Last visit was: 15 Dec 2024, 01:42 It is currently 15 Dec 2024, 01:42
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
User avatar
sunniboy007
Joined: 30 Aug 2003
Last visit: 29 May 2004
Posts: 252
Own Kudos:
Location: BACARDIVILLE
Posts: 252
Kudos: 167
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Paul
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,708
Own Kudos:
Posts: 2,708
Kudos: 1,578
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
kpadma
Joined: 03 Jul 2003
Last visit: 20 Oct 2004
Posts: 380
Own Kudos:
Posts: 380
Kudos: 502
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Paul
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,708
Own Kudos:
Posts: 2,708
Kudos: 1,578
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kpadma
Sunny!

Welcome back ! and Good luck with your preparations!!

Good question. As simple as it looks, but I need to use the calculator
to find out the results.

Any if the question is changed to

A = SQRT(20) - SQRT(15)
B = SQRT(17) - SQRT(12)

What is the simple method to solve this?


The way to look at this is when you add 1 to an integer, the difference in the square root of the two numbers will become increasingly smaller as the value of the original integer is increased. For instance, if you take SQRT(1000) - SQRT(999), the difference will be much smaller than SQRT(1) - SQRT(0).

Therefore, A - [sqrt12 - sqrt11] is less than B - [sqrt11 - sqrt10] and the square of A (a smaller decimal) is then less than the square of B. C is then wrong
User avatar
gmatblast
Joined: 11 Nov 2003
Last visit: 27 Dec 2004
Posts: 190
Own Kudos:
Location: Illinois
Posts: 190
Kudos: 64
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Paul
kpadma
Sunny!

Welcome back ! and Good luck with your preparations!!

Good question. As simple as it looks, but I need to use the calculator
to find out the results.

Any if the question is changed to

A = SQRT(20) - SQRT(15)
B = SQRT(17) - SQRT(12)

What is the simple method to solve this?

The way to look at this is when you add 1 to an integer, the difference in the square root of the two numbers will become increasingly smaller as the value of the original integer is increased. For instance, if you take SQRT(1000) - SQRT(999), the difference will be much smaller than SQRT(1) - SQRT(0).

Therefore, A - [sqrt12 - sqrt11] is less than B - [sqrt11 - sqrt10] and the square of A (a smaller decimal) is then less than the square of B. C is then wrong


That is a good observation Paul!! Thanks

Another simple way to look at this is:

A = sqrt12 - sqrt11 = 1 / (sqrt12 + sqrt11)

B = sqrt11 - sqrt10 = 1 / (sqrt11 + sqrt10 )

A is obviously less than B. So A < B is the answer. (choice B)

Paul,

In your first response to this problem you selected A as an answer. I assume that it is a typo.
User avatar
Paul
Joined: 15 Dec 2003
Last visit: 10 Nov 2012
Posts: 2,708
Own Kudos:
Posts: 2,708
Kudos: 1,578
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Yes, Gmatblast, I meant B. Thx for pointing it out. I'll edit my answer



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderator:
Senior Moderator - Masters Forum
3116 posts