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

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Director
Joined: 02 Mar 2006
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Location: France
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I had never troubles with fractions; now I have some doubts! [#permalink]  15 Nov 2006, 07:14
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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
Director
Joined: 28 Dec 2005
Posts: 921
Followers: 2

Kudos [?]: 38 [0], given: 0

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
Manager
Joined: 10 Jul 2006
Posts: 75
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Kudos [?]: 2 [0], given: 0

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
Director
Joined: 02 Mar 2006
Posts: 579
Location: France
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Kudos [?]: 47 [0], given: 0

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?
Manager
Joined: 29 Sep 2006
Posts: 96
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Kudos [?]: 12 [0], given: 0

E for me too ..picking number strategy is good here...
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
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Kudos [?]: 195 [0], given: 0

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
Director
Joined: 02 Mar 2006
Posts: 579
Location: France
Followers: 1

Kudos [?]: 47 [0], given: 0

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?
VP
Joined: 21 Aug 2006
Posts: 1026
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Kudos [?]: 21 [0], given: 0

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.
_________________

The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.

Director
Joined: 02 Mar 2006
Posts: 579
Location: France
Followers: 1

Kudos [?]: 47 [0], given: 0

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?
Senior Manager
Joined: 20 Feb 2006
Posts: 373
Followers: 1

Kudos [?]: 7 [0], given: 0

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
Director
Joined: 02 Mar 2006
Posts: 579
Location: France
Followers: 1

Kudos [?]: 47 [0], given: 0

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!
Senior Manager
Joined: 20 Feb 2006
Posts: 373
Followers: 1

Kudos [?]: 7 [0], given: 0

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))
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

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?

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

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