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Manager  Joined: 18 Feb 2015
Posts: 81
In the sequence above each term after  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 69% (01:38) correct 31% (01:54) wrong based on 373 sessions

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1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

Originally posted by HarveyKlaus on 10 Jul 2016, 02:35.
Last edited by abhimahna on 30 Jun 2017, 03:39, edited 1 time in total.
Edited the question
Retired Moderator G
Joined: 26 Nov 2012
Posts: 555
Re: In the sequence above each term after  [#permalink]

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1
HarveyKlaus wrote:
Hi all,
I am not able to solve this question by the normal sequence formula. Can someone please help? Thanks

1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. $$2^1$$, $$2^2$$,$$2^3$$.... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.
Intern  Joined: 02 Feb 2016
Posts: 2
Re: In the sequence above each term after  [#permalink]

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1
In the Sequence notice that the sequence is just the 1/(2^n) ...
so for 1st term=1/2^1=1/2
2nd term=1/(2^2)1/4, 3rd term=1/(2^3)=1/8 and so on...
Thus the 10th term will be 1/(2^10)=1/1024
Roughly, 1/1024 can be 1/1000=0.001 but since denominator is a bit more than 1000 therefore the actual value will be a bit less than 0.001.

thus the ans will lie btw. 0.0001 and 0.001.(D)
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3664
Re: In the sequence above each term after  [#permalink]

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msk0657 wrote:
HarveyKlaus wrote:

1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. $$2^1$$, $$2^2$$,$$2^3$$.... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.

Then 1/1024 = 0.024
msk0657 , how did you get this value?

On a calculator it comes out as .0009765625.
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Manager  B
Joined: 26 Mar 2016
Posts: 68
Location: Greece
GMAT 1: 710 Q51 V34 GPA: 2.9
Re: In the sequence above each term after  [#permalink]

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It's just about calculating the tenth term which is 2^10=1024.
Then you should be able to define where this number fits.
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+1 Kudos if you like the post Intern  Joined: 16 Jan 2017
Posts: 1
Re: In the sequence above each term after  [#permalink]

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msk0657 wrote:
Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. $$2^1$$, $$2^2$$,$$2^3$$.... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.

What is your approach to solve for 2^10 and 1/1024?
Manager  G
Joined: 16 Oct 2011
Posts: 109
GMAT 1: 570 Q39 V41 GMAT 2: 640 Q38 V31 GMAT 3: 650 Q42 V38 GMAT 4: 650 Q44 V36 GMAT 5: 570 Q31 V38 GPA: 3.75
Re: In the sequence above each term after  [#permalink]

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HarveyKlaus wrote:
1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

When comparing powers of some value to either powers of 10 or some decimal with a trailing 1 (which is also a power of 10), one approach is to rewrite our term, to get it as close to a power of 10 as possible. For (1/2)^10, a base of 1/8 is gonna be as close to 1/10 as we can get. Then (1/2)^10 = 1/(2^3 *2^3 *2^3 *2) = (1/8)^3*1/2 is approximately
(1/10)^3 *1/2 = .001*1/2 = approximately .0005, which falls between .0001 and .001. The answer is D
Senior Manager  S
Joined: 12 Sep 2017
Posts: 306
In the sequence above each term after  [#permalink]

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msk0657 wrote:
HarveyKlaus wrote:
Hi all,
I am not able to solve this question by the normal sequence formula. Can someone please help? Thanks

1/2, 1/4, 1/8, 1/16, 1/32, ....

In the sequence above each term after after the first one-half the previous term. If x is the tenth term of the sequence, then x satisfies which of the following inequalities?

A) 0.1 < x < 1
B) 0.01 < x < 0.1
C) 0.001 < x < 0.01
D) 0.0001 < x < 0.001
E) 0.00001 < x < 0.0001

Hi,

This is how I tried.

We are asked to find the tenth term of the series 1/2, 1/4, 1/8, 1/16, 1/32, ....

If you observe that denominator is power of 2 i.e. $$2^1$$, $$2^2$$,$$2^3$$.... then tenth term will be 2^10.

Here 2^10 = 1024

Then 1/1024 = 0.024. Now check where this number fits in and this value is greater than 0.001. So C is eliminated.

Then only D satisfies this condition.

Hello msk0657 !

How did you get this result?

Then 1/1024 = 0.024

Kind regards! In the sequence above each term after   [#permalink] 20 Jan 2019, 11:50
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