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Is Z Between X and Y?

This topic has 6 expert replies and 2 member replies
bml1105 Master | Next Rank: 500 Posts
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Is Z Between X and Y?

Post Tue Jun 03, 2014 2:24 pm
If x, y, and z are positive numbers, is z between x and y?

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


OA: C

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NandishSS Master | Next Rank: 500 Posts Default Avatar
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Post Tue Apr 25, 2017 5:40 am
ceilidh.erickson wrote:
NandishSS wrote:
How to quickly choose the numbers to satisfy

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

I'm Getting Stuck to choose the numbers. Please throw some light

Thanks
Nandish
Here's my general process: go BIG then small!

So for each statement, I think "is there an example in which the numbers are HUGE or FAR APART that keeps the statement true?"
ex: (1) x < 2z < y
x = 0
y = 1,000,000
2z = 100, z = 50
When x and y are far apart, it's easy to think of a number for z where both z and 2z are between x and y.

Next, I think: "is there an example in which the numbers are SMALL or very CLOSE TOGETHER that keeps the statement true?"
ex: (1) x < 2z < y
x = 3
y = 5
2z = 4, z = 2
When x and y are close together, it's possible for 2z to be in between x and y, but for 2z not to be.

Then repeat the same thought process of BIG then small (or FAR then CLOSE) for the 2nd statement.
Hi ceilidh,

Thanks A lot:-) You People made my day

Thanks
Nandish

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Post Tue Jun 03, 2014 2:38 pm
bml1105 wrote:
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

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Post Tue Jun 03, 2014 2:52 pm
bml1105 wrote:
If x, y, and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y
Statement 1: x < 2z < y
Let x=2 and y=10.
In this case:
2 < 2z < 10
1 < z < 5.
If z=4, then it is between x=2 and y=10.
If z=1.5, then it is not between x=2 and y=10.
INSUFFICIENT.

Statement 2: 2x < z < 2y
Let x=2 and y=10.
In this case:
2*2 < z < 2*10
4 < z < 20
If z=5, then it is between x=2 and y=10.
If z=19, then it is not between x=2 and y=10.
INSUFFICIENT.

Statements combined:
Inequalities can be ADDED TOGETHER.
Adding together x < 2z < y and 2x < z < 2y, we get:
(x+2x) < (2z+z) < (y+2y)
3x < 3z < 3y
x < z < y.
Thus, z is between x and y.
SUFFICIENT.

The correct answer is C.

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NandishSS Master | Next Rank: 500 Posts Default Avatar
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Post Mon Apr 24, 2017 3:54 am
bml1105 wrote:
If x, y, and z are positive numbers, is z between x and y?

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


OA: C
Hi GMATGuruNY And Brent@GMATPrepNow,

How to quickly choose the numbers to satisfy

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

I'm Getting Stuck to choose the numbers. Please throw some light

Thanks
Nandish

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Post Mon Apr 24, 2017 5:05 am
NandishSS wrote:
bml1105 wrote:
If x, y, and z are positive numbers, is z between x and y?

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


OA: C
Hi GMATGuruNY And Brent@GMATPrepNow,

How to quickly choose the numbers to satisfy

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

I'm Getting Stuck to choose the numbers. Please throw some light

Thanks
Nandish
The process of choosing numbers isn't always easy. In most cases, we are looking for numbers that meet the given information AND provide conflicting answers to the target question.

So, for example, for statement 1, we have x < 2z < y
Notice that this statement is similar to what the target question is asking, yet instead of z, we have 2z.
So, we might first make x and y spread out apart enough to allow us to play with the z values.
Let's start with x = 3 and y = 10
First value of z is kind of irrelevant (as long as it meets the given information).
Let's say z = 4.
This value of z meets the condition that x < 2z < y. Also, when x = 3, y = 10, and z = 4, the answer to the target question is "YES, z IS between x and y"

At this point, we'll try to find a value of z so that the answer to the target question is "NO, z is NOT between x and y."
This is where we need to play around with some z-values.
Let's see what happens if we keep x = 3 and y = 10, and work with different z-values.
In this case, if we make z small AND close to the x-value, it might be the case that doubling z (aka 2z) will make 2z big enough so that x < 2z < y, HOWEVER z itself will NOT be between x and y.
So, how about z = 2. When we double z, we get 2z = 6, so we meet the condition that x < 2z < y, however z itself (z = 2) leads us to a DIFFERENT answer to our target question: "NO, z is NOT between x and y."

As I said, choosing numbers isn't always easy. It can take a lot of playing around with numbers.

If you're interested, we have a video on choosing numbers to test when answer Data Sufficiency questions: https://www.gmatprepnow.com/module/gmat-data-sufficiency/video/1102

Cheers,
Brent

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Post Mon Apr 24, 2017 6:59 am
NandishSS wrote:
How to quickly choose the numbers to satisfy

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

I'm Getting Stuck to choose the numbers. Please throw some light

Thanks
Nandish
Here's my general process: go BIG then small!

So for each statement, I think "is there an example in which the numbers are HUGE or FAR APART that keeps the statement true?"
ex: (1) x < 2z < y
x = 0
y = 1,000,000
2z = 100, z = 50
When x and y are far apart, it's easy to think of a number for z where both z and 2z are between x and y.

Next, I think: "is there an example in which the numbers are SMALL or very CLOSE TOGETHER that keeps the statement true?"
ex: (1) x < 2z < y
x = 3
y = 5
2z = 4, z = 2
When x and y are close together, it's possible for 2z to be in between x and y, but for 2z not to be.

Then repeat the same thought process of BIG then small (or FAR then CLOSE) for the 2nd statement.

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EdM in Mind, Brain, and Education
Harvard Graduate School of Education


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Post Tue Apr 25, 2017 7:12 am
NandishSS wrote:
Hi ceilidh,

Thanks A lot:-) You People made my day

Thanks
Nandish
My pleasure! Very glad to help.

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EdM in Mind, Brain, and Education
Harvard Graduate School of Education


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