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

 It is currently 20 Mar 2019, 20:44

### 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

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

# (1001^2-999^2)/(101^2-99^2)=

Author Message
TAGS:

### Hide Tags

Intern
Joined: 11 Sep 2006
Posts: 44

### Show Tags

Updated on: 08 Jan 2014, 05:00
2
00:00

Difficulty:

5% (low)

Question Stats:

83% (01:29) correct 17% (01:38) wrong based on 129 sessions

### HideShow timer Statistics

$$\frac{1001^2-999^2}{101^2-99^2}=$$

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

Originally posted by Googlore on 20 Sep 2006, 14:35.
Last edited by Bunuel on 08 Jan 2014, 05:00, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
SVP
Joined: 01 May 2006
Posts: 1768

### Show Tags

20 Sep 2006, 14:38
1
1
We should apply : a^2-b^2 = (a+b)*(a-b)

[(1001^2) - (999^2)]/[(101^2)-(99^2)]
= (1001+999)*(1001-999) / [(101+99)*(101-99)]
= (2000)*2 / [200*2]
= 10
Director
Joined: 23 Jun 2005
Posts: 778
GMAT 1: 740 Q48 V42

### Show Tags

20 Sep 2006, 21:15
I arrived at the same answer as Fig in a different way.
Write it as
(a+b)^2 = a^2 +2ab_b^2 -----> 1
(a-b)^2 = a^2-2ab+b^2. -----> 2
When you subtract 1 from 2, you get 4ab. Applying this
(1000+1)^2 -(1000-1)^2/(100+1)^2 - (100-1)^2
4000/400 = 10
Manager
Joined: 29 Jul 2009
Posts: 157

### Show Tags

09 May 2010, 18:55
$$\frac{1001^2-999^2}{101^2-99^2}=$$

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

Q10.jpg [ 8.23 KiB | Viewed 3851 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 53738
Re: what is the strategy to be used?  [#permalink]

### Show Tags

09 May 2010, 19:05
1
Hi,

I'm sure there is a strategy to use in order to solve the question...I dont think that the GMAT expects me to do the calculation!!

You should apply the following formula: $$x^2-y^2=(x-y)(x+y)$$

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

_________________
Senior Manager
Joined: 08 Nov 2010
Posts: 329
Re: what is the strategy to be used?  [#permalink]

### Show Tags

12 Dec 2010, 01:09
1
damn. So simple. Thanks m8.
_________________
Intern
Joined: 27 Dec 2013
Posts: 32
Concentration: Finance, General Management
Schools: ISB '15
Re: Is there a faster way to calculate this that I'm missing...  [#permalink]

### Show Tags

07 Jan 2014, 23:49
Googlore wrote:
Is there a faster way to calculate this that I'm missing...

[(1001^2) - (999^2)]/[(101^2)-(99^2)]

Remember this formula a² - b² = (a+b) (a-b)

Applying this formula,

= (1001+999) x (1001-999)/ (101+99) x (101-99)

= 2000 x 2/ 200 x 2

= 10
_________________

Kindly consider for kudos if my post was helpful!

Non-Human User
Joined: 09 Sep 2013
Posts: 10151

### Show Tags

15 Jan 2019, 21:30
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.
_________________
Re: (1001^2-999^2)/(101^2-99^2)=   [#permalink] 15 Jan 2019, 21:30
Display posts from previous: Sort by