 Jun 2016, 07:06

18

13

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

Hi,

since we have ONLY 10s in denominator, the # of zeroes between the decimal point and the first nonzero digit in the decimal expansion of \((\frac{t}{1000})^4\) will depend on 't'...

ONE 10 will put the decimal point and the REMAINING 10s will result in " zeroes between the decimal point and the first nonzero digit in the decimal expansion of \((\frac{t}{1000})^4\)"..

so we have \(10^{12}\), so if we have t as a single digit, there will be 11 zeroes, that is12 - number of digits in t..... BUT 0s are <8, so t should have digits>12-8 or >4......

lets see the choices -

I. 3 -----\(3^4 = 81\)... two digits ..# of 0s \(= 12 - 2 = 10 > 8.\)...NO II. 5 ------\(5^4 = 625\)... three digits ..# of 0s = 12 - 3 = 9 > 8....NO III. 9------\(9^4 = 6561\)... four digits ..# of 0s = 12 - 4 = 8 = 8....NO, we are looking for <8

If there are fewer than 8 zeroes between the decimal point and the
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16 Aug 2016, 03:34

2

1

hey chetan thanks for the explanation but i don't get this one at all. is there another way possible and if so can u please post it ? thanks

chetan2u wrote:

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

Hi,

since we have ONLY 10s in denominator, the # of zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4 will depend on 't'...

ONE 10 will put the decimal point and the REMAINING 10s will result in " zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4"..

so we have \(10^{12}\), so if we have t as a single digit, there will be 11 zeroes, that is12 - number of digits in t..... BUT 0s are <8, so t should have digits>12-8 or >4......

lets see the choices -

I. 3 -----\(3^4 = 81\)... two digits ..# of 0s \(= 12 - 2 = 10 > 8.\)...NO II. 5 ------\(5^4 = 625\)... three digits ..# of 0s = 12 - 3 = 9 > 8....NO III. 9------\(9^4 = 6561\)... four digits ..# of 0s = 12 - 4 = 8 = 8....NO, we are looking for <8

If there are fewer than 8 zeroes between the decimal point and the
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16 Aug 2016, 08:08

9

1

shivmalhotra10 wrote:

hey chetan thanks for the explanation but i don't get this one at all. is there another way possible and if so can u please post it ? thanks

chetan2u wrote:

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

Hi,

since we have ONLY 10s in denominator, the # of zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4 will depend on 't'...

ONE 10 will put the decimal point and the REMAINING 10s will result in " zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4"..

so we have \(10^{12}\), so if we have t as a single digit, there will be 11 zeroes, that is12 - number of digits in t..... BUT 0s are <8, so t should have digits>12-8 or >4......

lets see the choices -

I. 3 -----\(3^4 = 81\)... two digits ..# of 0s \(= 12 - 2 = 10 > 8.\)...NO II. 5 ------\(5^4 = 625\)... three digits ..# of 0s = 12 - 3 = 9 > 8....NO III. 9------\(9^4 = 6561\)... four digits ..# of 0s = 12 - 4 = 8 = 8....NO, we are looking for <8

none A

Hi I'll try another way..

First just logic What is \(\frac{1}{100}=0.01\)... \(\frac{20}{100}=0.2\) and so on... So we have number of zeroes the way asked is the 0s in denominator - number of digits in numerator...

Let's see the Q now ... Denominator has \(1000^4\), so 12 Zeroes.... We are looking for 0s <8.... so numerator should have digits >12-8...

Now the best way to solve this as it does not really require to be made complicated A simple way which will work here is take t as smallest 2 digit number 10... so (10/1000)^4= (.01)^4= .00000001 so 7 zeroes... This should tell us that any 't' <10 will give us MORE than 7 Zeroes and 10 or more will give7 or less Zeroes...

Now the choices given are all less than 10, so all of them will give 8 or more Zeroes

Re: If there are fewer than 8 zeroes between the decimal point and the
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20 Aug 2016, 09:41

3

Another way is to understand that there will 12 zeros and some other number of other digits. Then just write down 12 zeros and substitute at the end 81, 625 and 6561. The only thing is left - to calculate the number of zeros left.

