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karlfurt
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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 08:14
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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



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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 10:40
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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
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enola
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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 14:09
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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
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karlfurt
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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 15:14
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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?
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ayushi
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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 15:58
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E for me too ..picking number strategy is good here...
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ywilfred
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I had never troubles with fractions; now I have some doubts! 15 Nov 2006, 20:23
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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
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I had never troubles with fractions; now I have some doubts! 16 Nov 2006, 01:37
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karlfurt
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?
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OA E.
Can please someone explain the point I have above?
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I had never troubles with fractions; now I have some doubts! 16 Nov 2006, 04:37
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karlfurt
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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?
Show more


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.
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karlfurt
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I had never troubles with fractions; now I have some doubts! 16 Nov 2006, 08:36
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karlfurt
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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.
Show more


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?
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MBAlad
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I had never troubles with fractions; now I have some doubts! 16 Nov 2006, 15:43
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karlfurt
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?
Show more


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
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karlfurt
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I had never troubles with fractions; now I have some doubts! 18 Nov 2006, 04:03
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MBAlad
karlfurt
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
Show more


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!
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MBAlad
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I had never troubles with fractions; now I have some doubts! 18 Nov 2006, 06:25
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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))
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I had never troubles with fractions; now I have some doubts! 18 Nov 2006, 06:37
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karlfurt
MBAlad
karlfurt
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!
Show more


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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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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