Last visit was: 24 Apr 2026, 03:14 It is currently 24 Apr 2026, 03:14
Home
Messages
Tests
Schools
Events
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
User avatar
GMAT1900
User avatar
Joined: 03 Mar 2007
Last visit: 25 Jul 2015
Posts: 14
Own Kudos:
5
Posts: 14
Kudos: 5
3 more tricky Math Questions 03 Mar 2007, 17:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have attached three questions. I cant seem to figure them out and I was wondering if anyone had any ideas or could tell me the best way to solve them and what was the main concept in each


[/img]



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 below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
Fig
User avatar
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
253
Posts: 1,031
Kudos: 253
3 more tricky Math Questions 03 Mar 2007, 17:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Question 3)

Have a look here : https://www.gmatclub.com/phpbb/viewtopic.php?t=37781

:)
User avatar
GMAT1900
User avatar
Joined: 03 Mar 2007
Last visit: 25 Jul 2015
Posts: 14
Own Kudos:
5
Posts: 14
Kudos: 5
3 more tricky Math Questions 03 Mar 2007, 17:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks for #3 FIG ...
User avatar
Fig
User avatar
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
253
Posts: 1,031
Kudos: 253
3 more tricky Math Questions 03 Mar 2007, 17:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Question 2)

(A) for me :)

1/(b-a) < a - b ?
<=> (a - b) - 1/(b-a) > 0 ?
<=> ((a - b)^2 + 1)/(a-b) > 0 ?

Since (a-b)^2 + 1 > 0, we have
=> a-b > 0 ?
<=> a > b ?

From 1
a < b

SUFF.

From 2
|a-b| > 1
<=> a-b > 1 or b-a > 1
<=> a > b+1 or b > a+1

We cannot conclude from which a and b is superior.

INSUFF.
User avatar
kyatin
User avatar
Joined: 29 Jan 2007
Last visit: 16 Oct 2016
Posts: 248
Own Kudos:
162
Location: Earth
Posts: 248
Kudos: 162
3 more tricky Math Questions 03 Mar 2007, 20:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Fig,

I dint get this..

Since (a-b)^2 + 1 > 0, we have
=> a-b > 0 ?
<=> a > b ?

If u simplified it to
Is (a-b)^2 + 1 > 0 ?

Then isnt that true always... I mean can we not answer it without any option. Because square is positive no matter what?

So really speaking any value of a and b will satisfy that relation.
avatar
xufan_dfdq
avatar
Joined: 03 Mar 2007
Last visit: 03 Mar 2007
Posts: 2
Posts: 2
Kudos: 0
3 more tricky Math Questions 03 Mar 2007, 20:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Question 1)

90/(v-3)-90/(v+3)=0.5
=>90(v+3)-90(v-3)=0.5(v^2-9)
=>v^2=1089
=>v=33

=> 90/(v+3)=90/36=2.5
User avatar
voltron
User avatar
Joined: 28 Dec 2006
Last visit: 30 Jun 2007
Posts: 6
Posts: 6
Kudos: 0
3 more tricky Math Questions 04 Mar 2007, 00:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
for #2, i would assume the first step is....

1/(a-b) < b-a
1/(a-b) < (-1)(a-b)

with (1) a < b
we know that a-b is negative so plug in with whole numbers and fractions
both should result in the statement being true

ex1. a=2, b = 4
1/(a-b) < (-1)(a-b)
1/(2-4) < (-1)(2-4)
1/ -2 < 2

ex2. a=-2/4 b=-1/4
1/(a-b) < (-1)(a-b)
1/(-2/4 + 1/4) < (-1) (-2/4 + 1/4)
1/(-1/4) < (-1) (-1/4)
-4 < 1/4
User avatar
Fig
User avatar
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
253
Posts: 1,031
Kudos: 253
3 more tricky Math Questions 04 Mar 2007, 02:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kyatin
Fig,

I dint get this..

Since (a-b)^2 + 1 > 0, we have
=> a-b > 0 ?
a > b ?

If u simplified it to
Is (a-b)^2 + 1 > 0 ?

Then isnt that true always... I mean can we not answer it without any option. Because square is positive no matter what?

So really speaking any value of a and b will satisfy that relation.
Show more


My objective was to simplify this expression :
((a - b)^2 + 1)/(a-b) > 0

As u also say, a square of a number + 1 is always positive. So, we could divid the left and the right of the inequation by (a - b)^2 + 1).

Then, we have:
1/(a-b) > 0
a - b > 0
a > b

All the process is to simplify what we are looking for :)



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!