Re: If there are fewer than 8 zeroes between the decimal point and the
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07 Nov 2016, 08:56

I got it wrong because it says " fewer than 8 zeroes between the decimal point and the first nonzero digit ....". So I did not count the first one. Shouldn't they be more precise ?
_________________

03/29/16 GmatPrep CAT1 600 (Q 47 V 26) + long break. 09/26/16 MGMAT CAT1 560 (Q41 V 27) 10/12/16 MGMAT CAT2 630 (Q46 V 31) 10/21/16 MGMAT CAT3 640 (Q42 V36) 10/28/16 GmatPrep Cat2 660 (Q49 V 31) 11/15/16 MGMAT Cat4 640 (Q41 V 36)

Re: If there are fewer than 8 zeroes between the decimal point and the
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07 Nov 2016, 17:26

13

4

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

We are given that the decimal expansion of (t/1000)^4 has fewer than 8 zeroes between the decimal point and the first nonzero digit. We are also given that 3, 5 and 9 are possible values of t. Let’s test each Roman numeral:

I. 3

If t = 3, then (t/1000)^4 = (3/1000)^4 = (.003)^4 has twelve decimal places with the digits 81 (notice that 3^4 = 81). So there must be 10 zeros between the decimal point and the first nonzero digit 8 in the decimal expansion. This is not a possible value of t.

II. 5

If t = 5, then (t/1000)^4 = (5/1000)^4 = (.005)^4 has twelve decimal places with the digits 625 (notice that 5^4 = 625). So there must be 9 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.

III. 9

If t = 9, then (9/1000)^4 = (9/1000)^4 = (.009)^4 has twelve decimal places with the digits 6561 (notice that 9^4 = 6561). So there must be 8 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.

Recall that we are looking for fewer than 8 zeros between the decimal point and the first nonzero digit in the decimal expansion. So none of the given numbers are possible values of t.

If there are fewer than 8 zeroes between the decimal point and the
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24 Mar 2017, 16:19

1

chetan2u wrote:

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

Hi,

since we have ONLY 10s in denominator, the # of zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4 will depend on 't'...

ONE 10 will put the decimal point and the REMAINING 10s will result in " zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4"..

so we have \(10^{12}\), so if we have t as a single digit, there will be 11 zeroes, that is12 - number of digits in t..... BUT 0s are <8, so t should have digits>12-8 or >4......

lets see the choices -

I. 3 -----\(3^4 = 81\)... two digits ..# of 0s \(= 12 - 2 = 10 > 8.\)...NO II. 5 ------\(5^4 = 625\)... three digits ..# of 0s = 12 - 3 = 9 > 8....NO III. 9------\(9^4 = 6561\)... four digits ..# of 0s = 12 - 4 = 8 = 8....NO, we are looking for <8

none A

Hi, do you agree with this rationale?

According to the stem, [(t/1000)^4]>1x10^-8 1x10^-8 has 7 0s between the decimal and first non-zero digit. 1x10^-8 is the smallest number with 7 0s between the decimal and first non-zero digit.

Re: If there are fewer than 8 zeroes between the decimal point and the
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25 Mar 2017, 05:32

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

The number of zeroes between 't' and the decimal point has to be <8. Then (t/10^(7 or lesser)) matches. Now we need to take into account the size of 't'. So (t/10^(7 or lesser+size of 't'))

The expression in question is t^4/10^12; to match this size of 't' should be anything above or equal to 5. t^4 with no option has the size>5, so option A
_________________

Re: If there are fewer than 8 zeroes between the decimal point and the
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29 Apr 2017, 00:16

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

(t^4)/(10^12) should have less then 8 zero between first non zero number

Re: If there are fewer than 8 zeroes between the decimal point and the
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01 May 2017, 19:34

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

My 2 cents.

I think it is easier if we change as this :

= (t x 10^-3)^4 = t^4 x 10^-12

So, in here, if t =1, we have 12 zeroes. So what we need to find is how many digits do t^4 contains.

I. 3 3^4 is two digit number, meaning, there will be 10 zeroes...so this is no good

II. 5 5^4 is three digit number, meaning, there will be 9 zeroes...so no good

III. 9 9^4 is four digit number, meaning, there will 8 zeroes...so no good.

Re: If there are fewer than 8 zeroes between the decimal point and the
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22 Jul 2018, 08:28

5

Top Contributor

1

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

First: (t/1000)^4 = (t^4)/(1000^4) Now recognize that 1000^4 = (10^3)^4 = 10^12 So, (t/1000)^4 = (t^4)/(1000^4) = (t^4)/(10^12)

IMPORTANT: When we divide a number by 10^12, we must move the decimal point 12 spaces to the left So, for example, 1234567/10^12 = 0.000001234567 Likewise, 8888/10^12 = 0.000000008888 And, 66666666666666/10^12 = 66.666666666666

