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If x, y, and z are positive numbers, Is z between x and y?
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10 Nov 2007, 21:51
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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
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Re: If x, y, and z are positive numbers, Is z between x and y?
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17 Jul 2014, 01:28
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. Answer: C.
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Re: If x, y, and z are positive numbers, Is z between x and y?
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Updated on: 10 Nov 2007, 22:34
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.



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Re: If x, y, and z are positive numbers, Is z between x and y?
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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.



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



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Re: If x, y, and z are positive numbers, Is z between x and y?
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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. Answer: C. 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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Re: If x, y, and z are positive numbers, Is z between x and y?
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19 Aug 2019, 05:57
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. Answer: C. 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?
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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 Posted from my mobile device
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Re: If x, y, and z are positive numbers, Is z between x and y?
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