NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 00:10 It is currently 26 Apr 2024, 00:10
Decision Tracker
My Rewards
New posts
New comers' posts

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

I had never troubles with fractions; now I have some doubts!

User avatar
karlfurt
Senior Manager
Senior Manager
Joined: 02 Mar 2006
Posts: 347
Own Kudos [?]: 536 [0]
Given Kudos: 0
Location: France
Send PM
I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 08:14
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hide Show timer Statistics
I had never troubles with fractions; now I have some doubts!

If 1<a<b<c and if x=1/a and y=1/1/b and z=1/1/1/c then which of the following is true?

z < y < x
x < y < z
x < z < y
y < z < x
z < x < y



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 Problem Solving (PS) 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
mbagal1
Director
Director
Joined: 28 Dec 2005
Posts: 696
Own Kudos [?]: 107 [0]
Given Kudos: 0
 Q49  V45
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 10:40
I get z < x< y


1<a<b<c


Given:
x=1/a and y=1/1/b =b and z=1/1/1/c=1/c

Since c is the largest, 1/c will be the lowest.

Given: b> a, b will also be greater than 1/a.

Therefore z < x< y
User avatar
enola
Intern
Intern
Joined: 10 Jul 2006
Posts: 34
Own Kudos [?]: 12 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 14:09
This problem could also be solved by picking number.

once we know x =1/a; y = b and z=1/c

Pick a=2, b=3, c=4. so we have x=1/2, y=3, z=1/4. Now all we have to do is to rearrange it 1/4<1/2<3. Thus z<x<y Answer E
User avatar
karlfurt
Senior Manager
Senior Manager
Joined: 02 Mar 2006
Posts: 347
Own Kudos [?]: 536 [0]
Given Kudos: 0
Location: France
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 15:14
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?
User avatar
ayushi
Manager
Manager
Joined: 29 Sep 2006
Posts: 74
Own Kudos [?]: 264 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 15:58
E for me too ..picking number strategy is good here...
User avatar
ywilfred
SVP
SVP
Joined: 07 Jul 2004
Posts: 2004
Own Kudos [?]: 1899 [0]
Given Kudos: 0
Location: Singapore
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   15 Nov 2006, 20:23
x = 1/a
y = 1/1/b = b
z = 1/1/1/c = 1/c

So 1/c is smallest, followed by 1/a then b.

So z < x < y

Ans E
User avatar
karlfurt
Senior Manager
Senior Manager
Joined: 02 Mar 2006
Posts: 347
Own Kudos [?]: 536 [0]
Given Kudos: 0
Location: France
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   16 Nov 2006, 01:37
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


OA E.
Can please someone explain the point I have above?
User avatar
ak_idc
Director
Director
Joined: 21 Aug 2006
Posts: 546
Own Kudos [?]: 70 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   16 Nov 2006, 04:37
karlfurt wrote:
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


OA E.
Can please someone explain the point I have above?


Karl, I really appreciate ur interst in understanding things thoroughly.

As you highlighted above, there is a confusion. I think, the guy setting the question should make it clear by putting brackets.
User avatar
karlfurt
Senior Manager
Senior Manager
Joined: 02 Mar 2006
Posts: 347
Own Kudos [?]: 536 [0]
Given Kudos: 0
Location: France
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   16 Nov 2006, 08:36
ak_idc wrote:
karlfurt wrote:
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


OA E.
Can please someone explain the point I have above?


Karl, I really appreciate ur interst in understanding things thoroughly.

As you highlighted above, there is a confusion. I think, the guy setting the question should make it clear by putting brackets.


Yes, but how every one did find the correct answer? :?:
How can you say that z=1/1/1/c = 1/c and not z=1/1/1/c= c?
User avatar
MBAlad
Manager
Manager
Joined: 20 Feb 2006
Posts: 184
Own Kudos [?]: 82 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   16 Nov 2006, 15:43
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


Karlfurt,

Dont panic, we're probably just all a bit more used to the way things are written on this forum than you are.

By the way, 1/2/3/4=(1/2) * (4/3) = 4/6 not 2
User avatar
karlfurt
Senior Manager
Senior Manager
Joined: 02 Mar 2006
Posts: 347
Own Kudos [?]: 536 [0]
Given Kudos: 0
Location: France
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   18 Nov 2006, 04:03
MBAlad wrote:
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


Karlfurt,

Dont panic, we're probably just all a bit more used to the way things are written on this forum than you are.

By the way, 1/2/3/4=(1/2) * (4/3) = 4/6 not 2


Don't worry for me, I don't panic at all. I just cannot see how you guys can say that 1/1/1/C =1/C. and you are right, I've made a silly mistake. So if you agree that

1/2/3/4 = 1/2 . 4/3 = 4/6, how can you say that 1/1/1/c = 1/c and not c!

Really, there is something here I cannot understand!
User avatar
MBAlad
Manager
Manager
Joined: 20 Feb 2006
Posts: 184
Own Kudos [?]: 82 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   18 Nov 2006, 06:25
I think if you look at the context of the question then you can see what is most likely being asked and the progression of x, y and z ie 1/a........1/1/b or 1/(1/b)........1/(1/(1/c))
User avatar
Fig
VP
VP
Joined: 01 May 2006
Posts: 1033
Own Kudos [?]: 251 [0]
Given Kudos: 0
Send PM
Re: I had never troubles with fractions; now I have some doubts! [#permalink] New post   18 Nov 2006, 06:37
karlfurt wrote:
MBAlad wrote:
karlfurt wrote:
but isn't 1/2/3/4=(1/2) * (4/3) =2 because 0,5/0,25=2...

so z=1/1/1/c= c?

and is more generally 1/1/b = 1/(1/b) = b OR (1/1)/(1/b)=(1/1)/b=1/b?


Karlfurt,

Dont panic, we're probably just all a bit more used to the way things are written on this forum than you are.

By the way, 1/2/3/4=(1/2) * (4/3) = 4/6 not 2


Don't worry for me, I don't panic at all. I just cannot see how you guys can say that 1/1/1/C =1/C. and you are right, I've made a silly mistake. So if you agree that

1/2/3/4 = 1/2 . 4/3 = 4/6, how can you say that 1/1/1/c = 1/c and not c!

Really, there is something here I cannot understand!


Hi kalfurt :)... Following your PM, I will try to give my view on it :)

To me, I prefer to see the problem with a function f(x) = 1/x.

So, we have :

o x = f(a) = 1/a
o y = f(f(b)) = 1/(1/b) = 1/1/b = b
o z = f(f(f(c))) = 1/(1/(1/c)) = 1/1/1/c = 1/c

As well, the bold section should be :

1/2/3/4 = 1/[2 * (4/3)] as 4/3 = 1/[3/4] = 1/3/4 and so 2*4/3 = 2*1/[3/4] = 2/3/4

:)



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 Problem Solving (PS) 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!
GMAT Club Bot
gmatclubot
Re: I had never troubles with fractions; now I have some doubts! [#permalink]
18 Nov 2006, 06:37
Moderators:
Bunuel
Math Expert
92915 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions