Last visit was: 23 Jan 2025, 14:00 It is currently 23 Jan 2025, 14:00
Home
GMAT
Messages
Tests
Schools
Reviews
Social
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
Sub 505 Level|   Exponents|      
User avatar
Googlore
User avatar
Joined: 11 Sep 2006
Last visit: 21 Mar 2007
Posts: 26
Own Kudos:
13
 [10]
Posts: 26
Kudos: 13
 [10]
(1001^2 - 999^2)/(101^2 - 99^2) = Updated on: 15 Oct 2019, 08:47
Originally posted by Googlore on 20 Sep 2006, 14:35.
Last edited by Bunuel on 15 Oct 2019, 08:47, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

5% (low)

Question Stats:

86% (01:24) correct 14% (01:43) wrong based on 529 sessions

History

Date
Time
Result
Not Attempted Yet
\(\frac{1001^2-999^2}{101^2-99^2}=\)

A. 10
B. 20
C. 40
D. 80
E. 100
ShowHide Answer
Official Answer
A
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Jan 2025
Posts: 98,904
Own Kudos:
696,121
 [4]
Given Kudos: 91,889
Products:
GMAT Club Tests Forum Quiz
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 98,904
Kudos: 696,121
 [4]
(1001^2 - 999^2)/(101^2 - 99^2) = 09 May 2010, 19:05
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
\(\frac{1001^2-999^2}{101^2-99^2}=\)

A. 10
B. 20
C. 40
D. 80
E. 100

\(\frac{1001^2-999^2}{101^2-99^2}=\frac{(1001-999)(1001+999)}{(101-99)(101+99)}=\frac{2*2000}{2*200}=10\)

Answer: A.
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 23 Jan 2025
Posts: 20,072
Own Kudos:
24,832
 [1]
Given Kudos: 289
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 20,072
Kudos: 24,832
 [1]
(1001^2 - 999^2)/(101^2 - 99^2) = 16 Oct 2019, 19:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Googlore
\(\frac{1001^2-999^2}{101^2-99^2}=\)

A. 10
B. 20
C. 40
D. 80
E. 100

Both the numerator and the denominator are a difference of squares. Using the given information, we have:

(1001 - 999)(999 + 1001)/[(101 - 99)(101 + 99)]

(2)(2000)/[(2)(200)] = 10

Answer: A
User avatar
BrentGMATPrepNow
User avatar
GMAT Club Legend
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,784
Own Kudos:
32,581
 [1]
Given Kudos: 799
Location: Canada
Expert reply
Posts: 6,784
Kudos: 32,581
 [1]
(1001^2 - 999^2)/(101^2 - 99^2) = 12 Feb 2021, 14:12
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Googlore
\(\frac{1001^2-999^2}{101^2-99^2}=\)

A. 10
B. 20
C. 40
D. 80
E. 100


Key Property: \(x^2 - y^2 = (x+y)(x-y)\) (i.e., factoring a difference of squares)

\(\frac{1001^2-999^2}{101^2-99^2}=\frac{(1001+999)(1001-999)}{(101+999)(101-99)}\)

\(=\frac{(2000)(2)}{(200)(2)}\)

\(=\frac{4000}{400}\)

\(= 10\)

Answer: A
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 36,165
Own Kudos:
950
Posts: 36,165
Kudos: 950
(1001^2 - 999^2)/(101^2 - 99^2) = 09 Jul 2023, 10:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
98904 posts
Krunaal
PS Forum Moderator
331 posts