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

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Manager
Joined: 02 Apr 2006
Posts: 83
If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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10 Nov 2007, 21:51
18
00:00

Difficulty:

55% (hard)

Question Stats:

65% (01:48) correct 35% (02:01) wrong based on 252 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 58311
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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17 Jul 2014, 01:28
4
6
pretttyune wrote:
If x, y, and z are positive numbers, Is Z between X and Y?

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

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.

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Intern
Joined: 04 Nov 2007
Posts: 36
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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Updated on: 10 Nov 2007, 22:34
1
Statement 1 & 2 will give YES & NO answers to the question when substituted with different values. So, they are not sufficient.

Take values x=1;y=6;z=2 & x=3;y=6;z=2 & you will see why statement 1 is not sufficient.

Take both the statements together.
4X<8Z<4Y
4X<Z<4Y
All values that satisfy these two statements satisfy the question with an answer YES.

Hope that helps

Originally posted by Vemuri on 10 Nov 2007, 22:30.
Last edited by Vemuri on 10 Nov 2007, 22:34, edited 1 time in total.
Manager
Joined: 25 Jul 2007
Posts: 94
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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10 Nov 2007, 22:35
Statement 1:

Assume x=1,y=2. z=2/3.
The given equation is satisfied but z is not between x & y.

Assume x=1,y=5,z=2
The given equation is satisfied and z is not between x & y.

Therefore Statement 1 is insufficient.

Statement 2:

Assume x=1,y=2,z=5.
The given equation is satisfied but z is not between x & y.

Assume x=1,y=6,z=5.
The given equation is satisfied and z is not between x & y.

Therefore Statement 2 is insufficient.

Taking both the statements together we have,

x<2z<y
x<z/4<y

You will find that for any values of x,y & z that satisfy both the equations, z will always be between x and y.

Therefore the answer is C.
Director
Joined: 29 Jun 2017
Posts: 929
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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02 Mar 2018, 08:50
1
pretttyune wrote:
If x, y, and z are positive numbers, Is z between x and y?

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

it is hard
from 1
2z<y , so z< y because all are positive
but
2z>x dose not mean z>x
1 is not sufficient

similarly we know 2 is not sufficient

from both 1 and 2
we have
z<y from 1 and z>x from 2. so both is sufficient.

we do not need to pick numbers, doing so is long
Director
Joined: 24 Oct 2016
Posts: 529
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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19 Aug 2019, 05:52
Bunuel wrote:
pretttyune wrote:
If x, y, and z are positive numbers, Is Z between X and Y?

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

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.

Bunuel VeritasKarishma I have a doubt regarding an alternate method mentioned below. I would really appreciate your help

Let's say that we divided inequalities in stmt 2 by 2.
2x < z < 2y => x < z/2 < y

Now if we add this modified stmt 2 inequalities with the one in stmt 1, we get

2x < 5z/2 < 2y => 4x < 5z < 4y

From 5z < 4y, we can infer that z < y.
From 4x < 5z, we can infer anything whether x < z or x > z.

How can we infer x < z in this case? Am I doing something wrong here?

Thanks!
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Posts: 58311
Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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19 Aug 2019, 05:57
1
dabaobao wrote:
Bunuel wrote:
pretttyune wrote:
If x, y, and z are positive numbers, Is Z between X and Y?

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

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.

Bunuel VeritasKarishma I have a doubt regarding an alternate method mentioned below. I would really appreciate your help

Let's say that we divided inequalities in stmt 2 by 2.
2x < z < 2y => x < z/2 < y

Now if we add this modified stmt 2 inequalities with the one in stmt 1, we get

2x < 5z/2 < 2y => 4x < 5z < 4y

From 5z < 4y, we can infer that z < y.
From 4x < 5z, we can infer anything whether x < z or x > z.

How can we infer x < z in this case? Am I doing something wrong here?

Thanks!

We know that x < z from (2): 2x < z. This means that x < z.
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Re: If x, y, and z are positive numbers, Is z between x and y?  [#permalink]

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19 Aug 2019, 06:35
pretttyune wrote:
If x, y, and z are positive numbers, Is z between x and y?

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

Given: x, y, and z are positive numbers.

Asked: Is z between x and y?

(1) x < 2z < y
NOT SUFFICIENT

(2) 2x < z < 2y
NOT SUFFICIENT

Combining (1) & (2)
3x<3z<3y
x<z<y
SUFFICIENT

IMO C

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Re: If x, y, and z are positive numbers, Is z between x and y?   [#permalink] 19 Aug 2019, 06:35
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