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# If x, y and z are positive numbers, is z between x and y?

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If x, y and z are positive numbers, is z between x and y? [#permalink]

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04 Jun 2009, 11:42
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If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Oct 2013, 00:15, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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04 Jun 2009, 14:23
x<(y/2)<z
case 1
x y z
1.0 1.1 1.2
TRUE

case 1
x y z
1.0 1.2 1.1
FALSE

INSUFFICIENT

x<2y<z amkes it necessary for y to be between x and z
SUFFICIENT

IMO B
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06 Jun 2009, 17:23
1
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Quote:
x<2y<z amkes it necessary for y to be between x and z
SUFFICIENT

Say,
x=5
y=3
z=10
x<2y<z is true --?5<6<10 but we can't say that y is b/w x and z..
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08 Jun 2009, 19:56
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1) Case 1: y/2=4, case 2: y/2=2, in two case x=1, z=6
In two case, we have different ans (not suf)
2) with x=3 and 2y = 10 and 4, we have different ans (not suf)
Together we have, x<y/2<y<2y<z => YES
Hence C

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Re: x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). [#permalink]

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15 Oct 2013, 16:30
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J

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Re: x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). [#permalink]

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16 Oct 2013, 00:24
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jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J

No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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16 Oct 2013, 04:46
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Hmm wonder where i went wrong in my reasonining for the second statement. Thanks Bunuel

PS. I think theres a small typo in the answer think it should be x, y and z only after divided by 3

Thanks again
Cheers!
J

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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10 Nov 2013, 15:52
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vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

question- is "x < z < y" or "y < z < x" ????

1) x < 2z < y
since 2z<y ( and all are positive) we can infer that z<y . But we do not know weather x<z
Insufficient
(2) 2x < z < 2y
since 2x<z , we can infer x<z , but we cant say if z<y
Insufficient

Combine both 1 and 2 - x<z<y ... sufficient
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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11 Nov 2013, 13:32
1
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vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 4, 2z = 5, y = 6 --> 4 < 5 < 6 BUT z < x < y -- NO
If x = 4, 2z = 10, y = 12 --> 4 < 10 < 12 and x < z < y -- YES

Not sufficient.

(2) 2x < z < 2y.

If 2x = 4, z = 5, 2y = 6 --> 4 < 5 < 6 BUT x < y < z --> NO
If 2x = 4, z = 5, 2y = 18 --> 4 < 5 < 9 and x < z < y --> YES

Not sufficient.

(1)+(2)

x < 2x (always, for positive numbers)
2x < z (from 2)
z < 2z (always, for positive numbers)
2z < y (from 1)

Which gives x < 2x < z < 2z < y

Therefore x < z < y

Sufficient.

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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11 Nov 2013, 21:24
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Question Basically asks whether x<z<y or y<z<x

from St 1 we have x<2z<y
Consider x=4, y= 9, 2z= 7.8 --->z= 3.9 Is Z between x and y : No
Consider x=4, y=9 and 2z= 8.4 ---->z =4.2. Is Z between x and y : yes

So A and D ruled out

Consider St 2, we have 2x<z<2y------> x<z/2< y

Consider x= 4, y=14, z/2= 6----> then z=12, Is Z between x and y : yes
Consider x=4, y=14, z/2=8 ----> then z=16, Is Z between x and y: no

So B ruled out.

Combining we get x<2z<y and x<z/2<y. Now if if both z/2 and 2z are in between x and y then z will also between X and Y.

Ans C
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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22 Feb 2014, 08:00
Hi Bunnel,

I have made mistakes in these questions a couple of times... I solve them incorrectly under time pressure as I jumble up the number cases to pick.

Do you have a Drill or a 'set of questions' for specifically this type of questions that test inequalities in multiple variables?
Do you have some theory tips / pointers? (For eg. - I finnd that consider very close values in Case 1 and considering extremely far values in Case 2 helps...)

THanks for the help

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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14 Apr 2014, 05:42
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

I want to know what is wrong with the following method:

1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes
x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No

Insufficient.

Statement 2 :
Since all are positive I can reduce by 2 , we get x<(z/2)<y

x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes

x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no

1+2

x < 2z < y
x<z/2<y

we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1

I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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14 Apr 2014, 06:58
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Expert's post
qlx wrote:
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

I want to know what is wrong with the following method:

1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes
x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No

Insufficient.

Statement 2 :
Since all are positive I can reduce by 2 , we get x<(z/2)<y

x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes

x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no

1+2

x < 2z < y
x<z/2<y

we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1

I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.

(1) x < 2z < y
(2) 2x < z < 2y

When you add the inequalities you get:

(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

Hope this helps.
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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16 Apr 2014, 13:05
Bunuel wrote:
qlx wrote:
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

I want to know what is wrong with the following method:

1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes
x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No

Insufficient.

Statement 2 :
Since all are positive I can reduce by 2 , we get x<(z/2)<y

x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes

x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no

1+2

x < 2z < y
x<z/2<y

we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1

I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.

(1) x < 2z < y
(2) 2x < z < 2y

When you add the inequalities you get:

(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

Hope this helps.

sorry I didn't get it. My question was since all are positive I can definitely reduce statement 2 by 2, right?
now statement 2 becomes x<(z/2)<y
why am I not getting the answer if I take statement 1 as x < 2z < y , and statement 2 as x<(z/2)<y

Adding 1 and 2 we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So how come this is not working out if after reducing stat.2 I add it to 1?

In the same way if we were to solve :

(Q2)
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So what is the answer to Q2, C or E

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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17 Apr 2014, 01:28
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qlx wrote:
Bunuel wrote:
qlx wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

I want to know what is wrong with the following method:

1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes
x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No

Insufficient.

Statement 2 :
Since all are positive I can reduce by 2 , we get x<(z/2)<y

x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes

x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no

1+2

x < 2z < y
x<z/2<y

we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1

I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.

(1) x < 2z < y
(2) 2x < z < 2y

When you add the inequalities you get:

(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

Hope this helps.

sorry I didn't get it. My question was since all are positive I can definitely reduce statement 2 by 2, right?
now statement 2 becomes x<(z/2)<y
why am I not getting the answer if I take statement 1 as x < 2z < y , and statement 2 as x<(z/2)<y

Adding 1 and 2 we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So how come this is not working out if after reducing stat.2 I add it to 1?

In the same way if we were to solve :

(Q2)
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So what is the answer to Q2, C or E

1. You can reduce 2x < z < 2y by 2 no matter whether x, y, and z are positive or negative. You can always reduce/multiply an inequality by a positive value (2 in our case).

2. Reducing by 2 and adding, though legit is not a correct way to solve. You can add the two inequalities without reducing and directly get (x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

3. When you reduce and then add you'll get 2x < 2.5z < 2y but you cannot plug arbitrary value of x, y, and z there. The values must satisfy all the inequalities, while x=1, z=2, y=3 and x=2, z=2, y=3 does NOT satisfy neither x < 2z < y nor 2x < z < 2y.

Hope it's clear.
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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17 Apr 2014, 06:11
Thank you , the fact that after addition of the 2 statements , the values we choose should satisfy all the inequalities , made it clear.

Now for the question below :
(Q2)
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Simply adding 1 and 2 gives 2x<2.5y<2z
for this we have to plug and test , we have to choose values that satisfy all the 3 inequalities , This way we get the answer as C .

But if we were to multiply statement 1 by 2 and then add to statement 2 we would get 3x<3y<3z
which is x<y<z , this way was much easier here.

so as we can see , in some cases it is easier to get the answer without any manipulation(reducing/ multiplication), as in first case
(1) x < 2z < y
(2) 2x < z < 2y

and in some cases it is easier to get the answer after manipulation( reducing/ multiplication) as in this case :
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

So are these deductions correct?

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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17 Apr 2014, 06:26
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Expert's post
qlx wrote:
Thank you , the fact that after addition of the 2 statements , the values we choose should satisfy all the inequalities , made it clear.

Now for the question below :
(Q2)
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Simply adding 1 and 2 gives 2x<2.5y<2z
for this we have to plug and test , we have to choose values that satisfy all the 3 inequalities , This way we get the answer as C .

But if we were to multiply statement 1 by 2 and then add to statement 2 we would get 3x<3y<3z
which is x<y<z , this way was much easier here.

so as we can see , in some cases it is easier to get the answer without any manipulation(reducing/ multiplication), as in first case
(1) x < 2z < y
(2) 2x < z < 2y

and in some cases it is easier to get the answer after manipulation( reducing/ multiplication) as in this case :
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

So are these deductions correct?

Yes, sometimes you need some kind of manipulations before adding/subtracting,
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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28 Sep 2014, 08:17
Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J

No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

Hi Bunuel,
It may be a dumb question but i am really not able to understand why Statement 1 is not sufficient

In a you have mentioned that
If x = 1, y = 10, and z = 1 --> answer NO.

But if i put the values in inequality i get
x < 2z < y= 1 < 2 < 10 which is absolutely fine. then why the answer is No

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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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28 Sep 2014, 08:38
kd1989 wrote:
Bunuel wrote:
jlgdr wrote:

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J

No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

Hi Bunuel,
It may be a dumb question but i am really not able to understand why Statement 1 is not sufficient

In a you have mentioned that
If x = 1, y = 10, and z = 1 --> answer NO.

But if i put the values in inequality i get
x < 2z < y= 1 < 2 < 10 which is absolutely fine. then why the answer is No

It seems that you don't understand how DS questions work.

The question asks whether z is between x and y.

(1) says that x < 2z < y.

If x = 1, y = 10, and z = 1 (x < 2z < y), then z is NOT between x and y --> answer NO.
If x = 1, y = 10, and z = 2 (x < 2z < y), then z IS between x and y --> answer YES.

We have two different answers to the question, which means that the statement is NOT sufficient.
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Re: If x, y and z are positive numbers, is z between x and y? [#permalink]

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20 Oct 2014, 22:08
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[quote="vcbabu"]If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

1 - This would conclude that y is greater than z i.e. y > z.
No information can be deduced about the relationship between x and z.

x can be 3 and z can be 2

or

x can be 2 and z can be 3

There are infinite options.

2 - Here, we can conclude that z > x

but nothing can be said of z and y.

Combining both

we get that y >z >x

Thus C
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Re: If x, y and z are positive numbers, is z between x and y?   [#permalink] 20 Oct 2014, 22:08

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