Now let's check each option

I. 3 It t = 3, then (t^4)/(10^12) = (3^4)/(10^12) = 81/(10^12) = 0.000000000081 There are 10 zeroes between the decimal point and the first nonzero digit Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement I

II. 5 It t = 5, then (t^4)/(10^12) = (5^4)/(10^12) = 625/(10^12) = 0.000000000625 There are 9 zeroes between the decimal point and the first nonzero digit Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement II

III. 9 It t = 9, then (t^4)/(10^12) = (9^4)/(10^12) = 6561/(10^12) = 0.000000006561 There are 8 zeroes between the decimal point and the first nonzero digit Since the question tells us that there are fewer than 8 zeroes between the decimal point and the first nonzero digit, we can ELIMINATE statement III

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18 Aug 2018, 00:12

JeffTargetTestPrep wrote:

Bunuel wrote:

If there are fewer than 8 zeroes between the decimal point and the first nonzero digit in the decimal expansion of (t/1000)^4, which of the following numbers could be the value of t?

I. 3 II. 5 III. 9

A) None B) I only C) II only D) III only E) II and III

We are given that the decimal expansion of (t/1000)^4 has fewer than 8 zeroes between the decimal point and the first nonzero digit. We are also given that 3, 5 and 9 are possible values of t. Let’s test each Roman numeral:

I. 3

If t = 3, then (t/1000)^4 = (3/1000)^4 = (.003)^4 has twelve decimal places with the digits 81 (notice that 3^4 = 81). So there must be 10 zeros between the decimal point and the first nonzero digit 8 in the decimal expansion. This is not a possible value of t.

II. 5

If t = 5, then (t/1000)^4 = (5/1000)^4 = (.005)^4 has twelve decimal places with the digits 625 (notice that 5^4 = 625). So there must be 9 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.

III. 9

If t = 9, then (9/1000)^4 = (9/1000)^4 = (.009)^4 has twelve decimal places with the digits 6561 (notice that 9^4 = 6561). So there must be 8 zeros between the decimal point and the first nonzero digit 6 in the decimal expansion. This is not a possible value of t.

Recall that we are looking for fewer than 8 zeros between the decimal point and the first nonzero digit in the decimal expansion. So none of the given numbers are possible values of t.

If there are fewer than 8 zeroes between the decimal point and the
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20 Sep 2018, 20:39

1

\((\frac{t}{1000})^4\) in decimal form is \((0.000t)^4\). This means that \(0.000t\) is multiplied to itself 4 times. It will be important to observe that:

These examples show that when a decimal has 3 places containing zeroes is raised to its second power, we’re expecting 3 x 2 = 6 digits after the decimal point. Similarly, we’re expecting 4 x 3 = 12 numbers after the decimal place when 0.0003, 0.0005, and 0.0009 are raised to the fourth power.

First, let’s check the values of\(3^4\),\(5^4\), and \(9^4\):

\(3^4 = 81\), \(5^4 = 625\), and \(9^4 = 6561\)

This means that for\((0.0003)^4\) or \((3/1000)^4\), there will be 12-2 = 10 zeroes and 2 nonzero digits after the decimal point.

Similarly, \((5/1000)^4\) will have 9 zeroes and 3 nonzero digits while \((9/1000)^4\) will have 8 zeroes and 4 nonzero digits after the decimal point.

We can see that for the three options, none of them had fewer than 8 zeroes. Thus, the final answer is

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28 Apr 2019, 13:21

(t/1000)^4 = t^4/10^12 = 0.000..x (with maximum 7 digit 0 after . and before x and x>=1) so t^4/10^12 >= 1/10^8 ==> t^4 >= 10^4 ==> t>=10 None of the answers is true => Option A

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14 Sep 2019, 07:17

Consider simple fractions \(\frac{1}{1000}\)=0.001 (1 digit give n-1 0s) \(\frac{11}{1000}\)=0.011 (1 digit give n-2 0s) \(\frac{111}{1000}\)=0.111 (1 digit give n-3 0s)

We clearly know that question has 12 0s in the denominator (\(10^3\)= 3 0s, \(10^12\)=12 0s)

So it must have less than 8 0s, which means it has at max 7 0s. Let us consider the case with 7 0s after the decimal, thus numerator must have n-7 digits i.e 12-7 = 5 digits

Not let us see the number of digits for 1,2,3 I. 3 => 3^4 =81 NO II. 5 => 5^4 =625 NO III. 9 => 9^4 =81*81=729*9=4 digits NO

Thus the correct answer is A. None _________________