GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Oct 2019, 13:47

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M26-16

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
M26-16  [#permalink]

Show Tags

New post 16 Sep 2014, 01:25
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

57% (01:22) correct 43% (01:40) wrong based on 153 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re M26-16  [#permalink]

Show Tags

New post 16 Sep 2014, 01:25
1
2
Official Solution:

If \(n\) is an integer and \(\frac{1}{10^{n+1}} \lt 0.00737 \lt \frac{1}{10^n}\), then what is the value of \(n\)?

A. 1
B. 2
C. 3
D. 4
E. 5


There is no need for algebraic manipulation to solve this question.

\(\frac{1}{10^{n+1}}\) is 10 times less than \(\frac{1}{10^n}\), and both when expressed as decimals are of a type 0.001 (some number of zeros before 1). Which means that the given expression to hold true we should have: \(0.001 \lt 0.00737 \lt 0.01\), which means that \(n=2\) \((\frac{1}{10^n}=0.01\), so \(n=2)\).


Answer: B
_________________
Intern
Intern
avatar
Joined: 26 Jul 2011
Posts: 26
GMAT ToolKit User
Re: M26-16  [#permalink]

Show Tags

New post 18 Sep 2014, 14:16
Hi Bunuel,

I am unable to understand the explanation. Please help me on this.

This is how is approached, but got stuck.

\(\frac{1}{10^{n+1}} \lt \frac{737}{10^5} \lt \frac{1}{10^n}\)

Approximate 737 to 1000

\(\frac{1}{10^{n+1}} \lt \frac{10^3}{10^5} \lt \frac{1}{10^n}\)

\(\frac{1}{10^{n+1}} \lt \frac{1}{10^2} \lt \frac{1}{10^n}\)

Thanks,
Gabriel
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M26-16  [#permalink]

Show Tags

New post 19 Sep 2014, 02:14
gabriel87 wrote:
Hi Bunuel,

I am unable to understand the explanation. Please help me on this.

This is how is approached, but got stuck.

\(\frac{1}{10^{n+1}} \lt \frac{737}{10^5} \lt \frac{1}{10^n}\)

Approximate 737 to 1000

\(\frac{1}{10^{n+1}} \lt \frac{10^3}{10^5} \lt \frac{1}{10^n}\)

\(\frac{1}{10^{n+1}} \lt \frac{1}{10^2} \lt \frac{1}{10^n}\)

Thanks,
Gabriel


For alternative solutions go through this topic: 12-easy-pieces-or-not-126366.html

As for your solution you cannot approximate the way you did, it won't give you a correct answer.
_________________
Intern
Intern
avatar
Joined: 09 Sep 2015
Posts: 5
Location: Grenada
GMAT 1: 630 Q46 V31
GPA: 3.53
WE: Education (Education)
Re M26-16  [#permalink]

Show Tags

New post 25 Jul 2016, 10:41
I think this is a high-quality question.
Intern
Intern
avatar
Joined: 02 Nov 2015
Posts: 20
Re: M26-16  [#permalink]

Show Tags

New post 26 Jul 2016, 05:22
What I did-

Approximate 737 to 700= 7 X 100.

Doing so will give you power of 3 to 10 in denominator. Now, this inequality becomes very simple as we need to have smaller powers than 3 (of 10) in denominator for the right most expression to have greater value. Also, please remember that you can not have n=1, else the left most expression would attain greater value than the middle expression

I hope this helps.

Regards
Yash
Intern
Intern
avatar
B
Joined: 19 Jul 2016
Posts: 49
Reviews Badge
Re: M26-16  [#permalink]

Show Tags

New post 26 Jan 2017, 10:38
hi Bunuel
please elaborate it further. i didn't understand why we are taking/considering n=2



thnx
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: M26-16  [#permalink]

Show Tags

New post 26 Jan 2017, 10:44
SVP
SVP
User avatar
V
Joined: 26 Mar 2013
Posts: 2345
Reviews Badge CAT Tests
Re: M26-16  [#permalink]

Show Tags

New post 27 Jan 2017, 02:54
4
The number is scary 0.00737. So try be simple. Put it as 0.007

Plugging in numbers: 0.001< 0.007 < 0.01..........n=2

Answer; B
Intern
Intern
avatar
B
Joined: 20 Nov 2017
Posts: 16
Re: M26-16  [#permalink]

Show Tags

New post 25 Feb 2018, 16:30
Hi Bunuel,

I solved this problem by multiplying the entire inequality by 10^5 and solved it. This should work all the time with such questions, am I right?

Thank you
Manager
Manager
avatar
B
Joined: 09 Oct 2015
Posts: 226
Re: M26-16  [#permalink]

Show Tags

New post 08 Aug 2018, 03:48
2
multiply every side by 10^5

As the number is positive, no change in sign will occur

10^4-n < 737 < 10^5-n

only n = 2 satisfies this
Intern
Intern
avatar
B
Joined: 23 Jun 2018
Posts: 24
Re: M26-16  [#permalink]

Show Tags

New post 08 Aug 2018, 06:40
1
Bunuel wrote:
If \(n\) is an integer and \(\frac{1}{10^{n+1}} \lt 0.00737 \lt \frac{1}{10^n}\), then what is the value of \(n\)?

A. 1
B. 2
C. 3
D. 4
E. 5


We can plug in the values and solve this question

When it comes to plugging in, i generally pick the middle option (option 3)

when n = 3

1/10^(3+1) = 1/10^4 = 0.0001. This is not less than 0.007.

Now, lets consider n=2
1/10^3 = 0.001 and 1/10^2 = 0.01
0.001< 0.007<0.01

Hence, option B
Intern
Intern
avatar
B
Joined: 01 May 2017
Posts: 10
Re M26-16  [#permalink]

Show Tags

New post 13 Aug 2018, 19:07
I think this is a high-quality question and I agree with explanation. Multiplied the entire inequality by 10^3 which makes the middle # 7.37. N = 2 will put the inequality between 1 and 10
Intern
Intern
avatar
B
Joined: 24 Feb 2014
Posts: 36
Location: United States (GA)
WE: Information Technology (Computer Software)
CAT Tests
Re M26-16  [#permalink]

Show Tags

New post 26 Sep 2019, 22:15
I think this is a high-quality question and I agree with explanation.
GMAT Club Bot
Re M26-16   [#permalink] 26 Sep 2019, 22:15
Display posts from previous: Sort by

M26-16

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne