Is Z between Z & Y?

[fourteenstix]

Anyone have a good approach to this? Plugging in numbers seemed to get pretty confusing.

[Brent@GMATPrepNow]

If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y
(2) 2x < z < 2y

Target question: Is z between x and y?

Statement 1: x < 2z < y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 3, y = 10, and z = 2, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 2x < z < 2y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 1, y = 2, and z = 3, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x < 2z < y
Statement 2: 2x < z < 2y
Since the two inequalities are facing the same direction, we can add them to get:
3x < 3z < 3y
Divide all three parts by 3 to get: x < z < y
As we can see, z IS definitely between x and y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

[visheshchadha]

There is no need to even put values for the First and Second Statement alone.

I) x < 2z < y => x/2 < z < y/2
Now, y/2 may be bigger than x and in that case x < z < y, else not. So, no unique solution.

II) 2x < z < 2y Now, 2x may be lesser than y and in that case x < z < y, else not. So, no unique solution.

Just visualize the 4 variables mentioned above on Number line and it becomes easy.

Combining I and II : 3x < 3z < 3y => x < z < y. Hence both statements are required.

[Java_85]

Hard question for me, But adding the both sides of in-equations sounds like a good idea to solve such a questions!

[a_new_start]

If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y
(2) 2x < z < 2y

Target question: Is z between x and y?

Statement 1: x < 2z < y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 3, y = 10, and z = 2, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 2x < z < 2y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 1, y = 2, and z = 3, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x < 2z < y
Statement 2: 2x < z < 2y
Since the two inequalities are facing the same direction, we can add them to get:
3x < 3z < 3y
Divide all three parts by 3 to get: x < z < y
As we can see, z IS definitely between x and y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

Dear Brent,

What if we are unable to think of the values of x, y and z under exam pressure.

Any other way to tackle this question.

[Brent@GMATPrepNow]

Dear Brent,

What if we are unable to think of the values of x, y and z under exam pressure.

Any other way to tackle this question.

When statements are not sufficient, it can often be difficult to show that they are not sufficient using means other than PLUGGING IN.
Here, each statement contains very little information, so I can't see any options other than plugging in.

Fortunately, with the statements combined, we have something to work with.

Cheers,
Brent

[a_new_start]

Dear Brent,

What if we are unable to think of the values of x, y and z under exam pressure.

Any other way to tackle this question.

When statements are not sufficient, it can often be difficult to show that they are not sufficient using means other than PLUGGING IN.
Here, each statement contains very little information, so I can't see any options other than plugging in.

Fortunately, with the statements combined, we have something to work with.

Cheers,
Brent

Yeah. True.

I agree with you.

Thank You

[GMATGuruNY]

If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y
(2) 2x < z < 2y

What if we are unable to think of the values of x, y and z under exam pressure.

Any other way to tackle this question.

One suggestion:
Since the question stem asks whether z is between x and y, test values for x and y that are SPREAD OUT.
Let x=10 and y=100.

Statement 1: x < 2z < y
Substituting x=10 and y=100, we get:
10 < 2z < 100
5 < z < 50.
If z=20, then z is between x and y.
If z=6, then z is NOT between x and y.
INSUFFICIENT.

Statement 2: 2x < z < 2y
Substituting x=10 and y=100, we get:
2*10 < z < 2*100
20 < z < 200.
If z=30, then z is between x and y.
If z=150, then z is NOT between x and y.
INSUFFICIENT.

Statements combined:
Adding together x < 2z < y and 2x < z < 2y, we get:
x+2x < 2z+z < y+2y
3x < 3z < 3y
x < z < y.
SUFFICIENT.

The correct answer is C.

[a_new_start]

If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y
(2) 2x < z < 2y

What if we are unable to think of the values of x, y and z under exam pressure.

Any other way to tackle this question.

One suggestion:
Since the question stem asks whether z is between x and y, test values for x and y that are SPREAD OUT.
Let x=10 and y=100.

Statement 1: x < 2z < y
Substituting x=10 and y=100, we get:
10 < 2z < 100
5 < z < 50.
If z=20, then z is between x and y.
If z=6, then z is NOT between x and y.
INSUFFICIENT.

Statement 2: 2x < z < 2y
Substituting x=10 and y=100, we get:
2*10 < z < 2*100
20 < z < 200.
If z=30, then z is between x and y.
If z=150, then z is NOT between x and y.
INSUFFICIENT.

Statements combined:
Adding together x < 2z < y and 2x < z < 2y, we get:
x+2x < 2z+z < y+2y
3x < 3z < 3y
x < z < y.
SUFFICIENT.

The correct answer is C.

Mitch,

You have really made it easy.

Kudos!

I'm satisfied now.

[DavidG@VeritasPrep]

Note that summing inequalities is often helpful in data sufficiency questions when you're testing the statements together. Another good thread here:

https://www.beatthegmat.com/is-m-z-0-t13539.html